I need to find the periodicity of the following signal:

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

Now I understand that the basic procedure to determine the periodicity is to find a $ N $ such that $$ x\left [ n \right ] = x\left [ n + N\right ] $$

I applied the procedure to the aforementioned signal and got the following results:
$$ \begin{aligned} x\left[ n\right] &= \cos \left(\frac{\pi n^{2}}{8}\right) \\ x\left[ n + N\right] &= \cos \left(\frac{\pi (n + N)^{2}}{8}\right)\\ &= \cos \left(\frac{\pi (n^{2} + 2nN + N^{2})}{8}\right)\\ &= \Re \left \{\exp \left(i \frac{\pi (n^{2} + 2nN + N^{2})}{8}\right) \right \}\\ &= \Re \left \{\exp \left(i \frac{\pi n^{2}}{8}\right)\exp \left(i \frac{\pi 2nN}{8}\right)\exp \left(i \frac{\pi N^{2}}{8}\right) \right \} \end{aligned} $$ Now for the signal to conform to $ x\left [ n \right ] = x\left [ n + N\right ]$: $$ \Re \left \{\exp \left(i \frac{\pi 2nN}{8}\right)\exp \left(i \frac{\pi N^{2}}{8}\right) \right \} = 1 $$ $ \exp(i\omega) = 1 $ only when $ \omega = 2\pi k $ where $k$ is an integer. Hence:
$$ \frac{\pi 2nN}{8} = \frac{\pi N^{2}}{8} = 2\pi k $$ Now logically I can see that if $ N = 8 $ the first term would be reduced to a multiple of $ 2\pi $ for all $ n $ and hence the fundamental period would be $ N = 8 $ but how would I go about proving this mathematically, that the minimum period is indeed 8? like what set of operations would I perform on:
$$ \frac{\pi 2nN}{8} = \frac{\pi N^{2}}{8} = 2\pi k $$
to obtain $ N = 8 $?


3 Answers 3


You have to prove that the period $N$ is the smallest number satisfying

$$\frac{\pi}{8}(n+N)^2=\frac{\pi n^2}{8}+2\pi k,\quad k\in\mathbb{N}\tag{1}$$

From (1) you get

$$\frac{\pi}{8}n^2+\frac{\pi}{8}2nN+\frac{\pi}{8}N^2=\frac{\pi n^2}{8}+2\pi k$$

which is equivalent to

$$\frac{\pi}{8}2nN+\frac{\pi}{8}N^2=2\pi k\tag{2}$$



for any value of $n$. Clearly, the smallest number $N$ for which the left-hand side of (3) is a multiple of $16$, regardless of the value of $n$, is $N=8$:

$$16n+16\cdot 4=16k$$


You want to prove that $\frac{nN}{8}$ is an integer for any integer $n$. Consider $n=1$. Clearly, $N$ cannot be less than 8.

You also need to prove that $\frac{N^2}{16}$ is an integer. This means that $N$ is a multiple of 4. So, the smallest $N$ that meets both conditions is 8.


For anyone interested in another way to solve this: The radial frequency $\omega$ is defined as

$\omega=\frac{d(\theta)}{dn}= \frac{d(\pi/8 n^2)}{dn}$


$\frac{2 \pi }{N} = 2 \pi/8 n $

Getting the N

$ N= \frac1n 8$

where n is an integer. For the lowest n, we get N=8

Note that although at n=2, you might get N=4, this will not satisfy periodicity at n=1. On the other hand, an N=8 works for n=1 and n=2 (two periods),...


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