I am having trouble in deriving a result, which asks me to find a relation between $X_{1}(k)$ and $X_{2}(k)$ where $X_{1}(k)$ are the DFS coefficients of single period of signal $x(n)$ and $X_{2}(k)$ are the DFS coefficients of two periods of the same $x(n)$. My approach is this.

$$X_{1}(k) = \sum_{n=0}^{N-1}x(n)W_{N}^{-kn}$$ and $$X_{2}(k) = \sum_{n=0}^{2N-1}x(n)W_{2N}^{-kn}$$ Since we know that $W_{N} = W_{2N}^{2}$. That means that, if I square $X_{2}(k)$, then I get this $$X_{2}^{2}(k)=\sum_{n=0}^{2N-1}x^{2}(n)W_{2N}^{-2kn}=\sum_{n=0}^{2N-1}x^{2}(n)W_{N}^{-kn}$$ Now this is where I can't move further. What should I do with the last expression. I mean what are the DFS coefficients of the square of a sequence? And also note that now we are over two periods ($2N-1$).

  • 1
    $\begingroup$ That squaring argument would not help you... and indeed wrong (square of $X_2[k]$ is not what you think). You better apply the steps of the solution algorithm pointed by Hilmar. $\endgroup$
    – Fat32
    Commented Oct 7, 2018 at 22:08
  • $\begingroup$ Do we really need to use $W_{2N}$ in the expression of $X_{2}(k)$. I think that would be true if it was a DFT vector but here we are talking of DFS vector, so shouldn't it be $W_{N}$ only due to the fundamental frequency? I know I am questioning my very own solution. $\endgroup$ Commented Oct 8, 2018 at 14:10
  • $\begingroup$ Hint: $W_{2N}^{kn} = e^{-j \frac{2\pi}{2N} k n } = e^{-j \frac{2\pi}{N} \frac{k}{2} n } = W_{N}^{ \frac{k}{2}n}$ $\endgroup$
    – Fat32
    Commented Oct 8, 2018 at 19:48

1 Answer 1


Here is an outline (in case it's homework)

  1. Use a different frequency index for $X_2(l)$
  2. Split the $X_2(l)$ into two sums: from $0:N-1$ and from $N-1:2N$
  3. Use the periodicity of x[n], i.e. $x(n) = x(n+N)$
  4. Substitute $m = n-N$ in the second sum
  5. Recombine the two sums again
  6. Stare what happens to even $l = 2k$ and odd $l = 2k+1$ terms
  • $\begingroup$ What I got from your algorithm is this $$ X_{2}(l) = 2 \sum_{n=0}^{N-1}x(n)W_{2N}^{-ln}$$. First of all, is it correct? If it is, I don't understand why you say put $l=2k$ and $l=2k+1$, because $l$ and $k$ are just the indexes of the DFS. We just replaced $k$ by $l$. Then why? $\endgroup$ Commented Oct 8, 2018 at 14:08
  • $\begingroup$ Your intermediate solution is wrong. There should be x[n] times the sum of two different W coefficients $ln$ and $l(n+N)$. Once you have that, pop the actual definition of $W$ and see if it simplifies further $\endgroup$
    – Hilmar
    Commented Oct 9, 2018 at 16:54
  • $\begingroup$ Ok I will try and notify you. $\endgroup$ Commented Oct 9, 2018 at 21:14

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