8. Z-TRANSFORM

99

8. Z-TRANSFORM

Historically, Laplace transforms were used to study signals defined by the solutions to

linear ordinary differential equations. Beginning in the 1950’s, discrete time signals began to

appear. Unfortunately, the Laplace transform is not well suited for the study of discrete time

signals and systems. Instead, another transform, called the Z-transform, is used.

8.1 Direct Z-transform

It is know that discrete-time signal can be obtained by multiplying continuous time

signal x(t) by impulse-train

n

)nTt()t(p

n

)nTt()nT(x)t(x (8.1)

Using the Laplace transform’s shifting property:

ksTe)kTt(L

and a short hand notation zesT , we get

k

kx(kT)zX(z) x(t)Z

if T=1 we get

k

kz)k(x)z(X (8.2)

Equation (8.2) is called bilateral or two sided Z-transform. For causal signals the one sided Z-

transform uses

(8.3)

Example 8.1

Find the Z-transform of the DTS shown in Figure 8.2

0k

kz)k(x)z(X

-3T -2T -T 0 T 2T 3T

t

x(t)

Figure8.1

x[2T]

x[T]

-2 -1 0 1 2 3

-2

2

-2

1

3

1 k

x[k]

X(z) = -2z-3

+2z-2

+z-1

+3+-2z1+z

2

Figure 8.2

8. Z-TRANSFORM

100

Example 8.2

Find the Z-transform of a causal step function show in Figure 8.3

........zzz1)z(X 321

0k

This infinitely long series can also be represented in closed form using :

x1

1.....xxx1x 32

if x <1

1z1)z(X 2z .........z 3 has a closed form

1z

z

z1

1)z(X

1

if 1z < 1 or z >1

Example 8.3

Find the Z-transform of the causal exponential function

te)t(x or kekx

kekx ka where a= e

1

k

0k

1

0k

kk

az1

1)az(za)z(X

=

az

z

if a 1z <1 or z>a

ez;

ez

z)z(X

Example 8.4

Find the Z-transform of an anticausal sequences shown in Figure 8.4

-1 k≤-1

kx

0 k≥ 0

0 T 2T 3T t

xs(t)

Figure 8.3

Figure 8.4

-4 -3 -2 -1 0

-1

x[k]

k

8. Z-TRANSFORM

101

1zif;z1

z

z1

z111

z1

1

1z1zzzkx)z(X

1

0k

k

0k

k1

k

kk1

z

Computer Study:

M - file ztrans.m is used to find Z-transform of the time domain function.

Consider the following two examples.

Example 8.5 x(t)= t2e

2

t2

ez

ze

Example 8.6 X(t)=t

8.2 Region of Convergence of Z-Transform

The domain of values of z guaranteeing that a Z-transform of kx exists is called the

region of convergence, or simply ROC. The ROC constrain of an annular ring in the Z-plane

that is centered around the origin. The ROC is an important concept for a variety of reasons.

As shown in the Examples 8.2 and 8.4. There is no unique relationship between the sequences

and their Z-transforms.

The Z-transform for both functions is 1z

z)z(X

, but ROC are different. Hence, the

Z-transform must always be specified with its ROC. The regions of convergence for the

Examples 8.2 and 8.4 are given in figures 8.5 (a) and (b) respectively.

» syms t

» x=exp(-2*t);

» ztrans(x)

ans =

z/(z-exp(-2))

» syms t

» x=t; ztrans(x)

ans =

z/(z-1)^2

a)

Figure 8.5

pole

z=1

ROC: z>1

zero

ROC: z<1

b)

X(z)=z/(z-1)

z=1

2)1(

z

z

for causal for anticausal

8. Z-TRANSFORM

102

We can readily verify that when 1z causal step sequences X(z) convergence where as when

1z X(z) diverges. For example, if we let z=2 we find that the series on the RHS of

equation 4.8 adds up to 2:

X(z)=1+1/0,5+ 25,0/1 + 35,0/1 +3... =1+2+4+8+...

Example

Find thez-transform and the region for convergence for each of the discrete-time sequences

given in figure 4.1.

