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I'm trying to understand how altering the frequency response of a H(z) low pass filter, will visually alter it's frequency response plot. For example: by doing H(z^2), would the frequency response represent the filter's cut out frequency happening much later than H(z)? If I multiply H(z) for something like 0.5(1-z^(-1)) as I have zeros in 0 and 0.5 a sort of band-pass will show up between 0 and 0.5?
Thank you
It is just like ordinary functions manipulation. You have to remember that z and the frequency are linked with $z = exp(j \omega \cdot T)$
$H(z^2)$ you are doing something equivalent to $H(2 \omega)$ on the continuous frequency domain, i.e. you are narrowing the filter band, and thus stretching its time response.
On the second example, you have to remember that $H_1(z) H_2(z)$ is the response of two cascading filter, that the product of the Z transforms is the convolution of the the time responses.
The filter you are suggesting $0.5 (1 - z^{-1})$ will be a sort of derivative estimate because a signal with response $x[n]$ will be transformed to $(x[n] - x[n-1])/2$ and this is a high-pass filter. You can also think of it as the inverse of the step response, you can spend some time contemplating a table of pairs of signals and their Z transform, for instance this and slowly but certainly you will build some intuition.
If you have a filter impulse response represented by H(z), replacing z with z^2 is equivalent to inserting a zero-valued sample in between each original impulse response sample. Such an operation compresses the frequency response of your original filter by a factor of two.
