# How to find inverse z transform

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?

• 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. – TimWescott Oct 6 '20 at 15:11
• 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. – Fat32 Oct 6 '20 at 16:18
• @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. – TimWescott Oct 6 '20 at 23:15

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.
• Nice. It may be worth mentioning that the given equation is true for $|z|>\frac12$. – Matt L. Oct 7 '20 at 6:57