How to get output from High Pass filter on this signal?

Question

I'm trying to solve this exercise:

Consider a High-Pass filter with a cutoff frequency of $W=2\pi$

Calculate the output signal if the input signal is $x(t)=\displaystyle \frac{\sin(4\pi t)}{\pi t}$

The fourier transform for the input signal is:

$X(\omega)= \left\{ \begin{array}{lcc} 1 & \text{if} & -4\pi \lt \omega \lt 4\pi \\[2ex] 0 & \text{otherwis}e \\ \end{array} \right.$

Now, the output signal is : $Y(\omega)=H(\omega)X(\omega)$

$Y(\omega)= \left\{ \begin{array}{lcc} 1 & \text{if} & 2\pi \le |\omega| \le 4\pi \\[2ex]0 & \text{otherwise} \\ \end{array} \right.$

I can get here, but the solution to the exercise converts this function with inverse fourier transform, giving the result of $x(t)=2\displaystyle \frac{\sin(\pi t)}{\pi t}\cos(3\pi t)$.

I don't know how did they get to that conclusion, I have tried to search for it, but i don't know how to look for that.

I hope that the question is clear. Sorry if I'm lacking some details, it's my first time asking this type of questions.

Answer 1

What you get as $x(\omega)$ is two $\mathrm{sinc}(\pi t) = \displaystyle\frac{\sin(\pi t)}{\pi t}$ shifted by $+\omega_0$ and $-\omega_0$ in frequency ($\omega_0 = 3 \pi$). By the frequency shift property of the FT:

$$ \mathcal F\left\{x(t) \cdot e^{i\omega_0 t}\right\} = X\left(e^{j(\omega - \omega_0)}\right) $$

Then, remember $e^{i\omega t} = \cos(\omega t) + i \cdot \sin(\omega t)$. Therefore:

$$ x(\omega) = \sqcap_{2\pi} (\omega - \omega_0) + \sqcap_{2\pi} (\omega + \omega_0) $$

Doing inverse: $$ \begin{align} x(t) &= \mathcal F^{-1}\left\{x(\omega)\right\} \\ &=\mathrm{sinc} (\pi t) \cdot \big(\cos(\omega_0 t) + i \cdot \sin(\omega_0 t)\big) + \mathrm{sinc} (\pi t) \cdot \big(\cos(-\omega_0 t) + i \cdot \sin(-\omega_0 t)\big) \\ &= 2 \cdot \mathrm{sinc} (\pi t)\cdot \cos(\omega_0 t) \\ &= 2 \cdot \frac{\sin(\pi t)}{\pi t}\cdot \cos(3 \pi t) \end{align} $$

since $\cos(-x)=\cos(x)$ and $\sin(-x)=-\sin(x)$

Answer 2

So you have to take the inverse Fourier transform of the output of the highpass filter which you have correctly defined. Instead of directly computing the result $y(t)$ by evaluating the inverse Fourier transform integral; $$y(t) = \frac{1}{2\pi} \int_{2\pi < \lvert \omega\rvert <4\pi} {1 e^{j\omega t} d\omega }$$ you can also use the linearity property of the Fourier transform to directly write the answer, without explicitly evaluating the integral.

More specifically, you can see by inspection of $Y(\omega)$ that $$ Y(\omega) = Y_{4\pi}(\omega) - Y_{2\pi}(\omega),$$ where $Y_{4\pi}(\omega)$ corresponds to the frequency response of an ideal lowpass filter with cutoff frequency $W_1 = 4\pi$ radians per second, and $Y_{2\pi}(\omega)$ corresponds to the same type of filter with a cutoff frequency $W_2 = 2\pi$ radians per second. Since you already know the inverse Fourier transform of the ideal zero phase lowpass filter with a cutoff frequency $W$ radians per second and a gain of unity as given by: $h_{W}(t) = \displaystyle \frac{\sin(W t)}{\pi t}$, therefore you conclude; $$ y(t) = h_{4\pi}(t) - h_{2\pi}(t) = \frac{\sin(4\pi t)}{\pi t} -\frac{\sin(2\pi t)}{\pi t} $$ What remains is trigonometric simplification; (At this point I've to note that from a number of possible trigonometric simplifications (which are all equivalent as a correct answer) I'll choose the one that yields the provided answer, and how do I decide on that is by inspection of the provided answer using educated guess again):

\begin{align} y(t) &= \frac{1}{\pi t} \big( {\sin(4\pi t) - \sin(2\pi t) } \big)\\ &= \frac{1}{\pi t}\big( {\sin(3\pi t + \pi t) - \sin(3\pi t -\pi t) }\big)\\ &= \frac{1}{\pi t}\bigg( {\sin(3\pi t) \cos(\pi t) + \cos(3\pi t) \sin(\pi t) - \big[ \sin(3\pi t) \cos(\pi t) - \cos(3\pi t) \sin(\pi t)\big]}\bigg)\\ &= \frac{1}{\pi t}\big( {\sin(3\pi t) \cos(\pi t) + \cos(3\pi t) \sin(\pi t) - \sin(3\pi t) \cos(\pi t) + \cos(3\pi t) \sin(\pi t) }\big)\\ & = \frac{1}{\pi t} {2\cos(3\pi t) \sin(\pi t) }\\ &= 2\frac{\sin(\pi t)}{\pi t} {2\cos(3\pi t) } \end{align}

Note that another perspective in the inspection of $Y(\omega)$ may result in the conlcusion as @oxuf did, and that interpretation can provide a more simpler and elegant procedure of obtaining the same result, I belive.