DTFS of sin(ω₀n) with N = 6

Since $x[n] = \sin(\omega_0 n)$ has period $N = 6$, the fundamental frequency is

$$\omega_0 = \frac{2\pi}{N} = \frac{2\pi}{6} = \frac{\pi}{3}$$

Fourier series coefficients

Using Euler's formula:

$$x[n] = \sin\frac{\pi n}{3} = \frac{1}{2j}e^{j(2\pi/6)n} - \frac{1}{2j}e^{-j(2\pi/6)n}$$

Comparing with $x[n] = \sum_{k=\langle 6\rangle}a_k e^{jk(2\pi/6)n}$:

$$a_1 = \frac{1}{2j} = -\frac{j}{2} = \frac{1}{2}e^{-j\pi/2}, \qquad a_{-1} = -\frac{1}{2j} = \frac{j}{2} = \frac{1}{2}e^{j\pi/2}$$

All other $a_k$ in one period are zero. The coefficients repeat every 6: $a_{-1} = a_5 = a_{11}$, $a_1 = a_7 = a_{-5}$, etc.

Spectra

 |a_k|  (repeats every 6)
   0.5  |           |     |           |
       -+--o--o--o--+--o--+--o--o--o--+--> k
       -5 -4 -3 -2 -1  0  1  2  3  4  5

 angle a_k (deg)
    90              |                 |
       -+--o--o--o--+--o--+--o--o--o--+--> k
   -90  |                 |
       -5 -4 -3 -2 -1  0  1  2  3  4  5

Lines of height $1/2$ at $k = \pm1, \pm5, \pm7, \dots$; phase $-90^\circ$ at $k = 1, 7, -5, \dots$ and $+90^\circ$ at $k = -1, 5, \dots$

Average power

By Parseval's relation:

$$P = \sum_{k=\langle 6\rangle}|a_k|^2 = |a_1|^2 + |a_5|^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$

Check in time: samples over one period are $0, \frac{\sqrt3}{2}, \frac{\sqrt3}{2}, 0, -\frac{\sqrt3}{2}, -\frac{\sqrt3}{2}$, so $P = \frac{1}{6}\left(4\times\frac{3}{4}\right) = 0.5$.

Answer: $a_1 = -j/2$, $a_{-1} = a_5 = j/2$, others 0; average power $P = 0.5$ W.