Note: This is not a homework problem. I'm just stalled at a point because I think I might be interpreting the duality property incorrectly.

If $x_1[n]$ and $x_2[n]$ are periodic with period N, then if I periodically convolve them in time domain as follows:

$$y[n]=\sum_{m=0}^{N-1} x_1[m]x_2[n-m] \iff Y[k]=X_1[k]X_2[k] $$

Then I want to prove the corresponding property wherein I multiply in time domain using duality property.

As per duality property if $x[n]$ is periodic in the time domain with period N with Fourier series coefficients as X[k]

$$x[n] \iff X[k]$$


$$X[n] \iff Nx[-k]$$

Then if I multiply in the time domain this is what I get

$$x_1[n]x_2[n] \iff N \sum_{m=0}^{N-1} X_1[m]X_2[k-m]$$

But textbook says

$$x_1[n]x_2[n] \iff {\frac{1}{N}} \sum_{m=0}^{N-1} X_1[m]X_2[k-m]$$

where am I going wrong?

  • 1
    $\begingroup$ I'm not a big fan of the "standard" DFT scaling convention. If you use a scale factor of $1/\sqrt{N}$ for both forward and backward transform, the duality simply becomes $x[n] \iff X[-k]$ and as an added benefit Parseval's Theorem hold as well, i.e. $\sum |x[n]|^2 = = \sum|X[k]|^2$ and the DFT conserves energy (or power). $\endgroup$
    – Hilmar
    Sep 8, 2021 at 11:23

1 Answer 1


well, the thrid equation should be $X[n] \iff Nx[-k]$

The DFS of $x_1[n]x_2[n]$ is $$ \mathrm{DFS}\{x_1[n]x_2[n]\} = \sum_{n=0}^{N-1} x_1[n]x_2[n]W_N^{nk} \tag{1} $$ where $W_N^{nk}=e^{-j\frac{2\pi}{N}nk}$.

Now recall the definition of IDFS $$ x_1[n] = \mathrm{IDFS}\{X_1[m]\} = \frac{1}{N}\sum_{m=0}^{N-1}X_1[m]W_{N}^{-mn} \tag{2} $$ Substituting Eq. (2) into Eq. (1) and swap the order of summation, we have $$ \begin{aligned} \mathrm{DFS}\{x_1[n]x_2[n]\} &= \sum_{n=0}^{N-1} \left(\frac{1}{N}\sum_{m=0}^{N-1}X_1[m]W_{N}^{-mn}\right)x_2[n]W_N^{nk}\\ &=\frac{1}{N}\sum_{m=0}^{N-1}X_1[m] \left( \sum_{n=0}^{N-1}x_2[n]W_N^{n(k-m)} \right)\\ &=\frac{1}{N}\sum_{m=0}^{N-1}X_1[m] X_2[k-m] \end{aligned}\tag{3} $$

  • $\begingroup$ I was able to prove it this way....but wanted to prove it using the duality theorem....using the duality theorem m not able to account for the \frac{1}{N} in the formula $\endgroup$
    – Orpheus
    Sep 8, 2021 at 9:46
  • $\begingroup$ @Orpheus I guess what you meant is $\mathrm{DFS}\{Y[n]\} = \mathrm{DFS}\{X_1[n]X_2[n] \} = N y[-k] = N \sum_{m=0}^{N-1} x_1[m]x_2[k-m]$ and then you have $\mathrm{DFS}\{ x_1[n]x_2[n]\} = N \sum_{m=0}^{N-1} X_1[m]X_2[k-m]$. The second step is incorrect. $x[n]$ and $X[k]$ are a DFS pair and $\iff$ is not a $=$. If you didn't mean it, you may provide a detailed proof in your original question. $\endgroup$
    – ZR Han
    Sep 8, 2021 at 10:31
  • $\begingroup$ if $y[n]=\sum_{m=0}^{N-1}x_1[m]x_2[n-m]$ (periodic convolution) then $DFS(y[n])=X_1[k]X_2[k]$....but if I do $Y[n]=X_1[n]X_2[n]$ then as per duality theorem that states $DFS(Y[n])=Ny[-k] $ what I get is $DFS(Y[n])=N\sum_{m=0}^{N-1}x_1[m]x_2[k-m]$...but that is not true...as you proved elegantly without duality theorem there should be a $\frac{1}{N}$...and I know m wrong....but don't realize where m wrong.... $\endgroup$
    – Orpheus
    Sep 8, 2021 at 10:54
  • $\begingroup$ @Orpheus What you have is $DFT(X_1[n]X_2[n])=N\sum x_1[m]x_2[k-m]$ which cannot derive that $DFS(x_1[n]x_2[n])=N\sum X_1[m]X_2[k-m]$. $\endgroup$
    – ZR Han
    Sep 8, 2021 at 12:52

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