z transform

7212019 Z Transform

httpslidepdfcomreaderfullz-transform-56de7166594ea 152

6003

Signals

and

Systems

ZTransform

September

22

2011

1

7212019 Z Transform


Block DiagramSystem Functional

Difference Equation System Function

Unit-Sample Response

+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2

y[n] = x[n] + y[nminus1] + y[nminus2] H (z) = Y (z)

X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

ConceptMap Discrete-TimeSystems

Multiple

representations

of

DT

systems

2

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R


Relations

among

representations

3

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R


Two

interpretations

of

ldquoDelayrdquo

4

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Rrarr 1

z


Relation

between

System

Functional

and

System

Function

5

7212019 Z Transform


Block Diagram System Functional



+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2

y[n] = x[n] + y[nminus1] + y[nminus2] H (

z) =

Y (z)

X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

CheckYourself

What

is

relation

of

System

Functional

to

Unit-Sample

Response

6

7212019 Z Transform


CheckYourself

Expand

functional

in

a

series

Y 1

X

=

H(R)

=

1

minus

R

minus

R2

1 +R +2R2 +3R3 +5R4 +8R5 + middot middot middot

1 minusRminusR2 1

1 minusR minusR2

R +R2

R minusR2 minusR3

2R2 +R3

2R2 minus2R3 minus2R4

3R3 +2R4

3R3

minus3R4

minus3R5

middot middot middot

H(R)

=

1

1

minus

R

minus

R2

=

1

+

R

+

2R2+

3R3+

5R4+

8R5+

13R6+

middot

middot

middot

7

7212019 Z Transform


1048573

CheckYourself

Coefficientsof series representationof H(R)

1

H(R) =

middot middot

1

minus R minus R2

= 1 +

R

+ 2R2+ 3R3+ 5R4+ 8R5+

13R6+

middot

are the successive samples in theunit-sample response

h[n] 11

23

58

1321

3455

If

a

system

is

composed

of

(only)

adders

delays

and

gains

then

H(R) =h[0]+h[1]R+h[2]R2+h[3]R3+h[4]R4+middot middot middot

= h[n]Rn

n

We

can

write

the

system

function

in

terms

of

unit-sample

response

8

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Functional

to

Unit-Sample

Response

9

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

10

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

11

7212019 Z Transform


Block Diagram

System Functional



+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H (z) =

h[n]zminusn

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

12

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

Today

thinking

about

a

system

as

a

mathematical

function H (z)

rather

than

as

an

operator

13

7212019 Z Transform


1048573

1048573

ZTransform

We call the relationbetween H (z)and h[n] the Z transform

minusn

H (z) =

h[n]z

n

Z

transform

maps

a

function

of

discrete

time n

to

a

function

of z

Although

motivated

by

system

functions

we

can

define

a

Z

trans

form forany signal

minusnX (z) =

x[n]z

n=minusinfin

Noticethatwe includen lt0aswellasn gt0rarrbilateralZtransform

(there isalsoaunilateralZ transformwith similarbutnot identical

properties)

14

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

the

unit-sample

signal

x[n] =

δ [n]

n

δ [n]

minusnX (z) = x[n]z =x[0]z

0 = 1

n=minusinfin

15

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

a

delayed

unit-sample

signal

n

x[n]

x[n

] =δ

[n

minus

1]

minusn

minus1

minus1X (z) = x[n]z =x[1]z =z

n=minusinfin

16

infin

Check Yo rself

7212019 Z Transform


CheckYourself

What is theZ transformof the following signal

minus4minus3minus2minus1 0 1 2 3 4n

x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

17

Check Yourself

7212019 Z Transform


1048573 1048573

CheckYourself

What

is

the

Z

transform

of

the

following

signal


x[n] =7

8

nu[n]

n

n7

7 1minusn

minusnX (z) = z u[n] = z =

8 8 1minus

7 minus18

zn=minusinfin n=0

18

infin infin

Check Yourself

7212019 Z Transform


CheckYourself

What is theZ transformof the following signal 2


x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

19

Z Transform Pairs

7212019 Z Transform


1048573 1048573

ZTransformPairs

Thesignalx[n]which isafunctionof timenmapstoaZtransform

X (z)which isa functionof z

n

x[n] =

7

u[n] harr X (z) =

1

minus18 1

minus

7

8

z

For

what

values

of z

does X (z)

make

sense

The

Z

transform

is

only

defined

for

values

of

z

for

which

the

defining

sum converges

infin n

infin n7 7 1minusn minusn

X (z) =

z u[n] =

z

=8 8 1minus

7

8

zminus1

n=minusinfin n=0

Therefore

1048608104860810486081048608 7

8z

minus1

1048608104860810486081048608 lt1 ie |z|gt

7

8

20

infin infin

Regions of Convergence

7212019 Z Transform



The Z transform X (z) is a function of z defined for all z inside a

Region of Convergence (ROC)

n7 1 7

x[n] = u[n] harr X (z) =

|z|gt

8 1

minus

7 88

zminus1

7

ROC

|z|gt

8

21

Z Transform Mathematics

7212019 Z Transform


1048573

1048573

1048573 1048573

ZTransformMathematics

Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2

then x1[n] +x2[n] harr

X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then

minusn Y (z) = y[n]z

n=minusinfin

minusn = (x1[n] +x2[n])z

n=minusinfin

minusn+

minusn = x1[n]z x2[n]z

n=minusinfin n=minusinfin

=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC

Wehavealready seenanexampleof thisproperty

δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin

Let y[n] =x[nminus1] then

infin

minusn minusnY (z) = y[n]z = x[n

minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin

Rational Polynomials

7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with

constantcoefficientscanalsobedescribedbyaZ transform that is

a

ratio

of

polynomials

in

z

b0y[n] +b1y[nminus1]+b2y[nminus2]+middot middot middot=a0x[n] +a1x[nminus1]+a2x[nminus2]+middot middot middot

Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y

(z)+middot middot middot

=

a0X (z)+a1z

minus1X (z)+a2z

minus2X (z)+middot middot middot

a0

+a1zminus1+a2zminus2+


H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+


a0zk+a1zkminus1+a2zkminus2+middot middot middot =

b0zk+b1zkminus1+b2zkminus2+


24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo

remwe canexpress thepolynomialsasaproductof factors

k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z



(zminusz0) (zminusz1) middot middot middot (zminuszk)

=

(zminus p0) (zminus p1) middot middot middot (zminus pk)

where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in

theZ-plane The edgesof the circlesareat thepoles

Example

x[n] =

αnu[n]

minusn minusn X (z) = αnu[n]z = αnz

n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform


WhatDT signalhas the followingZ transform

78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region

minus4minus3minus2minus1 0 1 2 3 4 n

x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0

7doesnotconverge if |z|lt

8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8

y[n] = (y[n+1]minusx[n+1])

7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7

minus2 0 minus 9830723

minus3 0 minus

middot middot middot middot middot middot

n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot

minus4 minus3 minus2 minus1 0 1 2n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8

minus4 minus3 minus2 minus1 0 1 2 n

y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8

minus4 minus3 minus2 minus1 0 1 2

n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2

u[nminus1] + 2 +2nminus1u[minusn]

n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=

minus22z2minus5z+ 2 2

minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic

ularly easy to solve

Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute

This equation canbeuseful toprove theorems

There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on

several importantproperties

Property

x[n]

X (z)

