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

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?

  • $\begingroup$ I don't really understand. If $x[4]$ only equal to $4i+1$, or all the signal values? $\endgroup$ Commented Nov 4, 2019 at 19:53
  • $\begingroup$ For any constant $c$, $M\ge 0$, $\sum_{-M}^{M} c = (2M+1)c$ $\endgroup$ Commented Nov 4, 2019 at 19:54
  • 2
    $\begingroup$ how does this sum function ? What do you mean by this phrase ? $\endgroup$
    – Fat32
    Commented Nov 4, 2019 at 20:37
  • $\begingroup$ @LaurentDuval That makes sense, but how can I derive that formula for $$(2M+1)c$$? The x[4] part was a mistake, corrected it. $\endgroup$ Commented Nov 4, 2019 at 21:14

2 Answers 2


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 $(2M+1)c$.


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