(1) The sequence of figure 4.1(a) is noncausal, since x(n) is not zero for n<0, but it is

of a finite duration. The sequence has values x(-6)=0, x(-5)=1, x(-4)=3, x(-3)=5, x(-2)=3, x(-

1)=1 and x(0)=0. from equation 4.6, the z-transform is given by

n

nznxzX1 = 2353 2345 zzzz

It is readily verified that the value of X (z) becomes infinite when z . Trus the ROC is

everywhere in the z-plane except at z= .

a) b) c)

(2) Again, the sequence in figure 4.1(b) is not causal. It is of a finite duration, and

double sided. The values of the sequence are x(3)=0, x(-2)=1, x(-1)=3, x(0)=5, x(1)=3,

x(2)=1 and x(3)=0. from equation 4.6, the z-transform is given by

n

nznxzX 2 = 212 353 zzzz

It is evident that the value of X(z) is infinite if z=0 or if z= . Therefore the region of

convergence is everywhere expect at z=0 and z= .

1

3

5

3

1

-6 -5 -4 -3 -2 -1 0

x(n)

n

1

3

5

3

1

-3 -2 -1 0 1 2 3

x(n)

n

1

3

5

3

1

0 1 2 3 4 5 6

x(n)

n

8. Z-TRANSFORM

103

(3) figure 4.1(c) represents a causal, finite duration sequence with values x(0)=0,

x(1)=1, x(2)=3, x(3)=5, x(4)=3, x(5)=1 and x(6)=0. the z-transform is given by

n

nznxzX 3 = 54321 353 zzzzz

In this case , X(z)= for z=0. Trus the region of convergence is everywhere except at z=0.

Example 8.7

ku)4/1()3/1(kx s

kk

Now the procedure continues as:

)z(X11 z

4

11

1

z3

11

1

)4

1z)(

3

1z(

)12

1z2(z

For convergence, the individual terms must converge. This means that:

1z

3

11

1

1z3

1 < 1

1z

4

11

1

1z4

1 < 1

The overlap of two ROC’s corresponds to the region of converges of X(z), that is z>1/3.

Example 8.8

Consider the two-sided sequence defined by:

nnu

where can be a real or complex number, does not have a Z-transform, regardless of the

absolute value . This followed by noting that the Z-transform expression can be rewritten as:

n1

n

nn

0n

n zzzU

The first term on the right-hand side of the equation converges for z > ,whereas the

second term converges for z < ,and hence, there is no overlap of two ROC’s.

Example 8.10

Determine the ROC of the Z-transform H(z) of the sequence nUs)6.0(nh n .

ROC: z>1/3

Figure 8.6

X(z)=z/(z-1)

1/3 1/4

8. Z-TRANSFORM

104

0n

n1nn

0n

)z6.0(z)6.0()z(H

which simplifies to 6.0z

z

z6.01

1)z(H

1

provided z0.6. This implies that the ROC is just outside the

circle going through the point z= -0.6 and extending all the

way to z=, as show in figure. Notice that H(Z) has a zero at z=0

and a pole at z= -0.6.

Method of Residue

This method of residue is used if original function is represented by s-

variable:

ii ss

1Tze1

)(Fresidues)z(F

The residue ror simple poles

1Ts

i

ii

ze1

1

)s('D

)s(N

i

For multiple poles

is

1T

n

i1n

1n

ize1

1)(Fs

d

d

!1n

1

Example 8.16

1ss

1)s(F

2

Using the residues:

0s1s s;s

1T2

2

121

)ze1)(1(

1residues)z(F

For simple pole s1=-1:

T1T

1

T211T1ez

z

ze1

1

e1

1

12

1

ze1

1

)('D

1

For double pole s1=0:

221

11

0

21T2

1T1T

0

T2

2

2

)1z(

Tz

1z

z

)z1(

Tz)z1(

)ze1()1(

zTe1ze1

e1

1

1

1

d

d

T22 ez

z

)1z(

Tz

1z

z

)1s(s

1z

z=-0.6

Figure 8.7

8. Z-TRANSFORM

105

Comparison between F(z) and F(s)