ROC

Linearity ax1[n] +bx2[n] aX 1(z) +bX 2(z) sup(R1

capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n

minusm] X 1(z)X 2(z)

sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R

852000 852000 = 1 +R+R2+R3+middot middot middot times 1 +R+R2+R3+middot middot middot

1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y

= 1 + 2R+ 3R2+ 4R3+middot middot middot= (n+1)Rn

X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms

Th f Z T f t l diff ti l ti d d

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems

Relations among representations

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu

6003 Signals and SystemsFall 2011

For information about citing these materials or our Terms of Use visit httpocwmiteduterms

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R


Multiple

representations

of

DT

systems

2

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R


Relations

among

representations

3

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R


Two

interpretations

of

ldquoDelayrdquo

4

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Rrarr 1

z


Relation

between

System

Functional

and

System

Function

5

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


z) =

Y (z)

X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

CheckYourself

What

is

relation

of

System

Functional

to

Unit-Sample

Response

6

7212019 Z Transform


CheckYourself

Expand

functional

in

a

series

Y 1

X

=

H(R)

=

1

minus

R

minus

R2


1 minusRminusR2 1

1 minusR minusR2

R +R2

R minusR2 minusR3

2R2 +R3


3R3 +2R4

3R3

minus3R4

minus3R5


H(R)

=

1

1

minus

R

minus

R2

=

1

+

R

+

2R2+

3R3+

5R4+

8R5+

13R6+

middot

middot

middot

7

7212019 Z Transform


1048573

CheckYourself


1

H(R) =

middot middot

1

minus R minus R2

= 1 +

R

+ 2R2+ 3R3+ 5R4+ 8R5+

13R6+

middot


h[n] 11

23

58

1321

3455

If

a

system

is

composed

of

(only)

adders

delays

and

gains

then


= h[n]Rn

n

We

can

write

the

system

function

in

terms

of

unit-sample

response

8

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Functional

to

Unit-Sample

Response

9

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

10

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

11

7212019 Z Transform


Block Diagram

System Functional



+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H (z) =

h[n]zminusn

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

12

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

Today

thinking

about

a

system

as

a

mathematical

function H (z)

rather

than

as

an

operator

13

7212019 Z Transform


1048573

1048573

ZTransform


minusn

H (z) =

h[n]z

n

Z

transform

maps

a

function

of

discrete

time n

to

a

function

of z

Although

motivated

by

system

functions

we

can

define

a

Z

trans

form forany signal

minusnX (z) =

x[n]z

n=minusinfin



properties)

14

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

the

unit-sample

signal

x[n] =

δ [n]

n

δ [n]


0 = 1

n=minusinfin

15

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

a

delayed

unit-sample

signal

n

x[n]

x[n

] =δ

[n

minus

1]

minusn

minus1


n=minusinfin

16

infin

Check Yo rself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

17

Check Yourself

7212019 Z Transform


1048573 1048573

CheckYourself

What

is

the

Z

transform

of

the

following

signal


x[n] =7

8

nu[n]

n

n7

7 1minusn


8 8 1minus

7 minus18

zn=minusinfin n=0

18

infin infin

Check Yourself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

19

Z Transform Pairs

7212019 Z Transform


1048573 1048573

ZTransformPairs



n

x[n] =

7

u[n] harr X (z) =

1

minus18 1

minus

7

8

z

For

what

values

of z

does X (z)

make

sense

The

Z

transform

is

only

defined

for

values

of

z

for

which

the

defining

sum converges

infin n


X (z) =

z u[n] =

z

=8 8 1minus

7

8

zminus1

n=minusinfin n=0

Therefore

1048608104860810486081048608 7

8z

minus1

1048608104860810486081048608 lt1 ie |z|gt

7

8

20

infin infin


7212019 Z Transform





n7 1 7


|z|gt

8 1

minus

7 88

zminus1

7

ROC

|z|gt

8

21


7212019 Z Transform


1048573

1048573

1048573 1048573


Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2


X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then


n=minusinfin


n=minusinfin

minusn+



=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R


Relations

among

representations

3

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R


Two

interpretations

of

ldquoDelayrdquo

4

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Rrarr 1

z


Relation

between

System

Functional

and

System

Function

5

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


z) =

Y (z)

X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

CheckYourself

What

is

relation

of

System

Functional

to

Unit-Sample

Response

6

7212019 Z Transform


CheckYourself

Expand

functional

in

a

series

Y 1

X

=

H(R)

=

1

minus

R

minus

R2


1 minusRminusR2 1

1 minusR minusR2

R +R2

R minusR2 minusR3

2R2 +R3


3R3 +2R4

3R3

minus3R4

minus3R5


H(R)

=

1

1

minus

R

minus

R2

=

1

+

R

+

2R2+

3R3+

5R4+

8R5+

13R6+

middot

middot

middot

7

7212019 Z Transform


1048573

CheckYourself


1

H(R) =

middot middot

1

minus R minus R2

= 1 +

R

+ 2R2+ 3R3+ 5R4+ 8R5+

13R6+

middot


h[n] 11

23

58

1321

3455

If

a

system

is

composed

of

(only)

adders

delays

and

gains

then


= h[n]Rn

n

We

can

write

the

system

function

in

terms

of

unit-sample

response

8

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Functional

to

Unit-Sample

Response

9

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

10

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

11

7212019 Z Transform


Block Diagram

System Functional



+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H (z) =

h[n]zminusn

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

12

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

Today

thinking

about

a

system

as

a

mathematical

function H (z)

rather

than

as

an

operator

13

7212019 Z Transform


1048573

1048573

ZTransform


minusn

H (z) =

h[n]z

n

Z

transform

maps

a

function

of

discrete

time n

to

a

function

of z

Although

motivated

by

system

functions

we

can

define

a

Z

trans

form forany signal

minusnX (z) =

x[n]z

n=minusinfin



properties)

14

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

the

unit-sample

signal

x[n] =

δ [n]

n

δ [n]


0 = 1

n=minusinfin

15

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

a

delayed

unit-sample

signal

n

x[n]

x[n

] =δ

[n

minus

1]

minusn

minus1


n=minusinfin

16

infin

Check Yo rself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

17

Check Yourself

7212019 Z Transform


1048573 1048573

CheckYourself

What

is

the

Z

transform

of

the

following

signal


x[n] =7

8

nu[n]

n

n7

7 1minusn


8 8 1minus

7 minus18

zn=minusinfin n=0

18

infin infin

Check Yourself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

19

Z Transform Pairs

7212019 Z Transform


1048573 1048573

ZTransformPairs



n

x[n] =

7

u[n] harr X (z) =

1

minus18 1

minus

7

8

z

For

what

values

of z

does X (z)

make

sense

The

Z

transform

is

only

defined

for

values

of

z

for

which

the

defining

sum converges

infin n


X (z) =

z u[n] =

z

=8 8 1minus

7

8

zminus1

n=minusinfin n=0

Therefore

1048608104860810486081048608 7

8z

minus1

1048608104860810486081048608 lt1 ie |z|gt

7

8

20

infin infin


7212019 Z Transform





n7 1 7


|z|gt

8 1

minus

7 88

zminus1

7

ROC

|z|gt

8

21


7212019 Z Transform


1048573

1048573

1048573 1048573


Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2


X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then


n=minusinfin


n=minusinfin

minusn+



=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R


Two

interpretations

of

ldquoDelayrdquo

4

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Rrarr 1

z


Relation

between

System

Functional

and

System

Function

5

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


z) =

Y (z)

