I have just started out learning the basics of discrete time. I would like some help to understand if cos2n is periodic or not.
I know the discrete time formula for periodicity: x[n] = x[n+N].
I Also know that ω=2πf, ω=2π/Ν.
I searched the web and i found a solution that goes like so:
x [ n + N ] = C o s ( 2 n + 2 N ) therefore 2 N = Ω = m 2π, m ∈ Z, (at quora.com).
I do not really understand it, i think that m integer does not fit in the standard formula i know of: ω=2πf. Can you solve it using a step by step guide for 'dummies'?
Best Answer
Consider a discrete-time sequence x[n] based on a sinusoid with angular frequency \$\omega _0\$: x[n]=cos[\$\omega _0\$n].
If this sequence is periodic with a period of N samples, then the following must be true: cos[\$\omega _0\$ (n + N)] = cos [\$\omega _0\$n] (Eq. 1). However, the left hand side can be expressed as cos [\$\omega _0\$(n + N)] = cos [\$\omega _0\$n + \$\omega _0\$N]. The cosine function is periodic with a period of 2\$\pi\$ and therefore cos [\$\omega _0\$n] = cos [\$\omega _0\$n + 2\$\pi\$r] (Eq.2), for integer values of r. Comparing Eq.1 and Eq.2 we get:
$$\omega _0 N =2\pi ; \ \text{(for r=1)}$$ $$2\pi f _0 N =2\pi$$ $$f _0 = \frac{1}{N}$$
Since N is integer the signal will be periodic if \$f_0\$ is a rational number. Otherwise, it will be non-periodic.
In your example, \$\omega _0\$ = 2 so \$f_0\$ = \$\frac{2}{2\pi}\$. But you cannot find an integer division (\$\frac{1}{N}\$) that results in an irrational number (\$f_0\$). So no N can be found to meet the criteria for this signal to be periodic.