# DTFT inconsistency

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.

• 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. – Matt L. Jan 6 '16 at 8:07
• Enough information to upvote or validate? – 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 $X^2(\omega)$).

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

$$\frac{\sin(2\omega)}{\sin(\omega/2)}=\frac{1}{2\pi}\int_0^{2\pi}\frac{\sin(2\theta)}{\sin(\theta/2)}\frac{\sin(2(\omega-\theta))}{\sin((\omega-\theta)/2)}d\theta\tag{5}$$

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.

• 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 $\omega$. – robert bristow-johnson Jan 6 '16 at 8:38