How can I prove this discrete signal is periodic?

[Boltzman Oscillation]

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 don't know what I need to do from here. The absolute value throws me off, I don't 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 doesn't 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.

 

[BvU]

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 !

 

[Boltzman Oscillation]

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]

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]

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

 

[Boltzman Oscillation]

And I wouldn't call it a brute force method. More: orientation phase

Well I try but I just don't understand this summation, maybe you can help me understand:

x[1] = Σ2^(-2m)
x[2] = ∑2^(-abs(1-2m))
x[3] = ∑2^(-abs(2-2m))
x[4] = ∑2^(-abs(3-2m))
x[5] = ∑2^(-abs(4-2m))

I don't even work with sums so I am totally confused.

 

[BvU]

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 ?

 

[Boltzman Oscillation]

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[0] = ∑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[1] = ∑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] = ∑2^(-abs(2(1-m)) i can get: 2^(-2) + 2^(0) + 2^(-2) + 2^(-6)

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

 

[BvU]

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 $1+ {1\over 2} + {1\over 4} + {1\over 8} + {1\over 16} + {1\over 32} + ... = 2$