This is a question related to continuous Time Fourier Transform.

I have attempted it this way:

$G(j\omega)$ is a rectangular pulse in frequency domain. Using results, inverse Fourier transform of $G(\omega)$ will be $$g(t)=\frac{\sin(2t)}{\pi t}$$

Now given that $g(t)=x(t)\cos(t)$

Using $\sin (2t)=2\cos(t)\sin(t)$, I got $$x(t)=2\frac{\sin(t)}{\pi t}$$

But, I am struggling to draw the parallel for part b of the question.

I was using the multiplication property for part b, and it turned out,

$$G(j\omega)=\frac{X\left(\omega-\frac 23\right)+X\left(\omega+\frac 23\right)}{2}$$

How to find $X(\omega)$ and proceed from here onwards?


1 Answer 1


Your procedure for the first exercise is correct. Note however that, as you may have noticed already, that approach is only useful if the cosine has unitary pulsatance. There is a generic way to do this, if we think of the problem in the frequency domain.

I'll apply this method to the exercise you already solved, so that you can work the second one on your own. In the frequency domain, your exercise can be stated as follows:

$$G(\omega)=\frac{1}{2\pi}X(\omega) *[\pi\delta(\omega-1)+\pi\delta(\omega+1)]$$

Remember that convolving signals with Dirac deltas can be easily made graphically. It's just centering the signal being convolved (in this case, $X(\omega$)) in the position where the impulses are.

Notice that when the impulses are centered in $\omega=\pm1$, if $X(\omega)$ was a rectangular pulse of width $2$ and amplitude $2$ as well, then the convolution would return a rectangular pulse of width $4$ (one for each impulse, and as they don't overlap they keep the rectangular shape) of amplitude $2\pi$. But because convolution in the frequency domain has to be divided by $2\pi$ when going back to the time domain, we get that if $X(\omega)=\mathrm{rect}(\omega/2)$, then the equality holds. That transform corresponds indeed to the signal $x(t)$ you found using the trigonometric identity.

See that this approach is useful for any position of the Dirac deltas. In the second case, they are at $\omega=\pm 2/3$. Can you solve it now?

  • $\begingroup$ Thanks Tendero for explaining it so tersely. Like I have already mentioned in my question, extrapolating this generic method to the part b.) gives G(w)=(X(w-2/3) + X(w+2/3))/2 , that is because of the impluses present at +-2/3, the center of spectrum X(w) shifts to these values. But, then I am not being able to visulaize X(w) that would after shifting give G(w). Like you visualized it to be a rectangular pulse of width 2. $\endgroup$ Jan 29, 2018 at 15:56
  • $\begingroup$ @YOGENDRASINGH When I wrote the answer, I thought it would be trivial to extend this to other cases. But it certainly isn't. I'm not being able to think of a solution for the problem. Even more, I found this is a problem in Oppenheim-Willsky, and the solution manual doesn't provide an answer either (ït says the solution is in a figure but it's not there). I believe there is some kind of typo or something, unless someone comes up with an idea... $\endgroup$
    – Tendero
    Jan 29, 2018 at 17:24

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