0
$\begingroup$

Objective : Compute $y(t)$ from $Y(j\omega)=H(j\omega)X(j\omega)$ where : $$ x(t)=\left(\frac{\sin(2t)}{\pi t}\right)^{2} $$ and $$ H(j\omega)=\begin{cases}e^{-j\omega}&\text{if $|\omega|<4$}\\0&\text{if otherwise}\end{cases} $$ what I did is find FT of $\left(\frac{\sin(2t)}{\pi t}\right)\left(\frac{\sin(2t)}{\pi t}\right)$ as follow : $$ \mathcal{F}\{s\cdot s\}(t)=\frac{1}{2\pi}[S*S](j\omega) $$

We have : $$S(j\omega)=\mathcal{F}\left\{\frac{\sin(2t)}{\pi t}\right\}=\begin{cases}1&\text{if $|\omega|<2$}\\0&\text{if otherwise}\end{cases}$$ I have computed the modulation : $$ X(j\omega)=\frac{1}{2\pi}[S(j\omega)*S(j\omega)]=\frac{1}{2\pi}\int_{-\infty}^{\infty}S(j\Omega)S(j(\omega-\Omega))\;\text{d}\Omega=\begin{cases}\frac{\omega}{2\pi}&\text{if $0<\omega<2$}\\\frac{2-\omega}{2\pi}&\text{if $2<\omega<4$}\\0&\text{if otherwise}\end{cases} $$

The first issue I have is multiplying $H(j\omega)$ with $X(j\omega)$ since I have two piecewise functions. The second issue is how to thus compute its inverse Fourier transform.

I would hope someone can please help me. Thank you

$\endgroup$
4
  • $\begingroup$ I assume is the homework ? If, yes, please the homework tag. $\endgroup$
    – Hilmar
    Mar 24 at 13:13
  • $\begingroup$ Hello, thank you I have edited $\endgroup$ Mar 24 at 13:14
  • $\begingroup$ On a completely unrelated note, shouldn't it be "Kuznetsova", if it's "Maria"? Not Russian, couldn't speak it to save my life, just curious. $\endgroup$ Mar 24 at 13:39
  • 1
    $\begingroup$ Great point $:$) I have a Syrian father and a Russian mother. Kuznetsova is considered as a middle name but I prefer it to be placed along with my name @aconcernedcitizen $\endgroup$ Mar 24 at 17:48
0
$\begingroup$

So you know that in the frequency domain the function

$$s(t)=\frac{\sin(2t)}{\pi t}\tag{1}$$

is a rectangle with height $1$ for $|\omega|<2$. Now think graphically about convolving $S(j\omega)$ with itself. Can your result be right? You found that $X(j\omega)$ is only non-zero for positive frequencies. That would correspond to a complex-valued time domain function. So check that result again. The result of the convolution of two rectangles is basic and you should be able to see the result graphically.

After having figured out the convolution of $S(j\omega)$ with itself, it should be easy to obtain the bandwidth, i.e. the frequency $\omega_c$ above which the spectrum of the result is zero. With that knowledge it's straightforward to apply $H(j\omega)$ if you know the time-shifting property of the Fourier transform.

$\endgroup$
3
  • $\begingroup$ Hello, thank you for your answer. But why do I have to convolve $s(t)$ and not $S(j\omega)$? Furthermore, indeed the convolution between two identical rectangles is a triangle with its frequency spectrum being square of the frequency spectrum of $S(j\omega)$ This can be seen in the piecewise function describing $X(j\omega)$ $\endgroup$ Mar 24 at 13:18
  • $\begingroup$ @MariaKuznetsov: You're right, I meant convolution in the frequency domain, will edit ... So if you know the result should be a triangle then you see that your result for $X(j\omega)$ must be wrong, don't you? $\endgroup$
    – Matt L.
    Mar 24 at 13:32
  • $\begingroup$ I see, the interval is wrong. I am really embarrassed for my stupid mistake. Thank you for pointing it :) $\endgroup$ Mar 24 at 13:34
0
$\begingroup$

The general solution

Your problem has a piecewise definition in the frequency domain. That means the solution in the frequency domain is also piecewise. The individual section of the solutions are given by the intersections of the sections for $H(\omega)$ and $X(\omega)$. Within each solution section there is no section boundary, so you can just multiply as usual.

The inverse Fourier Transform is an integral: To execute this over a piecewise definition you need to break down the integral into multiple integral using the section boundaries as the integration intervals.

In this case that's fairly tedious but as (unfortunately) typical with home work and book exercises there is a short cut solution here.

The easy solution

You still should find intersect the sections and determine the sections in the frequency domain where the solution is non zero. Then look at $H(\omega)$ and figure out what difference the piecewise definition actually makes. As @Matt already pointed out, the time shifting property should get you easily too a solution that doesn't require the full integral.

$\endgroup$
2
  • $\begingroup$ Ah I see from the fact that $e^{-j\omega}$ adjusts the timing of $x(t)$ in time-domain. I shall admit it as the "way it should be solved" but I am always curious with heavy mathematical computations to know where I will reach and hopefully not get disappointed if I get a different answer :) . Thank you for your answer! $\endgroup$ Mar 24 at 13:30
  • $\begingroup$ Also a little update. The good news is that the area of the isosceles triangle reduces the computations to just one integral $\text{:D}$ $\endgroup$ Mar 24 at 15:29

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.