# The sum part of the formula for a signals power

$$P_x=\lim_{M\rightarrow \infty }\frac{1}{2M+1}\sum_{n=-M}^{M}\left | x[n] \right |^2$$

I have used this formula to find out the powers for different signals, but the sum part of the formula still puzzles me. Lets take one example: $$x[n] = 4i+1$$ Simply calculating the absolute value, I got this: $$|x[n]|^2=|4i+1|^2=(\sqrt{4^2+1^2})^2=17$$ 17 seems to be the answer I got after going through the rest of the formula, but I am not sure how the sum works. By using Wolframa, I arrived at: $$\sum_{n=-M}^{M}17 =34M+17$$ Which would complete the limit part of the formula with the final result of 17, as discussed. How does this sum function, as described above?

• I don't really understand. If $x$ only equal to $4i+1$, or all the signal values? Nov 4 '19 at 19:53
• For any constant $c$, $M\ge 0$, $\sum_{-M}^{M} c = (2M+1)c$ Nov 4 '19 at 19:54
• how does this sum function ? What do you mean by this phrase ? Nov 4 '19 at 20:37
• @LaurentDuval That makes sense, but how can I derive that formula for $$(2M+1)c$$? The x part was a mistake, corrected it. Nov 4 '19 at 21:14

A more general expression states that for $$M \geq N$$:

$$\sum_{n= N}^{n = M} c = (M-N+1) \cdot c$$

where the derivation simply relies on fact that the epxression has (M-N+1) terms :

$$\sum_{n= N}^{n = M} c = \{ c + c + ... + c\} = (M-N+1) \cdot c$$

And when applied for your particular case (with $$N = -M$$) it becomes: $$\sum_{n= -M}^{n = M} c = (M-(-M)+1) \cdot c = (2M+1) \cdot c$$

For instance, from $$-3$$ to $$3$$, you have $$-3,\,-2\,-1,\,0\,1,\,2,\,3$$, hence $$2\times 3+1$$ terms. More generally, the sum from $$-M$$ to $$M$$ is composed of $$2M+1$$ terms:

• indices with $$m$$ strictly negative (a total of $$M$$),
• those which $$m$$ strictly positive (a total of $$M$$),
• plus one at zero ($$1$$).

If all terms are the same constant $$c$$, the total is $$(2M+1)c$$.