Fourier coefficients of a function

26 vues (au cours des 30 derniers jours)
SSBGH
SSBGH le 6 Mai 2022
Commenté : SSBGH le 6 Mai 2022
given this function with this interval is there any way to calculate the Fourier coefficients (ak, a0, bk) of this fuction in matlab?

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Paul
Paul le 6 Mai 2022
Hi SSBGH
If you want a closed form solution, use the symbolic toolbox to define f(x)
syms x
f(x) = ....
Once you have that you can use the symbolic toolbox to compute a0, ak, and bk in terms of their defining integrals using the int function
syms k integer positive
a0 = int(...)
ak = int(...)
bk = int(...)
The defining integrals should be in your notes or text book, or can be found here.
If you want to numerically compute a0, ak, and bk for given values of k, first define f(x) as an anonymous function
f = @(x) ....
Then use the integral function to compute a0, or ak, or bk for a specific value of k
Your're more likely to get more help if you post your code and explain where you think it might be having problems.
  1 commentaire
SSBGH
SSBGH le 6 Mai 2022
thanks alot man

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Plus de réponses (1)

Torsten
Torsten le 6 Mai 2022
Modifié(e) : Torsten le 6 Mai 2022
syms x omega n
f = 0.5*(sin(x)+abs(sin(x)));
F(omega) = int(f*exp(1i*omega*x),x,-pi,pi)
n = 10;
g = real(F(0))/(2*pi) + sum(real(F(1:n))/pi.*cos((1:n)*x)+imag(F(1:n))/pi.*sin((1:n)*x))
g = matlabFunction(g);
x = linspace(-pi,pi,100)
plot(x,g(x))
hold on
f = matlabFunction(f);
plot(x,f(x))
  2 commentaires
SSBGH
SSBGH le 6 Mai 2022
after writing your code im getting bunch of errors do you know why?
Torsten
Torsten le 6 Mai 2022
I don't know. Maybe MATLAB cannot evaluate F(1).
Try
g = real(F(0))/(2*pi) + 0.5*sin(x) + sum(real(F(2:n))/pi.*cos((2:n)*x)+imag(F(2:n))/pi.*sin((2:n)*x))

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