X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

CheckYourself

What

is

relation

of

System

Functional

to

Unit-Sample

Response

6

7212019 Z Transform


CheckYourself

Expand

functional

in

a

series

Y 1

X

=

H(R)

=

1

minus

R

minus

R2


1 minusRminusR2 1

1 minusR minusR2

R +R2

R minusR2 minusR3

2R2 +R3


3R3 +2R4

3R3

minus3R4

minus3R5


H(R)

=

1

1

minus

R

minus

R2

=

1

+

R

+

2R2+

3R3+

5R4+

8R5+

13R6+

middot

middot

middot

7

7212019 Z Transform


1048573

CheckYourself


1

H(R) =

middot middot

1

minus R minus R2

= 1 +

R

+ 2R2+ 3R3+ 5R4+ 8R5+

13R6+

middot


h[n] 11

23

58

1321

3455

If

a

system

is

composed

of

(only)

adders

delays

and

gains

then


= h[n]Rn

n

We

can

write

the

system

function

in

terms

of

unit-sample

response

8

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Functional

to

Unit-Sample

Response

9

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

10

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

11

7212019 Z Transform


Block Diagram

System Functional



+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H (z) =

h[n]zminusn

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

12

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

Today

thinking

about

a

system

as

a

mathematical

function H (z)

rather

than

as

an

operator

13

7212019 Z Transform


1048573

1048573

ZTransform


minusn

H (z) =

h[n]z

n

Z

transform

maps

a

function

of

discrete

time n

to

a

function

of z

Although

motivated

by

system

functions

we

can

define

a

Z

trans

form forany signal

minusnX (z) =

x[n]z

n=minusinfin



properties)

14

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

the

unit-sample

signal

x[n] =

δ [n]

n

δ [n]


0 = 1

n=minusinfin

15

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

a

delayed

unit-sample

signal

n

x[n]

x[n

] =δ

[n

minus

1]

minusn

minus1


n=minusinfin

16

infin

Check Yo rself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

17

Check Yourself

7212019 Z Transform


1048573 1048573

CheckYourself

What

is

the

Z

transform

of

the

following

signal


x[n] =7

8

nu[n]

n

n7

7 1minusn


8 8 1minus

7 minus18

zn=minusinfin n=0

18

infin infin

Check Yourself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

19

Z Transform Pairs

7212019 Z Transform


1048573 1048573

ZTransformPairs



n

x[n] =

7

u[n] harr X (z) =

1

minus18 1

minus

7

8

z

For

what

values

of z

does X (z)

make

sense

The

Z

transform

is

only

defined

for

values

of

z

for

which

the

defining

sum converges

infin n


X (z) =

z u[n] =

z

=8 8 1minus

7

8

zminus1

n=minusinfin n=0

Therefore

1048608104860810486081048608 7

8z

minus1

1048608104860810486081048608 lt1 ie |z|gt

7

8

20

infin infin


7212019 Z Transform





n7 1 7


|z|gt

8 1

minus

7 88

zminus1

7

ROC

|z|gt

8

21


7212019 Z Transform


1048573

1048573

1048573 1048573


Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2


X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then


n=minusinfin


n=minusinfin

minusn+



=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Rrarr 1

z


Relation

between

System

Functional

and

System

Function

5

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


z) =

Y (z)

X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

CheckYourself

What

is

relation

of

System

Functional

to

Unit-Sample

Response

6

7212019 Z Transform


CheckYourself

Expand

functional

in

a

series

Y 1

X

=

H(R)

=

1

minus

R

minus

R2


1 minusRminusR2 1

1 minusR minusR2

R +R2

R minusR2 minusR3

2R2 +R3


3R3 +2R4

3R3

minus3R4

minus3R5


H(R)

=

1

1

minus

R

minus

R2

=

1

+

R

+

2R2+

3R3+

5R4+

8R5+

13R6+

middot

middot

middot

7

7212019 Z Transform


1048573

CheckYourself


1

H(R) =

middot middot

1

minus R minus R2

= 1 +

R

+ 2R2+ 3R3+ 5R4+ 8R5+

13R6+

middot


h[n] 11

23

58

1321

3455

If

a

system

is

composed

of

(only)

adders

delays

and

gains

then


= h[n]Rn

n

We

can

write

the

system

function

in

terms

of

unit-sample

response

8

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Functional

to

Unit-Sample

Response

9

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

10

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

11

7212019 Z Transform


Block Diagram

System Functional



+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H (z) =

h[n]zminusn

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

12

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

Today

thinking

about

a

system

as

a

mathematical

function H (z)

rather

than

as

an

operator

13

7212019 Z Transform


1048573

1048573

ZTransform


minusn

H (z) =

h[n]z

n

Z

transform

maps

a

function

of

discrete

time n

to

a

function

of z

Although

motivated

by

system

functions

we

can

define

a

Z

trans

form forany signal

minusnX (z) =

x[n]z

n=minusinfin



properties)

14

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

the

unit-sample

signal

x[n] =

δ [n]

n

δ [n]


0 = 1

n=minusinfin

15

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

a

delayed

unit-sample

signal

n

x[n]

x[n

] =δ

[n

minus

1]

minusn

minus1


n=minusinfin

16

infin

Check Yo rself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

17

Check Yourself

7212019 Z Transform


1048573 1048573

CheckYourself

What

is

the

Z

transform

of

the

following

signal


x[n] =7

8

nu[n]

n

n7

7 1minusn


8 8 1minus

7 minus18

zn=minusinfin n=0

18

infin infin

Check Yourself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

19

Z Transform Pairs

7212019 Z Transform


1048573 1048573

ZTransformPairs



n

x[n] =

7

u[n] harr X (z) =

1

minus18 1

minus

7

8

z

For

what

values

of z

does X (z)

make

sense

The

Z

transform

is

only

defined

for

values

of

z

for

which

the

defining

sum converges

infin n


X (z) =

z u[n] =

z

=8 8 1minus

7

8

zminus1

n=minusinfin n=0

Therefore

1048608104860810486081048608 7

8z

minus1

1048608104860810486081048608 lt1 ie |z|gt

7

8

20

infin infin


7212019 Z Transform





n7 1 7


|z|gt

8 1

minus

7 88

zminus1

7

ROC

|z|gt

8

21


7212019 Z Transform


1048573

1048573

1048573 1048573


Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2


X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then


n=minusinfin


n=minusinfin

minusn+



=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


z) =

Y (z)

X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

CheckYourself

What

is

relation

of

System

Functional

to

Unit-Sample

Response

6

7212019 Z Transform


CheckYourself

Expand

functional

in

a

series

Y 1

X

=

H(R)

=

1

minus

R

minus

R2


1 minusRminusR2 1

1 minusR minusR2

R +R2

R minusR2 minusR3

2R2 +R3


3R3 +2R4

3R3

minus3R4

minus3R5


H(R)

=

1

1

minus

R

minus

R2

=

1

+

R

+

2R2+

3R3+

5R4+

8R5+

13R6+

middot

middot

middot

7

7212019 Z Transform


1048573

CheckYourself


1

H(R) =

middot middot

1

minus R minus R2

= 1 +

R

+ 2R2+ 3R3+ 5R4+ 8R5+

13R6+

middot


h[n] 11

23

58

1321

3455

If

a

system

is

composed

of

(only)

adders

delays

and

gains

then


= h[n]Rn

n

We

can

write

the

system

function

in

terms

of

unit-sample

response

8

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Functional

to

Unit-Sample

Response

9

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

10

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

11

7212019 Z Transform


Block Diagram

System Functional



+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H (z) =

h[n]zminusn

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

12

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

Today

thinking

about

a

system

as

a

mathematical

function H (z)

rather

than

as

an

operator

13

7212019 Z Transform


1048573

1048573

ZTransform


minusn

H (z) =

h[n]z

n

Z

transform

maps

a

function

of

discrete

time n

to

a

function

of z

Although

motivated

by

system

functions

we

can

define

a

Z

trans

form forany signal

minusnX (z) =

x[n]z

n=minusinfin



properties)

