Rayleigh energy theorem; power from Fourier series

Rayleigh energy theorem

For an energy signal $x(t)$ with Fourier transform $X(f)$:

$$E = \int_{-\infty}^{\infty}|x(t)|^2dt = \int_{-\infty}^{\infty}|X(f)|^2df$$

Derivation: substitute $x^*(t) = \int X^*(f)e^{-j2\pi ft}df$ into the energy integral and interchange the order of integration:

$$\begin{aligned} E &= \int_{-\infty}^{\infty}x(t)\,x^*(t)\,dt = \int_{-\infty}^{\infty}x(t)\left[\int_{-\infty}^{\infty}X^*(f)e^{-j2\pi ft}df\right]dt \\ &= \int_{-\infty}^{\infty}X^*(f)\left[\int_{-\infty}^{\infty}x(t)e^{-j2\pi ft}dt\right]df \\ &= \int_{-\infty}^{\infty}X^*(f)X(f)\,df = \int_{-\infty}^{\infty}|X(f)|^2df \end{aligned}$$

So energy can be found from the spectrum; $|X(f)|^2$ is the energy spectral density.

Power of a periodic signal from $C_n$

A periodic signal with period $T_0$ ($f_0 = 1/T_0$) can be written $x(t) = \sum_n C_ne^{j2\pi nf_0t}$. Its average power is

$$\begin{aligned} P &= \frac{1}{T_0}\int_{T_0}x(t)x^*(t)\,dt = \sum_n\sum_m C_nC_m^*\frac{1}{T_0}\int_{T_0}e^{j2\pi(n-m)f_0t}dt \\ &= \sum_{n=-\infty}^{\infty}|C_n|^2 \end{aligned}$$

since the integral is 1 for $m = n$ and 0 otherwise. This is Parseval's power theorem: the power is the sum of the squared magnitudes of the Fourier coefficients. For a real signal, $P = C_0^2 + 2\sum_{n=1}^{\infty}|C_n|^2$.

Example: $x(t) = 2 + 4\cos 2\pi f_0t$ has $C_0 = 2$, $C_{\pm1} = 2$, so $P = 4 + 4 + 4 = 12$ W.