1s

1

)s(

1

1s

1

)1s(s

1)p(F

22

T22 ez

z

)1z(

Tz

1z

z

)1s(s

1z)z(F

Example 8.17

Find the Z-transform of t2sintx

22 2s

2t2sinL

; j2s

1T2cosz2z

T2sinz

1)ee(z2

2z

)ee(z

j2

1

1zezez

zezzez

j2

1

)ez)(ez(

)ez(z)ez(z

j2

1

ez

z

ez

z

j2

1

ze1

1

ze1

1

j2

1

ze1

1

2

2

ze1

1

2

2)z(X

2jT2jT22

jT2jT2

jT2jT22

jT22jT22

jT2jT2

jT2jT2

jT2jT2

1jT21jT2

j2

1T

j2

1T

1T2cosz2z

T2sinzt2cosZ

2

Computer Study

MATLAB can be used to determine the ROC’s of a rational Z-transform. The M-file

z, p, k = tf2zp(num, den) determines the zeros, poles and the gain constant of a rational Z-

transform expressed as a ratio of polynomials in descending powers of z. The output files are

the column vectors z and p containing the zeros and poles of the rational Z-transform, and the

gain constant k. The statement num, den = tf2zp (z, p, k) is used in implement the reverse

process. From the zero-pole description, the factored form of the transfer function can

be obtained using the function sos = tf2sos(z, p, k). The statement computes the coefficients

of each second order factor given as an 6L matrix sos, where

01b 11b 21b 01a 11a 21a

02b 12b 22b 02a 12a 22a

sos = . . . . . .

oLb L1b L2b L0a L1a L2a

8. Z-TRANSFORM

106

where the k-th row contains the coefficients of the numerator and the denominator of the

k-th second order factor of the Z-transform G(z):

L

1k2

k2

1

k1k0

2

k2

1

k1k0

zazaa

zbzbb)z(G

The pole-zero plot of the rational Z-transform can also be plotted by using the M-files

zplane (zeros,poles), zplane (num,den)

It should be noted that the argument zeros and poles must be entered as column vectors,

whereas, the argument num and den needed to be entered as row vectors.

The following example illustrates the application of the above functions.

Example 8.11 Express the following Z-transform in factored form, plot its poles and zeros,

and the n determine its ROC’s.

423

2

z5z8z

6z5z2)z(G

The factored form of the Z-transform is given by:

321

213

z5531.0z7678.00000.1z2322.70000.1

z9627.0z8023.03209.0z232.6)z(G

» syms z

» G=(2*z^2+5*z-6)/(z^3+8*z^2+5*z-4);

» num=[2 5 -6];

» den=[1 8 5 -4];[z,p,k]=tf2zp(num,den)

z =

-3.3860

0.8860

p =

-7.2322

-1.2209

0.4530

k =

2

» sos=zp2sos(z,p,k)

sos =

6.2322 0 0 1.0000 7.2322 0

0.3209 0.8023 -0.9627 1.0000 0.7678 -

0.5531

» z=[ -3.3860;0.8860];

» p=[ -7.2322; -1.2209; 0.4530];

» k=2;

» [num,den]=zp2tf(z,p,k)

num =

0 2.00000000000000

5.00000000000000 -5.99999200000000

den =

1.00000000000000 8.00010000000000

5.00053868000000 -3.99989621994000

» printsys(num,den)

num/den =

2 z^2 + 5 z - 6

------------------------------------

z^3 + 8.0001 z^2 + 5.0005 z - 3.9999

The pole-zero configuration

developed by the program is

shown in Figure 8.9. From the

equation the ROC are:

1R : z 7.2323

2R : -7.23z 1.2209

3R : 1z 0.453

Figure 8.9

-7 -6 -5 -4 -3 -2 -1 0 1 -2

-1

0

1

2

8. Z-TRANSFORM

107

8.2 The properties of the Z-transform

The z-transforms can be represented by following shorthand notations:

X(z); txz ; sxz ; kxz

1. Linearity:

)z(bxzaxtbxtaxz 2121

2. Time shifting:

zXzkTtxz k

T1kzXTXz)T0(XzzXzkTtxZ 1kkk

3. Frequency shifting:

azeXasxZ

4. Multiplication by t n :

zXdz

dTZtxtZ 1

n

txtzzX 1n

1

zXdz

dzzX

dz

dTzttxZ

1T

Example 8.12

Find the Z-transform of x(t) = t

x(t) kt = kkx

0k

k

11z

zz)z(X

0k

k

0k

11k1 kzdz

)z(dXzzk

dz

zdX

22

1

)1z(

z

)1z(

z1zz

dz

)z(dXz)z(X

2

0k

k

)1z(

zkzzX

Example 8.13 Find the Z-transform of kukkx 2 .

