Question

9. The purpose of this problem is to use MATLAB to compute the magnitude, phase, and total
energy of a Fourier transform.
a) Develop a MATLAB routine to plot the magnitude and phase of a given Fourier
transform H(jω). The input part of your program will, of course, require that you
specify the particular H(jω) of interest, but, once this is done, your program should not
depend on the Fourier transform specified. You will need to select an appropriate
range of frequencies for these plots. Test your program using the following three
signals for a = 4 (plot all three using the same range of frequencies):
i) h(t) = e-at u(t)
ii) h(t) = e-a|t|
iii) h(t) = te-at u(t)
b) Extend the MATLAB routine that you developed in part a) to find the approximate
total energy in each of the signals, as well as to find the approximate frequency ω in
rad/sec below which 90% of the total signal energy is contained.
c) Suppose you wanted to design a simple low-pass filter with a specified cutoff
frequency. Assuming that you could choose the desired value of a and based on the
total energy criterion alone, which impulse response would you choose? Why?

Answer 1

%Part a Solution

clc;
clear all;
close all;
a=4;
F=50;
t = 0:1/F:10-1/F; %time instant
signal_1=exp(-a.*t);%signal 1
signal_2=exp(-a*(abs(t)));%signal 2
signal_3=t.*exp(-a.*t);%signal 3
Ts=mean(diff(t));
Fs=1/Ts;
%%Choose the signal
signal=signal_1;
a = fft(signal); %fourier transform of signal 1
freq = (0:length(a)-1)*50/length(a); %frequency
l = length(signal);
fshift = (-l/2:l/2-1)*(50/l);
yshift = fftshift(a);%for two sided spectrum
plot(fshift,abs(yshift))
title('Magnitude')
power = abs(yshift).^2/l;
plot(fshift,power)
title('Power')
phase=angle(yshift); %phase
%%Part b Solution
y_squared=abs(yshift);
Et=sum(y_squared)*Fs/l;%total energy
Eg=0.9*(Et);% 90 % of the total energy
df = Fs/l; % frequency increment
for i=1:500
E(i)=sum(y_squared(1:i))*Fs/l;
end
for i=1:500
if E(i)<=Eg;
out = i;
end
end
freq_Hz=(50/500)*out;%frequency in Hz.

%%Part c Solution

To design a simple low pass filter with a specified cut off frequency based on the total energy criterian alone we will  choose the third signal because the impulse response of third signal has the highest cut-off frequency