Question

Butterworth filter

a)Design a 5th order low pass Butterworth low-pass filter with a cut-off frequency of 1592 Hz and a dc gain of 3dB. Find and present the mathematical transfer function of the filter, showing all your steps.

b) Write a Matlab code to plot the magnitude of this function with a linear scale in dB units on the ordinate, and a log scale of frequency on the abscissa.
The plot range should be: ordinate- linear scale from -100dB to +10dB; abscissa- log scale from 1000 rads/sec to 100,000 radians/sec.

Answer 1

Question a

The normalized transfer function of a 5^th order Butterworth Low Pass filter with a cut off frequency of 1 rad/sec is given as

Let K denote the dc gain. So

We need a dc gain of 3 dB. So the gain should be

The dc gain is the gain of the transfer function when s = 0

So

The cut off frequency required is

So the cut off frequency in rad/sec will be

Now we can make the cut off frequency as by replacing s in the normalized transfer function with

Question b

MATLAB Script

clear;
close all;
clc;
format long

b = [1.41254e20]; % Numerator Coefficients
a = [1, 3.2369e4, 5.23897e8, 5.24045e12, 3.2397e16, 1e20]; % Denominator Coefficients

w = [1000:100000];

H = freqs(b, a, w);

H_dB = 20*log10(abs(H));

semilogx(w, H_dB, 'linewidth', 2);
grid;
ylim([-100, 10]);
xlabel('Frequency, \omega (rad/sec)');
ylabel('Magnitude in dB');
title('Magnitude Response');

After executing we get

We can see that the pass band gain is 2.9969 dB which is approximately equal to 3 dB .The gain at 10000 rad/sec is equal to 0 dB. So the cut off frequency is 10000 rad/sec. This frequency is equal to 1592 Hz as follows.