21

)1z(

zzX

Using the table

0k

kkzzX

8. Z-TRANSFORM

108

34

2

4

2

11z

1zz

1z

1zz

1z

z2z21z22zzzX

dz

dzzX

An extension to this result is given by kkakx for a 1. Letting kkx1 and

kxakx 1

k , and given knowledge of zX1 , it fallows that:

21az

az)a/z(X)z(X

5. The initial value theorem:

Lim 0k )kT(x =Lim z X(z)

6. The fınal value theorem:

Lim k )kT(x =Lim 1z (z-1)X(z)

7. Differentiation:

),z(Fd

d,kTx

d

dz

8. Integration:

1

0

1

0

d),z(xd),kT(xz

Example 8.14

Given atez

z

as

1z

; Find

2)as(

1z

2t

T

T2 )ez(

Te

ez

z

d

d

as

1z

d

d

as

1z

d

d

as

1

d

dz

as

1z

9. Convolution:

zxzxnxnx 2121 ; ROC: 21 RR

where: zxnx 11 ROC: 1R

zxnx 22 ROC: 2R

Example 8.15

Find the output signal of the system shown in the Figure 8.10

U(z)=z t2e=

2

ez

z

; H(z)=1z

z

Y(z)=)ez)(1z(

z

ez

z

1z

z2

2

2

1.4 Mapping the s-plane into z-plane

The connection between s-plane and z-plane is interpreted in the figure 8.11. A mapping

rule is based on z=cost + jsint. Some important mapping are listed below:

1. The points s= j2k (k=0, 1, 2. . .) is mapped to z=exp(j2k) cos2k =1.

2. The point s=j(2k+1) is mapped to z=expj((2k+1) cos(2k+1) =-1.

H(z) U(t)=e

-2t Y(z)

Figure 8.10

8. Z-TRANSFORM

109

3. The j axis (locus p6) maps onto z6 - the unit circle with the radius z =1. For the values

s=j, (0≤≤2) , if =0, z=exp(0)=1; if =/2sin /2=1.

4. The left half of s-plane, for values s=-j, such that ≤ ) maps to the interior of the

unit circle.

5. The right half of s-plane, for values s=j, such that ≤ ) maps to the exterior of the

unit circle.

6. The locus p1, for values s=-+j/2, (0≤≤-) maps onto z1. If =0,

z1=exp(j/2)sin(/2)=1, if =-, z1=exp(-)=0.

7. The locus p2, for values s=-, (0≤≤-) maps onto z2. If =0, z2=exp(j/4)

sin(/4)=0.707, if =-, z2=exp(-)=0.

8. The locus p3, for values s=--j/4, (0≤≤) maps onto z3. If =0, z3=exp(j/4)

jsin(/4)=0.707, if =-, z3=exp(-)=0.

9. The locus of p4, for values s=--j, (0≤≤-) maps onto z4. If =0, z4=1, if =-,

z4=exp(-)=0.

10. The locus p5, for values as s=-+j, such that 0≤≤-; maps onto z5. If =0,

z=-1; If =, z=0.

11. The locus p7=1- j), maps onto circle with radius z6=e1

. In Figure 8.11 1=-0.5.

