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I am trying to compute the trigonometric fourier series coefficients of a periodic square wave time signal that has a value of 2 from time 0 to 3 and a value of -12 from time 3 to 6. It then repeats itself. I am trying to calculate in MATLAB the fourier series coefficients of this time signal and am having trouble on where to begin.

The equation is x(t) = a0 + sum(bk*cos(2*pi*f*k*t)+ck*sin(2*pi*f*k*t))

The sum is obviously from k=1 to k=infinity.

a0, bk, and ck are the coefficients I am trying to find. Thanks for the help.

Youssef Khmou
on 6 Mar 2013

Edited: Youssef Khmou
on 6 Mar 2013

hi Jay , computing a0 bk and ck is bout theory i think, anyway try :

You have first to construct the original signal "Square(t)" so as to compare it with Fourier approximation :

clear , close all;

Fs=60;

t=0:1/Fs:20-1/Fs;

y=square(t,50);

y(y>0)=2;

y(y<0)=-12;

figure, plot(t,y);

axis ([0 20 -20 10])

% Fourier Series

a0=0;

Fy=zeros(size(t));

N=10;

for n=1:2:N

Fy=Fy+(4/n*pi)*sin(2*pi*n*t/(2*pi));

end

hold on,

plot(t,Fy,'r')

legend(' Square ','Fourier Approx');

Try now to to compute an, and bn and increase the number of iterations N and conclude

You have also to adjust the amplitudes

Rick Rosson
on 6 Mar 2013

Edited: Rick Rosson
on 6 Mar 2013

>> doc fft

>> doc real

>> doc imag

Kamal Kaushal
on 1 Mar 2020

clear , close all;

Fs=60;

t=0:1/Fs:20-1/Fs;

y=square(t,50);

y(y>0)=2;

y(y<0)=-12;

figure, plot(t,y);

axis ([0 20 -20 10])

% Fourier Series

a0=0;

Fy=zeros(size(t));

N=10;

for n=1:2:N

Fy=Fy+(4/n*pi)*sin(2*pi*n*t/(2*pi));

end

hold on,

plot(t,Fy,'r')

legend(' Square ','Fourier Approx');

Hemang Mehta
on 23 Oct 2020

clear , close all;

Fs=60;

t=0:1/Fs:20-1/Fs;

y=square(t,50);

y(y>0)=2;

y(y<0)=-12;

figure, plot(t,y);

axis ([0 20 -20 10])

% Fourier Series

a0=0;

Fy=zeros(size(t));

N=10;

for n=1:2:N

Fy=Fy+(4/n*pi)*sin(2*pi*n*t/(2*pi));

end

hold on,

plot(t,Fy,'r')

legend(' Square ','Fourier Approx');

Aaron Vargas
on 9 Dec 2020

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