# from fourier transform to fourier series

• Jan 31st 2008, 07:48 PM
from fourier transform to fourier series
HI,
I really need to find a relationship between the coefficients of the Fourier transform coefficients and those for Fourier series, especially how to find Fourier series coefficients from the Fourier transform coefficients.

The formula should be like this:
an= X1*(Fn)+X2*(F-n)
bn= X3*(Fn)+X4*(F-n)

Where an and bn are the Fourier series coefficients for cosine and sine respectively. Fn are the Fourier transform coefficients.
FinallyX1,X2,X3,X4 are some coefficients( I don't know if they are constant or not)

I could not derive it by myself and did some research with books and online but I have not found something good.

Thank you
B
• Feb 1st 2008, 06:35 AM
Opalg
Quote:

I really need to find a relationship between the coefficients of the Fourier transform coefficients and those for Fourier series, especially how to find Fourier series coefficients from the Fourier transform coefficients.

The formula for the n-th complex Fourier coefficient of the function f(x) is $\hat{f}(n) = \frac1{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx$. I'm assuming that a_n and b_n are defined by $a_n = \frac1\pi\int_{-\pi}^{\pi}f(x)\cos(nx)\,dx$ and $b_n = \frac1\pi\int_{-\pi}^{\pi}f(x)\sin(nx)\,dx$.
If in the formulas for the complex Fourier coefficients you use the facts that $e^{inx} = \cos(nx) + i\sin(nx)$ and $e^{-inx} = \cos(nx) - i\sin(nx)$ then you should easily be able to express a_n and b_n interms of $\hat{f}(n)$ and $\hat{f}(-n)$.