Question

Plot the magnitude of the and the pole- zero plot frequency response given below. using Hcz) Hcz) = (1-ovci :) 1-0 72 (Porto

please show steps and formulas.

Answer 1

(->_956o-1) (- 560-7) (-³-²560-7) (-³560 -1) - 2 -Z - I

To find and plot the Pole Zero Map

The zeros are at locations where the numerator polynomial is zero. So

1- -2 = 0

0 = (1-2 +1) G-Z - 1)

(z - 1)(2+1) = 0

(2-1) = 0 and (z+1) = 0

So the zeros are at

z= 1 and z= -1

The poles are at locations where the denominator polynomial is zero. So

0 = (-_9S60 -1) (-560 - 1) (-ಶ.S60 -1) (-560 - 1)

First Pole

1 -0.95e72-1=0

2 -0.95e= 0

z= 0.95ejz

== 0.95(cos(©)+ jsin()

(**)seo - -

z = 0.67175 + 70.67175

Another way to find the pole is

1 -0.95e72-1=0

So the pole is at

z= 0.95ejz

The pole is at the point at an angle at a distance 0.95.

+ rad

Second Pole

1 -0.95e-777-1 = 0

2 -0.95e1=0

z = 0.95e

z= 0.95(cos () – jsin()

(-)560 - -

z = 0.67175 -0.67175

Another way to find the pole is

1 -0.95e-777-1 = 0

So the pole is at

z = 0.95e

The pole is at the point at an angle at a distance 0.95.

pol

Third Pole

1 -0.95e-72-1=0

2 -0.95e1 = 0

z= 0.95e

(9) us: +(6)00) s60 = =

z= 0.95(0 + i)

z= +0.95

Another way to find the pole is

1 -0.95e-72-1=0

So the pole is at

z= 0.95e

The pole is at the point at an angle at a distance 0.95.

+ rad

Fourth Pole

1 -0.95e-777-1 = 0

2 -0.95e-13 = 0

z = 0.95e13

(9) us! - (so)s6o ==

z= 0.95(0-i)

z= -10.95

Another way to find the pole is

1 -0.95e-777-1 = 0

So the pole is at

z = 0.95e13

The pole is at the point at an angle at a distance 0.95.

rad

So the poles are at

z = 0.67175 + 70.67175, z= 0.67175 -0.67175, z= +10.95, z= -10.95

Hence the pole zero plot will be

Pole-Zero Map Imaginary Axis -0.8 -0.6 -0.4 -0.2 0.2 0.6 0.8 Real Axis

To find and plot the Magnitude Response

(->_956o-1) (- 560-7) (-³-²560-7) (-³560 -1) - 2 -Z - I

We know that

= +0= () ups{+ (so = ?

- = - 0 = () s[-(ဠ) ၀၁ = --

So

1-2-2 * *(1 -0.95€',-1) (1 -0.95e-42-1)(1 – j0.952-1)(1 +j0.952-1)

1-2-2 (1 -0.95 (ete-)2-1 +0.9527-2) (1 +0.9522-2)

Using

= = = (957 = ೪ * *

We get

1-2-2 H(2) G1 -0.95 x 22-1 +0.9522-2)(1 +0.952z-2)

(z-252060 + 1)(z-292060 +1-75€¥€'T – 1) = (2) 2-2-T

To get the frequency response put

لازم = =

1- e-j2w (1 - 1.3435e-jw +0.9025e-j2w) (1 + 0.9025e-j2w)

Using MATLAB

clc;
clear all;
close all;

b = [1, 0, -1];
a1 = [1, -1.3435, 0.9025];
a2 = [1, 0, 0.9025];

a = conv(a1, a2);

w = -pi:pi/1000:pi;

H = freqz(b, a, w);

plot(w/pi, abs(H), 'linewidth', 2);
grid;
xlabel('Normalized Frequency, \omega (\times \pi units)');
ylabel('|H(e^{j\omega})|');
title('Magnitude Response');

sys = tf(b,a,1)
figure
pzmap(sys)

After executing we get

Magnitude Response 1 ) -0.8 -0.6 -0.4 0.4 0.6 0.8 -0.2 0.2 Normalized Frequency, w (x units)

Pole-Zero Map Imaginary Axis -0.8 -0.6 -0.4 -0.2 0.2 0.6 0.8 Real Axis