Let $x[n] = u[n] - u[n-4]$ (a discrete pulse of length 4), and $X(\omega)$ is its DTFT.

Let $x_1[n] = x[n]*x[n]$. I expect DTFT of $x_1[n]$ to be same as $X(\omega)$, because $x_1[n]$ has the same sample values. But it is actually a convolution: $X(\omega)*X(\omega)$. This is puzzling. Please explain.

  • $\begingroup$ Your question is a bit difficult to understand because you use the symbol * for multiplication as well as for convolution. I think that in your first equation you actually mean multiplication ($x_1[n]=x^2[n]$), otherwise $x_1[n]$ wouldn't be equal to $x[n]$. If this is the case, then my answer below should make sense to you. $\endgroup$ – Matt L. Jan 6 '16 at 8:07
  • $\begingroup$ Enough information to upvote or validate? $\endgroup$ – Laurent Duval Nov 10 '19 at 18:05

There is a confusion: the convolution of $x$ by $x$, denoted $x*x$, yields a product in the Fourier domain: $X(\omega)\times X(\omega)$ (sometimes denoted $X(\omega). X(\omega)$ or simply $X(\omega) X(\omega)$, or $X^2(\omega)$).

  • $\begingroup$ and vice versa: product in the time domain is also equivalent to convolution on the frequency domain $\endgroup$ – Hilmar Jan 6 '16 at 15:56
  • $\begingroup$ Indeed, and since the signal is a unit discrete window, I assumed the question was not about its self-product $\endgroup$ – Laurent Duval Jan 6 '16 at 16:20

As you probably already guessed, there is no inconsistency. Since you can trust in the correctness of the correspondence

$$x_1[n]\cdot x_2[n]\longleftrightarrow \frac{1}{2\pi}(X_1\star X_2)(\omega)=\frac{1}{2\pi}\int_0^{2\pi}X_1(\theta)X_2(\omega-\theta)d\theta\tag{1}$$

where $\star$ denotes convolution, you know that if $x[n]=x^2[n]$ (as in your example) the following equation must hold:

$$X(\omega)=\frac{1}{2\pi}(X\star X)(\omega)\tag{2}$$

For the given sequence you have

$$X(\omega)=e^{-j3\omega /2}\frac{\sin(2\omega)}{\sin(\omega/2)}\tag{3}$$

So from $(2)$ we have the equality

$$e^{-j3\omega /2}\frac{\sin(2\omega)}{\sin(\omega/2)}=\frac{1}{2\pi}\int_0^{2\pi}e^{-j3\theta /2}\frac{\sin(2\theta)}{\sin(\theta/2)}e^{-j3(\omega-\theta) /2}\frac{\sin(2(\omega-\theta))}{\sin((\omega-\theta)/2)}d\theta\tag{4}$$

from which


follows. So you have actually proved that awfully looking equation given by $(5)$!

I assume that with a bit of effort you could actually prove $(5)$ directly, but your method is of course much more elegant.

  • $\begingroup$ just to dot the t's and cross the i's, we need to be clear that for both $X_1(\omega)$ and $X_2(\omega)$ (and $X(\omega)$) that they are all periodic with period $2\pi$. that is $$ X(\omega + 2\pi) = X_1(\omega) $$ for all real $\omega$. $\endgroup$ – robert bristow-johnson Jan 6 '16 at 8:38

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.