lecture4-2

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Z-Transform

( ) [ ]

=

=n

jwnjw

enxeX

Fourier Transform

z-transform

( ) [ ]

=

=

n

nznxzX

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Z-transform operator:

The z-transform operator is seen to transform thesequencex[n] into the functionX{z}, where zis a

continuous complex variable. From time domain (or space domain, n-domain) to the

z-domain

Z-Transform (continue)

{ }Z[ ]{ } [ ] ( )zXzkxnxZ

k

k ==

=

[ ] [ ] [ ] [ ]{ } ( )ZXnxZknkxnxz

k ==

=

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Two sided or bilateral z-transform

Unilateral z-transform

Bilateral vs. Unilateral

( ) [ ]

=

=n

nznxzX

( ) [ ]

=

=0n

nznxzX

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Example of z-transform

( ) 54321 24642 +++++= zzzzzzX

01246420x[n]

N>5543210n1n

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If we replace the complex variable zin the z-transform by ejw, then the z-transform reduces tothe Fourier transform.

The Fourier transform is simply the z-transformwhen evaluating X(z) in a unit circle in the z-plane.

Generally, we can express the complex variablez in the

polar form asz = rejw. Withz expressed in this form,

Relationship to the FourierTransform

( ) [ ]( ) [ ]( )

=

=

==

n

jwnn

n

njwjwernxrenxreX

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In this sense, the z-transform can be interpreted asthe Fourier transform of the product of the originalsequencex[n] and the exponential sequence rn.

For r=1, the z-transform reduces to the Fourier transform.

Relationship to the FourierTransform (continue)

The unit circle in thecomplexz plane

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Relationship to the FourierTransform (continue)

Beginning atz = 1 (i.e., w = 0) throughz =j (i.e., w =/2) toz = 1 (i.e., w = ), we obtain the Fouriertransform from 0 w .

Continuing around the unit circle in thez-planecorresponds to examining the Fourier transformfrom w = to w = 2. Fourier transform is usually displayed on a linear

frequency axis. Interpreting the Fourier transform as thez-transform on the unit circle in the z-plane correspondsconceptually to wrapping the linear frequency axisaround the unit circle.

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Region of convergence (ROC) Since the z-transform can be interpreted as the Fourier

transform of the product of the original sequencex[n] andthe exponential sequence rn, it is possible for the z-

transform to converge even if the Fourier transform doesnot.

Because

X(z) is convergent (i.e. bounded) i.e., x[n]rn 1. This meansthat the z-transform for the unit step exists with ROC |z|>1.

Convergence Region of Z-transform

( ) [ ] [ ]

=

=

=

n

n

n

nznxznxzX

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In fact, convergence of the power series X(z)depends only on |z|.

If some value ofz, sayz =z1, is in the ROC, then all

values ofz on the circle defined by |z|=|z1| will alsobe in the ROC.

Thus the ROC will consist of a ring in the z-plane.

ROC of Z-transform

[ ]

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ROC of Z-transform RingShape

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Analytic Function and ROC

The z-transform is a Laurent series ofz. A number of elegant and powerful theorems from the

complex-variable theory can be employed to study thez-transform.

A Laurent series, and therefore the z-transform,represents an analyticfunction at every point inside theregion of convergence.

Hence, the z-transform and all its derivatives exist andmust be continuous functions ofz with the ROC.

This implies that if the ROC includes the unit circle, the

Fourier transform and all its derivatives with respect tow must be continuous function ofw.

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Z-transform of a causal FIR system

The impulse response is

Take the z-transform on both sides

Z-transform and LinearSystems

[ ] [ ]=

=M

m

m mnxbny

0

[ ] [ ]=

=M

m

m mnbnh0

0bMb3b2b1b00h[n]

N>MM3210n

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Z-transform of Causal FIR

System (continue)

Thus, the z-transform of the output of a FIR system is

the product of the z-transform of the input signal andthe z-transform of the impulse response.

( ) [ ]{ } [ ] [ ]

[ ] [ ]( )

[ ]{ } [ ]{ } [ ] ( ) ( )zXzHzXnhZnxZzb

zmnxzbzmnxb

zmnxbmnxbZnyZzY

M

m

m

m

M

m

mn

n

m

m

M

m

n

nm

n

M

m

nm

M

m

m

==

==

=

==

=

=

=

=

=

= =

=

0

00

00

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Z-transform of Causal FIRSystem (continue)

H(z

)is called the system function(or transfer

function) of a (FIR) LTI system.

( ) ==

M

m

mmzbzH0

x[n] y[n]

h[n]

X(z) Y(z)

H(z)

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Multiplication Rule ofCascading System

X(z) Y(z)H1(z)

Y(z) V(z)H2(z)

X(z) Y(z)H1(z)

V(z)H2(z)

X(z) Y(z)H1(z)H2(z)

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Example

Consider the FIR systemy[n] = 6x[n] 5x[n1] +x[n2] The z-transform system function is

( )

( )( ) 211

21

2

1

3

1

623

56

z

zz

zz

zzzH

==

+=

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Delay of one Sample

Consider the FIR systemy[n] =x[n1], i.e., the one-sample-delay system.

