Inverse Fourier Transform of piecewise function

Question

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

Answer 1

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.

Answer 2

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.