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This seems a 2D filter. $A(z_\ell,z_b)=(1-\alpha{z_\ell^{-1}})\cdot\frac{1-{z_b^{-1}}}{1-\beta{z_b^{-1}}}$ Is a FIR on the $\ell$ dimension and recursive in the $b$ dimension. Let's express in terms of inputs $X(z_\ell,z_b)$ and $Y(z_\ell,z_b)$, $$Y(z_\ell,z_b) = (1-\alpha{z_\ell^{-1}})\cdot\frac{1-{z_b^{-1}}}{1-\beta{z_b^{-1}}} X(z_\ell,z_b)$$ Let's ...

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A type I (odd length $N$, even symmetry) FIR highpass filter with transfer function $$H_{HP}(z)=\sum_{n=0}^{N-1}h_{HP}[n]z^{-n}\tag{1}$$ has a real-valued amplitude function $$A_{HP}(\omega)=H_{HP}\big(e^{j\omega}\big)e^{j\omega (N-1)/2}\tag{2}$$ If the highpass filter is normalized such that it approximates unity in its passband, it can be transformed into ...

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You can model the effect of the various delays and still work in the s-domain. Your DAC can be modeled as a zero-order-hold with a T period. In the s-domain you could model the zero-order-hold by a $\frac{T}{2}$ delay or $e^{\frac{-Ts}{2}}$. While this is not technically accurate, it should be good enough. Secondly, your software must have some kind of ...

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I'd recommend to work in the time domain. The output $y[n]$ is given by the convolution of $x[n]$ and $h[n]$: \begin{align}y[n]&=\sum_{k=-\infty}^{\infty}x[n-k]h[k]\\&=u[n]\sum_{k=0}^n(0.4)^{n-k}(0.8)^k\cos\left(\frac{k\pi}{3}\right)\\&=u[n](0.4)^n\sum_{k=0}^n2^k\cos\left(\frac{k\pi}{3}\right)\tag{1}\end{align} It's clear that the first ...

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If you put the transfer function into the $z$ form, you get $$H(z) = \frac{Y(z)}{X(z)} = \frac{z - 0.5}{z}$$ Then you can immediately see that $Y(z) = z - 0.5$, and $X(z) = z$. Thus, by inspection, the transfer function has a pole at $z = 0$, and a zero at $z = 0.5$.

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There is already a good answer, but here is some extra explanation. This is called "pole/zero cancellation". That means you have a pole, say, $z_p$ but you also have a zero at $z_p$ so the zero cancels the pole. One particular example where this can be useful is a moving average filter of length $N$, The transfer function is H(z) = \frac{1}{N} \...

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You can simplify the z-transform further, $H(z) = \frac{Y(z)}{X(z)}= \frac{0.5-0.5z^{-2}}{1-z^{-1}}$ $H(z) = \frac{(1-z^{-1})(1+z^{-1})}{2(1-z^{-1})}$ canceling the common pole and zero $H(z) = \frac {1+z^{-1}}{2}$ take inverse z-transform $Y(n) = \frac{x(n)+x(n-1)}{2}$

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You can write your transfer function as the sum of three 2-nd order transfer functions (partial fractions), and then the impulse response will be the sum of the impulse response of each. To help you with the partial fractions you can use this calculator

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