Why is cos(n/6) aperiodic?

This is a very common example in most Signal Processing books I have come across.

x(n) = cos($\frac{n}{6}$) is a non-periodic discrete signal because it doesn't satisfy the periodicity condition for discrete time signals i.e, it is not of the form 2$\pi$($\frac{m}{N}$).

My question is :

the coefficient of n, i.e, $\Omega_0$=$\frac{1}{6}$ here can also be expressed as $\frac{1}{6}$ = $\frac{1}{6}$ * $\frac{2\pi}{2\pi}$ = 2$\pi$$\frac{1}{12\pi} Now, substituting for \pi = \frac{22}{7} in above, we get 2\pi$$\frac{7}{12*22}$. So, $\frac{1}{6}$ can be written as 2$\pi$($\frac{7}{264}$), which is in the form 2$\pi$($\frac{m}{N}$) with a period N=264.

I'm sure I'm missing something which may be obvious but it would be of great help if someone could point it out and explain.

• pi is NOT 22/7. It's an irrational number – Hilmar Feb 10 '16 at 19:25
• $x(0) = \cos(0) = 1$. Now find the next larger (integer) value of $n$ such that $\cos\left(\frac n6\right) = 1$. – Dilip Sarwate Feb 10 '16 at 20:11
• Which is not possible and hence is non-periodic (since the function cannot return to 1)? – skrowten_hermit Feb 11 '16 at 14:09

The problem with your reasoning is that $\pi \ne \frac{22}{7}$; $\pi$ is an irrational number. There is no period $N$ for which $x[n] = x[n+N] \ \forall \ n \in \mathbb{Z}$. Hence, the sequence is not periodic.
• I added that $n$ needs to be an integer. Otherwise the the statement can be true if $n \in \mathbb{R}$. – Peter K. Feb 10 '16 at 19:36
• Yes, I've just had too many people on this site not get that $n$ is an integer to want the grief of not specifying it. :-) – Peter K. Feb 10 '16 at 20:15
• Indeed, for all possible choices of nonzero integer $N$, there is no integer $n \in \mathbb Z$ for which $x[n] = x[n+N]$ holds, that is, we don't need to consider the possibility that $x[n] = x[n+N]$ holds for some, but not all, $n \in \mathbb Z$. – Dilip Sarwate Feb 10 '16 at 20:15
• Now I get it. It is okay to multiply and divide by $\pi$, but $\pi$ here is in radians (as multiple of $\pi$ in radians along the axis for time n) and not substitutable for $\frac{22}{7}$ in the denominator like I did, which leaves me with 2$\pi$($\frac{1}{12\pi}$). So, equating/comparing this with 2$\pi$($\frac{m}{N}$) gives me m=1 and N=12$\pi$ and since N is definitely not an integer, periodicity test fails.I got it right, didn't I? – skrowten_hermit Feb 11 '16 at 10:56
The periodicity of a signal holds if we can show $$x(n)=x(n+N)$$, otherwise, the signal is nonperiodic. Simply start with \begin{align} x(n+N) &= \cos( \frac{n}{6} + \frac{N}{6}) \\ &= \cos(\frac{n}{6})\cos(\frac{N}{6}) - \sin(\frac{n}{6})\sin(\frac{N}{6}) \end{align} In order for $$x(n)=x(n+N)$$ to hold, $$\cos(\frac{N}{6})=1$$ and $$\sin(\frac{N}{6})=0$$. We search for the smallest value of $$N$$. This is true if $$\frac{N}{6}=2\pi \implies N = 12\pi$$. Indeed, if $$N=12\pi$$, we get \begin{align} x(n+N) &= \cos( \frac{n}{6} + \frac{N}{6}) \\ &= \cos(\frac{n}{6})(1) - \sin(\frac{n}{6})(0) \\ &= \cos(\frac{n}{6}) \\ &= x(n) \end{align} But $$N$$ must be a positive integer, therefore, the signal is nonperiodic.