Question

could you please solve this problem using Matlab

4. Using the Fourier series coefficients {ak} obtained, construct the following signal N(t)>akekwot k=-N where ak is the kth Fourier series coefficients of r(t). Therefore, N(t) is the so-called 2N-term approximation of original signal r(t). In one figure, plot âN(t) versus t, for N 10, 20, 30, 40. In the same plot, also plot original signal (t). Visually compare N(t) and r(t) in the figure and comment on your observations

Answer 1

The following example is considered to demonstrate in MATLAB.

x(t) =1 for 0<=t<=1, 0 otherwise and T=2

MATLAB code:

clc;clear all;close all;
T=2%Fundamnetal period
w0=2*pi/T;%Angular frequency
t=0:T/500:T;%time range
x=1.*(t>=0 &t<=1)%Define square wave
figure;
for i=1:1:4
M=[10 20 30 40];
N=M(i);
%calculate ak-Fourier series coefficients
k=0;
for n=-N:1:N
k=k+1;
a(k)=(trapz(t,x.*exp(-j*n*w0*t)));
end
a=a/T
n=-N:1:N

%Fourier series approximation
k=0;
for t=0:T/500:T
k=k+1;
xN(k)=sum(a.*(exp(j*n*w0*t)));
end
t=0:T/500:T
subplot(2,2,i)
plot(t,xN,'g','LineWidth',3,t,x,'r','LineWidth',3);grid;
title(sprintf('Fourier seies approximation for N=%d',N))
xlabel('t')
legend('xN(t)','x(t)')

end

Plot:

Fourier seies approximation for N=10 Fourier seies approximation for N=20 1.4 xN(t) 1.2 XN(t) X(t) X(t) 0.6 0.4 0 0.2 0 0.5 1 1.5 2 0 0.5 1 1.5 2 Fourier seies approximation for N=30 Fourier seies approximation for N=40 1.4 1.2 1.2 xN(t) XN(t) X(t) X(t) 000 0.6 0.4 0.2 0.8 0.6 0.4 0.2 -0.2 -0.2 0.5 1.5 2 0.5 1 1.5 2