# Discrete-time signals

1. Sep 23, 2006

Folks,

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 $$x [n]$$ periodic? If so, what is the fundamental period?

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

Here is what I've got:

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

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

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

$$\exp \left( j \frac{\pi}{8} n^2 \right) = \exp \left( j\frac{\pi}{8} n^2 \right) \exp \left( j\frac{\pi}{8} 2n N_0 \right) \exp \left( j\frac{\pi}{8} N_0 ^2 \right)$$

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

$$\frac{\pi}{8} \left( N_0 ^2 + 2nN_0 \right) = 2\pi$$

However $$N_0$$ should be independent of $$n$$, and so $$2nN_0 = 0$$. Then $$N_0 = 8$$.

Last edited: Sep 23, 2006
2. Sep 24, 2006

### 0rthodontist

I assume you're restricting to integer values of n. It looks regular somehow but it doesn't appear periodic at first glance. The possible values of cos(pi/8 * n^2) depend on pi/8 * n^2 modulo 2pi. The potential modulos you could get are 0, pi/8, 2pi/8, ..., 15pi/8. If pi n^2 / 8 - 2 pi k = r pi/8 for 0 <= r <= 15, then you have
n^2 / 8 - 2 k = r/8
n^2 = 16 k - r
so looking at the remainders of n^2 mod 16 using Haskell
Code (Text):

Prelude> [(mod (n^2) 16) | n <- [1..30]]
[1,4,9,0,9,4,1,0,1,4,9,0,9,4,1,0,1,4,9,0,9,4,1,0,1,4,9,0,9,4]
It does look like they form a repeating pattern of length 8, namely 1, 4, 9, 0, 9, 4, 1, 0. You want to show that n^2 is congruent to (n+8)^2 mod 16, and that shows it is periodic.