# Determine filter given transfer function [closed]

I am given $$H(z) = 1 + \frac{\alpha}{1-\alpha z^{-1}}$$ where alpha is between $0$ and $1$. This is apparently is a low-pass filter with cutoff frequency $f_c$.

How can I see that? And how can I compute/plot the frequency response in Matlab? Neither freqz(1+alfa,[1 -alfa]) nor abs(H(z)) seems to yield a low-pass filter on the above transfer function in Matlab.

Furthermore, $1-H(z)$ should be a high-pass filter. But again, using freqz(-alfa,[1 -alfa]) in matlab it is clearly not a high-pass filter.

## closed as unclear what you're asking by Matt L., MBaz, Peter K.♦Dec 14 '17 at 13:11

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Should the numerator be $1+\alpha$ (as in the argument of freqz()), or is $H(z)$ correct as it stands? – Matt L. Dec 14 '17 at 10:57
• Mattl L: it is the same.. – Peter Alexander Dec 14 '17 at 11:00
• Either $H(z)$ is wrong, or your call to freqz is wrong. – Matt L. Dec 14 '17 at 11:11

No $$H(z) = 1 + \frac{\alpha}{1-\alpha z^{-1}}$$ is not a lowpass filter. It's a lowpass filter in parallel with an all pass filter (kind of low-boost shelving filter). If you want a lowpass filter which rejects the high frequencies then you may take the following:
$$H_{lp}(z) = \frac{\alpha}{1-\alpha z^{-1}}$$ which is a lowpass filter for the range of parameter $\alpha$. And therefore $1 - H_{lp}(z)$ will be the complementary highpass filter in simplest terms.
freqz(alfa, [1 -alfa])

Note that the gain of the filter will be dependent on $\alpha$! Note also that as Matt.L commented you might have $1 + \alpha$ in the numerator instead...