Fourier coefficients of a function

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SSBGH
SSBGH am 6 Mai 2022
Kommentiert: SSBGH am 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?

Akzeptierte Antwort

Paul
Paul am 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.

Weitere Antworten (1)

Torsten
Torsten am 6 Mai 2022
Bearbeitet: Torsten am 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 Kommentare
SSBGH
SSBGH am 6 Mai 2022
after writing your code im getting bunch of errors do you know why?
Torsten
Torsten am 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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