# How can I prove this discrete signal is periodic?

## Homework Statement

Prove the discreet signal is periodic: ## Homework Equations

for periodic funtions: x[n] = x[n + N]

## The Attempt at a Solution

I made an equality (im going to leave the sigma out for simplicity):

2^(-abs(n-2m)) = 2^(-abs(n+N-2m))
I dont know what I need to do from here. The absolute value throws me off, I dont know what to do with it. My guess would be that the only thing that matters is what is inside the abs since everything else is the same so I really only need for the follwing to be true:

abs(n-2m) = abs(n+N-2m).......................................................(1)
but then N = 0

I guess I could make N = (-2n +4m) since using this in equation (1) will get me:

abs(n-2m) = abs(-n+2m) = abs(-(n-2m)) which is true. Then N can equal (-2n + 4m). Now doesnt N have to be independent of n? Did I do this right? I sort of started getting the idea as I went along here so sorry if I solved it.

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BvU
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Why don't you work out a few ##x(n)## to get a feeling for what this is about ? big N pops up in no time at all !

Why don't you work out a few ##x(n)## to get a feeling for what this is about ? big N pops up in no time at all !
I understand what you mean but I am not a big fan of this brute force method :( is this really the best way?

BvU
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If you don't see an alternative ...

I note that you ran into trouble with your approach, so at least one of the choices you made was not such a good idea. My bet is that leaving out the summation is the one BvU
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And I wouldn't call it a brute force method. More: orientation phase.
Who knows you don't even have to do the complete working out And I wouldn't call it a brute force method. More: orientation phase
Well I try but I just dont understand this summation, maybe you can help me understand:

x = Σ2^(-2m)
x = ∑2^(-abs(1-2m))
x = ∑2^(-abs(2-2m))
x = ∑2^(-abs(3-2m))
x = ∑2^(-abs(4-2m))

I dont even work with sums so I am totally confused.

BvU
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dont even work with sums
Well, there is a first time for everything ... And I venture to assume that you do know ##\sum 2^{-|m|}## ?

Fortunately you will accept that a sum yields the same result if all the terms are exactly the same, so we can reduce your problem to showing that that indeed is the case for a certain big N

rings a bell ? Well, there is a first time for everything ... And I venture to assume that you do know ##\sum 2^{-|m|}## ?

Fortunately you will accept that a sum yields the same result if all the terms are exactly the same, so we can reduce your problem to showing that that indeed is the case for a certain big N

rings a bell ? I agree with you on "a sum yields the same result if all the terms are exactly the same" but I am not familiar with that sum. No bells have been rung sir. It is over for me.

but let me try reasoning
at x = ∑2^-abs(-2m) I can get numbers 2^0 + 2^(-2) + 2^(-4) + 2^(-6)+ ... + 2^(-2n) where n = 0,1,2,3,4,5...
at x = ∑2^(-abs(1-2m)) i can get the following: 2^-1 + 2^-3 + 2^-5 + .... +2^(-(2n+1)) where n = 0,1,2,3,4...
at x = ∑2^(-abs(2(1-m)) i can get: 2^(-2) + 2^(0) + 2^(-2) + 2^(-6)

hmm it looks like x is just a shifted form of x, then x at m = -1 should equal to 2^(-2)
errrr it equals 2^(-2). So ima go ahead and goes that N = 2 since x = x = x[0 + 2]. The fundamental period is 2. Man integrals are so much easier. Is there a faster way of doing this?

BvU
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2019 Award
Is there a faster way of doing this?
Yes: you conclude that the exponents go through the same set of values for all even ##n## and also through a same set of values (but different from the even one) for all odd ##n##. Here's a picture of |n-2m| as a function of ##m## for a few values of ##n## There are only two distinct sets of values if ##m## runs from ##-\infty## to ##+\infty##

Furthermore you should know that ## 1+ {1\over 2} + {1\over 4} + {1\over 8} + {1\over 16} + {1\over 32} + .... = 2##

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