Lowpass filter impulse response in frequency domain

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?

• It's OK to post homework questions here, but it's not enough to just post the exercise. It's important for us to understand what you've done and why and where you're stuck. People will not do your homework for you. – Matt L. Feb 29 '16 at 6:24

1 Answer

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...