Difficult discrete-time periodicity problem

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SUMMARY

The discussion focuses on determining the periodicity of the discrete-time signal x[n] = cos(π/6 * n^2). The user seeks to establish whether this function is periodic and, if so, to find its fundamental period. The key equation derived is (2*n*N + N^2) = 12 * k, which indicates that for periodicity, this must hold true for all n. The user emphasizes the need to find the lowest N that satisfies this condition.

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Homework Statement



Determine whether x[n] is periodic, if so find the fundamental period.

ps forgive my notation, I'm new to physics forums and haven't had a chance to figure out the exact notation syntax yet.x[n]=cos(π/6 * n^2)

The Attempt at a Solution



x(n) is periodic when x(n) = x(n +N)x(n+N) = cos(π/6 * (n+N)^2)

π/6(2*n*N + N^2) = k*2*π

this is kinda where i got stuck

(2*n*N + N^2) = 12 * k
 
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hi jti5017! welcome to pf! :smile:

try using the X2 button just above the Reply box :wink:

is it true for the same N, for all n ? :wink:
 
thanks for the reply!

it has to be true for all n to be periodic by definition i believe.
i mean, I've personally never heard of a sinusoidal period being a function of time (in this case discrete time)

And there are an infinite number of 'N's (provided it is periodic) but I am specifically looking for the lowest N, the fundamental period
 

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