The z-transform system function is

( ) 1=zzH

z1

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Delay of k Samples

( ) kzzH =

Similarly, the FIR systemy[n] =x[nk], i.e., the k-sample-delay system, is the z-transform of theimpulse response [n k].

zk

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System Diagram of A CausalFIR System

The signal-flow graph of a causal FIR system

can be re-represented by z-transforms.

x[n]

TD

TD

TD

x[n-2]

x[n-1]

x[n-M]

+

+

+

+

b1

b0

b2

bM

y[n]

x[n]

z1

z1

z1

x[n-2]

x[n-1]

x[n-M]

+

+

+

+

b1

b0

b2

bM

y[n]

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Z-transform of GeneralDifference Equation

Remember that the general form of a linearconstant-coefficient difference equation is

When a0 is normalized to a0 = 1, the systemdiagram can be shown as below

[ ] [ ]==

=M

m

m

N

k

k mnxbknya00

for all n

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Review of Linear Constant-coefficient Difference Equation

x[n]

TD

TD

TD

x[n-2]

x[n-1]

x[n-M]

+

+

+

+

b1

b0

b2

bM

+

+

+

+

y[n]

TD

TD

TD

a1

a2

aN y[n-N]

y[n-2]

y[n-1]

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Z-transform of Linear Constant-coefficient Difference Equation

The signal-flow graph of difference

equations represented by z-transforms.

X(z)

z1

z1

z

1

+

+

+

+

b1

b0

b2

bM

+

+

+

+

Y(z)

z1

z1

z1

a1

a2

aN

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From the signal-flow graph,

Thus,

We have

Z-transform of DifferenceEquation (continue)

( ) ( ) ( )=

=

=N

k

kk

mM

m

m zzYazzXbzY10

( ) ( ) mM

m

m

N

k

kk zzXbzzYa

==

=00

( )( )

=

==N

k

k

k

mM

m

m

za

zb

zX

zY

0

0

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Let

H(z) is called the system functionof the LTI system defined

by the linear constant-coefficient difference equation. The multiplication rule still holds: Y(z) =H(z)X(z), i.e.,

Z{y[n]} =H(z)Z{x[n]}.

The system function of a difference equation is a rationalformX(z) = P(z)/Q(z).

Since LTI systems are often realized by difference equations,

the rational form is the most common and useful for z-transforms.


[ ] =

==

N

k

kkm

M

m

m zazbzH00

/

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When ak= 0 for k= 1 N, the difference equationdegenerates to a FIR system we have investigatedbefore.

It can still be represented by a rational form of thevariablez as


( ) mM

m

mzbzH == 0

( )M

mMM

m

m

z

zb

zH

=

= 0

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When the inputx[n] = [n], the z-transform ofthe impulse response satisfies the followingequation:

Z{h[n]} =H(z)Z{[n]}. Since the z-transform of the unit impulse [n] is

equal to one, we have

Z{h[n]} =H(z)

That is, the system functionH(z) is the z-

transform of the impulse response h[n].

System Function and ImpulseResponse

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Generally, for a linear system,y[n] = T{x[n]}

it can be shown that

Y{z} =H(z)X(z).whereH(z), the system function, is the z-transform of theimpulse response of this system T{}.

Also, cascading of systems becomes multiplication ofsystem function under z-transforms.

System Function and ImpulseResponse (continue)

X(z) Y(z) (=H(z)X(z))

H(z)/H(ejw

)X(ejw) Y(ejw) (=H(ejw)X(ejw))

Z-transform

Fourier transform

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Pole: Thepoleof a z-transformX(z) are the values ofz for

whichX(z)= .

Zero: The zeroof a z-transformX(z) are the values ofz for

whichX(z)=0.

WhenX(z) = P(z)/Q(z) is a rational form, andboth P(z) and Q(z) are polynomials ofz, thepoles of are the roots ofQ(z), and the zeros are

the roots ofP(z), respectively.

Poles and Zeros

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Zeros of a system function The system function of the FIR system y[n] = 6x[n]

5x[n1] +x[n2] has been shown as

The zeros of this system are 1/3 and 1/2, and thepole is 0.

Since 0 and 0 are double roots ofQ(z), the poleis a second-order pole.

Examples

( )( )( )zQzP

z

zz

zH =

=2

21

31

6

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Right-sided sequence: A discrete-time signal is right-sided if it is nonzero

only for n0.

Consider the signalx[n] = an

u[n].

For convergentX(z), we need

Thus, the ROC is the range of values ofz for which

|az1| < 1 or, equivalently, |z| > a.

Example: Right-sidedExponential Sequence

( ) [ ] ( )

=

=

==0

1

n

n

n

nnazznuazX

( )

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By sum of power series,

There is one zero, atz=0, and one pole, atz=a.

Example: Right-sidedExponential Sequence (continue)

( ) ( ) azaz

z

azazzX

n

n>

=

==

=

1

11

0

1 ,

: zeros : poles

Gray region: ROC

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Left-sided sequence:

A discrete-time signal is left-sided if it is nonzero onlyfor n 1.

