# Power of periodic signal

Given a signal $s(t) = e^{j(t+\pi)}$, I conclude that the signal is periodic with period $T=2\pi$, so its power should be $$P = \frac{1}{T}\int_{-T/2}^{T/2}|s(t)|^2dt= \frac{1}{2\pi}\int_{-\pi}^{\pi}|1|^2dt = \frac{1}{2\pi}t|^{\pi}_{-\pi}=\frac{\pi-(-\pi)}{2\pi}=1$$

Is the formula for the power of a periodic signal correct?

• Welcome to DSP.SE! What is $P$? – Peter K. Nov 15 '15 at 12:45
• @PeterK. Power, my textbook said for periodic signal I can instead of $\lim\int_{T/2}^{T/2}|s(t)|^2dt = \frac{1}{T}\int_{-T/2}^{T/2}|s(t)|^2dt$ – temp8jfhfhf Nov 15 '15 at 12:51
• I edited. My question is if I can use that equality i wrote in my other comment for periodic signals – temp8jfhfhf Nov 15 '15 at 12:53

## 1 Answer

The average power of a signal $s(t)$ is given by

$$\overline{s^2(t)}=\lim_{T\rightarrow\infty}\frac{1}{T}\int_{-T/2}^{T/2}|s(t)|^2dt\tag{1}$$

If $s(t)$ is periodic, $(1)$ is equivalent to the average power in one period:

$$\overline{s^2(t)}=\frac{1}{T}\int_{-T/2}^{T/2}|s(t)|^2dt\tag{2}$$

where $T$ now denotes the period of $s(t)$.