Z-transform & Inverse Z-transform Examples
Example :01
x(z) = _______18z^3_______
18z^3 + 3z^2 -4z – 1
CODE:
b = [18 0 0 0];
a = [18 3 -4 -1];
[r p k] = residuez(b,a);u = ones(1,10);
n = 0:9;
f = (0.3600.*0.5000.^n).*u + (0.2400.*-0.3333.^n).*u + (0.4000.*-0.3333.^n).*u;
subplot(2,1,1)
stem(n,f);
subplot(2,1,2)
impz(b,a);
OUTPUT:
Example :02
x(z) = _______1 – 2z^-1_______
(1 – 0.2z^-1)( 1+0.6z^-1)
CODE:
b = [1 -2 0];
a = [1 0.4 -0.12];
[r p k] = residuez(b,a)u = ones(1,18);
n = 0:17;
f = (3.2500*-0.6000.^n).*u + (–2.2500.*0.2000.^n).*u
subplot(2,1,1)
stem(n,f);
subplot(2,1,2)
impz(b,a);
OUTPUT:
Example :03
Determine the Inverse z-transform of the following sequences, using partial fraction expansion method.
x(z) = _______1 – z^-1 – 4z^-2 + 4z^-3_______
1 – 2.75z^-1 + 1.625z^- 2 – 0.25z^-3
CODE:
b = [1 -1 -4 4];
a = [1 -2.75 1.625 -0.25];
[r p k] = residuez(b,a)u = ones(1,18);
n = 0:17;
f = (0*2.^n).*u + (-10.*0.5.^n).*u + (27.*0.25.^n).*u;
subplot(2,1,1)
stem(n,f);
subplot(2,1,2)
impz(b,a);
OUTPUT:
Example :04
A digital filter is described by the difference equation,
Y(n) = x(n) + x(n-1) + 0.9y(n-1) – 0.81y(n-2)
Using the freqz function, plot the magnitude and phase of the frequency response of the filter.
CODE:
b = [1,1];
a = [1, -1, -0.9, 0.81];
w = [0:1:500]*pi/500;
H = freqz(b,a,w);
magH = abs(H);
phaH = angle(H)*180/pi;
subplot(2,1,1)
stem(w/pi,magH);
title(‘Magnitude’);
subplot(2,1,2)
stem(w/pi,phaH);
title(‘Phase’);
OUTPUT:
Example :05
x(z) = _______z^-1_______
1 – 2z^-1 + 3z^ – 2
CODE:
b = [0 1 0];
a = [1 -2 3];
[r p k] = residuez(b,a)u = ones(1,30);
n = 0:29;
f = (0-0.356j.*(1.000+1.4142j).^n).*u + (0+0.356j.*(1.000-1.4142j).^n).*u
subplot(2,1,1)
stem(n,f);
subplot(2,1,2)
impz(b,a);
OUTPUT:
Example :06
x(z) = _______2 – 3z^-1_______
1 – 5z^-1 + 6z^ – 2
CODE:
b = [2 -3 0];
a = [1 -5 6];
[r p k] = residuez(b,a)u = ones(1,10);
n = 1:10;
f = (3.*3.^n).*u + (-1.*2.^n).*u;
subplot(2,1,1)
stem(n,f);
subplot(2,1,2)
impz(b,a);
OUTPUT:
