Period of cos(πn/5)·sin(πn/3)

Energy signal: $0 < E < \infty$, $P = 0$, where $E = \sum_{n}|x[n]|^2$. Power signal: $0 < P < \infty$, $E = \infty$, where $P = \lim_{N\to\infty}\frac{1}{2N+1}\sum_{-N}^{N}|x[n]|^2$. Periodic sequences are power signals.

Periodicity of $x[n] = \cos\left(\frac{\pi n}{5}\right)\sin\left(\frac{\pi n}{3}\right)$

Use $\cos A\sin B = \frac{1}{2}[\sin(A+B) - \sin(A-B)]$:

$$\begin{aligned} x[n] &= \frac{1}{2}\left[\sin\left(\frac{\pi}{5}+\frac{\pi}{3}\right)n - \sin\left(\frac{\pi}{5}-\frac{\pi}{3}\right)n\right] \\ &= \frac{1}{2}\sin\left(\frac{8\pi}{15}n\right) + \frac{1}{2}\sin\left(\frac{2\pi}{15}n\right) \end{aligned}$$

• $\omega_1 = 8\pi/15$: $\omega_1/2\pi = 4/15$ (rational), $N_1 = 15$.
• $\omega_2 = 2\pi/15$: $\omega_2/2\pi = 1/15$ (rational), $N_2 = 15$.

$$N = \text{LCM}(15, 15) = 15$$

Check: $x[n+15] = \cos(\frac{\pi n}{5}+3\pi)\sin(\frac{\pi n}{3}+5\pi) = (-\cos\frac{\pi n}{5})(-\sin\frac{\pi n}{3}) = x[n]$.

(Taking the LCM of the periods of the two factors, 10 and 6, gives 30, which is a period but not the fundamental one.)

Since it is periodic, it is a power signal: $P = \frac{1}{4}\left(\frac{1}{2}+\frac{1}{2}\right) = \frac{1}{4}$.

Answer: periodic, fundamental period $N = 15$ samples ($\omega_0 = 2\pi/15$ rad/sample).