# Complex form of Fourier series

1. Sep 29, 2014

### etf

Let function $f(t)$ is represented by Fourier series,
$$\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n\cos{\frac{2n\pi t}{b-a}}+b_n\sin{\frac{2n\pi t}{b-a}}),$$
$$a_0=\frac{2}{b-a}\int_{a}^{b}f(t)dt,$$
$$a_n=\frac{2}{b-a}\int_{a}^{b}f(t)cos\frac{2n\pi t}{b-a}dt,$$
$$b_n=\frac{2}{b-a}\int_{a}^{b}f(t)sin\frac{2n\pi t}{b-a}dt,$$
where $$a$$ and $$b$$ are lower and upper boundary.

Here is how I transformed it in order to get complex form:

But here is what I find on web:
$$f(t)=\sum_{-\infty}^{+\infty}Fne^{jnw0t}$$, where $$Fn=\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}f(t)e^{-jnw0t}dt$$

If I put in my solution a=-T/2, b=T/2, I will get $$Fn=\frac{2}{b-a}\int_{a}^{b}f(t)e^{-j\frac{2n\pi t}{b-a}}dt=\frac{2}{T/2-(-T/2))}\int_{-T/2}^{T/2}f(t)e^{-j\frac{2n\pi t}{T/2-(-T/2)}}dt=\frac{2}{T}\int_{-T/2}^{T/2}f(t)e^{-j\frac{2n\pi t}{T}}dt= \frac{2}{T}\int_{-T/2}^{T/2}f(t)e^{-jnw0t}dt$$

You can see that I get (2/T)*integral but it should be (1/T)*integral. Whats wrong?

2. Sep 29, 2014

### etf

Mistake found :)