Using Matlab 03

5/27/2018 Using Matlab 03

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DIGITAL SIGNAL PROCESSING USING

MATLAB

LECTURE 3: Z-TRANSFORM

Pn. Aqilah Baseri Huddin

Prof. Dr. Salina A. Samad


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DEFINITION OF Z-TRANSFORM

The z-transform of a discrete-time signal x[n] is defined as thepower series

In MATLAB:

>> help ztrans

Example: Find z-transform of =

>> syms n>> f = cos((pi*n)/2);

>> ztrans(f)

ans =

z^2/(z^2 + 1)

n

znxnxZzX ][)}({)(


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PROPERTIES OF Z-TRANSFORM

Property

1 Linearity 2 Time-shifting ( ) ()3 Scaling

() (

)

4 Time reversal () ()5 Differentiation () 6 Multiplication

()() 12 ()(

)

7 Convolution () ()()


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Example 1:

Let = 2 3 4, = 3 4 5 6 Find = ()()

From definition of the z-transforms,

Hence

}6,5,4,3{)(}4,3,2{)( 21 nxnx

x1 = [2,3,4]; x2 = [3,4,5,6];

x3 = conv(x1,x2)

x3 =

6 17 34 43 38 24

543213

24384334176

zzzzz)z(X

PROPERTIES OF Z-TRANSFORM


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PROPERTIES OF Z-TRANSFORMExample 2:

We can use the conv_m function in to multiply two z-domainpolynomials corresponding to noncausal sequences:

Let = 2 3, = 2 4 3 5 Find = ()()

From definition of the z-transforms

Therefore X3(z) = 2z3+ 8z2+ 17z + 23 + 19z-1+ 15z-2

}5,3,4,2{)(}3,2,1{)( 21 nxnx

x1 = [1,2,3]; n1 = [-1:1];

x2 = [2,4,3,5]; n2 = [-2:1];

[x3,n3] = conv_m(x1,n1,x2,n2)x3 =

2 8 17 23 19 15

n3 =

-3 -2 -1 0 1 2


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PROPERTIES OF Z-TRANSFORMTo divide one polynomial by another, we need an inverse operation called

deconvolution.In Matlab we use deconv, i.e. [p,r] = deconv(b,a) where we are dividingb by a in a polynomial part p and a remainder r.

Example,

We divide polynomial X3(z) in Example 1 by X1(z):

x3 = [6,17,34,43,38,24]; x1 = [2,3,4];

[x2, r] = deconv(x3,x1)

x2 =

3 4 5 6

r =

0 0 0 0 0 0


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Z-TRANSFORM PAIRS


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Z-TRANSFORM INVERSIONA Matlab function residuez is available to compute the residue

part and the direct (or polynomial) terms of a rational function in z-1.Let

Therefore we can find the Residues (R), Poles (p) and Direct terms

(C) of X(z).

NM

k

kk

N

k k

k

xxNN

MM

zCzp

R

RzR;za....za

zb....zbb)z(X

01

1

1

1

1

10

1

1

[R, p, C] = residuez(b,a)


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Z-TRANSFORM INVERSION

Lets look at an example:

Consider the rational function:

First rearrange X(z) so that it is a fucntion in ascending powers of z-1:

Use matlab:

143)( 2

zz

z

zX

21

1

22

2

143)143()(

zz

z

zzz

zzzX

b = [0,1]; a = [3,-4,1];

[R,p,C] = residuez(b,a)

R =

0.5000

-0.5000

p =

1.0000

0.3333

C =

[]


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Z-TRANSFORM INVERSIONWe obtain:

To convert back to the rational function form:

So that we get:

11

311

21

1

21

zz)z(X

[b,a] = residuez(R,p,C)

b =

0.0000

0.3333

a =

1.0000

-1.3333

0.3333

21

1

31

341

31

zz

z)z(X


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Another example, compute the inverse z-transform:90

901901

1

121 .z,

)z.()z.()z(X

b = 1; a = poly([0.9,0.9,-0.9])

a =

1.0000 -0.9000 -0.8100 0.7290

[R,p,c] = residuez(b,a)

R =

0.2500 + 0.0000i

0.5000 - 0.0000i

0.2500

p =

0.9000 + 0.0000i0.9000 - 0.0000i

-0.9000

c = []


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Using table 4.1:

)()9.0(25.0)1()9.0)(1(9.0

5.0)()9.0(25.0)( 1 nununnunx nnn


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Matlab verification:

[delta,n] = impseq(0,-1,7);

x = filter(b,a,delta) % check sequence

x =

Columns 1 through 7

0 1.0000 0.9000 1.6200 1.4580 1.9683 1.7715Columns 8 through 9

2.1258 1.9132

x = (0.25)*(0.9).^n + (0.5)*(n+1).*(0.9).^n + (0.25)*(-0.9).^n % answersequence

x =Columns 1 through 7

0 1.0000 0.9000 1.6200 1.4580 1.9683 1.7715

Columns 8 through 9

2.1258 1.9132


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Another Example. Determine the inverse z-transform of

so that the resulting sequence is causal and contains no complex

numbers.

