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.

  • 4
    $\begingroup$ pi is NOT 22/7. It's an irrational number $\endgroup$
    – Hilmar
    Feb 10 '16 at 19:25
  • 2
    $\begingroup$ $x(0) = \cos(0) = 1$. Now find the next larger (integer) value of $n$ such that $\cos\left(\frac n6\right) = 1$. $\endgroup$ Feb 10 '16 at 20:11
  • $\begingroup$ Which is not possible and hence is non-periodic (since the function cannot return to 1)? $\endgroup$ 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.

  • 1
    $\begingroup$ I added that $n$ needs to be an integer. Otherwise the the statement can be true if $n \in \mathbb{R}$. $\endgroup$
    – Peter K.
    Feb 10 '16 at 19:36
  • $\begingroup$ Indeed that is true; that condition is the typical assumption for discrete-time signals. $\endgroup$
    – Jason R
    Feb 10 '16 at 19:58
  • $\begingroup$ 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. :-) $\endgroup$
    – Peter K.
    Feb 10 '16 at 20:15
  • $\begingroup$ 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, $n \in \mathbb Z$. $\endgroup$ Feb 10 '16 at 20:15
  • $\begingroup$ 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? $\endgroup$ 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.

  • $\begingroup$ I like this approach. "But N must be a positive integer". Is this a precondition? Could you cite a source? $\endgroup$ Mar 14 '19 at 19:21
  • $\begingroup$ @skrowten_hermit, yes it is. This is because the signal is discrete. $\endgroup$
    – CroCo
    Mar 15 '19 at 11:47

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