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Consider a discrete transfer function that represents an anti causal filter such as a derivative filter: $$H(z) = (-z^{-2} -2z^{-1} +2z +z^2) (1/8T)$$
Where T is the sampling period.
Normally in MATLAB, I use fvtool and enter coefficients but these coefficients are relevant to negative exponentials only, i.e. for a causal filter.
I want to view the magnitude response, phase response, pole zero map of an anti-causal filter in MATLAB. How is that possible?
The transfer function $$H(z)=-z^{-2} -2z^{-1} +2z +z^2$$ can be written as $$\begin{align} H(z)&=z^2z^{-2}\left(-z^{-2} -2z^{-1} +2z +z^2\right)\\ &=z^2\left(-z^{-4} -2z^{-3} +2z^{-1} +1\right)\\ &=H_1(z)H_2(z) \end{align}$$ It can now be seen more easily that
$H_1(z)$ has two zeros at $z=0$ (and two poles at $\pm\infty$) .
$H_2(z)$ has four poles at $z=0$ and four zeros: one at $z=1$ and three at $z=-1$. What can be said about pole-zero map of $H(z)$ overall?
Since $|H_1(e^{j\omega})|=1$, we cnclude that $|H(e^{j\omega})|=|H_2(e^{j\omega})|$
Since $\angle H_1(e^{j\omega})=2\omega$, we cnclude that $\angle H(e^{j\omega})=\angle H_2(e^{j\omega})+2\omega$