Consider the signalx[n] = anu[n1].

If|az1| < 1 or, equivalently, |z| < a, the sum converges.

Example: Left-sided ExponentialSequence

( ) [ ]

( )

=

=

=

=

==

==

0

1

1

1

1

1

n

n

n

nn

n

nn

n

nn

zaza

zaznuazX

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By sum of power series,

There is one zero, atz=0, and one pole, atz=a.

Example: Left-sided ExponentialSequence (continue)

( ) azaz

z

za

za

zazX

zznuzn

,

( ) 31

311

1

3

1

1>

+

zznuzn

,

Thus

( ) ( )2

1

3

11

1

2

11

1

3

1

2

1

11>

+

+

+

z

zz

nunuznn

,

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Example: Sum of TwoExponential Sequences (continue)

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Example: Two-sided ExponentialSequence

( ) ( ) ( )121

3

1

= nununx

nn

Given

Since ( )

3

1

311

1

3

1

1

>

+

zz

nuzn

,

and by the left-sided sequence example

( )2

1

2

11

11

2

1

1

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Some Common Z-transform

Pairs (continue)

[ ]

( )[ ] [ ]

[ ][ ]

[ ] [ ] [ ][ ]

[ ] [ ] [ ]

[ ][ ] [ ] [ ]

[ ]

.0:ROC1

1

0

10

.:ROC21

.:ROC

21

1

1.:ROC21

1.:ROC21

1

.:ROC

1

1

1

2210

10

0

221

0

10

0

210

10

0

210

10

0

21

1

>

>+

>+

>+

>+

=

zzz

zX//

( ) ( )

( )( ) ( ) 2211

1411

2112

411

1

==

==

=

=

/

/

/

/

z

z

zXzA

zXzA

( )( )

( )( )

( )1

21

1

211411 +

=

z

A

z

AzX

//

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Example (continue)

Thus

and so

( )( )( ) ( )( )11 211

2

411

1

+

=

zzzX

//

[ ] [ ] [ ]nununxnn

= 4

1212

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Find the inverse z-transform of

Since both the numerator and denominator are ofdegree 2, a constant term exists.

B0 can be determined by the fraction of the coefficients ofz2,B

0= 1/(1/2) = 2.

Another Example

( )( )

( )( )( )1

1211

111

21

>

+=

zzz

zzX

/

( ) ( )( ) ( )12

1

1

0 1211 +

+=

z

A

z

ABzX

/

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Therefore

Another Example (continue)

( ) ( )( ) ( )

( )( )( )( )( )

( )( )( )

( ) 811211

512

9211

1211

512

12112

11

11

1

2

211

11

1

1

1

2

1

1

=

++=

=

++=

+

+=

=

=

z

z

zzz

zA

z

zz

zA

z

A

z

A

zX

/

/

/

/

/

( )( )( ) ( )

[ ] [ ] ( ) [ ] [ ]nununnx

zzzX

n 82192

1

8

211

92

11

+=

+

=

/

/

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We can determine any particular value of the

sequence by finding the coefficient of theappropriate power ofz1 .

Power Series Expansion

( ) [ ]

[ ] [ ] [ ] [ ] [ ] ...... ++++++=

=

=

212 21012 zxzxxzxzx

znxzXn

n

Example: Finite-length

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By directly expandX(z), we have

Thus,

Example: Finite-length

Sequence

( ) 1112 11501 += zzzzzX .

( ) 12 50150 += zzzzX ..

[ ] [ ] [ ] [ ] [ ]1501502 +++= nnnnnx ..

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Using the power series expansion for log(1+x) with|x|+= 1 1log

( )( )

=

+=

1

11

n

nnn

n

zazX

[ ] ( )

=

+

00

111

n

nnanx

nn /

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Suppose

Linearity

Z-transform Properties

[ ] ( )[ ] ( )

[ ] ( ) 2

1

ROC

ROC

ROC

22

11

x

z

x

z

x

z

RzXnx

RzXnx

RzXnx

=

=

=

[ ] [ ] ( ) ( )21

ROC2121 xxz

RRzbXzaXnb xnax =++


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Time shifting

Multiplication by an exponential sequence

Z transform Properties

(continue)

[ ] ( ) xn

z

RzXznnx ROC00 = (except for the

possible addition

or deletion ofz=0orz=.)

[ ] ( ) xz

nRzzzXnxz 000 ROC = /


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Differentiation ofX(z)

Conjugation of a complex sequence


(continue)

[ ]( )

x

z

Rdz

zdXznnx ROC =

[ ] ( ) xz

RzXnx ROC = ***


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Time reversal

If the sequence is real, the result becomes

Convolution


(continue)

[ ] ( )x

z

RzXnx

1ROC1 = /

[ ] ( )x

z

RzXnx

1ROC1 = */**

[ ] [ ] ( ) ( ) 21containsROC

2121 xx

z

RRzXzXnxnx


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Initial-value theorem: Ifx[n] is zero for n

lecture4-2

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