Signals and Systems Homework: MATLAB Analysis and Solutions

This document presents a comprehensive solution to a Signals and Systems assignment, employing MATLAB for signal analysis and processing. The assignment covers various aspects, including the calculation of DC and AC components, plotting signals, and determining parameters like amplitude, period, and fundamental frequency. The solution delves into the effects of different frequency components, visualizing them through plots and stem diagrams. It also includes the implementation of a low-pass filter and analyzes the filtered signal's characteristics. Furthermore, the assignment calculates power metrics, such as total power and the dynamic power ratio, providing detailed insights into the signal's behavior. The MATLAB code is provided, along with the corresponding plots and numerical results, offering a complete and practical approach to solving the assignment problems. The document includes references to relevant textbooks. This assignment is available on Desklib, a platform providing AI-based study tools.

ASSIGNMENT
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Question a...............................................................................................................................1
Question b...............................................................................................................................2
Question c...............................................................................................................................3
Question d...............................................................................................................................4
Question e...............................................................................................................................5
clear;
clc;
close all;
T=1e-3;
A=2;
Question a
Fs=1e6;
t=-T:1/Fs:T;
d_c=2*A/pi; %Dc component of the signal
a_c=0;%Ac component calculation
for n=2:200
a_c=a_c+((-1).^(1+n/2))/(n.^2-1)*cos(2*pi*n*t/T);
end
s_t=d_c+4*A./pi.*a_c;
figure;
plot(t,abs(s_t));
grid on;
xlabel('Time in seconds');
ylabel('Amplitude');
amp=max(abs(s_t));% Amplitude calculation
per=T;% Period calculation
fund=1/T;% Fundamental frequency is obtained when n=1
L=length(s_t);
poww=(norm(s_t).^2)/L; % power calculation in watts
[R, SPURPOW, SPURFREQ] = sfdr(abs(s_t));
powed=SPURPOW; % power calculation in dBW
fprintf('Amplitude: %f volts\n Period: %f seconds\n Fundamental frequency: %f Hz\

n',amp,per,fund);
fprintf('Total power in watts: %f watts\n Total power in dBW: %f dBW\n',poww,powed);
Amplitude: 2.015035 volts
Period: 0.001000 seconds
Fundamental frequency: 1000.000000 Hz
Total power in watts: 2.059740 watts
Total power in dBW: -29.002412 dBW
Question b
k=[0 2 4 6];
figure;
grid on;
xlabel('Time in seconds');
ylabel('Amplitude');
hold on;
for i=1:4

n=k(i);
s_e=d_c+((-1).^(1+n/2))/(n.^2-1)*cos(2*pi*n*t/T);
plot(t,abs(s_e));
end
legend('n=0','n=2','n=4','n=6')
hold off;
Question c
figure;
hold on;
k=0:10;
for i=1:11;
n=k(i);
s_t=d_c+((-1).^(1+n/2))/(n.^2-1)*cos(2*pi*n*t/T);
subplot(3,4,i);
stem(abs(s_t));
grid on;

xlabel('frequency');
ylabel('Power spectral density');
end
hold off;
Question d
fcut=7000*pi*2;
sot=lowpass(abs(s_t),fcut/(2*pi*Fs));
figure;
plot(t,sot);
grid on;
xlabel('Time in seconds');
ylabel('Amplitude');
L=length(sot);
poww1=(norm(sot).^2)/L; % power calculation in watts
powrat=poww1/poww;
f=-1/T:(200/T-7000);

highest_freq=max(f);
hh=highest_freq/fund;
fprintf('The highets frequency of So(t) is as multiple of Fo: %f \n',hh);
fprintf('The highets frequency of So(t) in Hz is: %f \n',highest_freq);
fprintf('The total power of So(t) is: %f \n',poww1);
fprintf('The total power of So(t) as a ratio of power of s(t): %f \n',powrat);
% Comment
% The signal resembles the original signal s(t).
The highets frequency of So(t) is as multiple of Fo: 193.000000
The highets frequency of So(t) in Hz is: 193000.000000
The total power of So(t) is: 1.617594
The total power of So(t) as a ratio of power of s(t): 0.785339
Question e
figure;
sfdr(sot);

r=sfdr(sot);
hhh=1/T;
a=(200*hhh-7000);
f=-hhh:1e4:a;
pww=[];
for j=1:length(f);
ff=f(j);
for n=2:200;
ss_e=d_c+((-1).^(1+n/2))/(n.^2-1)*cos(2*pi*n*t*ff);
end
L=length(ss_e);
pw=(norm(ss_e).^2)/L; % power calculation in watts
pww=[pww pw];
end
pwwr=pww./poww;
pww_db=10*log10(pww);
q=[pww;pwwr;pww_db];
fprintf('power in watts Normalised power Normalised watts in dB')
q'
fprintf('Dynamic power ratio is: %f\n',r);
power in watts Normalised power Normalised watts in dB
ans =
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982

1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
1.6211 0.7871 2.0982
Dynamic power ratio is: 257.678527

References
[1] M. J. Roberts, Signals and systems Analysis Using Transform methods and MATLAB,
New York: McGraw-Hill, 2012.
[2] D. Silage, Digital communication Systems using Matlab and Simulink, Bookstand
Publishiing, 2009.