14

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

the

unit-sample

signal

x[n] =

δ [n]

n

δ [n]


0 = 1

n=minusinfin

15

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

a

delayed

unit-sample

signal

n

x[n]

x[n

] =δ

[n

minus

1]

minusn

minus1


n=minusinfin

16

infin

Check Yo rself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

17

Check Yourself

7212019 Z Transform


1048573 1048573

CheckYourself

What

is

the

Z

transform

of

the

following

signal


x[n] =7

8

nu[n]

n

n7

7 1minusn


8 8 1minus

7 minus18

zn=minusinfin n=0

18

infin infin

Check Yourself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

19

Z Transform Pairs

7212019 Z Transform


1048573 1048573

ZTransformPairs



n

x[n] =

7

u[n] harr X (z) =

1

minus18 1

minus

7

8

z

For

what

values

of z

does X (z)

make

sense

The

Z

transform

is

only

defined

for

values

of

z

for

which

the

defining

sum converges

infin n


X (z) =

z u[n] =

z

=8 8 1minus

7

8

zminus1

n=minusinfin n=0

Therefore

1048608104860810486081048608 7

8z

minus1

1048608104860810486081048608 lt1 ie |z|gt

7

8

20

infin infin


7212019 Z Transform





n7 1 7


|z|gt

8 1

minus

7 88

zminus1

7

ROC

|z|gt

8

21


7212019 Z Transform


1048573

1048573

1048573 1048573


Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2


X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then


n=minusinfin


n=minusinfin

minusn+



=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


CheckYourself

Expand

functional

in

a

series

Y 1

X

=

H(R)

=

1

minus

R

minus

R2


1 minusRminusR2 1

1 minusR minusR2

R +R2

R minusR2 minusR3

2R2 +R3


3R3 +2R4

3R3

minus3R4

minus3R5


H(R)

=

1

1

minus

R

minus

R2

=

1

+

R

+

2R2+

3R3+

5R4+

8R5+

13R6+

middot

middot

middot

7

7212019 Z Transform


1048573

CheckYourself


1

H(R) =

middot middot

1

minus R minus R2

= 1 +

R

+ 2R2+ 3R3+ 5R4+ 8R5+

13R6+

middot


h[n] 11

23

58

1321

3455

If

a

system

is

composed

of

(only)

adders

delays

and

gains

then


= h[n]Rn

n

We

can

write

the

system

function

in

terms

of

unit-sample

response

8

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Functional

to

Unit-Sample

Response

9

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

10

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

11

7212019 Z Transform


Block Diagram

System Functional



+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H (z) =

h[n]zminusn

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

12

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

Today

thinking

about

a

system

as

a

mathematical

function H (z)

rather

than

as

an

operator

13

7212019 Z Transform


1048573

1048573

ZTransform


minusn

H (z) =

h[n]z

n

Z

transform

maps

a

function

of

discrete

time n

to

a

function

of z

Although

motivated

by

system

functions

we

can

define

a

Z

trans

form forany signal

minusnX (z) =

x[n]z

n=minusinfin



properties)

14

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

the

unit-sample

signal

x[n] =

δ [n]

n

δ [n]


0 = 1

n=minusinfin

15

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

a

delayed

unit-sample

signal

n

x[n]

x[n

] =δ

[n

minus

1]

minusn

minus1


n=minusinfin

16

infin

Check Yo rself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

17

Check Yourself

7212019 Z Transform


1048573 1048573

CheckYourself

What

is

the

Z

transform

of

the

following

signal


x[n] =7

8

nu[n]

n

n7

7 1minusn


8 8 1minus

7 minus18

zn=minusinfin n=0

18

infin infin

Check Yourself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

19

Z Transform Pairs

7212019 Z Transform


1048573 1048573

ZTransformPairs



n

x[n] =

7

u[n] harr X (z) =

1

minus18 1

minus

7

8

z

For

what

values

of z

does X (z)

make

sense

The

Z

transform

is

only

defined

for

values

of

z

for

which

the

defining

sum converges

infin n


X (z) =

z u[n] =

z

=8 8 1minus

7

8

zminus1

n=minusinfin n=0

Therefore

1048608104860810486081048608 7

8z

minus1

1048608104860810486081048608 lt1 ie |z|gt

7

8

20

infin infin


7212019 Z Transform





n7 1 7


|z|gt

8 1

minus

7 88

zminus1

7

ROC

|z|gt

8

21


7212019 Z Transform


1048573

1048573

1048573 1048573


Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2


X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then


n=minusinfin


n=minusinfin

minusn+



=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


1048573

CheckYourself


1

H(R) =

middot middot

1

minus R minus R2

= 1 +

R

+ 2R2+ 3R3+ 5R4+ 8R5+

13R6+

middot


h[n] 11

23

58

1321

3455

If

a

system

is

composed

of

(only)

adders

delays

and

gains

then


= h[n]Rn

n

We

can

write

the

system

function

in

terms

of

unit-sample

response

8

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Functional

to

Unit-Sample

Response

9

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

10

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

11

7212019 Z Transform


Block Diagram

System Functional



+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H (z) =

h[n]zminusn

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

12

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

Today

thinking

about

a

system

as

a

mathematical

function H (z)

rather

than

as

an

operator

13

7212019 Z Transform


1048573

1048573

ZTransform


minusn

H (z) =

h[n]z

n

Z

transform

maps

a

function

of

discrete

time n

to

a

function

of z

Although

motivated

by

system

functions

we

can

define

a

Z

trans

form forany signal

minusnX (z) =

x[n]z

n=minusinfin



properties)

14

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

the

unit-sample

signal

x[n] =

δ [n]

n

δ [n]


0 = 1

n=minusinfin

15

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

a

delayed

unit-sample

signal

n

x[n]

x[n

] =δ

[n

minus

1]

minusn

minus1


n=minusinfin

16

infin

Check Yo rself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

17

Check Yourself

7212019 Z Transform


1048573 1048573

CheckYourself

What

is

the

Z

transform

of

the

following

signal


x[n] =7

8

nu[n]

n

n7

7 1minusn


8 8 1minus

7 minus18

zn=minusinfin n=0

18

infin infin

Check Yourself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

19

Z Transform Pairs

7212019 Z Transform


1048573 1048573

ZTransformPairs



n

x[n] =

7

u[n] harr X (z) =

1

minus18 1

minus

7

8

z

For

what

values

of z

does X (z)

make

sense

The

Z

transform

is

only

defined

for

values

of

z

for

which

the

defining

sum converges

infin n


X (z) =

z u[n] =

z

=8 8 1minus

7

8

zminus1

n=minusinfin n=0

Therefore

1048608104860810486081048608 7

8z

minus1

1048608104860810486081048608 lt1 ie |z|gt

7

8

20

infin infin


7212019 Z Transform





n7 1 7


|z|gt

8 1

minus

7 88

zminus1

7

ROC

|z|gt

8

21


7212019 Z Transform


1048573

1048573

1048573 1048573


Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2


X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then


n=minusinfin


n=minusinfin

minusn+



=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Functional

to

Unit-Sample

Response

9

7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

10

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

11

7212019 Z Transform


Block Diagram

System Functional



+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H (z) =

h[n]zminusn

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

12

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

Today

thinking

about

a

system

as

a

mathematical

function H (z)

rather

than

as

an

operator

13

7212019 Z Transform


1048573

1048573

ZTransform


minusn

H (z) =

h[n]z

n

Z

transform

maps

a

function

of

discrete

time n

to

a

function

of z

Although

motivated

by

system

functions

we

can

define

a

Z

trans

form forany signal

minusnX (z) =

x[n]z

n=minusinfin



properties)

