How to find Fourier series for the periodic function?

Hello, I need to find the the Fourier series for the periodic function f(t) = 0 for -pi < t <= 0, and f(t) = t for 0 < t <= pi, and f(t) = f(t + 2pi). I can do the calculation by hand, but I want to ask if Matlab can help me solve this problem. If yes, could you provide me some instruction of how to write codes into Matlab to get the series that I want? Thank you so much!

p/s: This question has nothing to do with my homework, I just simply want to know if Matlab can help me solve this kind of problem so that I can save a lot of time and increase the accurateness of my answer if I encounter this type of problem in the future.

[Torsten]

You know the integral formulae for the Fourier coefficients of a function f ? Then use MATLAB's "int" to evaluate them, and you'll get the formal Fourier series for f.

[Kieth Duong]

I see, so does it mean that Matlab can't help me directly transform some functions into Fourier series and I must write down clearly the formula of coefficients of Fourier series then Matlab can help me do the calculation stuff?

[Torsten]

You don't need to write down the coefficients. You need to write down the formula by which the coefficients are computed.

Don't you have a textbook which states

a_i = 1/pi * integral_{t=-pi}^{t=pi} f(t) * cos(i*t) dt
b_i = 1/pi * integral_{t=-pi}^{t=pi} f(t) * sin(i*t) dt

?

You can compute the values for these integrals using MATLAB's "int".

Or you go the "lazy way" Matt J suggested in his answer. I did not yet know of this MATLAB function.

[Kieth Duong]

okay I understand now, thank you so much!

Answer 1

[Matt J]

You can use fourier, e.g.

syms x n T

f=rectangularPulse(x);

series = simplify( fourier(f,2*pi*n/T)/T   )

series = 

[Paul]

Shouldn't that be 2*pi*n/T in the argument to fourier() (no negative sign) ?

[Matt J]

The documentation was a little hard to read, but yes.

[Paul]

Assuming the default symprefs ....

Also, any approach using fourier or int still needs to do a little bit of work to account for cases when the denominator goes to zero if we are going to use series in subsequent computations. So in this case we'd need

syms x n T
f=rectangularPulse(x);
series(n) = simplify( fourier(f,2*pi*n/T)/T   );
series(n) = piecewise(n == 0, 1, series(n))

series(n) = 

[Matt J]

I think rather than series(0)=1 it should be 1/T.

syms x n T
f=rectangularPulse(x);
series(n) = simplify( fourier(f,2*pi*n/T)/T   );

a0=limit(series(n),n,0)

a0 = 

[Paul]

Agreed.