# How's Fourier series modified for function f(t)= f(2Pi t)?

• Poirot

## Homework Statement

How are the coefficients of the Fourier series modified for a function with a period 2πT?

## Homework Equations

a0 = 1/π ∫π f(x) dx
an = 1/π ∫π f(x) cos(nx) dx
bn = 1/π ∫π f(x) sin(nx) dx

## The Attempt at a Solution

I tried letting x= t/T
so dx = dt/T and the limits x = ± π, x = ± πT
and I was going to plug this into the integrals but I don't think it's right.

Any help would be greatly appreciated, thanks!

Usually the function f(t) is periodic over period T. (not ## 2 \pi ## T). The integration over t will go from t=0 to t=T. You can readily google the topic to get the precise form of the coefficients, etc.

For general period ##2p## the Fourier coefficients are$$b_n = \frac 1 p \int_{-p}^p f(x) \sin(\frac{n\pi x}{p})~dx$$ $$a_n =\frac 1 p \int_{-p}^p f(x) \cos(\frac{n\pi x}{p})~dx$$ $$a_0 = \frac 1 {2p}\int_{-p}^p f(x)~dx$$
##a_0## may or may not have the ##2## in the denominator depending on whether you start the FS with ##a_0## or ##\frac {a_0} 2##.