14

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

the

unit-sample

signal

x[n] =

δ [n]

n

δ [n]


0 = 1

n=minusinfin

15

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

a

delayed

unit-sample

signal

n

x[n]

x[n

] =δ

[n

minus

1]

minusn

minus1


n=minusinfin

16

infin

Check Yo rself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

17

Check Yourself

7212019 Z Transform


1048573 1048573

CheckYourself

What

is

the

Z

transform

of

the

following

signal


x[n] =7

8

nu[n]

n

n7

7 1minusn


8 8 1minus

7 minus18

zn=minusinfin n=0

18

infin infin

Check Yourself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

19

Z Transform Pairs

7212019 Z Transform


1048573 1048573

ZTransformPairs



n

x[n] =

7

u[n] harr X (z) =

1

minus18 1

minus

7

8

z

For

what

values

of z

does X (z)

make

sense

The

Z

transform

is

only

defined

for

values

of

z

for

which

the

defining

sum converges

infin n


X (z) =

z u[n] =

z

=8 8 1minus

7

8

zminus1

n=minusinfin n=0

Therefore

1048608104860810486081048608 7

8z

minus1

1048608104860810486081048608 lt1 ie |z|gt

7

8

20

infin infin


7212019 Z Transform





n7 1 7


|z|gt

8 1

minus

7 88

zminus1

7

ROC

|z|gt

8

21


7212019 Z Transform


1048573

1048573

1048573 1048573


Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2


X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then


n=minusinfin


n=minusinfin

minusn+



=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

10

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

11

7212019 Z Transform


Block Diagram

System Functional



+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H (z) =

h[n]zminusn

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

12

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

Today

thinking

about

a

system

as

a

mathematical

function H (z)

rather

than

as

an

operator

13

7212019 Z Transform


1048573

1048573

ZTransform


minusn

H (z) =

h[n]z

n

Z

transform

maps

a

function

of

discrete

time n

to

a

function

of z

Although

motivated

by

system

functions

we

can

define

a

Z

trans

form forany signal

minusnX (z) =

x[n]z

n=minusinfin



properties)

14

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

the

unit-sample

signal

x[n] =

δ [n]

n

δ [n]


0 = 1

n=minusinfin

15

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

a

delayed

unit-sample

signal

n

x[n]

x[n

] =δ

[n

minus

1]

minusn

minus1


n=minusinfin

16

infin

Check Yo rself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

17

Check Yourself

7212019 Z Transform


1048573 1048573

CheckYourself

What

is

the

Z

transform

of

the

following

signal


x[n] =7

8

nu[n]

n

n7

7 1minusn


8 8 1minus

7 minus18

zn=minusinfin n=0

18

infin infin

Check Yourself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

19

Z Transform Pairs

7212019 Z Transform


1048573 1048573

ZTransformPairs



n

x[n] =

7

u[n] harr X (z) =

1

minus18 1

minus

7

8

z

For

what

values

of z

does X (z)

make

sense

The

Z

transform

is

only

defined

for

values

of

z

for

which

the

defining

sum converges

infin n


X (z) =

z u[n] =

z

=8 8 1minus

7

8

zminus1

n=minusinfin n=0

Therefore

1048608104860810486081048608 7

8z

minus1

1048608104860810486081048608 lt1 ie |z|gt

7

8

20

infin infin


7212019 Z Transform





n7 1 7


|z|gt

8 1

minus

7 88

zminus1

7

ROC

|z|gt

8

21


7212019 Z Transform


1048573

1048573

1048573 1048573


Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2


X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then


n=minusinfin


n=minusinfin

minusn+



=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

11

7212019 Z Transform


Block Diagram

System Functional



+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H (z) =

h[n]zminusn

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

12

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

Today

thinking

about

a

system

as

a

mathematical

function H (z)

rather

than

as

an

operator

13

7212019 Z Transform


1048573

1048573

ZTransform


minusn

H (z) =

h[n]z

n

Z

transform

maps

a

function

of

discrete

time n

to

a

function

of z

Although

motivated

by

system

functions

we

can

define

a

Z

trans

form forany signal

minusnX (z) =

x[n]z

n=minusinfin



properties)

14

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

the

unit-sample

signal

x[n] =

δ [n]

n

δ [n]


0 = 1

n=minusinfin

15

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

a

delayed

unit-sample

signal

n

x[n]

x[n

] =δ

[n

minus

1]

minusn

minus1


n=minusinfin

16

infin

Check Yo rself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

17

Check Yourself

7212019 Z Transform


1048573 1048573

CheckYourself

What

is

the

Z

transform

of

the

following

signal


x[n] =7

8

nu[n]

n

n7

7 1minusn


8 8 1minus

7 minus18

zn=minusinfin n=0

18

infin infin

Check Yourself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

19

Z Transform Pairs

7212019 Z Transform


1048573 1048573

ZTransformPairs



n

x[n] =

7

u[n] harr X (z) =

1

minus18 1

minus

7

8

z

For

what

values

of z

does X (z)

make

sense

The

Z

transform

is

only

defined

for

values

of

z

for

which

the

defining

sum converges

infin n


X (z) =

z u[n] =

z

=8 8 1minus

7

8

zminus1

n=minusinfin n=0

Therefore

1048608104860810486081048608 7

8z

minus1

1048608104860810486081048608 lt1 ie |z|gt

7

8

20

infin infin


7212019 Z Transform





n7 1 7


|z|gt

8 1

minus

7 88

zminus1

7

ROC

|z|gt

8

21


7212019 Z Transform


1048573

1048573

1048573 1048573


Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2


X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then


n=minusinfin


n=minusinfin

minusn+



=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


Block Diagram

System Functional



+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z)

= z2

z2

minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

H (z) =

h[n]zminusn

H(R) =

h[n]Rn

CheckYourself

What

is

relation

of

System

Function

to

Unit-Sample

Response

12

7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

Today

thinking

about

a

system

as

a

mathematical

function H (z)

rather

than

as

an

operator

13

7212019 Z Transform


1048573

1048573

ZTransform


minusn

H (z) =

h[n]z

n

Z

transform

maps

a

function

of

discrete

time n

to

a

function

of z

Although

motivated

by

system

functions

we

can

define

a

Z

trans

form forany signal

minusnX (z) =

x[n]z

n=minusinfin



properties)

14

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

the

unit-sample

signal

x[n] =

δ [n]

n

δ [n]


