Transform Domain Representation of Discrete

Time Signals

The Z Transform

Yogananda Isukapalli

1

• The z-transform for discrete time signals is the counterpart of the Laplace transform for the continuos-time signals.

• The z-transform is a generalization of the Fourier transform, the principal motivation for introducing the generalization is that the Fourier transform does not converge for all sequences, which limits the class of signals that could be transformed using the Fourier transform.

• The other advantage of the z-transform is that it allows us to bring in the power of complex variable theory to bear on the problems of discrete time signals and systems.

Introduction

2

Discrete-time signals and systems

• Given an analog signal x(t), it could be represented as discrete time signal by a sequence of weighted & delayed impulses.

… -2T -T 0 T 2T …..

{x[t]}t=nT n= …-2, -1, 0, 1, 2, ..

Note: We represent the discrete time signal as x[nT] = x[n] where T is the sampling period and is omitted in the representation.

• A discrete-time system is essentially a mathematical algorithm that takes an input sequence, x[n], and produces an output sequence, y[n].

• Linear time-invariant (LTI) systems form an important class of discrete systems used in DSP.

3

Discrete-time signals and systems contd…

• The input-output relationship of an LTI system is given by the convolution sum

å¥

-¥=

-=k

knxkhny )()()( (1)

where h(k) is the impulse response of the system.

• The values of h(k) completely define the discrete-time system in the time domain.

• An LTI system is stable if its impulse response satisfies the condition

å¥

-¥=

¥<k

kh )( (2)

and causal if

h(k) = 0, k<0 (3)

Definitionof

Z-Transform

• For any given sequence x[n], its z-transform is defined as :

.....]2[]1[]0[]1[]2[.....

][)(

21012 ++++-+-+=

=

--

¥

-¥=

-åzxzxzxzxzx

znxzXn

n

• If we let z = rejw then the expression reduces to :

å¥

-¥=

--=n

jwnnjw ernxreX ][)(

which can be interpreted as the Fourier transform, for r=1 (i.e.,|z|=1), the z-transform of x[n] reduces to the Fourier transform provided it exists.

• Example : x[n] = an n�0= 0 n< 0 Find the z-transform

Answer: {an}, n � 0zz a-

5

where z is a complex variable.

(4)

(5)

Region of Convergence

Whenever an infinite series converges, the z-transform X(z) has a finite value in some region A of the complex plane. This region is termed as the Region of Convergence (ROC).

A

Complex z -PlaneRe(z)

For z=rejw & r=1 thus making |z|=1, this contour in the z-plane is a circle of unity radius and is termed as the unit circleUnit Circle

|z|=1

Im(z)

Re(z)

Im(z)

w

6

Region of Convergencecontd...

Thus for any given sequence, the set of values of z for which the z-transform converges is called the region of convergence, which we abbreviate as ROC.

Convergence of the Fourier transform requires the sequence to be absolutely summable but in the case of the z-transform applying the above condition we get : for absolute convergence.

It is clear from the above equation that because of the multiplication of the sequence by the real exponential r-n, it is possible for the z-transform to converge even if the Fourier transform does not. Thus for the Fourier transform to converge it must include the unit circle in its ROC.

| [ ] |x n r n

n

- < ¥=-¥

¥

å

7

Types of Sequences

Right-handed sequences : x[n]¹0 n³0=0 n<0

Example x[n]=Aanej�n n³0, =0 n<0

A z

z a ejROC z a.

.. . :| | | |

->

qAnswer: X(z)=

Left-handed sequences : x[n] =0 n³0¹ 0 n<0

Example x[n]=-anu[-n-1]

zz a

ROC z a-

<. . :| | | |Answer: X(z)=

a

ROC

aROC

8

Types of Sequencescontd..