Computer study

M-file zplane(pc,T) convert the points (xc- in column vectors) in the s-plane into their

locations in the z-plain with sampling period (T) the unit circle for reference. Each point is

represented with a 'o' in the z-plane. The results of computer mapping ofr the values of xc =

, for the interval 0:2*pi are represented below.

z-plane Im

s-plane

Figure 8.11

/2

-/4

P1

P2

P3

P6

P4

Re 1;0

j

P5

-j

z1

z2

z3

-1;0

0

z5

z6

-0.5

P7

z4

z7

Im

Re

z7 =e-

z=1

cost

sint

8. Z-TRANSFORM

110

» sigma=0:pi/2:2*pi;

» p1=- sigma+sqrt(-1)*pi/2;

» z1=(exp(p1))'

z1 =

0.0000 - 1.0000i

0.0000 - 0.2079i

0.0000 - 0.0432i

0.0000 - 0.0090i

0.0 - 0.0019i

» zplane(z1,1)

» p2=-sigma;

» z2=(exp(p1))'

z2 =

1.0000

0.2079

0.0432

0.0090

0.0019

» zplane(z2,1)

» p3=- sigma-sqrt(-1)*pi/4;

» z3=(exp(p3))'

z3 =

0.7071 + 0.7071i

0.1470 + 0.1470i

0.0306 + 0.0306i

0.0064 + 0.0064i

0.0013 + 0.0013i

» zplane(z3,1)

» p4=- sigma-sqrt(-1)*sigma;

» z4=(exp(p4))'

z4 =

1.0000

0.0000 + 0.2079i

-0.0432 + 0.0000i

0.0000 - 0.0090i

0.0019 - 0.0000i

» zplane(z4,1)

» p5=- sigma+sqrt(-1)*pi;

» z5=(exp(p5))

» z5=(exp(p5))'

z5 =

8. Z-TRANSFORM

111

-1.0000 - 0.0000i

-0.2079 - 0.0000i

-0.0432 - 0.0000i

-0.0090 - 0.0000i

-0.0019 - 0.0000i

» zplane(z1,1)

» p6=sqrt(-1)*sigma;

» z6=(exp(p6))'

z6 =

1.0000

0.0000 - 1.0000i

-1.0000 - 0.0000i

0.0000 + 1.0000i

1.0000 + 0.0000i

» zplane(z6,1)

» p7=-0.5-sqrt(-1)*sigma;

» z7=(exp(p7))'

z7 =

0.6065

0.0000 + 0.6065i

-0.6065 + 0.0000i

0.0000 - 0.6065i

0.6065 - 0.0000i

» zplane(z7,1)

8.5 Partial Fraction Expansion.

8. INVERSE Z-TRANSFORM

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

z-transform. The IZT is particularly useful in DSP work, for example in finding the impluse

response of digital filters. Symbolically, the inverse z-transform may be defined as

)()( 1 zXZnx (4.9)

whre X(z) is the z-transform of x(n) and 1Z is the symbol for the inverse z-transform.

8. Z-TRANSFORM

112

Assuming a causal sequence, the z-transform, X(z), in equation 4.7 can be expanded

into a power series as

...)3()2()1()0()()( 321

0

1

zxzxzxxznxzXn

(4.10)

It is seen that the values of x(n) are the cofficients of ,...)1,0( nz n and so can be obtained

directly by inspection. In practice, X(z) is often expressed as a ratio of two polynomials in 1z

or equivalently in z:

M

M

N

N

zazazaa

zbzbzbbzX

...

...)(

2

2

1

10

2

2

1

10 =i

M

i

i

iN

i

i

Za

Zb

0

0 (4.11)

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 methode.

Each method has its own merits and demerits. In terms of mathematical rigor, the residue

method is perhaps the most elegant. The power series method, however, lends it self most

easily to computer implementation.

Computer study

M-file iztrans.m is used to find inverse Z-transform.

Example 8.1

6z5z

z)z(X

2

kU3kU2kx s

k

s

k

8.1 Inverse Z-transforms via long division

For causal sequences, the z-transform X(z) can be expended into a power serises in 1z .

For a rational X(z), a convenient way to determine the power serise is an expansion by long

division.

» syms z k

» x=z/(z^2+5*z+6);

» iztrans(x)

ans =

(-2)^k-(-3)^k

8. Z-TRANSFORM

113

.......zczccaza.......zaz

bzb.......zbzb)z(X 2

2

1

10

01

1n

1n

n

01

1n

1n

n

n

where .....c,c,c 210 are power serise coefficients.