0 = 1

n=minusinfin

15

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

a

delayed

unit-sample

signal

n

x[n]

x[n

] =δ

[n

minus

1]

minusn

minus1


n=minusinfin

16

infin

Check Yo rself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

17

Check Yourself

7212019 Z Transform


1048573 1048573

CheckYourself

What

is

the

Z

transform

of

the

following

signal


x[n] =7

8

nu[n]

n

n7

7 1minusn


8 8 1minus

7 minus18

zn=minusinfin n=0

18

infin infin

Check Yourself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

19

Z Transform Pairs

7212019 Z Transform


1048573 1048573

ZTransformPairs



n

x[n] =

7

u[n] harr X (z) =

1

minus18 1

minus

7

8

z

For

what

values

of z

does X (z)

make

sense

The

Z

transform

is

only

defined

for

values

of

z

for

which

the

defining

sum converges

infin n


X (z) =

z u[n] =

z

=8 8 1minus

7

8

zminus1

n=minusinfin n=0

Therefore

1048608104860810486081048608 7

8z

minus1

1048608104860810486081048608 lt1 ie |z|gt

7

8

20

infin infin


7212019 Z Transform





n7 1 7


|z|gt

8 1

minus

7 88

zminus1

7

ROC

|z|gt

8

21


7212019 Z Transform


1048573

1048573

1048573 1048573


Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2


X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then


n=minusinfin


n=minusinfin

minusn+



=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


1048573

1048573

CheckYourself

Start

with

the

series

expansion

of

system

functional

H(R) =

h[n]Rn

n

1

Substitute R rarr z

minusnH (z) = h[n]z

n

Today

thinking

about

a

system

as

a

mathematical

function H (z)

rather

than

as

an

operator

13

7212019 Z Transform


1048573

1048573

ZTransform


minusn

H (z) =

h[n]z

n

Z

transform

maps

a

function

of

discrete

time n

to

a

function

of z

Although

motivated

by

system

functions

we

can

define

a

Z

trans

form forany signal

minusnX (z) =

x[n]z

n=minusinfin



properties)

14

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

the

unit-sample

signal

x[n] =

δ [n]

n

δ [n]


0 = 1

n=minusinfin

15

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

a

delayed

unit-sample

signal

n

x[n]

x[n

] =δ

[n

minus

1]

minusn

minus1


n=minusinfin

16

infin

Check Yo rself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

17

Check Yourself

7212019 Z Transform


1048573 1048573

CheckYourself

What

is

the

Z

transform

of

the

following

signal


x[n] =7

8

nu[n]

n

n7

7 1minusn


8 8 1minus

7 minus18

zn=minusinfin n=0

18

infin infin

Check Yourself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

19

Z Transform Pairs

7212019 Z Transform


1048573 1048573

ZTransformPairs



n

x[n] =

7

u[n] harr X (z) =

1

minus18 1

minus

7

8

z

For

what

values

of z

does X (z)

make

sense

The

Z

transform

is

only

defined

for

values

of

z

for

which

the

defining

sum converges

infin n


X (z) =

z u[n] =

z

=8 8 1minus

7

8

zminus1

n=minusinfin n=0

Therefore

1048608104860810486081048608 7

8z

minus1

1048608104860810486081048608 lt1 ie |z|gt

7

8

20

infin infin


7212019 Z Transform





n7 1 7


|z|gt

8 1

minus

7 88

zminus1

7

ROC

|z|gt

8

21


7212019 Z Transform


1048573

1048573

1048573 1048573


Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2


X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then


n=minusinfin


n=minusinfin

minusn+



=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


1048573

1048573

ZTransform


minusn

H (z) =

h[n]z

n

Z

transform

maps

a

function

of

discrete

time n

to

a

function

of z

Although

motivated

by

system

functions

we

can

define

a

Z

trans

form forany signal

minusnX (z) =

x[n]z

n=minusinfin



properties)

14

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

the

unit-sample

signal

x[n] =

δ [n]

n

δ [n]


0 = 1

n=minusinfin

15

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

a

delayed

unit-sample

signal

n

x[n]

x[n

] =δ

[n

minus

1]

minusn

minus1


n=minusinfin

16

infin

Check Yo rself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

17

Check Yourself

7212019 Z Transform


1048573 1048573

CheckYourself

What

is

the

Z

transform

of

the

following

signal


x[n] =7

8

nu[n]

n

n7

7 1minusn


8 8 1minus

7 minus18

zn=minusinfin n=0

18

infin infin

Check Yourself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

19

Z Transform Pairs

7212019 Z Transform


1048573 1048573

ZTransformPairs



n

x[n] =

7

u[n] harr X (z) =

1

minus18 1

minus

7

8

z

For

what

values

of z

does X (z)

make

sense

The

Z

transform

is

only

defined

for

values

of

z

for

which

the

defining

sum converges

infin n


X (z) =

z u[n] =

z

=8 8 1minus

7

8

zminus1

n=minusinfin n=0

Therefore

1048608104860810486081048608 7

8z

minus1

1048608104860810486081048608 lt1 ie |z|gt

7

8

20

infin infin


7212019 Z Transform





n7 1 7


|z|gt

8 1

minus

7 88

zminus1

7

ROC

|z|gt

8

21


7212019 Z Transform


1048573

1048573

1048573 1048573


Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2


X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then


n=minusinfin


n=minusinfin

minusn+



=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

the

unit-sample

signal

x[n] =

δ [n]

n

δ [n]


0 = 1

n=minusinfin

15

infin

Si l Z t f

7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

a

delayed

unit-sample

signal

n

x[n]

x[n

] =δ

[n

minus

1]

minusn

minus1


n=minusinfin

16

infin

Check Yo rself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

17

Check Yourself

7212019 Z Transform


1048573 1048573

CheckYourself

What

is

the

Z

transform

of

the

following

signal


x[n] =7

8

nu[n]

n

n7

7 1minusn


8 8 1minus

7 minus18

zn=minusinfin n=0

18

infin infin

Check Yourself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

19

Z Transform Pairs

7212019 Z Transform


1048573 1048573

ZTransformPairs



n

x[n] =

7

u[n] harr X (z) =

1

minus18 1

minus

7

8

z

For

what

values

of z

does X (z)

make

sense

The

Z

transform

is

only

defined

for

values

of

z

for

which

the

defining

sum converges

infin n


X (z) =

z u[n] =

z

=8 8 1minus

7

8

zminus1

n=minusinfin n=0

Therefore

1048608104860810486081048608 7

8z

minus1

1048608104860810486081048608 lt1 ie |z|gt

7

8

20

infin infin


7212019 Z Transform





n7 1 7


|z|gt

8 1

minus

7 88

zminus1

7

ROC

|z|gt

8

21


7212019 Z Transform


1048573

1048573

1048573 1048573


Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2


X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then


n=minusinfin


n=minusinfin

minusn+



=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


1048573

SimpleZ transforms

Find

the

Z

transform

of

a

delayed

unit-sample

signal

n

x[n]

x[n

] =δ

[n

minus

1]

minusn

minus1


n=minusinfin

16

infin

Check Yo rself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

17

Check Yourself

7212019 Z Transform


1048573 1048573

CheckYourself

What

is

the

Z

transform

of

the

following

signal


x[n] =7

8

nu[n]

n

n7

7 1minusn


8 8 1minus

7 minus18

zn=minusinfin n=0

18

infin infin

Check Yourself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

19

Z Transform Pairs

7212019 Z Transform


1048573 1048573

ZTransformPairs



n

x[n] =

7

u[n] harr X (z) =

1

minus18 1

minus

7

8

z

For

what

values

of z

does X (z)

