Matlab Exercise Question Answer 2022

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Matlab Exercise
Students Name
Institutional Affiliation
Date

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Question 1
a)
f ( t )=t ( π 2−t2 ) ,−π ≤ t ≤ π
ω0=2 π f 0
Given that,
T =2 π
f = 1
T = 1
2 π
ω0=2 π f 0=2 π × 1
T =2 π × 1
2 π =1
b)
clear all
clc
t = -pi:0.1:pi;
f_t = t.*((pi.*pi)-(t.*t));
plot(t,f_t)
xlabel('Time, t')
ylabel('Magnitude')
title('f(t)=t(pi^2-t^2)')
%hold on
%grid on
%y = sin(t);
%plot(t,y)
grid on

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Published with MATLAB® R2019a
c)
For good approximation of this signal using fourier series, the time t shoul be very small. It was
observed that only values of t about 0.1 and lower gave a good approximation. For larger values of t ,
the signal was distorted. The plot below is for t=0.5. As it can be observed, it is not smooth.

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d)
The signal is periodic and odd as observed from the figure above. The signal can also be represented
by the function,
f ( t ) =sin (t )
The complex fourier series coefficients are given by,
cn= 1
T ∫
−T / 2
T /2
f (t )e− jnωt dt
cn= 1
2 π ∫
−2 π / 2
2 π /2
sin ( t)e− jnωt dt
cn= 1
2 π ∫
−π
π
sin (t) e− jnωt dt
cn= 1
2 π [ e¿ π −e−¿ π
n2−1 ]=0 for n≠ 1∨−1
For these two cases, we set n=1+ε and let ε approach zero,
c1= 1
2 π [ ei π (1+ ε )−e−¿ π (1+ε )
(1+ε )2−1 ]
¿ 1
2 π [ −ei π ε + e−i π ε
1+ 2 ε −1 ]
c1= [−1−i π ε+ 1−i π ε
1+2 ε−1 ]= 1
2 π [−2 i π ε
2 ε ]=−i
2 = 1
2 i
Similarly,
c−1=−1
2 i