# How to calculate the power of a finite length signal?

I am confused with these concepts. If the signal is expressed as r(t), I know the power of the signal is given by:

But if the the length of signal T is finite and cannot approach infinity, how can I calculate the power? Can I say the power is approximately equal to the following equation?

If the signal $$r(t)$$ is of finite length (and if integral of its square during its definition interval is also finite) then its average power given by $$\bar{P_x} = \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} |r(t)|^2 dt = \lim_{T \to \infty} \frac{1}{T} \int_{0}^{L} |r(t)|^2 dt = \lim_{T \to \infty} \frac{[\text{constant}]}{T} = 0$$ will be zero...

Btw, you may compute an instantaneous power of the signal, simply as the square of it $$|r(t)|^2$$...

• You are right. But in practical, signals often have finite length, energy and power, right? If I want to calculate the signal power, does it mean that the power of a signal is equal to the average of instantaneous power in a period of time? – Berman Song Oct 4 '18 at 6:26
• @BermanSong the power of a signal $x(t)$ is $|x(t)|^2$ whether it's finite length or infinite length. That's the instantaneous power. On the other hand $P_x$ is the average power and by default it's averaged over $t \in [-\infty,\infty]$ and defined to be the total energy over total time. However, you can also define an average over whatever time interval you wish. Then your average over that interval is what you mean... So you would be defining an average power over signal duration but be careful to interpret it though. A periodic sine was has nonzero $P_x$ do not mix them... – Fat32 Oct 4 '18 at 9:13
• @Fat32 by using $T \rightarrow \infty$ you're making the mistake that the OP was asking for a finite signal. See my answer below. – Dr. Dan Apr 12 at 1:23
• @Dr.Dan in fact my very comment above yours simply explains that alternate definition too ;-) ... – Fat32 Apr 12 at 7:45

Yes, you can say the average power is

$$\bar{P}_r(t) = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} \left| x(t) \right| ^2 dt$$

Where I've just used $$T = t_2-t_1$$ in the first term and limits are adjusted accordingly. Your version has $$t_1 = 0$$ and $$t_2 = T$$.

The instantaneous power is:

$$P_r(t_i) = \left| x(t_i) \right| ^2$$