Signal power

The average signal power of $x(t)$ is the time-average of $|x(t){|}^2$:

$P_x={\lim }_{T\to \infty }\frac{1}{T}{\int }_{-T/2}^{T/2}|x(t){|}^{2}dt.$

For a signal with infinite energy but a well-defined average squared value, this limit exists and gives a finite, useful number. Sinusoids and other persistent signals fall into this category.

Periodic signals: shortcut

If $x(t)$ is periodic with fundamental period $T$, then the average power computed over one period equals the average power over the whole real line:

$P_x=\frac{1}{T}{\int }_T|x(t){|}^{2}dt,$

where the subscript $T$ means “any interval of length $T$” — pick whichever is most convenient, typically $(-T/2,T/2)$ or $(0,T)$.

Average power of a sinusoid

The most useful fact in this note: for any real sinusoid $x(t)=A\cos (2\pi f_{0}t+\theta )$,

$P_x=\frac{A^2}{2},$

independent of frequency and phase. Derivation: using the periodic-signal formula with $T=T_0$ and the half-angle identity ${\cos }^{2}u=(1+\cos 2u)/2$,

$P_x=\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}{A}^2{\cos }^2(2\pi f_{0}t+\theta )dt=\frac{A^2}{2T_0}{\int }_{-T_0/2}^{T_0/2}\left[1+\cos (4\pi f_{0}t+2\theta )\right]dt.$

The first piece integrates to $T_0$; the second is a cosine integrated over exactly two of its own periods (the new angular frequency is $4\pi f_0$, twice the original), which gives zero. So $P_x=A^2/(2T_0)\cdot T_0=A^2/2$.

This $A^2/2$ shortcut shows up constantly. Worth committing to memory.

A non-sinusoid worked example

Suppose $x(t)$ is periodic with fundamental period 12, and one period (on $-4<t<8$) is $\mathrm{ramp}(-t/5)$. Find the average power.

$\mathrm{ramp}(-t/5)$ is zero for $t>0$ (ramp is zero for negative argument) and equals $-t/5$ for $t<0$. So on $(-4,8)$, the integrand is $(-t/5{)}^2=t^2/25$ for $t\in (-4,0)$, and zero for $t\in (0,8)$. Hence

$P_x=\frac{1}{12}{\int }_{-4}^0\frac{t^2}{25}dt=\frac{1}{300}{\left[\frac{t^3}{3}\right]}_{-4}^0=\frac{1}{300}\cdot \frac{64}{3}=\frac{16}{225}\approx \mathrm{0.0711.}$

The structural point: for periodic signals, integrate over one period, divide by the period length. That’s it.

Energy vs power

A signal is an energy signal if it has finite nonzero energy (and therefore zero average power); it’s a power signal if it has infinite energy and finite nonzero average power. The two categories don’t overlap. Some signals are neither (an exponentially growing signal has both infinite energy and infinite power), but for everything in this course one of the two applies.