Example 8.2 1z414.1z

z)z(X

12

Computer study

The inverse of a rational z-transform can also be readıly calculated using MATLAB. The

function impz can be utilized for this purpose. Three versions of this function are as follows:

[h,t]=impz(num,den)

[h,t]=impz(num,den, L)

[h,t]=impz(num,den, L,

FT)

Where the input data consists of the vector num and den

containing the coefficients of the numerator and the

denominator polynomials of the z-transform given in the

descending powers of z, the output impulse response vector h,

and the time index vector t. The first form, the length L of h is

determined automatically by the computer with t=0:L-1,

whereas in the remaining two forms it is supplied by the user

through the input data L. In the last form, the sampling interval

is FT

1. The default value of FT is 1. The following two

examples show application dennumimpzth ,, file to and

plot power.

Example 8.3

1414.1

)(2

zz

zzX

z z2-1.414z+1

z-1.414+z-1

z-1

+1.414z-2

+z-3

-z-5

. . .

1.414-z-1

1.414-2z-1

+1.414z-2

z

-1-1.414z

-2

z-1

-1.414z-2

+z-3

-z-3

-z-3

+1.414z-4

-z-5

X[k]=[k-1]+1.414[k-2]+ [k-3]+0-[k-5]. . .

» num=[1 0];

» den=[1 -1.414 +1];

» L=8;

» [x,k]=impz(num,den,L)

x =

1.0000

1.4140

0.9994

-0.0009

-1.0006

-1.4140

-0.9988

0.0017

k =

0

1

2

3

4

5

6

7

» stem(k,x,’fill’,’k’)

Power series coefficiens for

1414.1)(

2

zz

zzX

8. Z-TRANSFORM

114

132

1)(

2

2

zz

zzzX

Example 8.4

Power series coefficients for 132

1)(

2

2

zz

zzzX

0 1 2 3 4 5 6 7 -1.5

-1

-0.5

0

0.5

1

1.5

k

» num=[1 1 -1];

» den= [2 3 1];

» L=10;

» [x,t]=impz(num,den,L)

x =

0.5000

-0.2500

-0.3750

0.6875

-0.8438

0.9219

-0.9609

0.9805

-0.9902

0.9951

» stem(k,x,’fill’,’k’)

1 2 3 4 5 6 7 8 9

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

8. Z-TRANSFORM

115

8.2 The Inverse Z-Transform Using Partial Fractions

We now derive the expression for the inverse z-transform and outline the two

methods for its computation.

Recall that, for jrez , the z-transform G(z) given by the equation is merely the Fourier

transform of the modified sequence krkg . Accordingly, by the inverse Fourier transform,

we have:

de)re(G2

1rkg kjjk

(8.2)

By making the change of variable jrez , the above equation can be converted into a

contour integral given by :

Cinsidepolestheatz)z(Gofresidueskg 1k

(8.3)

Note that theequation mentioned above needs to be equated at all values of k which can be

quite complicated in most cases.

A rational G(z) can be expressed as:

)z(D

)z(P)z(G

where P(z) and D(z) are the polynomials in 1z . If the degree M of the numerator

polynomial P(z) is grester than or equal to the degree N of the denominator polynomial D(z),

we can divide P(z) by D(z) and re-express G(z) as:

)(

)()( 1

0 zD

zPzazG

NM

i

k

i

where the degree of the polynomial )z(P1 is less that that of D(z). The rational function

)z(D

)z(1P is called a proper fraction.

The expression of Eq (8.2) can be computed in a number of ways. Consider the following

cases:

Case 1: G(z) is a proper fraction with simple poles. Let the poles of G(z) be at kpz ,

k=1,2,3,……,N, where pk are distinct. A partial-fraction expansion of G(z) then is of the form

:

N

1k i

i

pz

za)z(G

(8.4)

where the constants l in the above expression, called the residues, are given by:

ipz

iiz

)z(G)pz(a

(8.5)

Each term of the sum on the right-hand side of Eq.(8.4) has an ROC given by z pk,

therefore, the inverse transform g[k] of G(z) is given by

kUpakg s

k

i

N

1k

i

8. Z-TRANSFORM

116

Note that the above approach with slight modifications can also be used to determine the

inverse z-transform of a noncausal sequence with a rational z-transform.