Mixed sequences : x[n]¹0 n³0¹ 0 n<0

Example x[n]=anu[n] - bnu[-n-1]

zz a

zz b

ROC z a z b-

+-

> <. . :| | | |&| | | |Answer: X(z)=

If |b| < |a| ROC : empty setIf |b| > |a| ROC : exists

ROC: a Çb

|a|<|z|<|b|

9

Z-transforms of some common sequences

10

Examples

1. x[n] = b|n| for all ‘n’

))(1(

)1(

1

111

11

)(1)(

)(

2

1

1

00

01

0

1

bzbzbz

bzz

bzbz

bzbz

zbzb

zbzb

zbzbzX

n

n

n

n

n

n

nn

n

n

n

n

nn

n

n

---

=

-+

-=

-+-

-=

+-=

+=

+=

-

-¥

=

¥

=

-¥

=

¥

=

-¥

=

--

-¥=

-

åå

åå

åå

ROC: b Ç1/b

11

|b|<|z|<|1/b|

The ROC is (1 – bz) > 0 and (z – b) >0, i.e.

| b | < | z | < 1 / | b |

å¥=

-¥=

-=

å¥=

-¥=

-=

=

n

n

nznn

n

n

nznxzX

nx

nnnx

][)3.0(

][1

][1

bygiven is ][1 of transform- ZThe

ROC. theDetermine][)3.0(][1Given

µ

µ

3.0:1by given is ROC The13.01

1][1

toconverges seriespower This0

)3.0(][1

0100][ that gConsiderin

>Â--

=

å¥=

=

-=

³<=

ïî

ïíì

zz

zX

n

n

nznzX

nnnµ

2.(a)(i)

12

Examples contd…

5.0:2 ROC and 15.011][2 Similarly,

][)5.0(][2For )(

>Â-+

=

-=

zz

nX

nnnxii µ

5.0:2 2 1by given is ][1 ofROC

5.0:2by given is][2 of ROC

3.0:1by given is][1 of ROC Now

][2][1][1

sequence theof ROC theDetermine To

)(

>Â=ÂÇÂ\

>Â

>Â

+=

zzY

zzX

zzX

nxnxny

ib

13

Examples contd…

Properties of ROC

1. The ROC is a ring or disc in the z-plane centered at the origin i.e.,0 �rR<|z|<rL �¥

2. The Fourier Transform of x[n] converges absolutely if and only if the ROC of the z-transform includes the unit circle

3. ROC cannot contain any poles

4. If x[n] is a finite duration sequence, then the ROC is the entire z-plane except possibly z=0 or z=¥

5. If x[n] is a right sided sequence, the ROC extends outward from the outermost finite pole in X(z) to z=¥

6. If x[n] is a left sided sequence, the ROC extends inward from the innermost finite pole in X(z) to z= 0

14

4. Differentiation of X(z) : ROC =Rx1(except for the possible addition or

deletion of z=0 or z=¥)

Properties of the Z-Transform

Let : x1[n] z X1(z), ROC = Rx1

x2[n] z X2(z), ROC = Rx2

1. Linearity: ax1[n]+bx2[n] z aX1(z)+bX2(z) ROC contains Rx1�Rx2

2. Time Shifting: x[n-k] z z-kX(z)ROC =Rx1(except for the possible addition ordeletion of z=0 or z=¥)

3. Multiplication by an exponential Sequence: z0

n x[n] z X(z/z0) ROC =|z0|Rx1

nx n zdX zdz

[ ]( )

« -

15

7. Convolution of Sequences :x1[n]�x2[n] z X1(z).X2(z), ROC=Rx1 Ç Rx2

6. Time Reversal : x[-n] z X(1/z),ROC=1/Rx1

5. Conjugation of a complex Sequence: x*[n] z X*(z*), ROC =|z0|Rx1

Properties of the Z-Transform contd…

8. Relationship with the Laplace transform:

Let z = esT, where s is the complex variable given by s = d + jw, then z = e(d+jw)T = edTejwT

Thus ssdT FfTzez wpwpw /2/2 and ===Ð=

where (rad s-1) is the sampling frequency.

As w varies from -µ to µ the s-plane is mapped to the z-plane as shown in the following

sw

figure.16

Properties of the Z-Transform contd…

Figure: Mapping of frequencies from the s-plane to the z-plane (wT is also referred to as q)

17

Example

Solution for Differential Equations:

y n akk

Ny n k bkx n k

k

L[ ] [ ] [=

=- + -

=å å1 0

y n y n x n x n

Y z z Y z X z z X z

z Y z z X z

H zY zX z

z

z

[ ] [ ] [ ] [ ]

( ) ( ) ( ) ( )

[ ] ( ) [ ] ( )

( )( )( )

[ ]

[ ]

= - - + + -

= - - + + -

+ - = + -

= =+ -

+ -

12

1 1

12

1 1

112

1 1 1

1 1

1 12

1

Thus here the transfer function H(z) is specified by the above equation.

