DTFS of a DT pulse and e^(j4πn/7)x[n]

The signal is $x[n] = 2$ for $n = -1, 0, 1$ and $x[n] = 0$ for $n = \pm2, \pm3$. One period runs over $n = -3$ to $3$, so $N = 7$ and $\omega_0 = 2\pi/7$.

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

Fourier series coefficients of x[n]

$$a_k = \frac{1}{N}\sum_{n=-1}^{1} 2\,e^{-jk(2\pi/7)n} = \frac{2}{7}\left[1 + e^{-j2\pi k/7} + e^{j2\pi k/7}\right] = \frac{2}{7}\left[1 + 2\cos\frac{2\pi k}{7}\right]$$

In closed form (the standard result for a DT rectangular pulse with $N_1 = 1$):

$$a_k = \frac{2}{7}\cdot\frac{\sin(3\pi k/7)}{\sin(\pi k/7)}, \quad k \ne 0, \pm7, \dots; \qquad a_0 = \frac{2(2N_1+1)}{N} = \frac{6}{7}$$

The coefficients are real and even ($a_k = a_{-k} = a_{7-k}$), because $x[n]$ is real and even. They repeat with period 7.

Coefficients of y[n] = e^{j4πn/7} x[n]

$$e^{j4\pi n/7} = e^{j2(2\pi/7)n} = e^{jM\omega_0 n} \quad \text{with } M = 2$$

By the frequency shifting property, $e^{jM\omega_0 n}x[n] \xleftrightarrow{FS} a_{k-M}$, so

$$b_k = a_{k-2} = \frac{2}{7}\cdot\frac{\sin\left(3\pi(k-2)/7\right)}{\sin\left(\pi(k-2)/7\right)}, \qquad b_2 = \frac{6}{7}$$

Answer: $a_k = \frac{2}{7}\frac{\sin(3\pi k/7)}{\sin(\pi k/7)}$ with $a_0 = 6/7$, and $b_k = a_{k-2}$. The spectrum is the same shape, moved 2 places to the right (peak at $k = 2$).