Energy of e^(jω₀n)u[n]; period of cos(41πn/7)

Energy and power of $x[n] = e^{j\omega_0 n}u[n]$

Formulas for a discrete-time signal:

$$E = \lim_{N\to\infty} \sum_{n=-N}^{N} |x[n]|^2, \qquad P = \lim_{N\to\infty} \frac{1}{2N+1}\sum_{n=-N}^{N} |x[n]|^2$$

Here $|x[n]|^2 = |e^{j\omega_0 n}|^2 \, u[n] = 1$ for $n \ge 0$ and $0$ for $n < 0$.

Energy:

$$E = \lim_{N\to\infty} \sum_{n=0}^{N} 1 = \lim_{N\to\infty}(N+1) = \infty$$

Power:

$$\begin{aligned} P &= \lim_{N\to\infty} \frac{1}{2N+1}\sum_{n=0}^{N} 1 \\ &= \lim_{N\to\infty} \frac{N+1}{2N+1} = \lim_{N\to\infty}\frac{1 + 1/N}{2 + 1/N} = \frac{1}{2} \end{aligned}$$

Answer: $E = \infty$, $P = 0.5$ W. Since the power is finite and non-zero, $x[n]$ is a power signal.

Period of $x[n] = \cos(41\pi n/7)$

A discrete-time sinusoid $\cos(\omega_0 n)$ is periodic only if $\omega_0/2\pi$ is rational, and then $N = 2\pi m/\omega_0$ with the smallest integer $m$ that makes $N$ an integer.

$$\begin{aligned} \omega_0 &= \frac{41\pi}{7} \\ \frac{\omega_0}{2\pi} &= \frac{41}{14} = \frac{m}{N} \quad \text{(rational, already in lowest terms)} \end{aligned}$$

So $m = 41$ and $N = 14$. Check: $\omega_0 N = \frac{41\pi}{7}\times 14 = 82\pi = 41 \times 2\pi$.

Answer: the signal is periodic with fundamental period $N = 14$ samples.