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

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

\begin{align*} x[n] &= x_c\left(\frac{n}{6000}\right) \\ &= \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 &= e^{-j(2\pi/N)} \\ X[k] &= \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] &= \sum\limits_{n=0}^{N-1} \sum\limits_{m=-9}^9 a_m e^{j(\pi m n/3)} W_N^{kn} \\ X[k] &= \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] &= 2\pi \begin{cases} a_0 + a_6 + a_{-6} & k = 0 \\ a_1 + a_7 + a_{-5} & k = 1 \\ a_2 + a_8 + a_{-4} & k = 2 \\ a_3 + a_9 + a_{-3} + a_{-9} & k = 3 \\ a_4 + a_{-2} + a_{-8} & k = 4 \\ a_5 + a_{-1} + a_{-7} & k = 5 \\ \end{cases} \end{align*}

  • $\begingroup$ Can you double-check if it is not $N$ instead of $2\pi$ in the $X[k]$ equation ? $\endgroup$
    – Gilles
    Commented Oct 21, 2015 at 12:15
  • $\begingroup$ The textbook solution for $X[k]$ absolutely has $2\pi$ not $N$. $N=6$, btw, so the variable wouldn't be necessary. $\endgroup$
    – clay
    Commented Oct 21, 2015 at 13:18

1 Answer 1


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


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.


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