Question

Part II: Design of Butterworth Filters Butterworth filters, described in a paper by Stephen Butterworth in 1930, are widely used for CT frequency-selective filtering. Butterworth filters have a simple analytic form and are designed to have a magnitude response that is maximally flat in the passband. In this section, you will use the Laplace transform to design and analyze Butterworth filters in the frequency domain. The textbook has some useful information about Butterworth filters, so check it out to help you with this section! An Nth order Butterworth lowpass filter has a frequency response whose magnitude satisfies 1 |H(jw)/2 = — For a Butterworth filter with a real-valued impulse response h(t), the transfer function satisfies the following equation: H(s)H(-) = 7 1+ (s/jw.)2N

Answer 1

Question (1)

H(s)H-s)= 2N 1+

So the poles will be at

2N S 1+ = 0 jac

2N S = -1 jwc.

-1 = ei(n+2nk) k 1,2,3,4,5,....

2N = ei(x+2nk) jae

1а+2лк) 2N е

j(+2k s j.wc.e 2N

So the poles are at

j(+2nk) k 1,2,3,4,5,6,....... S we.j.e 2N

Question (2)

N = 1

So, there will be two poles of H(s) H(-s)

j(T+2Tk) k 1,2 S = wej.e

j(+2n j(+2TX2) S1wej.e S2 = wj.e" 2

ј3л Ј5л 1 = wej.e 2 S2 = wej.e 2

1Зд е 2 Зп. j. sin = 0 + j(-1) =-j 2 cos 2

5T = COS 2 j5T e 2 = 0 + j(1) j

s1 = wj.(-) S2 = w.j.(j)

S2 =-we we

So the pole which will result in a stable and causal filter will be . Its because for a stable system, poles should lie in the left half of the s plane

S2

So

К H(s) с

Given H(0) =1

So

К H(0) 1

К - Ф.

So the transfer function of the first order Butterworth filter will be

H(s) s+w

Question (3)

10т rad/s

N = 1

So the transfer function of the filter will be

10л H(s) s+ 10л

The MATLAB Code

clc;
clear all;
close all;

wc = 10*pi;

b = [wc];
a = [1 wc];

w = linspace(0,1000,100);

H = freqs(b, a, w);

magH = abs(H);
HdB = 20*log10(magH);

figure
plot(w, magH, 'linewidth', 2);
grid on;
xlabel('\omega (rad/s)');
ylabel('|H(\omega)|');
title('Magnitude Response of the Filter');

figure
plot(w, HdB, 'linewidth', 2);
grid on;
xlabel('\omega (rad/s)');
ylabel('20.log(|H(\omega)|) (dB)');
title('Magnitude Response of the Filter[Magnitude in db]');

The plots obtained

Magnitude Response of the Filter 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 100 200 300 400 500 600 700 800 900 1000 w (rad/s) CO (m)H

Magnitude Response ofthe Filter[Magnitude in db] O -5 -10 -15 -20 -25 -30 -35 1000 100 200 300 400 500 600 700 800 900 w (rad/s) (IC)HD6or0z (dB)

Question (4)

The poles are given by

j(+2nk) k 1,2,3, ... (2N) S we.j.e 2N

N = 3

So

j(+2mk) k 1,2, 3, 4,5,6 S we.j.e 6

j(+2m) 6 we.j.e = wc.j.e 2

e(cos ).sin G))

1 = cj.(0 +j(1))

we.j.j

S1

5(л+4п) Ј5л = @c.j.e 6 S2w.j.e 6

5T (cos j. sin 6 S2 6 Ln

3 1 S2 2

1 -je S2

= @c.j.e 6 S3w.j.e 6

3j.cos) +j. sin ()) j. sin COs

3 = @e-j. 1 S3 2

1 3 j 2

j(+8) јЗл = @c. j.e_2 we.j.e S4 6

Зп j.sin (cos S4

S4 wj. (0-1)

SA s

j(+10 j11T = @e.j.e 6 6

11T 117T e(cos( j. sin 6 6

V3 2 2

V3 1 wej.w 2

j(+12T) j13m = @e.j.e 6 Sw.j.e 6

13T 13 e-1-(cos j. sin 6 S6 6

S6 = @e.j. 2

V3 1 +j.we 2 S6

So the poles are at

S1

V3 1 S2 2 2

V3 1 -jw $32jc-

S4

V3 1 wej.w 2

V3 1 +j.we 2 S6

Let us consider that the cut off frequency is

w 1 rad/s

Then the poles are

= -1

3 S2 2 2

3 S3 2 2

= 1 S4

3- 1 2

3 1 j. 2 2 S6

MATLAB Code

clc;
clear all;
close all;

s = tf('s');

sp = [-1, -1/2-j*sqrt(3)/2, 1/2-j*sqrt(3)/2, 1, 1/2+j*sqrt(3)/2, -1/2+j*sqrt(3)/2];
b = 1;

a = (s - sp(1))*(s - sp(2))*(s - sp(3))*(s - sp(4))*(s - sp(5))*(s - sp(6));

sys = b/a;

pzplot(sys);

The pole zero plot obtained

Pole-Zero Map 1 X X 0.8 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 X X 1 -1 0.5 1.5 -0.5 0 1 Real Axis (seconds) CO Co Imaginary Axis (seconds)