EN5101 Digital Control Systems

z-Transforms

1

Prof. Rohan Munasinghe

Dept of Electronic and Telecommunication Engineering

University of Moratuwa

Contents

• Forward z-transform

• Inverse z transform methods

– Power series, Recursive, Partial Fraction expansion

• z-transform Applications

• Applications of z-transform

– Pole-zero description of discrete-time systems

– Stability consideration

– Difference Equations

2

Discrete- time or sampled signal

n

x(n)

-1

1

2

1 2 3 4 5 6 7 8-2 9 10 11

3

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

• Example of discrete time systems are digital controllers, digital spectrum analyzers, and digital filters.

Discrete time systemx(n) y(n)

Analysis of Sampled Data Systems

4

Problem with the Laplace in discrete systems

5

Laplace (Continuous) to z (Discrete)

6

The z-transform

• The z-transform of a sequence, x(n), which is valid for all n, is defined as power series

where X(z) = Z{x(n)} and z=rejω is a complex variable

∞

−∞=

−=n

nznxzX )()(

7

• In causal system x(n) = 0 for n < 0, x(n) may be nonzero

value only in the interval 0< n< ∞, so one sided z-transform can be written as follow.

• Clearly, the z-transform is a power series with an infinite number of terms and so may not converge for all values of z.

∞

=

−=0

)()(n

nznxzX

Region of convergence (ROC) of Z- transform

• The region of convergence (ROC) of X(z) is the set of all

values of z for which X(z) attains a finite value.

• The region where the z-transform converges is known as

the region of convergence (ROC) and in this region, the

values of X(z) are finite.

• Thus, any time we cite a z-transform we should also

indicate its ROC.

8

Poles & Zeros of X(z)

• Values of z for which X(z) attains ∞ are referred to as poles of X(z).

• Values of z for which X(z) attains 0 are referred to as the zeroes of X(z).

For example

82

10133)(

2

2

−+

−−=

zz

zzzX

Find the poles and zeroes of X(z).

9

Solution:

• To find poles and zeroes of X(z), denominator and numerator of

the X(z) are required to factorize first. Then,

• To find poles, set the denominator of X(z) = 0

( z + 4) (z - 2) = 0

(z + 4) = 0 or (z - 2) = 0

Thus, z = - 4 and z = 2 are poles of X(z).

• To find zeroes X(z) = 0 set the numerator of X(z)=0

i.e (3z + 2) (z - 5) = 0,

(3z + 2) = 0 or (z - 5) = 0

Thus, z = -2/3 and z = 5 are zeros of the X(z).

)2)(4(

)5)(23(

82

10133)(

2

2

−+−+

=−+

−−=

zz

zz

zz

zzzX

10

Example

Determine the z-transform of the following finite-duration

signals.

• (a) x1(n) = [1,2,5,7,0,1]

• (b) x2(n) = [1,2,5,7,0,1]

↑

• (c) x3(n) = [0,0,1,2,5,7,0,1]

• (d) x4(n) = [2,4,5,7,0,1]

↑

• (e) x5(n) = δ(n)

• (f) x6(n) = δ(n - k), k >0

• (g) x7(n) = δ(n + k), k >0

11

Solution

x1(n) = [1,2,5,7,0,1]

x1(n) is finite length sequence. It is causal case and having

n = 0 to 5. Therefore,

x1(0)=1, x1(1)=2, x1(2)=5, x1(3)=7, x1(4)=0, x1(5)=1

5321

54321

5

1

4

1

3

1

2

1

1

1

0

1

5

0

11

7521

107521

)5()4()3()2()1()0(

)()(

−−−−

−−−−−

−−−−−−

=

−

++++=

+++++=

+++++=

=

zzzz

zzzzz

zxzxzxzxzxzx

znxzXn

n

The ROC is entire z-plane except at z = 0.

12

X2(n) = [1,2, 5,7,0,1]

X2(n) is finite length sequence and double sided (non-

causal case) having n=-2 to 3. Therefore,

x2(-2)=1, x2(-1)=2, x2(0)=5, x2(1)=7, x2(2)=0, x2(3)=1

312

321012

3

2

2

2

1

2

0

2

)1(

2

)2(

2

3

2

22

752

107521

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

)()(

−−

−−−

−−−−−−−−

−=

−

++++=

+++++=

++++−+−=

=

zzzz

zzzzzz

zxzxzxzxzxzx

znxzXn

n

The ROC is entire z-plane except at z = 0 and z = ∞.

