Power of periodic signal

Thread Starter

xxxyyyba

Joined Aug 7, 2012
289
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 \(\left | F_n_=_1 \right |^{2}+\left | F_n_=_2 \right |^{2}+\left | F_n_=_3 \right |^{2}\)?
 

WBahn

Joined Mar 31, 2012
24,850
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?
 

Thread Starter

xxxyyyba

Joined Aug 7, 2012
289
If you sum up all the power in all the harmonics, does it equal the power in the signal?
It must equal power in signal.
So what about F_(n-1), F_(n-2), and F_(n-3)?
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):
\(2(\left | F_n_=_1 \right |^{2}+\left | F_n_=_2 \right |^{2}+\left | F_n_=_3 \right |^{2})\)
 
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