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I am given the z-transform of a filter $$H(z) = \frac{z^2 - 1}{(z-z_1)(z-z_2)(z-z_3)(z-z_4)}$$

where $z_1^* = z_2$, and $z_3^* = z_4$. I'm interested in the operation of this filter, and what happens when its input is gaussian noise. Now, taking the inverse z-transform of $H(z)$ gives the impulse response. The output of the filter can then be determined via convolution between the noise input and the impulse response. I observe that the effect of this filter is to decay the noise as time moves forward. That is, when the input is gaussian noise, the output will eventually go to zero as $t\to \infty$.

Given the z-transform, I can also determine the frequency response of the filter via substitution $z = e^{j\omega}$, then plotting $\vert H(\omega) \vert$. I get the following graph representing the frequency response of this filter enter image description here

Question : Can the operation of the filter ( when the input is Gaussian noise ) be determined from this graph? This graph seems to indicate that this is a high-pass filter, and so it should have an attenuating effect on noise.

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  • $\begingroup$ "That is, when the input is gaussian noise, the output will eventually go to zero as $t \to \infty$." How do you arrive at this conclusion, and how do you define Gaussian noise? Typically Gaussian noise is defined as noise that never stops, with each $x_n \in N(0, \sigma)$. Because the noise never stops, the filter output never settles out to zero. $\endgroup$
    – TimWescott
    Oct 12, 2022 at 15:03

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Notice that $H(z)$ has two zeros, at $z=1$ (i.e $\omega = 0)$ and $z=-1$ (i.e $\omega=\pi$). Thus, it attenuates both high and low frequencies. Your graph is correct, and indeed shows that this is a band-pass filter (not solely a high-pass).

Yes, the operation of the filter is fully determined (at least in magnitude) from this graph. Any bias will be eliminated since there is a zero at $\omega=0$, and the components near $\omega=1$ will be amplified up to the peak level shown in the graph. Higher frequencies will be further attenuated, up to $\pi$ (i.e $F_s/2)$.

Specifically for Gaussian noise as input, you can know that no matter the value of the noise average, the output will eliminate it. When it comes to the rest of the spectrum, well, you may multiply the frequency response of Gaussian noise with the frequency response of your filter, and obtain the frequency response of the output.

Lastly, you said that:

when the input is gaussian noise, the output will eventually go to zero as t→∞

This is not, strictly speaking, correct. Every delta input eventually decays to zero if the filter is stable, but the output will actually still be a noise signal, only that their main frequency components will be around $\omega=1$.

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