We will have to find the poles of X(z) in the polar form to determine theROC of the causal sequence.

21

1

6402801

2401

z.z.

z.)z(X

b = [1,0.4*sqrt(2)]; a=[1,-0.8*sqrt(2),0.64];

[R,p,C] = residuez(b,a)

R = 0.5000 - 1.0000i

0.5000 + 1.0000i

p = 0.5657 + 0.5657i0.5657 - 0.5657i

C = []

Mp=abs(p) % pole magnitudes

Ap=angle(p)/pi % pole angles in pi units

Mp = 0.8000 0.8000Ap = -0.2500 0.2500


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From the Matlab calculation, we have:

= .+|.|

+ .

|.|

And from table 4.1 we have:

sequencesidedrighttoduez 8.0

)n(u)]nsin()n[cos(.

)n(u)]ee(j)ee(.[.

)n(ue.)j.()n(ue.)j.()n(x

n

njnjnjnjn

njnnjn

42

480

5080

80508050

4444

44


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Matlab Verification:

[delta,n] = impseq(0,0,6);

x = filter(b,a,delta) % check sequence

x =

1.0000 1.6971 1.2800 0.3620 -0.4096 -0.6951 -0.5243

x = ((0.8).^n).*(cos(pi*n/4)+2*sin(pi*n/4)) %answer sequence

x =

1.0000 1.6971 1.2800 0.3620 -0.4096 -0.6951 -0.5243


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SYSTEMS IN THE Z-DOMAINTo determine zeros and poles of a rational H(z), we can use

roots for both the numerator and denominator. (poly is theinverse of root)

We can plot these roots in a pole-zero plot using zplane(b,a).

This will plot poles and zeros given the numerator row/columnvector b and the denominator row/column vector a.

(H(z) = B(z)/A(z)) We can calculate the magnitude and the phase responses of

our system using freqz:


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SYSTEMS IN THE Z-DOMAIN

Example:

Given a causal system

y(n) = 0.9y(n-1) + x(n)

a. Find H(z) and sketch its pole-zero plot

b. Plot |H(ejw)| and


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Solution using Matlab:

a. use zplane function - >> b = [1,0]; a = [1,-0.9];>> zplane(b,a);

>> title(Pole-Zero Plot)

-1 -0.5 0 0.5 1

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

Real Part

ImaginaryPart

Pole-Zero Plot

0 0.9


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B. To plot the magnitude and phase response, we usefreqz function:

>> [H,w] = freqz(b,a,100);

>> magH = abs(H); phaH = angle(H);

>>

>> subplot(2,1,1); plot(w/pi,magH); grid>> xlabel('frequency in pi units'); ylabel('Magnitude');

>> title('Magnitude Response')

>> subplot(2,1,2); plot(w/pi,phaH/pi); grid

>> xlabel('Frequency in pi units'); ylabel('Phase in pi

units');

>> title('Phase Response')


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B. The plots:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

2

4

6

8

10

12

frequency in pi units

Magnitude

Magnitude Response

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-0.4

-0.3

-0.2

-0.1

0

Frequency in pi units

Phaseinpiunits

Phase Response


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B. Points to note:

We see the plots, the points computed is between 0 < < 0.99 Missing point at = To overcome this, use the second form of freqz:

>> [H,w] = freqz(b,a,200,whole);

>> magH = abs(H(1:101)); phaH = angle(H(1:101));Now the 101stelement of the array H will correspond to =

Similar result can be obtained using the third form:

>> w = [0:1:100]*pi/100;

>> H = freqz(b,a,w);

>> magH = abs(H); phaH = angle(H);

Try this


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B. Points to note:

Also, note that in the plots we divided the w and phaH

arrays by so that the plot axes are in the units of and easier toread

This is always a recommended practice!

Date post:	17-Oct-2015
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