Example 8.5

Let the z-transform of a causal sequence g[k] be given by :

)6.0z)(2.0z(

)0.2z(z)z(G

6.0z

a

2.0z

za)z(G z21

2.0z

1)6.0z)(2.0z(

0.2z)2.0z(a

75.2

8.0

2.2

75.18.0

4.1

)6.0z)(2.0z(

)0.2z()6.0z(a 6.0z2

kU)6.0(75.1)2.0(75.2kg s

kk

Example 8.6

Using MATLAB determine the partial fraction expansion of X(z):

5.1z5.3z2

12z3)z(X

23

3

5.0z

625.2

1z

6.3

5.1z

47.2)z(X

75.0z75.1z

z6z5.1)z(X

23

23

Multiplying the numerator and the denominator by 2.

num=[3 0 0 12];den=[2 -3.5 0 -1.5];

» [r,p,k]=residuez(num,den)

r =

1.9219

3.7891 - 0.3013i

3.7891 + 0.3013i

p =

1.9477

-0.0989 - 0.6126i

-0.0989 + 0.6126i

k =

-8

» r=[ 1.9219 3.7891 - 0.3013i 3.7891 + 0.3013i];

» p=[ 1.9477 -0.0989 - 0.6126i -0.0989 +

0.6126i];

» [num,den]=residuez(r,p,k)

num =

1.5001 -0.0000 0.0008 5.9999

den =

1.0000 -1.7499 -0.0000 -0.7500

8. Z-TRANSFORM

117

5.1z5.3z2

12z3)z(G

23

3

Case 2. G(z) has multiple poles, for example, if the pole at z is of multiplicity r and the

remaining N-r poles are simple and at ,rN,.....3,2,1k,pz k then the general partial-

fraction expansion of G(z) takes the form

r

1ii1ri

rN

ik k

kNM

0k

k

k)z1(

za

pz

zaza)z(G

where the constant ari (no longer called the residues for i1) are computed using the formula:

z

r

ir

ir

riz

)z(G)z(

)z(d

d

)!ir(

1a , i=1,2,3,…..,r

Example 8.7

2

2

)1z(

z)z(X

;

2

2221

)1z(

za

)1z(

za)z(X

2

2221

)1z(

a

)1z(

a

z

)z(X

; 0

z

)z(zXa 0 ; 1

z

)z(X)1z(a

1z

2

21

1z

dz

d

z

)z(X1z

dz

da

1z1z

2

22

Which results in the following time-series

kuk1kx s

Example 8.8 2)1z)(5.0z(

z)z(G

4z

)z(G)5.0z(a

5.0z

1

; 2z

)z(G)1z(a

1z

2

22

;

4z

)z(G)1z(

dz

da

1z

2

21

2)1z(

z2

1z

z4

5.0z

z4)z(G

Consider the following three cases:

1) z1

g[k] kkU2kU4kU)5.0(4 k

2) z2

1

1kUkU ss

g[k] 1kkU21kU41kU)5.0(4 sss

k

3)2

1z1

g[k] 1kkU21kU4kU)5.0(4 sss

k

z1

z2

1

2

1z1

0.5 1.0

0.5 1.0

0.5 1.0

8. Z-TRANSFORM

118

Example 8.9

)5.0z()1z(

z3z5z3)z(X

2

23

2

222111

)1z(

za

1z

za

5.0z

za)z(X

where )5.0z/(za11 an exponential, )1z/(za 21 a step function, and 2

22 )1z/(za a ramp

function. What is desired, however, is the partial fraction expansion of X(z)/z, where:

2

222111

)1z(

a

1z

a

5.0z

a

z

)z(X

where

5)1z(z

z3z5z3

z

)z(X)5.0z(a

5.0z

2

23

5.0z

1

2)2/1z(z

z3z5z3

z

)z(X)1z(a

1z

23

1z

2

22

2))2/1z(z(

)4/1z)(3z5z3(2

)2/1z(z

310z9

z

)z(X)1z(

dz

da

1z

2

23

1z

2

21

which results in

kuk22)5.0(5kx s

k

Example 8.10

Solve using Matlab: 1z4z3z18

z18)z(H

23

3

5.0z

36.0

)z33.01(

4.0

z33.01

24.0)z(H

211

» num=[18]; den=[18 3 -4 -1];

