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Suppose

$$Y(z) = \frac{\frac 12 z + 1}{z+\frac 12} \cdot \frac{z}{z-\frac12}\text.$$

According to Wolfram Alpha the inverse transform is, $2^{-n - 2} \cdot(5 - 3 \cdot (-1)^n)$. However, I cannot show why this is true. Could someone please guide me?

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  • $\begingroup$ Homework? Generally you transform the right hand side into a sum of terms that match a table of inverse z transforms (like this one: pfister.ee.duke.edu/courses/ece485/z_trans_table.pdf) using partial fraction expansion, then you write out the time-domain expression by inspection. $\endgroup$
    – TimWescott
    Commented Oct 6, 2020 at 15:11
  • $\begingroup$ You cannot find an inverse Z-transform unless the Region of Convergence (ROC) is also specified. There are causal and non-causal discrete-time sequences that has the same given Z-transform Y(z). So you must also know whether y[n] is causal or not before getting the inverse Z-transform. $\endgroup$
    – Fat32
    Commented Oct 6, 2020 at 16:18
  • $\begingroup$ @Fat32 if it's a practical application, and the poles are inside the unit circle, it's a causal system and stable. I suppose there are systems that are reasonable to model as both unstable and noncausal, but I can't even think of a silly example. $\endgroup$
    – TimWescott
    Commented Oct 6, 2020 at 23:15

1 Answer 1

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Using binomial formulas one easily gets $$ Y(z)=\frac{\frac12+z^{-1}}{1-\frac14z^{-2}} =(\tfrac12+z^{-1})\sum_{k=0}^\infty 2^{-2k}z^{-2k}. $$ Now compare to the coefficients in $Y(z)=\sum_{n=0}^\infty y_nz^{-n}$ for $n=2k$ and $n=2k+1$, even and odd indices.

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  • $\begingroup$ Nice. It may be worth mentioning that the given equation is true for $|z|>\frac12$. $\endgroup$
    – Matt L.
    Commented Oct 7, 2020 at 6:57

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