# formula derivation for periodic signal power

Studying DSP on my own. Taking an introductory class. Was given a formula for signal power

$$P = \lim_{M\to\infty} {\frac{1}{2M+1} \sum_{n=-M}^M} \lvert x[n] \rvert^2$$

I do understand why the formula looks the way it does.

But then we were given a formula for a power of a periodic signal, where $N$ is the period,

$$P = {\frac{1}{N} \sum_{n=0}^{N-1}} \lvert x[n] \rvert^2$$

I can't figure out how this formula is derived from the original one. How do I do it?

• If any of the answers below answers your question, please mark it as accepted by clicking on the check mark. – Matt L. Jan 11 '18 at 9:31

I'd like to show you a more formal derivation. Note that the first formula for arbitrary (non-periodic) signals could be rewritten as

$$P_x=\lim_{M\rightarrow\infty}\frac{1}{(2M+1)N}\sum_{n=-MN}^{(M+1)N-1}|x[n]|^2\tag{1}$$

for some integer $N\ge 1$. For $N=1$, Eq. $(1)$ is identical with the first formula in your question. If you choose $N>1$ you simply add more chunks of length $N$ (instead of length $1$) to your sum if you increase $M$.

Now assume that the signal $x[n]$ is periodic with period $N$. This means that the sum over $2M+1$ periods equals $(2M+1)$ times the sum over a single period:

$$\sum_{n=-MN}^{(M+1)N-1}|x[n]|^2=(2M+1)\sum_{n=0}^{N-1}|x[n]|^2\tag{2}$$

Plugging $(2)$ into $(1)$ gives

$$P_x=\lim_{M\rightarrow\infty}\frac{1}{(2M+1)N}(2M+1)\sum_{n=0}^{N-1}|x[n]|^2=\frac{1}{N}\sum_{n=0}^{N-1}|x[n]|^2\tag{3}$$

• Can you explain what does $M$ stand for? – flashburn Jan 9 '18 at 19:54
• @flashburn: $2M+1$ is the number of chunks of length $N$ in the sum. This number needs to get (infinitely) large, but it cancels out in the second formula due to the periodicity of $x[n]$. – Matt L. Jan 9 '18 at 21:17

The first definition you wrote corresponds to the average power of a discrete signal $x[n]$. If the signal is periodic, that means that

$$\exists N\in\mathbb{Z}:x[n] = x[n+N] \ \forall n$$

So if you find the period of the discrete signal, $N$, then if you use your first formula you would be doing the average between a lot of quantities that are all exactly the same. If we define $P_{interval}$ as the average power for a given interval, you would be doing something like (without being too formal in order to make the explanation clear and intuitive):

$$\frac{P_\mathrm{{period\ 1}}+P_\mathrm{{period\ 2}}+...+P_\mathrm{{period\ k}}}{k} \ , \ k\to\infty$$

But due to the periodicity of the signal, all the $P_\mathrm{{period\ i}}$ are the same.

When calculating the average of $k$ things that are equal, the result is equal to any of those things. In this case, as all the average powers for each period is equal to the rest, one can say that the total average equals the average power of (any) single period. And that corresponds to the second formula in your question.

The formula for signal power is the total energy of the signal, divided by the length of the signal: $$P = \frac{E}{N},$$ where $E$ is the energy of the signal, and $N$ is the length in samples.

The total power of the signal is given in both of your equations in the summation. The summation adds together the squared values of all the samples. That value is then divided by the total number of samples in the signal, which gives you the power.

In your upper equation, the signal is assumed to span the samples from $-N$ to $N$, which gives you a total number of $2N+1$ samples. Additionally, the upper equation assumes that the signal is of infinite length, and thus, $N$ approaches infinity in the limit.

The second equation assumes that the signal period is of length $N$, and thus the total energy given by the summation is divided by $N$. I am not sure why the summation goes from 0 to $N$ (N+1 samples), instead of 0 to $N-1$ (N samples), but that might just be a small mistake in the equation. Otherwise it is the same as the other equation.

Note that the $N$ for the periodic signal refers to the length of one period, instead of the length of the whole signal. For a periodic signal, dividing the total energy of one period with the length of that period gives you the same result as dividing the total energy of the whole signal with the length of the whole signal.