13

(e) x5(n) = δ(n)

By definition,

111

)0()(

)()(

0

=×=

==

=

−∞

−∞=

−

∞

−∞=

−

zzn

znxzX

n

n

n

n

δδ

=

=else

nforn

0

01)(δ

By using general formula,

14

z Transform of a sampled unit step

15

Example

Determine the z-transform of the signal

x(n) = (1/2)n u(n).

Solution:

The signal x(n) consists of an infinite number of non-zero

samples for n> 0.

x(n) = [1, (1/2), (1/2)2, (1/2)3,…………, (1/2)n,….]

The z-transform of the x(n) is the infinite power series

X(z) = 1 + ½ z-1 + (1/2)2 z -2 + ……(1/2)n z -n +….

( )

<

>=

0

0

0

1

nfor

nfornu

16

1||1

1.........1

32 <−

=++++ aifa

aaa

∞

=

−∞

=

− =

=0

1

0

)2

1(

2

1)(

n

n

n

n

n

zzzX

• This is an infinite geometric series, we recall that

where ‘a’ is the common ratio of the series. Consequently,

2/1|| :ROC

1)2/1()2/1(1

1)(

1

1

>

<−

= −−

z

zifz

zX

17

Example

Find the z-transform and the region of convergence for

each of the discrete-time sequence

18Fig 1 sample sequences

(a) The z-transform is given by

• It is readily verified that the value of X(z) becomes

infinite when z = ∞. Thus the ROC is everywhere in

the z- plane except at z = ∞.

12345

1

0)1()2()3()4()5()6(

1

0)1()2(

)3()4()5()6(

1

0

6

6

7

353)(

0135310)(

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

)3()4()5()6()(

)()()()(

zzzzzzX

zzzzzzzzX

zxzxzx

zxzxzxzzX

znxznxznxzXn

n

n

n

n

n

++++=

++++++=

+−+−+

−+−+−+−=

===

−−−−−−−−−−−−

−−−−

−−−−−−−−

−=

−

−=

−∞

−∞=

−

19

212

2

3210123

2

321

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

2

3

3

5

4

2

353)(

.0.1.35.3.1.0)(

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

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

)()()()(

−−

−−−

−−−

−−−−−−−

−=

−

−=

−∞

−∞=

−

++++=

++++++=

+++

+−+−+−=

===

zzzzzX

zzzzzzzzX

Zxzxzx

zxzxzxzxzX

znxznxznxzXn

n

n

n

n

n

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

z = ∞. So ROC is everywhere except z =0 and z = ∞.

20

(b) Again, the sequence in figure 1(b) is not causal. It is of a

finite duration, and double sided.

(d) It is a causal sequence of infinite duration. The z-transform of the

sequence is given by:

• This is a geometric series with a common ratio of Z-1.

• Generally, geometric series with a common ratio of ‘a’ can be

expressed as

• Which is a closed form of power series with common ratio ‘a’ and

the series converges if

......1)(

)()(

)()(

21

4

0

1

0

4

+++=

==

=

−−

∞

=

−∞

=

−

∞

−∞=

−

zzzX

zzzX

znxzX

n

n

n

n

n

n

.1<aaa

n

n

−=

∞

= 1

1

0

21

• Thus, power series of X4(z) converges if

• Thus, we may express X (z) in closed form provided that region of convergence ( ROC)

• In this case, the z-transform is valid everywhere outside a circle of unit radius (ie |z| =1) whose centre is at the origin. The exterior of the circle is the region of convergence.

:1>z

1)1(

1)(

......1)(

1

21

−=

−=

+++=

−

−−

z

z

zzX

zzzX

.111 ><− ziflyequivalentandz

22

The Unit Circle in the Complex z- Plane

23

The ROC of sequence (d), the shaded area is ROC

Unit Circle

Z Transform of a sampled exponential

24

25 26

The inverse z transform

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

• The inverse z-transform (IZT) is particularly useful in DSP work for example in finding the impulse response of digital filters.

