Frequency translation of an ideal low pass filter

Question

I am trying to create a discrete-time filter with a Fourier transform as follows.

$X(\omega) = \begin{cases} 1, & T - W \leq \omega \leq T + W\\ 0, & \text{all other values of } \omega \end{cases}, $

Looking at the provided solution, I have understood that the idea is to create a low-pass filter and then translate it in the frequency domain by $T$.

For the low-pass filter I've got the following time-domain function.

$x(n) = \frac{1}{\pi}\frac{\sin(Wn)}{n}$

To translate a function in frequency, it can be multiplied in the time domain by $e^{jTn}$. In the solution, they have first stated that this is correct, but in the final answer they have instead multiplied it by $2\cos(Tn)$. They also wrote that $2\cos(Tn) = e^{jTn}+e^{-jTn}$, which I have no problem with, but I don't understand why they went for this instead of $e^{jTn}$.

So to be clear, my question is: why did they multiply the low-pass filter by $2\cos(Tn)$ instead of $e^{jTn}$ to translate it in the frequency domain?

Answer 1

Multiplying the impulse response by $e^{jTn}$ would give a complex impulse response and a complex-valued filter output. Seems that they were aiming for a real-valued output. They did this by summing two filters, one with the pass band at $T - W \le \omega \le T + W$ and the other with the pass band at $-T - W \le \omega \le -T + W$. If there is overlap between the two pass bands then the frequency response of the composite filter will have a weird shape like:

Seems that they did not intend there to be overlap, but that $T \gt W$.