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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.

skrowten_hermit
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2 Answers2

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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.

Peter K.
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Jason R
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    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
  • Indeed that is true; that condition is the typical assumption for discrete-time signals. – Jason R Feb 10 '16 at 19:58
  • 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
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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.

CroCo
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