• Symbolically, the inverse z-transform may be defined as:

x(n) = Z -1 [X(z)]

27

• where X(z) is the z-transform of x(n) and Z -1 is the

symbol for the inverse z-transform.

• Assuming a causal sequence, the z-transform, X(z)

can be expanded into a power series as follow.

• It is seen that the values of x(n) are the coefficients of

z –n ( for n= 0, 1,2, ……) and so can be obtained

directly by inspection.

......)4()3()2()1()0()(

)()(

4321

3

0

+++++=

=

−−−−

∞

=

−

zxzxzxzxxzX

znxzXn

n

28

• In practice, X (z) is often expressed as a ratio of two

polynomials in z -1 or equivalently in z,

• In this form, the inverse z-transform, x (n), may be

obtained using one of several methods including the

following three:

(a) Power series expansion method

(b) Recursive method

(c) Partial fraction expansion method

)1.4........(......

......)(

2

2

1

10

2

2

1

10

M

N

N

N

zbzbzbb

zazazaazX

−−−

−−−

++++++++

=

29

Power Series Method

• Given the z-transform X(z) of a causal sequence, it can

be expanded into an infinite series in z -1 or z by long

division ( sometimes called synthetic division):

• In this method, the numerator and denominator of X(z)

are first expressed in either descending powers of z or

ascending powers of z -1 and the quotient is then

obtained by long division.

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

......

......)(

321

2

2

1

10

2

2

1

10

++++=

++++++++

=

−−−

−−−

−−−

zxzxzxxzX

zbzbzbb

zazazaazX

M

N

N

N

30

Example

• Given the following z-transform of a causal LTI system, obtain its IZT (Inverse Z- Transform) by expanding it into a power series using long division:

First, we expand X(z) into a power series with the numerator and denominator polynomials in ascending powers of z -1

and then perform the usual long division.

21

21

3561.01

21)(

−−

−−

+−++

=zz

zzzX

31

+ 3.6439

32

33

Recursive Method

M

N

N

N

zbzbzbb

zazazaazX

−−−

−−−

++++++++

=......

......)(

2

2

1

10

2

2

1

10

...3,2,1,.../)()( 0

1

=

−−=

=

nbbinxanxn

i

in

x(0) = a0/b0

x(1) = [a1-x(0)b1]/b0

x(2)=[a2 – x(1)b1 – x(0)b2 ]/b0

Generally,

where x(0) = a0/b0

We will repeat the previous example to illustrate the

recursive approach.

34

Example

Find the first four terms of the inverse z-transform, x (n),

using the recursive approach. Assume that the z-transform,

X(z), is the same as in example 4.5, that is:

SOLUTION:

Comparing the coefficients of X(z) above with those of

the general transform, we have,

a0= 1, a1 = 2, a2 = 1,

b0= 1, b1 = -1, b2 = 0.3561,

N = M = 2

21

21

3561.01

21)(

−−

−−

+−++

=zz

zzzX

35

By using general equation, we have,

x(0) = a0/ b0 = 1/1 = 1

x(1) = [ a1 - x(0) b1]/ b0 = [2-1 × (-1)]/1 = 3

x(2) = [ a2 - x(1) b1- x(0)b2]/b0 = [1- 3 ×(-1) -1 ×0.3561]/1= 3.6439

x(3) = [ a3 - x(2) b1- x(1)b2 + x(0) b3]/ b0

= 0 – x(2) b1 – x(1)b2

= [0 – 3.646439 × (-1) - 3 × 0.3561]/1 = 2.5756X(4) = [ a4 - x(3) b1- x(2)b2 + x(1) b3 + x(0) b4 ]/ b0

= [0 -2.5756 × (-1) – 3.6439 × 0.356 + 3 × 0 + 1 × 0]/1= 2.5756 - 1.2972 + 0+0 = 1.2784

Thus the first four values of the inverse z-transform

are:

X(0) =1, x(1) = 3, x(2) = 3.6439, x(3) = 2.5756

It is seen that both the recursive and direct, long division

methods lead to identical solutions.

