0

If I am given a function

$$ f(x) = \left\{ \begin{array}{ll} 2 & \quad x \in (0,6) \\0 & \quad x\in(0,-6) \end{array} \right. $$

$I=(-6,6)$

and I want to find the complex series representation, is it correct that

$\mathbb{C}_n=\frac{1}{12} \int_0^6(2*\exp(-\frac{-xni\pi}{12})) dx$?

Because the first half of interval contribution would just be $0$, Right?

To The MODERATOR .. I am not fully sure about it, but hope it improves upon earlier..

Narasimham
  • 40,495
Akitirija
  • 435

1 Answers1

1

The complex Fourier series is defined as: \begin{equation} f(x)=\sum_{-\infty}^{+\infty} c_n e^{i n x \frac{2 \pi}{L}} \end{equation} where \begin{equation} c_n=\frac{1}{L} \int_{0}^{L} f(x) e^{-i n x \frac{2 \pi}{L}} dx \end{equation} In your case $L=3$. Of course the Fourier series is periodic.

Upax
  • 2,073
  • 13
  • 18
  • Sorry, I am new to this, I thought $c_n=\frac{1}{2L}...$. How do you calculate L to begin with? – Akitirija Mar 14 '15 at 18:01
  • f(x) is different from zero only between 0 and 3. If you have 2L instead of L, your integral should be between -L and L, and you should have f(x) defined between -L and L. – Upax Mar 15 '15 at 08:59
  • 1
    Not the definition of f(x) is much better. With this definition your integral computed between -6 and +6 reduces to an integral between 0 and 6, since your function is 0 between -6 and 0. Your $c_n$ is correct. – Upax Mar 15 '15 at 10:26
  • OK, I think I understood. Thank your for taking the time, Upax! – Akitirija Mar 15 '15 at 13:25