Calculating Fourier Series Coefficients
198 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
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.
1 commentaire
omar seraj
le 1 Nov 2021
How to plot in Fourier series given a time interval -1.5 to 2.5I need a step by step answer to this problem please
Réponses (5)
Rick Rosson
le 6 Mar 2013
Modifié(e) : Rick Rosson
le 6 Mar 2013
>> doc fft
>> doc real
>> doc imag
0 commentaires
Youssef Khmou
le 6 Mar 2013
Modifié(e) : Youssef Khmou
le 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
0 commentaires
Kamal Kaushal
le 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');
0 commentaires
Hemang Mehta
le 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');
2 commentaires
Aaron Vargas
le 9 Déc 2020
hello , is there any chance that you can explain to me the variables and how it works ? im stuck in a homework for college
Rik
le 31 Mar 2022
Comment posted as flag by Hariharan Hariharan:
need the flowof code for case study
Voir également
Catégories
En savoir plus sur Spectral Measurements dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!