36

Partial fraction expansion method

• In this method, the z-transform is first into a sum of simple partial fractions. The inverse z-transform of each partial fraction is then obtained from z transform tables and then summed to give the overall inverse z-transform.

• In many practical cases, the z-transform is given as a ratio of polynomials in z or z-1 and has the now familiar form.

37

M

N

N

N

zbzbzbb

zazazaazX

−−−

−−−

++++

++++=

......

......)(

2

2

1

10

2

2

1

10

)2.4.........(..................

......)(

1......

11)(

1

0

2

2

1

10

11

2

2

1

1

10

=

−−−

−+=

−++

−+

−+=

−++

−+

−+=

M

k k

k

M

M

M

M

pz

zCB

pz

zC

pz

zC

pz

zCBzX

zp

C

zp

C

zp

CBzX

If the poles of X(z) are of first order, X(z) can be expanded as:

where pk are the poles of X(z) (assumed distinct), Ck are the partial fraction coefficients and

B0 = aN/bN

38

• Case 1: If N ≤ M ie the order of the numerator is less

than that of the denominator in X(z) expression, then B0

will be zero.

• Case 2: If N>M, then X(z) must be reduced first to make

N ≤ M by long division.

• The coefficient, Ck, associated with the pole pk may be

obtained by multiplying both sides of the equation by (z -

pk) /z and then letting z = pk.

kpzkk pz

z

zXC

=−= )(

)(

39

• Case 3: If X(z) contains one or more multiple-order poles ( that is poles that are coincident) then extra terms are required in the partial fraction equation.

• For example, if X(z) contains an mth-order pole at

z = pk the partial fraction expansion must include terms of the form

40

The coefficients, Di, may be obtained from the relationship

Evaluation of inverse z-transforms by the partial fraction

expansion method is best illustrated by examples.

41

Find the inverse z-transform of the following transfer function H(z).

)61(

)1()(

21

1

−−

−

−−−

=zz

zzH

Example

Solution:

Since N < M, case 1.

First, H(z) is converted for positive power of z.

42

)21()31()(

11 −− ++

−=

z

B

z

AzH

To find A

( ) ( )

5

2

3/5

3/2

)3/121(

3/11

)21(

131

)21(31

1

|)31)((

3/1

1

1

3/1

1

11

1

3/1

1

11

1

==×+

−=

+−

=−+−

−=

−=

=−

−

=

−−−

−

=

−

−−

−

A

z

zz

zz

zA

zzHA

zz

z

43

( ) ( )

5/32/5

2/3

2/31

2/11

)2/1(31

)2/1(1

)31(

121

)21(31

1

|)21)((

2/1

1

1

2/1

1

11

1

2/1

1

11

1

==++

=−×−

−−=

−−

=++−

−=

+=

−=−

−

−=

−−−

−

−=

−

−−

−

B

z

zz

zz

zB

zzHB

zz

z

To find B

By comparing to the z transform

11

1)(

)()(

−−=

=

zzX

nunx n

α

α

)21(

5/3

)31(

5/2)(

11 −− ++

−=

zzzH

The unit sample response of given system is:

h(n) = (2/5)(3)nu(n) + 3/5 (-2)nu(n)

= [(2/5)(3)n + 3/5 (-2)n]u(n)

44

Example

Find the inverse z-transform of the following X(z):

Solution:

we get, N =1, M =2,

Thus, N<M Case 1: B0 =0

21

1

375.025.01)(

−−

−

−−=

zz

zzX

45

• To change positive power of z by multiplying all terms

with z2 we get,

46

375.025.0)(

2 −−=

zz

zzX

)5.0)(75.0()(

+−=

zz

zzX

X(z) contains first-order poles at z= 0.75 and at z= -0.5.

