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Your diagram looks correct. Let's call the transfer function in the feedback loop $G(z)$. Consider the signal $w[n]$ at the input to the delay line. Its $\mathcal{Z}$-transform satisfies $$W(z)=X(z)+G(z)Y(z)\tag{1}$$ where $X(z)$ and $Y(z)$ are the $\mathcal{Z}$-transforms of the input and output sequences, respectively. The output is just a delayed ...


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Q1: You should ask yourself if $$\frac{z}{3z^2 - 4z + 1}\stackrel{?}{=}\frac{\frac 3 2}{z-1} - \frac{\frac 1 2 }{z-\frac 1 3}$$ really holds. You'll find out that you forgot to scale correctly. Q2: There is no reason to consider $X(z)/z$ instead of $X(z)$ in this case. And concerning your question about the anti-causal sequence, a multiplication by $z^{-...


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Robert's explanation is nice. Although, for finite difference, 𝑥[𝑛]−𝑥[𝑛−1] translates to 𝑥[𝑛+1]−𝑥[𝑛], which translates to X(z).z^(-1) - X(z) = X(z).{(1-z)/z}. And for the case of actual differentiation, the finite difference will generally be divided by sampling period T. Those other expressions from text books eg. z/(z-1) tend to get lost in ...


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