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I think I got to the right answer for the wrong reason. Could you please verify my approach? Any help is highly appreciated.

Problem:

Is [tex] x [n] [/tex] periodic? If so, what is the fundamental period?

[tex] x [n] = \cos \left( \frac{\pi}{8} n^2 \right) [/tex]

Here is what I've got:

If [tex] x [n] = x [n + N_0][/tex], then it is periodic. Let's check:

[tex] \cos \left( \frac{\pi}{8} n^2 \right) = \cos \left[ \frac{\pi}{8} \left( n + N_0 \right) ^2 \right] [/tex]

[tex] \exp \left( j\frac{\pi}{8} n^2 \right) = \exp \left[ j\frac{\pi}{8} n^2 + j\frac{\pi}{8} 2n N_0 + j \frac{\pi}{8} N_0 ^2 \right] [/tex]

[tex] \exp \left( j \frac{\pi}{8} n^2 \right) = \exp \left( j\frac{\pi}{8} n^2 \right) \exp \left( j\frac{\pi}{8} 2n N_0 \right) \exp \left( j\frac{\pi}{8} N_0 ^2 \right) [/tex]

[tex] \exp \left[ j \frac{\pi}{8} \left( N_0 ^2 + 2nN_0 \right) \right] = 1[/tex]

[tex] \frac{\pi}{8} \left( N_0 ^2 + 2nN_0 \right) = 2\pi [/tex]

However [tex] N_0 [/tex] should be independent of [tex] n [/tex], and so [tex] 2nN_0 = 0 [/tex]. Then [tex]N_0 = 8 [/tex].

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# Homework Help: Discrete-time signals

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