# Power spectrum of periodic signal

Discussion in 'Homework Help' started by xxxyyyba, Oct 31, 2015.

1. ### xxxyyyba Thread Starter Member

Aug 7, 2012
249
2
This is not really homework question, but this category looks most appropriate to me
We know that periodic function can be writen in terms of complex Fourier coefficients:
$f(t)=F_n_=_0+\sum_{n=-\infty,n\neq 0}^{n=\infty}F_ne^{jnw_0t}$, where $F_n=\frac{1}{T}\int_{\tau}^{\tau+T}f(t)e^{-jnw_0t}dt$ and $F_n_=_0$ is DC component. Power spectrum of signal is defined as $S_1_1(nw_0)=\left | F_n \right |^{2}$, where $\left | F_n \right |$ is modulus of complex Fourier coefficient $F_n$.
We can draw power spectrum as function of both negative and positive discrete frequencies nw0. My question is, if I want to draw it only for positive discrete frequencies, I should multiply $\left | F_n \right |^{2}$ by 2, right? Should $\left | F_n_=0 \right |^{2}$ be also multiplied by 2?

Last edited: Oct 31, 2015
2. ### WBahn Moderator

Mar 31, 2012
17,715
4,788
Yes and no, respectively.

Having said that, I don't understand your first expression (not that it isn't even an equation).

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3. ### xxxyyyba Thread Starter Member

Aug 7, 2012
249
2
I edited my post, I hope it is clear now

4. ### WBahn Moderator

Mar 31, 2012
17,715
4,788
Why exclude n=0 from the sum? Why not just get rid of the first term and let it appear naturally as part of the sum?

5. ### xxxyyyba Thread Starter Member

Aug 7, 2012
249
2
Because for n=0 Fn is usually undefined (something/0). For n=0 we calculate integral separately...