Since N< M, the partial fraction expansion has the form,

)5.0()75.0()5.0)(75.0(

1)( 21

++

−=

+−=

z

C

z

C

zzz

zX

5/4)5.075.0(

1

)5.0(

1

)5.0)(75.0(

1)75.0()()75.0(

75.0

1

75.075.0

1

=+

=+

=

+−−

=−

=

=

==

z

zz

zC

zz

z

z

zXzC

To find C1

47

5/4)75.05.0(

1

)75.0(

1

)5.0)(75.0(

1)5.0()()5.0(

5.0

2

5.05.0

2

−=−−

=−

=

+−+

=+

=

−=

−=−=

z

zz

zC

zz

z

z

zXzC

To find C2

)5.0(

.5/4

)75.0(

.5/4)(

)5.0(

5/4

)75.0(

5/4

)5.0()75.0(

)( 21

+−

−=

+−

−=

++

−=

z

z

z

zzX

zzz

C

z

C

z

zX

Then

48

From the z-transform table k an ⇔ k z /(z-a), therefore

nn

z

zZ

z

zZ )5.0(5/4

)5.0(

.5/4 and )75.0(5/4

)75.0(

.5/4 11 −−=

+−

=

−−−

The desired inverse z-transform x(n) is then

x(n) = 4/5 [(0.75)n –(-0.5) n], n>0

Example

2

2

)1)(5.0()(

−−=

zz

zzX

Find the discrete-time sequence x(n) with the following z-

transform

Solution:

X(z) has a first-order pole at z = 0.5 and a second order pole at z =1. (Case 3)

49

11

2

2

1

2

15.0)1)(5.0(

)1()()1(

===

−=

−−−

=

−=

zzzz

z

dz

d

zz

zz

dz

d

z

zXz

dz

dD

2

21

2)1()1()5.0()1)(5.0(

)(

−+

−+

−=

−−=

z

D

z

D

z

C

zz

z

z

zX

2)15.0(

5.0

)1()1)(5.0(

)5.0(2

5.0

2

5.0

2=

−=

−=

−−−

=== zz

z

z

zz

zzC

The partial fraction expansion has the form,

To obtain D1 ,

501

2

1

2

1

21

2

1

1

)5.0(

5.0

)5.0(

5.0

)5.0(

1.1).5.0(

5.0

===

=

−−

=−

−−=

−−−

=

−=

→−→

=

zzz

z

zz

zz

z

zzD

v

udvvdu

vz

uz

dz

dD

25.01

1

5.0

)()1(

11

2

2 =−

=−

=−

=== zz

z

z

z

zXzD

2)1(

2

)1(

2

)5.0(

2)(

−+

−−

−=

z

z

z

z

z

zzX

To obtain D2,

Combining the results, X(z) becomes,

51

The inverse z-transform of each term on the right hand

side is obtained from table 1 and summed to give x(n) as

follows.

x(n) = 2(0.5)n – 2+ 2n=2[(n-1) + (0.5) n] , n ≥0

52

Example

Properties of the z-transform

(1) Linearity

If the sequences x1(n) and x2(n) have a z-transform X1(z)

and X2(z), then the z-transform of their linear combination

is:

ax1(n) + bx2(n) ⇔ aX1(z) + bX2(z)

and the ROC will include the intersection of Rx and Ry, that

is Rx ∩ Ry..

53

Shifting a sequence (delaying or advancing) multiplies

the z-transform by a power of z, that is to say, if x(n)

has a z-transform X(z)

x(n – k) ⇔ z-kX(z)

Shifting does not change the region of convergence. Therefore, the z-transform of x(n) and x(n – k) have the same region of convergence.

Eg. x(n-2) ⇔ z-2X(z)

Shifting Property

54

55

z Transform of a shifted sequence

Properties: Time Reversal

If x(n) has a z-transform X(z) with a region of

convergence Rx that is the annulus α < |z| < β, the

z-transform of the time-reversal sequence x (-n) is

x (-n) ⇔ X(z -1)

and has a region of convergence 1/β < |z| < 1/α,

which is denoted by 1/Rx.

56

Multiplication by an Exponential

If a sequence x(n) is multiplied by a complex exponential αn,

αnx(n) ⇔ X(α-1z)

This corresponds to a scaling of the z-plane. If the region of

convergence of X(z) is r- < |z| < r+, which will be denoted by Rx,

the ROC of X(α-1z) is |α|r- < |z| < |α|r+, which is denoted by |α|Rx. As a special case, note that if x(n) is multiplied by a complex

exponential, ejnω0,

ejnω0x(n) ⇔ X(e-jω0z)

which corresponds to a rotation of the z-plane.

