Azzera filtri
Azzera filtri

Calculating a Fourier series with MATLAB manually problem

35 visualizzazioni (ultimi 30 giorni)
Hi there, I'm trying to transfer this academic terms learnings on the fourier series to MATLAB, and running into difficulties.
I want to be able to plot a fourier series approximation against the original function but when I try and calculate the fourier series to 5 steps of n in the final line I get a division by zero error.
I can't see where this division by zero occurs, the only division I am using is dividing by the period, T, which is just pi.
I understand this is probably a bad way to do things! And I don't think I understand the symsum enough, but any help gratefully recieved.
clear all
close all hidden
clc
syms t % time
syms n % number of terms to calculate the sum to
syms T % period
syms a_n(t, n, T) % fourier co-effecients
syms b_n(t, n, T)
syms a_0(t, n, T)
syms f(t)
f = sin(t)^2
tvec = -3 : 0.01 : 3;
xvec = sin(tvec).^2;
figure
plot(tvec, xvec)
title('Plot of f(t)')
T = sym(pi)
a_0 = (2/T)*int(f, t, 0, T)/2
a_n = (2/T)*int(f*cos(((2*pi)/T)*n*t), t, 0, T)
b_n = (2/T)*int(f*sin(((2*pi)/T)*n*t), t, 0, T)
exp = a_0 + symsum(a_n*cos(((2*pi)/T)*n*t) + b_n*sin(((2*pi)/T)*n*t), n, 1, 5)
% division by zero error
  10 Commenti
Torsten
Torsten il 31 Mar 2024
I'm interested in why for this specific function we get an a_n that is invalid for n=1.
But I alrady answered it - it's a MATLAB bug.
Owen
Owen il 31 Mar 2024
Thank you, I had misread that it was specifically a bug that b_n is not also invalid, but I see you meant the other way round, thanks!

Accedi per commentare.

Risposta accettata

Paul
Paul il 31 Mar 2024
Modificato: Paul il 1 Apr 2024
I don't think anything is wrong with the code in the Question, at least based on visual inspection (though it's always best to use sym(pi) rather than pi in symbolic expressions). Rather, int often doesn't recognize special cases when evaluating parameterized, Fourier integrals. The user needs to identify those cases and manually process them.
syms t % time
syms n % number of terms to calculate the sum to
f = sin(t)^2;
T = sym(pi);
a_0 = 2/T*int(f,t,0,T)/2
a_0 = 
a_n = 2/T*int(f*cos(2*sym(pi)/T*n*t),t,0,T)
a_n = 
b_n = 2/T*int(f*sin(2*sym(pi)/T*n*t),t,0,T)
b_n = 
I don't know why int() captures the special cases of n for b_n but does not for a_n. So we have to handle the special case for a_n manually (keeping in mind that we only care about n >= 1)
a_n = piecewise(n==1,limit(a_n,n,1),a_n)
a_n = 
%b_n = piecewise(n==1,limit(b_n,n,1),b_n)
We can simplify things by placing appropriate assumptions on n. This step isn't really necessary, but adds some clarity IMO.
assume(n,'integer')
assumeAlso(n,'positive'); % n >= 1 for the sin/cos Fourier series
a_n = simplify(a_n)
a_n = 
b_n = simplify(b_n)
b_n = 
0
The reconstruction yields the expected result
fr = a_0 + symsum(a_n*cos(2*sym(pi)/T*n*t),n,1,5)
fr = 
However, an incorrect result for a_n ensues if evaluating the integrals with the assumptions on n.
a_n = 2/T*int(f*cos(2*sym(pi)/T*n*t),t,0,T)
a_n = 
0
b_n = 2/T*int(f*sin(2*sym(pi)/T*n*t),t,0,T)
b_n = 
0
That looks like a bug.
Alternatively, we can Hold the integrals and then release() them after evaluating them at specified values of n (the @doc functionality isn't finding symbolic release() ?)
a_n(n) = 2/T*int(f*cos(2*sym(pi)/T*n*t),t,0,T,'Hold',true)
a_n(n) = 
b_n(n) = 2/T*int(f*sin(2*sym(pi)/T*n*t),t,0,T,'Hold',true)
b_n(n) = 
a_n = release(a_n(1:5))
a_n = 
b_n = release(b_n(1:5))
b_n = 
fr = a_0 + sum(a_n.*cos(2*sym(pi)/T*(1:5)*t))
fr = 
  1 Commento
Owen
Owen il 31 Mar 2024
Thanks Paul this has cleared everything up and thank you for the work-arounds, great to have some suggestions of new functrionality for me to learn and glad to know that my code was not the problem!

Accedi per commentare.

Più risposte (0)

Prodotti


Release

R2023b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by