make

sense

The

Z

transform

is

only

defined

for

values

of

z

for

which

the

defining

sum converges

infin n


X (z) =

z u[n] =

z

=8 8 1minus

7

8

zminus1

n=minusinfin n=0

Therefore

1048608104860810486081048608 7

8z

minus1

1048608104860810486081048608 lt1 ie |z|gt

7

8

20

infin infin


7212019 Z Transform





n7 1 7


|z|gt

8 1

minus

7 88

zminus1

7

ROC

|z|gt

8

21


7212019 Z Transform


1048573

1048573

1048573 1048573


Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2


X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then


n=minusinfin


n=minusinfin

minusn+



=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

17

Check Yourself

7212019 Z Transform


1048573 1048573

CheckYourself

What

is

the

Z

transform

of

the

following

signal


x[n] =7

8

nu[n]

n

n7

7 1minusn


8 8 1minus

7 minus18

zn=minusinfin n=0

18

infin infin

Check Yourself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

19

Z Transform Pairs

7212019 Z Transform


1048573 1048573

ZTransformPairs



n

x[n] =

7

u[n] harr X (z) =

1

minus18 1

minus

7

8

z

For

what

values

of z

does X (z)

make

sense

The

Z

transform

is

only

defined

for

values

of

z

for

which

the

defining

sum converges

infin n


X (z) =

z u[n] =

z

=8 8 1minus

7

8

zminus1

n=minusinfin n=0

Therefore

1048608104860810486081048608 7

8z

minus1

1048608104860810486081048608 lt1 ie |z|gt

7

8

20

infin infin


7212019 Z Transform





n7 1 7


|z|gt

8 1

minus

7 88

zminus1

7

ROC

|z|gt

8

21


7212019 Z Transform


1048573

1048573

1048573 1048573


Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2


X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then


n=minusinfin


n=minusinfin

minusn+



=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


1048573 1048573

CheckYourself

What

is

the

Z

transform

of

the

following

signal


x[n] =7

8

nu[n]

n

n7

7 1minusn


8 8 1minus

7 minus18

zn=minusinfin n=0

18

infin infin

Check Yourself

7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

19

Z Transform Pairs

7212019 Z Transform


1048573 1048573

ZTransformPairs



n

x[n] =

7

u[n] harr X (z) =

1

minus18 1

minus

7

8

z

For

what

values

of z

does X (z)

make

sense

The

Z

transform

is

only

defined

for

values

of

z

for

which

the

defining

sum converges

infin n


X (z) =

z u[n] =

z

=8 8 1minus

7

8

zminus1

n=minusinfin n=0

Therefore

1048608104860810486081048608 7

8z

minus1

1048608104860810486081048608 lt1 ie |z|gt

7

8

20

infin infin


7212019 Z Transform





n7 1 7


|z|gt

8 1

minus

7 88

zminus1

7

ROC

|z|gt

8

21


7212019 Z Transform


1048573

1048573

1048573 1048573


Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2


X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then


n=minusinfin


n=minusinfin

minusn+



=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


CheckYourself



x[n] =7

8

nu[n]

1

1

1

minus

78

z

2

1

1

minus

78

zminus1

3

z

1

minus

78

z

4

zminus1

1

minus

78

zminus1

5

none

19

Z Transform Pairs

7212019 Z Transform


1048573 1048573

ZTransformPairs



n

x[n] =

7

u[n] harr X (z) =

1

minus18 1

minus

7

8

z

For

what

values

of z

does X (z)

make

sense

The

Z

transform

is

only

defined

for

values

of

z

for

which

the

defining

sum converges

infin n


X (z) =

z u[n] =

z

=8 8 1minus

7

8

zminus1

n=minusinfin n=0

Therefore

1048608104860810486081048608 7

8z

minus1

1048608104860810486081048608 lt1 ie |z|gt

7

8

20

infin infin


7212019 Z Transform





n7 1 7


|z|gt

8 1

minus

7 88

zminus1

7

ROC

|z|gt

8

21


7212019 Z Transform


1048573

1048573

1048573 1048573


Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2


X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then


n=minusinfin


n=minusinfin

minusn+



=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


1048573 1048573

ZTransformPairs



n

x[n] =

7

u[n] harr X (z) =

1

minus18 1

minus

7

8

z

For

what

values

of z

does X (z)

make

sense

The

Z

transform

is

only

defined

for

values

of

z

for

which

the

defining

sum converges

infin n


X (z) =

z u[n] =

z

=8 8 1minus

7

8

zminus1

n=minusinfin n=0

Therefore

1048608104860810486081048608 7

8z

minus1

1048608104860810486081048608 lt1 ie |z|gt

7

8

20

infin infin


7212019 Z Transform





n7 1 7


|z|gt

8 1

minus

7 88

zminus1

7

ROC

|z|gt

8

21


7212019 Z Transform


1048573

1048573

1048573 1048573


Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2


X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then


n=minusinfin


n=minusinfin

minusn+



=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform





n7 1 7


|z|gt

8 1

minus

7 88

zminus1

7

ROC

|z|gt

8

21


7212019 Z Transform


1048573

1048573

1048573 1048573


Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2


X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then


n=minusinfin


n=minusinfin

minusn+



=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


1048573

1048573

1048573 1048573


Based

on

properties

of

the

Z

transform

Linearity

if

x1[n]

harr

X 1(z)

for

z

in

ROC1

and x2[n] harr

X 2(z)

for z

in

ROC2


X 1(z) +X 2(z)

for z

in (ROC1

cap

ROC2)

Let

y[n] =

x1[n] +

x2[n]

then


n=minusinfin


n=minusinfin

minusn+



=

X 1(z) +

X 2(z)

22

infin

infin

infin infin

Delay Property

7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


1048573

1048573 1048573

1048573

DelayProperty

If

x[n]harr

X (z)

for

z

in

ROC

then

x[n

minus1]harr

zminus1X (z)

for

z

in

ROC


δ [n]

harr

1

δ [n

minus1] harr

z

minus1

More

generally

minusn X (z) =

x[n]z

n=minusinfin


infin


minus1]z

n=minusinfin

n=minusinfin

Substitute m=n

minus1

Y (z) = x[m]z

minusmminus1 =z

minus1X (z)

m=minusinfin

23

infin

infin

infin


7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


RationalPolynomials

A

system

that

can

be

described

by

a

linear

difference

equation

with


a

ratio

of

polynomials

in

z


Taking

the

Z

transform

of

both

sides

and

applying

the

delay

property

b0Y

(z)+b1z

minus1Y

(z)+b2z

minus2Y


=

a0X (z)+a1z

minus1X (z)+a2z


a0



H (z) =

Y (z)

=

X (z)

b0

+

b1z

minus1

+

b2z

minus2

+





24


7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


RationalPolynomials

Applying

the

fundamental

theorem

of

algebra

and

the

factor

theo


k+

a1zkminus1+

a2zkminus2+


H (z) =

a0z




=


where

the

roots

are

calledpoles

andzeros

25

RationalPolynomials

7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


1048573 1048573

y

Regions

of

convergence

for

Z

transform

are

delimited

by

circles

in


Example

x[n] =

αnu[n]


n=minusinfin n=0

1

=

αzminus1

lt

11

minusαzminus1

=

z

zminusα

|z|gt

|α|


x[n] = αnu[n]

α

z-planeROC

z

z minus α

26

infin infin

CheckYourself

7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform



78

z-plane

ROCz

z minus 7

8

|z| lt 7

8

27

CheckYourself

7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


1048573

Recall

that

we

already

know

a

function

whose

Z

transform

is

the

outer region


x[n] =7

8

nu[n]

