# Power of periodic signal

Discussion in 'Homework Help' started by xxxyyyba, Nov 1, 2015.

1. ### xxxyyyba Thread Starter Member

Aug 7, 2012
249
2
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 $F_n$.
In book, they gave us some periodic signal to write it in terms of complex Fourier coefficients and calculate power of first three harmonics. What is power of first three harmonics? Is it $\left | F_n_=_1 \right |^{2}+\left | F_n_=_2 \right |^{2}+\left | F_n_=_3 \right |^{2}$?

Mar 31, 2012
17,777
4,805

3. ### xxxyyyba Thread Starter Member

Aug 7, 2012
249
2
There is no such a information in book... But I think it is like I wrote

4. ### WBahn Moderator

Mar 31, 2012
17,777
4,805
So what about F_(n-1), F_(n-2), and F_(n-3)?

If you sum up all the power in all the harmonics, does it equal the power in the signal?

5. ### xxxyyyba Thread Starter Member

Aug 7, 2012
249
2
It must equal power in signal.
Here is formula I found in book:
$P=\frac{1}{T}\int_{\tau}^{\tau+T}f^{2}(t)dt=F^{2}_n_=_0+\sum_{n=-\infty,n\neq 0}^{n=\infty}\left | Fn \right |^{2}$
It equals $F^{2}_n_=_0+\sum_{n=1}^{n=\infty}2\left | Fn \right |^{2}$
So power of first three harmonics should be (hopefully):
$2(\left | F_n_=_1 \right |^{2}+\left | F_n_=_2 \right |^{2}+\left | F_n_=_3 \right |^{2})$