Energy or power: 3cos(2πt), cos(n/6)

A signal is an energy signal if $0 < E < \infty$ ($P = 0$) and a power signal if $0 < P < \infty$ ($E = \infty$).

(a) $f(t) = 3\cos(2\pi t)$

This is periodic with $T = 1$ s, so its energy over all time is infinite. Average power over one period:

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

$E = \infty$, $P = 4.5$ W, so $f(t)$ is a power signal (in general, $A\cos\omega_0 t$ has $P = A^2/2$).

(b) $x[n] = \cos(n/6)$

Here $\omega_0 = 1/6$ and $\omega_0/2\pi = 1/(12\pi)$ is irrational, so $x[n]$ is not periodic. Its power must be found with the limit:

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

The sum of $\cos(n/3)$ stays bounded, so divided by $2N+1$ it goes to 0. The energy $\sum \cos^2(n/6)$ grows without limit, so $E = \infty$.

Answer: $E = \infty$, $P = 0.5$ W, so $x[n]$ is a power signal (even though it is aperiodic).