Energy and power: 4sin(2πt), cos(3n/4)

(i) $f(t) = 4\sin(2\pi t)$

Periodic with $T = 1$ s.

Energy:

$$E = \int_{-\infty}^{\infty}16\sin^2(2\pi t)\,dt = \infty$$

(each period contributes the same positive amount, and there are infinitely many periods).

Power:

$$\begin{aligned} P &= \frac{1}{T}\int_0^T 16\sin^2(2\pi t)\,dt = \int_0^1 8\left[1 - \cos(4\pi t)\right]dt \\ &= 8 - 0 = 8\ \text{W} \end{aligned}$$

Answer: $E = \infty$, $P = 8$ W, a power signal ($A^2/2 = 16/2$).

(ii) $x[n] = \cos(3n/4)$

$\omega_0 = 3/4$, $\omega_0/2\pi = 3/(8\pi)$ is irrational, so $x[n]$ is not periodic; use the limit definitions.

Energy:

$$E = \sum_{n=-\infty}^{\infty}\cos^2\frac{3n}{4} = \infty$$

Power:

$$\begin{aligned} P &= \lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^{N}\cos^2\frac{3n}{4} \\ &= \lim_{N\to\infty}\frac{1}{2N+1}\sum_{n=-N}^{N}\frac{1}{2}\left[1 + \cos\frac{3n}{2}\right] \\ &= \frac{1}{2} + \lim_{N\to\infty}\frac{1}{2(2N+1)}\sum_{n=-N}^{N}\cos\frac{3n}{2} = \frac{1}{2} + 0 \end{aligned}$$

(the cosine sum is bounded, so the second term goes to zero).

Answer: $E = \infty$, $P = 0.5$ W, a power signal.