Average power of sin(πn/3) via DTFS

Parseval's relation for the DTFS

For a periodic $x[n]$ with period $N$ and coefficients $a_k$:

$$\frac{1}{N}\sum_{n=\langle N\rangle}|x[n]|^2 = \sum_{k=\langle N\rangle}|a_k|^2$$

The average power over one period equals the sum of the powers $|a_k|^2$ of the $N$ harmonic components.

Average power of $x[n] = \sin(\pi n/3)$

Period: $\omega_0 = \pi/3$, $\frac{\omega_0}{2\pi} = \frac{1}{6}$, so $N = 6$ and the fundamental is $2\pi/6 = \pi/3$.

Coefficients:

$$x[n] = \frac{1}{2j}e^{j(2\pi/6)n} - \frac{1}{2j}e^{-j(2\pi/6)n}$$ $$a_1 = \frac{1}{2j} = -\frac{j}{2}, \qquad a_{-1} = a_5 = -\frac{1}{2j} = \frac{j}{2}$$

and $a_0 = a_2 = a_3 = a_4 = 0$ in one period.

Power by Parseval:

$$P = \sum_{k=0}^{5}|a_k|^2 = |a_1|^2 + |a_5|^2 = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{1}{2}$$

Check in the time domain: $x[n]$ for $n = 0 \dots 5$ is $0, \frac{\sqrt3}{2}, \frac{\sqrt3}{2}, 0, -\frac{\sqrt3}{2}, -\frac{\sqrt3}{2}$:

$$P = \frac{1}{6}\left(4\times\frac{3}{4}\right) = \frac{3}{6} = 0.5$$

Answer: average power $P = 0.5$ W.