# Average power of finite length discrete signal

I know that the power of a discrete signal is given by:

$$P(x) = \lim_{N \rightarrow \infty}\frac{1}{2N + 1}\sum ^N _{n = -N} | x[n]|^2$$

However, I would like to know whether the power of a discrite time finite length signal is defined as:

$$P(x) = \frac{1}{N - 1}\sum ^N _{n = 1} | x[n]|^2$$

Or as,

$$P(x) = \frac{1}{N}\sum ^{N - 1} _{n = 0} | x[n]|^2$$

Being $$N$$ the length of the signal, the number of samples of it.

## 1 Answer

A finite length signal has a power of zero. This becomes obvious if you just use the first formula in your question and apply it to a signal of finite length. Finite length signals are so-called energy signals, and their energy is given by

$$E_x=\sum_{n=0}^{N-1}\big|x[n]\big|^2\tag{1}$$

where $$N$$ is the length of the signal (and we assume that the signal is defined in the index range $$n\in[0,N-1]$$).

The expression

$$P_x=\frac{1}{N}\sum_{n=0}^{N-1}\big|x[n]\big|^2\tag{2}$$

is the power of an $$N$$-periodic signal, as explained in this answer.