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

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• 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...