» [r,p,k]=residuez(num,den)

r =

0.2400

0.4000

0.3600

p =

-0.3333

-0.3333

0.5000

k =[]

» [num,den]=residuez(r,p,k)

num =

1.0000 0.0000 0.0000

den =

1.0000 0.1667 -0.2222 -0.0556

8. Z-TRANSFORM

119

Using the numerator and the denominator coefficients we have:

0556.0z2222.0z1667.0z

z)z(X

23

3

It can be seen that the coefficients will be same as in the equation of the question if we

multiply each coefficient by 18.

Example 8.11. Find the inverse Z-transform of

2

3

)5.0z)(5.0z(

)1z()z(X

2

222110

)5.0z(

a

)5.0z(

a

)5.0z(

a

z

a

z

)z(X

8z

)z(zXa 0z0

4

27

)5.0z(

)1z(

dz

d

z

)z(X)5.0z(

dz

da

4

27

)5.0z(

)1z(

z

)z(X)5.0z(a

4

1

)5.0z(

)1z)(5.0z(

z

)z(X)5.0z(a

5.0z

3

5.0z

2

21

5.0z

3

5.0z

2

22

5.0z2

3

5.0z1

2)5.0z(

z

4

27

)5.0z(

z

4

27

)5.0z(

z

4

18

z

)z(X

kU)5.0(k4

27)5.0(

4

27)5.0(

4

1k8kx s

kkk

» num=[1 3 3 1]; den=poly([0 -0.5 0.5 0.5])

den =

1.0000 -0.5000 -0.2500 0.1250 0

» [r,p,k]=residue(num,den)

r =

-0.2500

-6.7500

6.7500

8.0000

p =

-0.5000

0.5000

0.5000

0

k = []

8. Z-TRANSFORM

120

Case 3. X(z) has a complex pole

Example 8.12. The second-order )4/1z)6

cos(z/()z5.1z3()z(X 22

has non

repeated complex roots. The partial expansion of X(z) is defined by:

)z(

za

)z(

za)z(X

*21

)z(

a

)z(

a

z

)z(X*

21

where =0.433 j0.25 and

z

*

2

z

1)z(z

)z5.1z3(

z

)z(X)z(a

*

1

*z*z

*

2 a)z(z

)z5.1z3(

z

)z(X)z(a

Also note that

kuz

1Zkx k1

1

and ku)(z

1Zkx k*

*

1

2

Where =0.433013-j0.25=0.5exp(-j/6). Therefore,

)6/jexp(5.0z

za

)6/jexp(5.0z

za)z(X *

11

which corresponds to a time-series, for k0

)12/6/kcos(2

11.3

))12/6/kjexp()12/j6/kj(exp(2

155.1

)6/kjexp(2

1)12/jexp(55.1)6/kjexp(

2

1)12/jexp(55.1kx

k

k

kk

which is seen to be a causal phase-shifted cosine wave with an exponentially descending

envelope. Also observe that 3)12/cos(1.30x , which can be verified using the initial

value theorem.

8. Z-TRANSFORM

121

8.3 Difference Equations

Long division can be intensive and tedious computational process. If a computer-based signal

processing is desired, the use of difference equation is generally more efficient. Assume that

the Z-transform of a time series is kx is X(z), where

N

0i

i

i

M

0i

i

i

za

zb

)z(X

Recall that the 1kZ and nznkZ . Therefore, it follows that:

Nkb)1N(kb....1kbkb

Mkxa)1M(kxa....1kxakxa

N1N10

M1M10

The response kx can be simulated by implementing the difference equation.

Example 8.13

Consider causal )5.0z)(1z(

z3z5z3)z(X

23

from example 11.

321

21

121

21

23

23

z5.0z2z5.21

z3z53

)z5.01()z1(

z3z53

5.0z2z5.2z

z3z5z3)z(X

Which produces a time-serises

ku)k22)5.0(5(kx s

k

Then kx , for 0k , can be simulated using

2k31k5k33kx5.02kx21kx5.2kx