18

example)

Inverse Z - Transform

• The inverse z-transform (IZT) allows us to recover the discrete-time sequence x(n), given its z-transform.

• In practice, X(z) is often expressed as a ratio of two polynomials in z-1 or equivalently in z:

MM

NN

zazazaazbzbzbb

zX---

---

++++

++++=

...

...)(

22

110

22

110 (6)

In this form, the inverse z-transform, x(n), may be obtained using one of several methods including the following three:

1. Power series expansion method

2. Partial fraction expansion method

3. Residue method

19

1. Power series method

• Given the z-transform, X(z), of a causal sequence as in eqn (6), it can be expanded into an infinite series in z-1 or z by long division or synthetic division:

.....)3()2()1()0(

...

...)(

321

22

110

22

110

++++=

++++

++++=

---

---

---

zxzxzxx

zazazaa

zbzbzbbzX

MM

NN

• The numerator and denominator are first expressed in either descending powers of z or ascending powers of z-1 and the quotient is then obtained by long division.

• The coefficients of z-n are the values of x(n).

20

Power series method examples

1.

Sol)

21

Either way, the z-transform is now expanded into the familiar power series, that is

.....5756.26439.331)( 321 ++++= --- zzzzX

Power series method examples contd…

The inverse z-transform can now be written directly:

;...5756.2)3(;6439.3)2(;3)1(;1)0( ==== xxxx

• The values of x(n) are also obtained recursively:

00

0

n

1

0212

011

00

/)0( where

,....2,1 ,a

)( )(

/])0()1([)2(

/])0([)1(/)0(

abx

n

ainxbnx

aaxaxbxaaxbx

abx

iin

=

=úúû

ù

êêë

é--

=

--=-=

=

å=

!!!!! (7)

22

Power series method examples contd…

2. We now redo the previous example using the recursive approach.

21

21

3561.0121)(

--

--

+-

++=

zzzzzX

Comparing the coefficients of X(z) with those of the general transform in eqn (6),

,2;3561.0,1,1,1,2,1 210210 ===-===== MNaaabbb

From eqn (7), we have

23

2. Partial fraction expansion method

MM

NN

zazazaazbzbzbb

zX---

---

++++

++++=

...

...)(

22

110

22

110

If the poles of X(z) are first order and N=M, then X(z) can be expanded as:

å=

---

-+=

-++

-+

-+=

-++

-+

-+=

M

k k

k

M

M

M

M

pzzC

B

pzzC

pzzC

pzzC

B

zpC

zpC

zpC

BzX

10

2

2

1

10

112

21

1

10

....

1....

11)(

(8)

• Consider eqn (6) again

where B0 = bN / aN

pk are the poles of X(z) (assumed distinct)

Ck are the partial fraction coefficients, also known as residues of X(z).

24

Partial fraction expansion method…

• If N<M, B0=0

• If N>M, X(z) must be first reduced, to make N£M, by long division and the remainder can be expressed as in eqn (8).

• The coefficient Ck associated with the pole pk can be obtained as:

kpzkk pz

zzXC

=

-= )()( (9)

• The inverse z-transform of each partial fraction is then obtained from z-transform tables.

25

Partial fraction expansion method…

• Multiple-order poles: If X(z) contains an mth-order pole, the partial fraction expansion must include terms of the form:

å= -

m

ii

k

i

pzD

1 )((10)

The coefficients, Di, may be obtained from the relation

kpz

mkim

im

i zzXpz

dzd

imD

=-

-

úû

ùêë

é --

=)()(

)!(1 (11)

26

Partial fraction expansion method examples

1. X(z) contains simple, first order poles:

21

1

375.025.01)(

--

-

--=

zzzzX Find the inverse

z-transform.

Soln)

54

5.01

)5.0)(75.0()75.0(

)75.0()(

)()(

(9),eqn From5.0p and 75.0p poles have We

5.075.0)5.0)(75.0(1)(

5.075.0)5.0)(75.0()(

M,N Since)5.0)(75.0(375.025.0

)(

:zby r denominato andnumerator thegMultiplyin

75.0

75.0

75.011

21

21

21

2

2

1

=

+=

+--

=

-=-=

-==

++

-=

+-=Þ

++

-=

+-=

<+-

=--

=

=

=

==

z

z

zpz

z

zzz

zzzXpz

zzXC

zC

zC

zzzzX

zzC

zzC

zzzzX

zzz

zzzzX

27

Partial fraction method examples….

