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I am trying to find the average signal power of some discrete functions.

For a finite interval, I can use the formula:

[1 / (n_2 - n_1)] * (SUM n_1 to n_2) of |x[n]|^2

For an infinite interval, I can use the formula:

[1 / (2N+1)] limit(n->infinity) of (SUM -N to N) of |x[n]|^2

I am not sure how to apply these formulas to various discrete functions. However, I was assigned this problem (homework), and this is my interpretation of it:

x[n] = (-1)^n

Squaring it becomes:

[x[n]]^2 = (-1)^(2n)

Now, because of the 2, I see that the power will always be even. I see that each term of the summation will be a positive one for n>=0:

n=1:   (-1)^(2*1) = (-1)^2 = 1
n=2:   (-1)^(2*2) = (-1)^4 = 1
n=3:   (-1)^(2*3) = (-1)^6 = 1
...

So, it seems that the summation of this discrete function, if my limits are from n=0 to n=N, is just 1+1+1+1+1... for N+1 number of terms (because the first n is zero). That is, it seems that the answer, for a sum from zero to N is just N+1. Is this correct?

I see that for values of n<0, each term will also be a positive 1.

(-1)^(2(negative-n)) = 1/(-1^(2(positive-n)).

n = -1:   (-1)^(-2) = (1/(-1^(2)) = 1/1 = 1
n = -2:   (-1)^(-4) = (1/(-1^(4)) = 1/1 = 1

So, if I have a summation of a finite interval from n1 to 2, it seems that the answer would be:

(n2 - n1) + 1

Is this correct? For example:

Sum from n = -10 to 10 of (-1)^n = [10 -(-10)]+1 = 21

Now, in this problem, I am not given any values for N. I am just supposed to find the average signal power of that discrete function. I see that the formula for infinite interval includes the limit n->infinity. So, is the answer just infinity?

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    \$\begingroup\$ I think you are mixing power and energy. Energy of the signal for \$n\to \infty\$ is \$\infty\$, but power is 1. \$\endgroup\$ – venny Sep 11 '14 at 20:24
  • \$\begingroup\$ Is this an EE question or a maths question? Before anyone starts jumping all over me I've asked this as a question. \$\endgroup\$ – Andy aka Oct 1 '14 at 19:09
  • \$\begingroup\$ EE because it's from signals and systems. \$\endgroup\$ – asdf Oct 27 '14 at 17:42
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The second formula you have posted gives the average power for an interval from -N to N as N tends to infinity. Using this formula, as you have stated the summation of individual 1's from -N to N, whatever N is,will just be 2N+1, and so 2N+1/2N+1=1 and so the limit is just 1,as N has been eliminated from the expression.

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  • \$\begingroup\$ Thanks! I figured it out but haven't logged on in a while and just saw your answer. \$\endgroup\$ – asdf Oct 27 '14 at 17:41

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