57

Properties: Convolution Theorem

Perhaps the most important z-transform property is the

convolution theorem, which states that convolution in the time

domain is mapped into multiplication in the frequency domain,

that is,

The region of convergence of Y(z) includes the interaction of Rx

and Ry, Rw contains Rx∩Ry However, the region of convergence of Y(z) may be larger, if there is a pole-zero cancellation in the

product X(z)H(z).

58

59

For a given n -

convolution

length

Real Convolution Theorem

60

Generalized

for any n

Example

Consider the two sequences

x(n) = αnu(n) and h(n) = δ(n) - αδ(n – 1)If y(n) is convolution of x(n) and h(n), find the sequence y(n)

Solution:The z-transform of x(n) is

X(z) = 1/(1 - αz-1) |z| > |α| and the z-transform of h(n) is

H(z) = 1 - αz-1 |z| > 0

• However, the z-transform of the convolution of x(n) with h(n) is Y(z) = X(z)H(z) = 1/(1 - αz -1)(1 - αz-1) = 1

• which, due to a pole-zero cancellation, has a region of convergence, that is the entire z-plane. By taking Inverse z-transform, y(n) = δ(n)

61

Properties: Derivative

If X(z) is the z-transform of x(n), the z-transform of nx(n) is

Repeated application of this property allows for the evaluation of the z-transform of nkx(n) for any integer k.

62

( )zXzn0−

( )dz

zdXz−

Property Sequence z-Transform RoC

Linearity ax(n) + by(n) aX(z) + bY(z) Contains Rx ∩ Ry

Shift x(n – n0) Rx

Time Reversal x(-n) X(z-1) 1/Rx

Exponential αnx(n) X(α-1z) αRx

Convolution x(n) ∗ y(n) X(z) Y(z) Contains Rx ∩ Ry

Conjugation x*(n) X*(z*) Rx

Derivative nx(n) Rx

Table 2: Properties of z- transform

63

A property that may be used to find the initial value of a causal

sequence from its z-transform is the initial value theorem.

If x(n) is causal, ie x(n)=0 for n<0], then

X(z) = x(0) + x(1)z-1 + x(2)z-2 + …

and

( ) ( )zXxz ∞→

= lim0

Initial Value Theorem

64

Example

Let x(n) be a left-sided sequence that is equal to zero for n > 0. If X(z) = (3z-1 + 2z-2)/(3 – z-1 + z-2), find x(0).

Solution:

For a left-sided sequence that is zero for n>0, the

z-transform is X(z) = x(0) + x(-1)z + x(-2)z2 + ………..

Therefore, it follows that

For the given z-transform, we see that

)(lim)0(0

zXxz →

=

( ) 213

23lim

3

23limlim)0(

202

2

21

21

00

=+−

+=×

+−

+==

→−−

−−

→→ zz

z

z

z

zz

zzzXx

zzz

65

Example

Generalize the initial value theorem to find the value of

a causal sequence x(n) at n = 1, and find x(1) when

X(z) = (2 + 6z-1)/(4 – 2z-2 + 13z -3).

Solution:

If x(n) is causal,

X(z) = x(0) + x(1)z-1 + x(2)z-2 + …

Therefore, note that if we subtract x(0) from X(z),

X(z) – x(0) = x(1)z-1 + x(2)z-2 + …

By multiplying both sides of this equation by z, we have,

z[X(z) – x(0)] = x(1) + x(2)z-1 + …

66

If we let z → ∞, we obtain the value for x(1), therefore,

x(1) = {z[X(z) – x(0)]}

Example: For the z-transform32

1

1324

62)(

−−

−

+−

+=

zz

zzX

∞→zlim

( )2

1

1324

62limlim)0(

32

1

=+−

+==

−−

−

∞→∞→ zz

zzXx

zz

67

Then, X(z) – x(0) = X(z) - ½ is:

)1324(

2/136

)1324(2

13212

)1324(2

1324124

2

1

1324

62

2

1)(

32

321

32

321

32

321

32

1

−−

−−−

−−

−−−

−−

−−−

−−

−

+−

−+=

+−

−+=

+−

−+−+=

−+−

+=−

zz

zzz

zz

zzz

zz

zzz

zz

zzX

68

32

21

32

321

1324

2/136)]0()([

1324

2/136)]0()([

−−

−−

−−

−−−

+−

++=−

+−

++×=−

zz

zzxzXz

zz

zzzzxzXz

Then,

( )

