Question

Question 2 [30pts]: By using a 3x1 subplot, plot x(t) signal in the first row. Take t between 0 s and 10 s with an increment of 1 ms. Furthermore, calculate and plot the Fourier series representation of x(t) = Ek--Nakejkwot for N = 10 and N = 100 in the second and third rows of the subplot. Note that, x(t) = Na elkwot should be calculated by MATLAB. Please comment on increasing N on your results. Do not forget axis labels.

Answer 1

Question (1)

I think the signal given is

3 S = 0)

Exponential Fourier Series

Any continuous time periodic signal, with a fundamental period can be represented as

x(t) = م والمعالجة | k=-00

where

<T>

For the signal x(t) the fundamental time period is given as .

To = 35

The angular frequency of the signal will be

2π 2π ωο Τ, 3

The signal is given as

3 S = 0)

So

Az = = x(t)e-jw kt dit

******+/+*.00 [ +Pamt-074) ***

e-jw kt.dt w

[D--13-

ax = [-[e-juok – 1] + [e-jag k3 – e-jwgk2]]

[Proxer-8 + omxz_– 04_– 1] + =

ਅੱਜ ਦੇ ਦਲ - - - - ੧੦

(ਸ ) + , - " " " - 10

( 3) cos.2 = ع -ج + عواره

ejkomen te komen = 2.cos kwo 2

So

[( ਸ ) 02:-ਅਜ਼) 12 ਅੰਕ : - - -

Substitute

ax = 7e- jkn [2.cos(tk) – 2.cos(k)]

as a pieska (cos(uk) – cos (5x)]

Using

e jkn = cos(Tk) - j.sin(Tk) = (-1){

We get

[(*)so» – 461–)](1–) **= 30

[(1.350,(7-) – *(1-))*7 = *0

[() s(-) - 1부 - 이

1-(-1)* cos(K) ax = ? jik

For k = 0, the above will be indeterminate. So we can find the value using L Hospitals rule

[1-(-1)* cos(k) Okno ao = lim jak

[o+k(-1)k-1 sin (5k) x 1 ao = lim jn

ao = 0

So the Fourier Coefficients are

k=0 on = {{– (-1)+cos[:)] kto jik

Question (2)

MATLAB Code

clc;
clear all;
close all;

T = 3;
t = 0:0.001:10;
t1 = [0:0.001:1];
t2 = [1.001:0.001:2];
t3 = [2.001:0.001:3-0.001];
x1 = ones(1,length(t1));
x2 = zeros(1,length(t2));
x3 = -1*ones(1,length(t3));

x_org = [x1, x2, x3, x1, x2, x3, x1, x2, x3, x1];

subplot(3,1,1);
plot(t, x_org, 'linewidth',2);
grid
xlabel('Time, t (s)');
ylabel('Amplitude ');
title('The original signal x(t)');

w0 = 2*pi/3;

a0 = 0;
x = a0;

for k = -10:10
if k ~= 0
ak = (1 - (-1)^k*cos(pi/3*k))/(j*pi*k);
x = x + ak*exp(j*k*w0*t);
end
end

subplot(3,1,2);
plot(t, x, 'linewidth',2);
grid
xlabel('Time, t (s)');
ylabel('Amplitude ');
title('The signal x(t) for N = 10');

a0 = 0;
x = a0;

for k = -100:100
if k ~= 0
ak = (1 - (-1)^k*cos(pi/3*k))/(j*pi*k);
x = x + ak*exp(j*k*w0*t);
end
end

subplot(3,1,3);
plot(t, x, 'linewidth',2);
grid
xlabel('Time, t (s)');
ylabel('Amplitude ');
title('The signal x(t) for N = 100');

The plots

The original signal x(t) Amplitude 5 6 Time, t(s) The signal x(t) for N = 10 Amplitude 6 7 8 Time, t(s) The signal x(t) for N = 100 Amplitude Time, t(s)

As N increases, the signal becomes more accurate to the original signal