Energy or power: 3cos(4πt), sin(n/6)

A signal with finite non-zero energy is an energy signal; one with finite non-zero average power (and infinite energy) is a power signal.

(i) $f(t) = 3\cos(4\pi t)$

Periodic with $T = 2\pi/4\pi = 0.5$ s, so $E = \infty$.

$$\begin{aligned} P &= \frac{1}{T}\int_0^{T}9\cos^2(4\pi t)\,dt = \frac{1}{0.5}\int_0^{0.5}\frac{9}{2}\left[1 + \cos(8\pi t)\right]dt \\ &= 2\times\frac{9}{2}\times 0.5 + 0 = 4.5\ \text{W} \end{aligned}$$

Answer: $E = \infty$, $P = 4.5$ W, so $f(t)$ is a power signal.

(ii) $x[n] = \sin(n/6)$

$\omega_0/2\pi = 1/(12\pi)$ is irrational, so $x[n]$ is aperiodic; its samples never die out, so $E = \sum \sin^2(n/6) = \infty$.

$$\begin{aligned} P &= \lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^{N}\sin^2\frac{n}{6} \\ &= \lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^{N}\frac{1}{2}\left[1 - \cos\frac{n}{3}\right] = \frac{1}{2} - 0 = 0.5\ \text{W} \end{aligned}$$

(the sum of $\cos(n/3)$ is bounded, so after dividing by $2N+1$ it tends to 0).

Answer: $E = \infty$, $P = 0.5$ W, so $x[n]$ is a power signal.