7

8

z-planeROC

z

z minus 7

8

What

changes

if

the

region

changes

The

original

sum

7

n

minusnX (z) =

z8

n=0


8

28

infin

CheckYourself

7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


The

functional

form

is

still

the

same

z

H (z) =

Y (z)

=

X (z)

z

minus

78

Therefore

the

difference

equation

for

this

system

is

the

same

y[n+1]minus

8

7

y[n] =x[n+1]

Convergence

inside

|z|

=

response

Solve

by

iterating

backwards

in

time

7

8 corresponds

to

a

left-sided

(non-causal)

8


7

29

CheckYourself

7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


Solve

by

iterating

backwards

in

time

8


7

Start

ldquoat

restrdquo

n x[n] y[n]

gt0 0 0

0 1 0 983072 minus1 0

minus

9830722

8

7

8

7

8

7


minus3 0 minus


n

minus 8

7983072minusn

8

minusn 7

n

y[n] =

minus n lt0 =

minus u[minus1

minusn]

7 830

CheckYourself

7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


Plot


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

31

CheckYourself

7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


What

DT

signal

has

the

following

Z

transform

7

8

z-plane

ROCz

z minus 7

8

|z| lt 7

8


y[n] = minus7

8

nu[minus1 minus n]

32

CheckYourself

7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


Two

signals

and

two

regions

of

convergence


x[n] =7

8

n

u[n]

7

8

z-planeROC

z

z minus 7

8


n

y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

33

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


Find

the

inverse

transform

of

X (z)=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

34

CheckYourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2

minus

5z

+ 2

2z

minus

1

minus

z

minus

2

Not

a

standard

form

35

Check

Yourself

7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


Standard

forms


x[n] = 7

8n

u[n]

7

8

z-planeROC

zz minus 7

8


y[n] = minus7

8

nu[minus1 minus n]

7

8

ROC

z-plane

z

z minus 7

8

36

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


Find

the

inverse

transform

of

minus3z

X (z) =

2z2minus5z+ 2

given

that

the

ROC

includes

the

unit

circle

Expand

with

partial

fractions

minus3z 1 2

X (z) = =

2z2minus5z

+ 2

2zminus1

minus

zminus2

Not

a

standard

form

Expand

it

differently

as

a

standard

form

minus3z 2z z z z

X (z) = = =

minus2z2minus

5z

+ 2

2z

minus1

minus

z

minus2 zminus

1

2

zminus2

Standard

form

a

pole

at

1

2

andapoleat 2

37

Check

Yourself

7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


z

minus

and

a

pole

at 2

1

22

z-plane

ROC

Region

of

convergence

is

ldquooutsiderdquo

pole

at 1

2

Ratio

of

polynomials

in

z

minus3z z z

= minus X (z) =

2minus5z

+ 2

1

22z

zminus2

1

2ndash

a

pole

at

n

u[n] + 2nu[minus1

minusn]

but

ldquoinsiderdquo

pole

at 2

1

x[n] =

2

38

Check

Yourself

7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


Plot

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

39

Check

Yourself

7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


Alternatively

stick

with

non-standard

form

minus3z 1 2

X (z) = =

2z2minus5z+ 2

2z

minus1

minus

z

minus2

Make

it

look

more

standard

1

minus1

z

minus1

z

X (z) =

minus2z

2z

zminus

1

2

z

minus2

Now

n 1 1

x[n] =

2 R u[n] + 2R +2n

u[minus1

minus

n]2 nminus1 1 1

=2 2


n

n

x[n]

1

= u[n

minus

1] +

+2

n

u[minusn]2

40

Check

Yourself

7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


Alternative

3

expand

as

polynomials

in

zminus1

minus1minus3z

minus3zX (z) =

=


minus5zminus1+ 2z

2 1

1 1= =

2

minuszminus1

minus

1

minus2zminus1 1

minus

1

2

zminus1

minus

1

minus2zminus1

Now

n1

x[n] =

2

u[n]+2nu[minus1

minusn]

n

x[n]

41

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


Find

the

inverse

transform

of

X (z)

=

minus3z

2z2minus5z+2

given

that

the

ROC

includes

the

unit

circle

n

x[n]

42

Solving

Difference

Equations

with

Z

Transforms

7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


Start

with

difference

equation

y[n]minus

2

1

y[n

minus1]=δ [n]

Take

the

Z

transform

of

this

equation

Y (z)

minus

1

2z

minus1Y (z) = 1

Solve for Y (z)1

Y (z) =

1

minus

12

zminus1

Take

the

inverse

Z

transform

(by

recognizing

the

form

of

the

trans

form)n1

y[n] =

u[n]

2

43

Inverse

Z

transform

7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


The

inverse

Z

transform

is

defined

by

an

integral

that

is

not

partic


Formally

x[n] =

1

X (z)z

nminus1dz2πj C

whereC represents

a

closed

contour

that

circles

the

origin

by

running

in

a

counterclockwise

direction

through

the

region

of

convergence

This

integral

is

not

generally

easy

to

compute


There

are

better

ways

(eg

partial

fractions)

to

compute

inverse

transforms

for

the

kinds

of

systems

that

we

frequently

encounter

44

Properties

of

Z

Transforms

7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

45

infin

Check

Yourself

7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


Find

the

inverse

transform

of Y (z)=

z

z

minus

1

2

|z|gt1

46

Check

Yourself

2

7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


1048573

2

z

Find

the

inverse

transform

of Y (z) = |z|

gt1z

minus1

y[n]

corresponds

to

unit-sample

response

of

the

right-sided

system

2 2 2Y z 1 1

= = =

minus1X z

minus1 1

minusz 1

minus R


1

R

R2

R3

middot

middot

middot

1

R

R2

R3

1

R

R2

R3

R

R2

R3

R4

R2

R3

R4

R5

R3

R4

R5

R6

middotmiddotmiddot

middotmiddotmiddot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

middot

infin

Y


X

n=0

y[n] =

h[n] = (n

+

1)u[n]

47

Check

Yourself

T bl l k th d

7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


Table

lookup

method

2

z

Y (z) =

harr

y[n]=

z

minus1

z

harr

u[n]

z

minus1

48

Properties

of

Z

Transforms


7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


1048573

The

use

of

Z

Transforms

to

solve

differential

equations

depends

on


Property

x[n]

X (z)

ROC


capR2)

Delay x[nminus1] z

minus1X (z) R

dX (z)Multiply

by

n

nx[n]

minusz dz

R

Convolve

in n x1[m]x2[n


sup(R1

capR2)

m=minusinfin

49

infin

Check

Yourself

Table look p method

7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


Table

lookup

method

Y (z)=

z

z

minus

1

2

harr

y[n]=

z

zminus1

harr u[n]

minusz

d

dz

z

z

minus1

=

z

1

z

minus1

2

harr nu[n]

z

times

minusz

d

dz

z

z

minus

1

=

z

zminus1

2

harr (n

+1)u[n+1]=(n

+1)u[n]

50

Concept

Map

Discrete-Time

Systems


7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform





+

Delay

+

Delay

X Y Y

X = H(R) =

1

1 minusRminusR2


X (z) =

z2

z2 minus z minus 1

h[n] 1 1 2 3 5 8 13 21 34 55 index shift

Delay rarr R

Z transform

Relations

among

representations

51

MIT OpenCourseWare

7212019 Z Transform


httpocwmitedu



7212019 Z Transform


httpocwmitedu



z transform

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