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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

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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.

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  • $\begingroup$ Thank you! So in the cascading filter I can assume that the 0.5(1-z^-1) is a high pass, so by cascading ill have a sort of band pass between H(z) and 0.5(1-z^-1)? $\endgroup$ – HelpMeBro Feb 25 at 18:05
  • $\begingroup$ if H(z) is a low pass filter, yes $\endgroup$ – Bob Feb 25 at 18:21
  • $\begingroup$ Thank you very much! By stretching the time response do you mean for example if I have the H(z) normalized frequency response plot null at 0.4pi rad, when I do H(z^2) it will go to 0.8? $\endgroup$ – HelpMeBro Feb 25 at 19:15
  • $\begingroup$ I expect it to go to 0.2*pi, because the zero is in z = exp(0.4j * pi), to determine where it goes you need to find z, such that z^2 = exp(0.4j * pi) for the modified transfer function. Thus z = exp(0.2j * pi). $\endgroup$ – Bob Feb 26 at 5:47
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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.

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