Lowpass filter impulse response in frequency domain

Question

A Linear Time-Invariant system with impulse response $h_1[n]$ is an ideal lowpass filter with cutoff frequency $\omega_c =\pi/2$. The frequency response of the system is $H_1\left(e^{j\omega}\right)$. Suppose a new LTI system with impulse response $h_2[n]$ is obtained from $h_1[n]$ by $h_2[n] = (-1)^n\cdot h_1[n]$. Sketch the frequency response $H_2\left(e^{j\omega}\right)$.

In this problem I know what a lowpass filter looks like in frequency domain but I don't know what $(-1)^n$ is. If I could just get some help with that, I could then convolve the two to get the frequency response. Or should I find the lowpass filter in time domain multiply it by $(-1)^n$ and then find its frequency response? Either way I would still need to know what to do with the $(-1)^n$.

Ok, so now I know that $H_1\left(e^{j\omega}\right)$ has a magnitude of $1$ and $0$ phase when $\omega <\omega_c$ and $0$ when $\omega>\omega_c$ and that $(-1)^n$ has a DTFT of: $$\frac{2}{1-e^{-j\omega}}$$ because $(-1)^n$ is not causal.

I then multiply the two and I get the same system as $h_1[n]$ but shifted by $1$ radian and its magnitude doubled? I am not totally sure, can someone please help?

Answer 1

Discrete time domain comes with its magics and one of the funniest of them out there is the result of the multiplication of an impulse response $h[n]$ by $(-1)^n$, which is equivalent to an alternating version of the original response.

The key to its understanding comes from a very little effort of converting $(-1)^n$ into a complex exponential and then invoking a fundamental DTFT property, namely:

if $x[n]$ has the DTFT of $X(e^{j\omega})$ , then $x[n]e^{j\omega_0 n}$ has the DTFT of $X(e^{j(\omega-\omega_0)})$

Therefore, since $(-1)^n = e^{j\pi n}$, then DTFT of $$h_2[n]=(-1)^n h_1[n]$$ is $$H_2(e^{j\omega}) = H_1(e^{j(\omega - \pi)})$$

Finally the interpretation of $H_1(e^{j(\omega - \pi)})$ is as follows, given that $H_1(e^{j\omega})$ is an ideal lowpass filter of cutoff frequency $\omega_c = \pi/2$ then $H_2(e^{j\omega})$ is a shifted by $\pi$ to the right version of $H_1(e^{j\omega})$ which will be an ideal highpass filter with a cutoff frequecy of $\omega_c = \pi /2$. And the best way to see this will come from drawing the magnitude reponses on the paper...