# Discrete Fourier Series of square of a signal

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 ($$2N-1$$).

• 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. Commented Oct 7, 2018 at 22:08
• 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. Commented Oct 8, 2018 at 14:10
• Hint: $W_{2N}^{kn} = e^{-j \frac{2\pi}{2N} k n } = e^{-j \frac{2\pi}{N} \frac{k}{2} n } = W_{N}^{ \frac{k}{2}n}$ Commented Oct 8, 2018 at 19:48

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
• 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? Commented Oct 8, 2018 at 14:08
• 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 Commented Oct 9, 2018 at 16:54
• Ok I will try and notify you. Commented Oct 9, 2018 at 21:14