2

3

1324

2/136lim)1(

)]}0([{lim)1(

32

21

=+−

++=

−=

−−

−−

∞→

∞→

zz

zzx

xzXzx

z

z

Consequently,

The One-Sided z-Transform

• The z-transform defined in previous section is the two-

sided, or bilateral, z-transform. The one-sided, or

unilateral, z-transform is defined by:

• The primary use of the one-sided z-transform is to solve

linear constant coefficient difference equations that have

initial conditions.

( ) ( )∞

=

−=0

1n

nznxzX

69

• Most of the properties of the one-sided z-transform are the same as those for the two-sided z-transform. One that is different, however, is the shift property.

• Specifically, if x(n) has a one-sided z-transform X(z), the one-sided z-transform of x(n – k) where k > 0 is:

• Then, one-sided z-transform of x(n – 1) and x(n-2) are:

−+−

=

−k

n

nk znxzXzknx1

1 )()()(

70

)2()1()()2(

)1()()1(

1

1

2

1

1

−+−+−

−+−−−

−

xxzzXznx

xzXznx

Example

Find the solution of the linear constant coefficient difference equation

y(n)=0.25 y(n–2) + x(n) for x(n)=δ(n–1) with y(-1)=y(-2)=1.

Solution:

The given equation has initial conditions. Thus, We need to apply one-sided z-transform to find its solution. If the one-sided z-transform of y(n) is Y1(z), the one-sided z-transform of y(n–2) is

)2()1()()2(1

1

2

0

−+−+=− −−∞

=

− yzyzYzznyn

n

71

Therefore, taking the z-transform of both sides of the

difference equation,

• y(n) = 0.25y(n – 2) + x(n), we have

• Y1(z) = 0.25[y(-2) + y(-1)z -1 + z-2Y1(z)] + X1(z)

For x(n) = δ(n – 1)

• X1(z) = z-1 × 1 = z-1

Substituting for y(-1) and y(-2), and solving for Y1(z), we have

Y1(z) = 0.25[1 + z -1 + z-2Y1(z)] + z -1

Y1(z) = ¼ + ¼ z -1 + ¼ z-2Y1(z)+ z -1

Y1(z) - ¼ z-2Y1(z) = ¼ + 5/4 z -1

Y1(z)[1 - ¼ z-2] = ¼ [(1 + 5z-1)]

Y1(z) = ¼ (1 + 5z-1)]/ [1 - ¼ z-2] 72

• Y1(z) = ¼ (1 + 5z-1)]/ [1 - ¼ z-2] and it is equivalent to

• Y1(z) may be expanded by using partial fraction

expansions as follow.

)2/11)(2/11(

)51(4/1

)4/11(

)51(4/1)(

11

1

2

1

1 −−

−

−

−

+−

+=

−

+=

zz

z

z

zzY

)2/11()2/11()(

)2/11)(2/11(

)51(4/1)(

111

11

1

1

−−

−−

−

++

−=

+−+

=

z

B

z

AzY

zz

zzY

73

And y(n) = [(11/8)(1/2)n – (9/8)(-1/2)n]u(n)

( )

( )

8

11

2

4/11

22/11

)251(4/1

)2/11(

)51(4/12/11

)2/11)(2/11(

)51(4/1

2/11)(

2

1

1

2

1

11

1

2

1

11

1

==×+×+

=

++

=−+−

+=

−=

=−

−

=

−−−

−

=

−

−−

−

A

z

zz

zz

zA

zzHA

zz

z

8/92

4/9

)2(2/11

)]2(51[4/1

)2/11(

)51(4/1)2/11)((

2

1

1

2

1

1

1

−=−

=

−×−−+

=−

+=+=

−=−

−

−=

−

−

−

B

z

zzzHB

zz

)2/11(

8/9

)2/11(

8/11)(

111 −− +−

−=

zzzY

74

Then

75

Application of z Transforms in Discrete

Systems (Difference Equations)

HW: Solve for y(k)

76

Car Loan Problem

HW: Solve for y(k)