[ ] 0 ,)5.0()75.0(54

)(

,5.075.0

)5/4(0.5z

(4/5)z0.75-z

(4/5)zX(z)

Therefore,54

75.01

)5.0()(

)()(

C

Similarly,

5.0

5.022

2

>--=

úûù

êëé

+-

-=

+-=

-=

-=

+=-=

-=

-==

nnx

Sozz

zz

z

zzzX

pzzzX

nn

z

zpz

2. X(z) contains a second order pole:

2

2

)1)(5.0()(

--=

zzzzX

Find the inverse z-transform.

28

Partial fraction method examples….

2

)1(

)1)(5.0()5.0(

)5.0()()1(15.0

)1)(5.0(

)(

5.02

5.02

25.0

221

2

2

=

-=

--

-=

-=

-+

-+

-=

--=

=

=

=

z

z

z

zz

zzzzz

zzzXC

zD

zD

zC

zzzzXSoln)

To obtain D1 we use eqn (11), with i=1, m=2.

2

)5.0(5.0

5.0

)1)(5.0()1()()1(

12

1

12

22

1

2

1

-=

-

--=

÷øö

çèæ

-=

úúû

ù

êêë

é

--

-=

úúû

ù

êêë

é -=

=

=

==

z

z

zz

zzz

zz

dzd

zzzzz

dzd

zzXz

dzdD

29

Partial fraction method examples….

Similarly, for D2 we use eqn (11), with i=2, m=2.

2 5.01

1

)1)(5.0()1()()1(

12

22

1

2

2

=-

=

--

-=

-=

== zz zzzzz

zzXzD

Combining the results, X(z) becomes

2)1(2

12

5.02)(

-+

--

-=

zz

zz

zzzX

Therefore,

[ ] 0 ,)5.0()1(2

22)5.0(2)(

³+-=

+-=

nn

nnxn

n

30

3. Residue method

• The inverse z-transform is obtained by evaluating the contour integral

ò -=c

n dzzXzj

nx )(21)( 1

p(12)

where C is the path of integration enclosing all the poles of z.

• Eqn (12) is evaluated using Cauchy’s residue theorem:

ò -=c

n dzzXzj

nx )(21)( 1

p

= sum of the residues of zn-1X(z) at all the poles inside C

31

Residue method…

• The residue of zn-1X(z) at the pole pk is given by:

[ ] [ ]kpzkm

m

k zFpzdzd

mpzFs =-

-

--

= )()()!1(

1),(Re1

1

(13)

where F(z)= zn-1X(z),

m is the order of the pole at pk,

Res[F(z),pk] is the residue of F(z)at z= pk.

• For a simple (distinct) pole, eqn (13) reduces to

[ ]kpz

nkkk zXzpzzFpzpzFs

=

--=-= )()()()(),(Re 1(14)

32

Residue method examples

1. )5.0)(75.0()(

+-=

zzzzX

Find x(n). Assume C is the circle |z|=1.

Soln) This problem is the same as example 1 under z-transform examples:

If we let F(z)= zn-1X(z) then

)5.0)(75.0(

)5.0)(75.0()(

1

+-=

+-=

-

zzz

zzzzzF

nn

F(z) has poles at z = 0.75 and z = -0.5.

Fig: A sketch of the contour of integration with the position of poles

33

Residue method examples…

x(n) = Res[F(z),0.75] + Res[F(z),-0.5] Since the poles are first order, eqn (14) will be used. Thus,

n

nz

nz

zzzz

zFzzFs

)75.0(54

5.075.0)75.0(

)5.0)(75.0()75.0(

)()75.0(]75.0),([Re

75.0

75.0

=

+=

+--

=

-=

=

=

n

z

nz

zzzz

zFzzFs

)5.0(54

)5.0)(75.0()5.0(

)()5.0(]5.0),([Re

5.0

5.0

--=

+-+

=

+=-

-=

-=

Thus, ])5.0()75.0)[(5/4()( nnnx --=

which is identical to the result obtained by the partial fraction expansion method.

34

Residue method examples…

2.

35