1497-Chapter3

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Chapter 3

The z-Transform

Der-Feng Tseng

Department of Electrical EngineeringNational Taiwan University of Science and Technology

(through the courtesy of Prof. Peng-Hua Wang of National Taipei University)

February 19, 2015

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Outline

1 3.1 z-Transform

2 3.2 Properties of the ROC

3 3.3 The Inverse z-transform

4 3.4 z-transform Properties

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3.1 z-Transform

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Definition

Definition (z-transform) The z-transform ofx[n] is defined as

Z{x[n]} =X(z)

n=

x[n]zn

The z-transform evaluated on the unit circle corresponds to the

Fourier transform. Letz=rej, we have

X(rej) =

n=

(x[n]rn)ejn.

Fourier transform ofx[n]rn.


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ROC

Now we consider the region in which the z-transform is converge. It is

calledregion of convergence (ROC).z-transform converges:

|X(z)|= |X(rej)|

n=|x[n]rn| < .

Fourier transform converges:

|X(ej)|

n=

|x[n]| < .

It is possible that X(z) converges but X(ej) does not. For example,thez transform of the unit step function converges for |z| >1.

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Examples 3.1 3.6

Example 3.1 x[n] =anu[n], X(z) = 11az1 ,|z| > |a|.

Example 3.2 x[n] = anu[n 1], X(z) = 11az1

,|z| < |a|.

Example 3.3 x[n] = (12)nu[n] + (13)

nu[n],

X(z) = 2z(z 112)

(z 12)(z+ 13)

, |z| > 12

.

Example 3.5 x[n] = (13)nu[n] (12 )

nu[n 1]

Example 3.6 x[n] =an(u[n] u[n N])


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3.2 Properties of the ROC


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Summary of Properties

1 The ROC is a ring or disk centered at the origin;

0 rR < |z| < rL 2 The Fourier transform ofx[n] exists if and only if the ROC includes

the unit circle.

3 The ROC cannot contain any poles.

4

Ifx[n] is a finite-duration sequence, then the ROC is the entirez-plane; except possible z= 0 or z = .

5 Ifx[n] = 0, n < N1< is a right-side sequence, then the ROCextends outward from the outermost finite pole to z= .

6 Ifx[n] = 0, N2 < n is a left-side sequence, then the ROC extendsinward from the innermost nonzero pole to z= 0.

7 A two-side sequence is an infinite-duration sequence. The ROC of a

two-side sequence will consist of a ring in the z-plane.

8 The ROC must be a connected region.


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Properties 1 & 2

Property 1. The ROC is a ring or disk centered at the origin.

The convergence of

X(z)

n=

x[n]zn

depends on only |z|.

Property 2. The Fourier transform ofx[n] exists if and only if the ROCincludes the unit circle.

X(z) reduces to the Fourier transform when |z| = 1.

System is stable. h[n] is absolutely summable. Fouriertransform exists. ROC includes the unit circle.


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Properties 3 & 4

Property 3. The ROC cannot contain any poles.

X(z) does not converge at poles.

Property 4. Ifx[n] is a finite-duration sequence, then the ROC is theentirez-plane; except possible z = 0 orz = .

Ifr= 0,,

|X(z)|= |X(rej)| M2

n=M1

|x[n]rn| < .

For example, let X1(z) = 1 + z1

, X2(z) = 1 + z, andX3(z) =z + 1 + z1. We have ROC1 = Z {0}, ROC2= Z {},

ROC3= Z {0,}.


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Property 5

Property 5. Ifx[n] = 0 forn < N1 is a right-side sequence, then the

ROC extends outward from the outermost finite pole to z= .Suppose x[n] =

Nk=1 Akd

nk , we need

n=N1

|x[n]rn| =

n=N1

N

k=1

Akdnkrn <

or

n=N1

|dnkrn| <

for k= 1, 2, . . . , N . Thus,|dkr1| |dk|


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Example 3.7

If a system has the zero-pole plot shown as follows. There are three

possible ROCs.

System 1: |z| >2: not stable!, causal!

System 2: |z|

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3.3 The Inverse z-transform


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Method 1: Inspection Method

Find the sequence x[n] with the following z-transform

X(z) = 11 12z

1, |z| > 1

2.

Solution:

|z| > 12 X(z) =

n=0

12

nzn x[n] =

12

nu[n]


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Method 2: Partial Fraction Expansion

Let

X(z) =

Mk=0

bkzk

N

k=0akz

k

Case 1. M < N. All poles are simple.

X(z) =N

k=1Ak

1 dkz1, Ak = (1 dkz

1)X(z)

z=dk


M h d 2 P i l F i E i

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Method 2: Partial Fraction Expansion

Case 2. M N. All poles are simple.

X(z) =MNr=0

Brzr +

Nk=1

Ak1 dkz1

Case 3. M N. X(z) has a s-order pole di

X(z) =MNr=0

Brzr +

Nk=1k=i

Ak1 dkz1

+

sm=1

Cm(1 diz1)m

[n r] zr


E l 3 8

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Example 3.8


X(z) = 1(1 14z

1)(1 12z1)

, |z| > 12

Solution:

X(z) = A1

1 1

4z1

+ A2

1 1

2z1

where A1= 1, A2= 2.

x[n] = (2 2n 4n)u[n].


E l 3 9

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Example 3.9


X(z) = (1 + z1)2

(1 12z1)(1 z1)

, |z| >1

Solution:

X(z) =B0+ A1

1 12z1

+ A2

1 z1

where B0= 2, A1= 9, A2 = 8.

x[n] = 2[n] (9 2n + 8)u[n].


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Method 3: Power Series Expansion

X(z) = + x[2]z2 + x[1]z+ x[0] + x[1]z1 + x[2]z2 +

x[n] = + x[2][n+ 2] + x[1][n+ 1] + x[0][n]

+ x[1][n 1] + x[2][n 2] +

Example 3.10

X(z) =z2(11

2z1)(1 + z1)(1 z1)

=z2 1

2

z 1 +1

2

z1

x[n] =[n+ 2] 1

2[n+ 1] [n] +

1

2[n 1]


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Method 3: Power Series Expansion

Example 3.11

X(z) = log(1 + az1) =n=1

(1)n+1anzn

n

x[n] = (1)n+1zn

n , n 1

0, n 0.


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3.4 z-transform Properties


Properties
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Properties

Linearity.

Z{ax1[n] + bx2[n]} =aZ{x1[n]} + bZ{x2[n]} ROC= Rx1Rx2

Time-Shifting.

Z{x[n n0]} =zn0Z{x[n]} ROC=Rx {z = or0}

Multiplication. IfZ{x[n]} =X(z), then

Z{zn0 x[n]} =X(z/z0)

with ROC= |z0|Rx.

Differentiation. IfZ{x[n]}=X(z), then

Z{nx[n]} = z d

dzX(z)

with ROC= Rx.


Properties

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Properties

Complex Conjugation. IfZ{x[n]} =X(z), then

Z{x[n]} =X(z)

with ROC= Rx.

Time Reversal. IfZ{x[n]} =X(z), then

Z{x[n]} =X(1/z)

with ROC= 1/Rx.

Time Reversal. Ifx[n] is real, then

Z{x[n]} =X(1/z)

with ROC= 1/Rx.


Properties

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Properties

Convolution.

Z{x1[n] x2[n]} = Z{x1[n]}Z{x2[n]} ROC= Rx1Rx2

Initial-Value Theorem. Ifx[n] is causal, then

lim

z

X(z) =x[0]


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