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DIGITAL SIGNAL PROCESSING

Z-TRANSFORM AND ITSAPPLICATIONS TO THE

ANALYSIS OF LTI SYSTEMS

Lectured by: Assoc. Prof. Dr. Thuong Le-Tien

National Distinguished Lecturer

September, 20111

Lecturer by Assoc.Prof.Dr. Thuong Le-Tien

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What the chapter can be presented.

1. Basis properties

2. Region of Convergence (ROC)3. Causality and Stability

4. Frequency spectrum

5. Inverse Z-Transform


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1. Basis Properties

Z-transform is basically as a tool for theanalysis,Design and implementation of digitalfilters. Z transform of a discrete time signalx(n)

X(z) = +x(-2)z2 + x(-1)z + x(0) + x(1)z-1 +x(2)z-2 +

if x(n) is causal, only negative power z-n,n 0 appear in the expansion.

n

n

nznxzX


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Z-transform of the transfer function h(n):

Example:

(a) h = {h0, h1, h2, h3} = {2,3,5,2}(b) h = {h0, h1, h2, h3, h4} = {1,0,0,0,-1}

Their Z-transform

a) H(z)= h0 + h1z-1 + h2 z

-2 + h3 z-3

= 2 + 3z-1 + 5z-2 + 2z-3

b) H(z)= h0 + h1z-1 + h2 z

-2 + h3 z-3 + h4 z

-4 =1 - z-4

n

n

nznhzH


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Z-transform has the three most important

properties that facilitate the analysis andsynthesis of linear systems

* Linearity property

* Delay property

* Convolution property

zXazXanxanxa 2211Z

2211

zXzDnxzXnx

DZZ

zHzXzYnx*nhny


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Example: Two filters of the above filters can be written

in the following closed forms(a) h(n) = 2(n) + 3(n-1) + 5(n-2) + 2(n-3)(b) h(n) = (n) - (n-4)Their transfer functions can be obtained using the

linearity and delay properties.z-transfrom of(n) is unity.

1z0znn 0n

n

nZ

,...1.3

,1.2

,1.1

33

22

11

zzn

zzn

zzn

Z

Z

Z


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Example: using the unit step identity u(n)-u(n-1)=(n),valid for all n, and the z-transform properties, determinethe z-transforms of two signals:

(a) x(n) = u(n) (causal) (b) x(n) = -u(-n-1) (anticausal)Solve:

(a) x(n) - x(n-1) = u(n) - u(n-1) = (n)

11Z

z1

1zX1zXzzXn1nxnx

321Z z2z5z323n22n51n3n2

4Z z1zH4nnnh

(b) x(n)-x(n-1)=-u(-n-1)+u(-(n-1)-1)= u(-n)-u(-n-1)=(-n)

1

1Z

z1

1zX1zXzzXn1nxnx

7Lecturer by Assoc.Prof.Dr. Thuong Le-Tien

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Example: determine the output by carrying out the

convolution operation as multiplication in z-domainh={1,2,-1,1}, x={1,1,2,1,2,2,1,1}

SolveZ-transform

H(z)= 1 + 2z-1 - z-2 + z-3

X(z)= 1 + z-1 + 2z-2 + z-3 + 2z-4 + 2z-5 + z-6 + z-7

Y(z) = X(z)H(z)Y(z)= 1 + 3z-1 + 3z-2 + 5z-3 + 3z-4 + 7z-5 + 4z-6 + 3z-7 + 3z-8 +z-10

The coefficients of the powers of z are the convolutionoutput samples:

y=h*x={1, 3, 3, 5, 3, 7, 4, 3, 3, 3, 0, 1}


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2. Region of Convergence (ROC)

ROC of X(z) is defined to be that subset of thecomplex z-plane C for which the series of the

formula converges, that is

The ROC is an important concept in manyrespects: It allows the unique inversion of theZ-transform and provides convenient

characterizations of the causality and stabilityproperties of a signal or system.

n

n

nznxzXCzROC


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Example, a causal signal:

x(n)=(0.5)nu(n)={1,0.5,0.52,}

Using the infinite geometric series formula

Which is valid for x < 1 and diverges otherwiseThe convergence of the geometric series requires:

Then, ROC={zCz>0.5} outside the circle of radius 0.5

x1

1xz5.0zX

0n

n

0n

n1

5.0zz

z5.01

1zX 1

5.0z1z5.0x 1

x1

1

x...xxx1 0n

n32


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Example for an anticausal signal x(n)=-(0.5)nu(-n-1)

Convergence with x < 1 and diverges otherwiseLet x=0.5z-1,

The same result as the causal case except the ROC

x1

xx...xxx1m

m32

1m1

1m

1m

m1

z5.01

z5.0

x1

xxz5.0zX

1z5.01

1

5.0z

zzX

1m

m11

n

n11

n

nn z5.0z5.0z5.0zX


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5.0z1z5.0x 1

5.0zCzROC

5.0z:ROCwhere

,5.01

115.0

5.0z:ROCwhere

,5.01

15.0

1

1

znu

znu

Zn

Zn

az:ROCwhere,az1

11nua

az:ROCwhere,az1

1nua

1

Zn

1

Zn

To summarize, the z-transform

Generally


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Example:

1zwhere,1

111

1zwhere,

1

11

1zwhere,111

1zwhere,1

1

1

1

1

1

znu

z

nu

znu

znu

Zn

Zn

Z

Z


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Example: determine z-transforms and ROCs

,...}0,1,0,1,0,1,0,1,0,1,0,1{

2

1

,...0,1,0,1,0,1,0,11

nu2

ncosx(n)6.

u(n)(-0.8)u(n)(0.8)x(n)5.

1)-u(nu(n)2

1x(n)4.

10)-u(n-u(n)(-0.8)x(n)3.

u(n)(-0.8)x(n)2.

10)-u(nx(n)1.

nn

n

n

n


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Solve

(1) Delay property: ROC z > 1.

(2)

(3) x(n) = (-0.8)nu(n) - (-0.8)10(-0.8)n -10 u(n-10))

Using the finite geometric series

1

1010

1

1010

1z8.01

z8.01

z8.01

z8.0

z8.01

1zX

x1

x1

x...xx1

N1N2

110

10

z1zzUzzX

8.08.0z:ROCwith,8.01

11

zX

1

1010

1

101099221

z8.01

z8.01

az1

za1za...zaaz1zX


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(4) ROC z > 1.

(5)

ROC z>0.8

(6)

211 z64.01

1

z8.01

1

z8.01

1

2

1zX

)n(ua)n(ua2

1)n(ue)n(ue

2

1nu

2

ncos)n(x n*n2/nj2/nj

211

1

1

1

1

1

1

2

1

zzzzX

a=ej/2=j and a*=e-j/2=-j.

211 z1

1

jz1

1

jz1

1

2

1zX


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3. Causality and Stability*Causal case:

The common ROC of all terms:

Anticausal case

ROC

...1nupA1nupAnx n22n11

21 pz,pz

...nupAnupAnx n22n11

...zp1

Azp1

AzX 12

21

1

1

21 pz,pz ii

pmaxz

ii pz min


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Example:

Find the Z-Transform and possible convergence regionx(n) = (0.8)nu(n) + (1.25)nu(n)

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4 Frequency Spectrum

Discrete time Fourier transform - DTFT

The evaluation of the z-transform on the unit circle:

Frequency response H() of a linear system h(n) withtransfer function H(z):

n

n

njenxX

0j

ez

XenxznxzXn

n

jn

n

n

ez j

n

n

njenhH


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Digital frequency:

Nyquist interval [-fs/2, fs /2] - < <

Fourier spectrum of signal x(nT) periodic

replication of the original analog spectrum atmultiples of fs.

jezzHH

sf

f2

n

f/jfn2^

senTxfX


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Example

deX2

1nx

nj

n

22dX

2

1nx

dfefXf1

nx

S

S

S

f

f

f/jfn2

S

n-,enx nj 0

m 0 m22X

INVERSE DTDT

Parseval


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Zeros and Poles of X(z) or H(z), on the z-plane, effect on the spectrum of X() or H().Example, consider a function has one pole

z = p1 and one zero z = z1.

1

1

11

11

pz

zz

zp1

zz1zX

1

j

1j

1j

1j

pe

zeX

pe

zeX


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GEOMETRIC INTERPRETATION OF FREQUENCY SPECTRUM


Example: A causal complex sinusoid

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1-M

M1-

2

21-

1

1

1-M

1-2

1-1

zp-1A...

zp-1A

zp-1A

)zp-(1)zp-(1)zp-(1

zN

zD

zNzX

1

1

pzij

1j

pz

1

ii zp1

zN

zXzp1A

Inverse Z-transform


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11

1

21

1

z25.11z8.01

z05.22

zz05.21

z05.22zX

1211211

z25.11A

z8.01A

zz05.21z05.22zX

1z8.01

z05.22zXz25.11A

18.0/25.11

8.0/05.22

z25.11

z05.22zXz8.01A

25.1z

1

1

25.1z1

2

8.0z

1

1

8.0z1

1

Example:


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Partial Fraction (PF)

1-M

1-2

1-1

1-M

1-2

1-1

zp-1zp-1zp-1

)zp-(1)zp-(1)zp-(1

M210

A...

AAA

zN

zD

zNzX

0z0

zXA

zDzR

zQzD

zRzDzQ

zD

zNzX


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Example: Compute all possible inverse Z-transform and stability

feature of the function

Inverted causal, stable:

Inverted anti-causal, unstable:

Solution

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Example:

Find possible

inverse Z-transform

Solution:

Four poles devide the z-plane into

four ROC regions

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