DTFS of 1 + sin + 3cos + cos(4πn/N)

Fundamental frequency $\omega_0 = 2\pi/N$. Rewrite each term with Euler's formula:

$$\begin{aligned} x[n] &= 1 + \frac{1}{2j}\left[e^{j\frac{2\pi}{N}n} - e^{-j\frac{2\pi}{N}n}\right] + \frac{3}{2}\left[e^{j\frac{2\pi}{N}n} + e^{-j\frac{2\pi}{N}n}\right] + \frac{1}{2}\left[e^{j\frac{4\pi}{N}n} + e^{-j\frac{4\pi}{N}n}\right] \\ &= 1 + \left(\frac{3}{2} + \frac{1}{2j}\right)e^{j(2\pi/N)n} + \left(\frac{3}{2} - \frac{1}{2j}\right)e^{-j(2\pi/N)n} + \frac{1}{2}e^{j2(2\pi/N)n} + \frac{1}{2}e^{-j2(2\pi/N)n} \end{aligned}$$

Comparing with $x[n] = \sum_{k=\langle N\rangle}a_k e^{jk(2\pi/N)n}$ (and using $\frac{1}{2j} = -\frac{j}{2}$):

Working: $|a_1| = \sqrt{1.5^2 + 0.5^2} = \sqrt{2.5} = 1.581$; $\angle a_1 = -\tan^{-1}(0.5/1.5) = -18.43^\circ$.

The coefficients are periodic with period $N$: $a_{k+N} = a_k$, so for example $a_{N-1} = a_{-1}$ and $a_{N-2} = a_{-2}$ (assuming $N \ge 5$ so that the harmonics do not overlap).

 |a_k|  (repeats every N)
 1.581     |     |
     1     |  |  |
   0.5  |  |  |  |  |
       -+--+--+--+--+--> k
       -2 -1  0  1  2

Since $x[n]$ is real, $a_{-k} = a_k^*$: the magnitude is even and the phase is odd.