Fourier Transform/Series DFT/DFS textbook problem (simple?) Suppose $x_c(t)$ is a periodic continuous time signal with period 1 ms and for which the Fourier series is

\begin{align*}
  x_c(t) &amp;= \sum\limits_{k=-9}^9 a_k e^{j(2000 \pi k t)} \\
\end{align*}

The Fourier series coefficients $a_k$ are zero for $|k| &gt; 9$. $x_c(t)$ is discretely sampled such that:

\begin{align*}
  x[n] &amp;= x_c\left(\frac{n}{6000}\right) \\
  &amp;= \sum\limits_{k=-9}^9 a_k e^{j(\pi k n/3)} \\
\end{align*}

$x[n]$ is periodic with $N=6$

Question: Find the DFS coefficients of $x[n]$ in terms of $a_k$.

My Work: The DFS coefficients of a periodic signal are:

\begin{align*}
  W_N &amp;= e^{-j(2\pi/N)} \\
  X[k] &amp;= \sum\limits_{n=0}^{N-1} x[n] W_N^{kn} \\
\end{align*}

Changing the variable $k$ in $x[n]$ to $m$ to avoid conflict and combining yields:

\begin{align*}
  X[k] &amp;= \sum\limits_{n=0}^{N-1}
    \sum\limits_{m=-9}^9 a_m e^{j(\pi m n/3)}
 W_N^{kn} \\
  X[k] &amp;= \sum\limits_{n=0}^{N-1}
  \sum\limits_{m=-9}^9 a_m e^{j(\pi n/3 (m - k))} \\
\end{align*}

I'm stumped on how to simplify or process this further.

I suspect this is the wrong approach. The problem gives Fourier series coefficients of the continuous function, there should be a direct way to convert them to the discrete Fourier series coefficients.

Textbook Answer: The answer given by the textbook is as follows. I am trying to figure out how to get to this answer.

\begin{align*}
  X[k] &amp;= 2\pi \begin{cases}
    a_0 + a_6 + a_{-6} &amp; k = 0 \\
    a_1 + a_7 + a_{-5} &amp; k = 1 \\
    a_2 + a_8 + a_{-4} &amp; k = 2 \\
    a_3 + a_9 + a_{-3} + a_{-9} &amp; k = 3 \\
    a_4 + a_{-2} + a_{-8} &amp; k = 4 \\
    a_5 + a_{-1} + a_{-7} &amp; k = 5 \\
  \end{cases}
\end{align*}
Your result

$$x[n]=\sum_{k=-9}^9a_ke^{j2\pi kn/N}\tag{1}$$

with $N=6$ is correct. If you compare $(1)$ to the IDFT (or DFS)

$$x[n]=\frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j2\pi kn/N}\tag{2}$$

you'll notice that in $(2)$ each term $e^{j2\pi kn/N}$ occurs only once for each value of $k$ (for fixed $n$), whereas in $(1)$, due to the periodicity of the complex exponential, each term $e^{j2\pi kn/N}$ occurs several times. E.g., for $k=0$ in $(2)$, in $(1)$ you get contributions for $k=0$, for $k=6$, and for $k=-6$ (because $N=6$). Consequently, comparing $(1)$ and $(2)$ gives

$$X[0]=N(a_0+a_6+a_{-6})$$

For all other values of $k$, the procedure is completely analogous: search for indices $k$ in $(1)$ which have the same complex exponential term $e^{j2\pi kn/N}$. You basically have to keep adding and subtracting the number $N=6$ to the indices, as long you remain in the range $|k|\le 9$, so e.g. for $X[2]$ you get contributions for $k=2$, $k=2+6=8$, and $k=2-6=-4$, because they all have the same exponential term $e^{j4\pi n/N}$.

PS: The term $2\pi$ in your textbook result doesn't make sense to me, it should be $N$, as already pointed out in a comment.