0
$\begingroup$

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]

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

then

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

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?

$\endgroup$
1
  • 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

1
$\begingroup$

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 $$ \begin{aligned} \mathrm{DFS}\{x_1[n]x_2[n]\} &= \sum_{n=0}^{N-1} \left(\frac{1}{N}\sum_{m=0}^{N-1}X_1[m]W_{N}^{-mn}\right)x_2[n]W_N^{nk}\\ &=\frac{1}{N}\sum_{m=0}^{N-1}X_1[m] \left( \sum_{n=0}^{N-1}x_2[n]W_N^{n(k-m)} \right)\\ &=\frac{1}{N}\sum_{m=0}^{N-1}X_1[m] X_2[k-m] \end{aligned}\tag{3} $$

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.