DTFS of sin ω₀n, ω₀ = 2π/N

Given $x[n] = \sin \omega_0 n$ with $\omega_0 = 2\pi/N$, the signal is periodic with fundamental period $N$. Its DTFS coefficients can be found by inspection, using Euler's formula.

Fourier series coefficients

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

Compare this with the synthesis equation $x[n] = \sum_{k=\langle N\rangle} a_k e^{jk(2\pi/N)n}$:

$$a_1 = \frac{1}{2j} = -\frac{j}{2}, \qquad a_{-1} = -\frac{1}{2j} = \frac{j}{2}$$

All other coefficients in one period are zero. DTFS coefficients repeat with period $N$, so $a_{k+N} = a_k$:

• $a_1 = a_{1+N} = a_{1-N} = \dots = -j/2$
• $a_{-1} = a_{N-1} = a_{2N-1} = \dots = j/2$

Over $k = 0, 1, \dots, N-1$: $a_1 = -j/2$, $a_{N-1} = j/2$, and every other $a_k = 0$.

For N = 5: $a_1 = -j/2$, $a_4 = j/2$, and $a_0 = a_2 = a_3 = 0$, repeating every 5 values of $k$. In magnitude form $|a_1| = |a_4| = 1/2$, with $\angle a_1 = -\pi/2$ and $\angle a_4 = +\pi/2$.

Plot of the signal for N = 5

$x[n] = \sin(2\pi n/5)$:

 x[n]
  0.95 |     *              *
  0.59 |     |  *           |  *
       |     |  |           |  |
  0  --*-----+--+--+--+--*--+--+--+--+-- n
       0     1  2  3  4  5  6  7  8  9
 -0.59 |           *  |           *  |
 -0.95 |              *              *

The pattern repeats every 5 samples.