Q3Signals and Systems

Given $x(n)=a^{n}u(n)+b^{n}u(-n-1)$.

The Z-transform is linear, so we evaluate the terms separately:

1. The right-sided sequence $x_1(n)=a^{n}u(n)$: $X_1(z)={\sum }_{n=0}^{\infty }{a}^n{z}^{-n}=\frac{1}{1-a{z}^{-1}}=\frac{z}{z-a}$. For convergence, $|a{z}^{-1}|<1⟹|z|>|a|$.

2. The left-sided sequence $x_2(n)=b^{n}u(-n-1)$: $X_2(z)={\sum }_{n=-\infty }^{-1}{b}^n{z}^{-n}$. Let $m=-n$. $X_2(z)={\sum }_{m=1}^{\infty }{b}^{-m}{z}^m={\sum }_{m=1}^{\infty }(b^{-1}z{)}^m=\frac{b^{-1}z}{1-b^{-1}z}=\frac{z}{b-z}=\frac{-z}{z-b}$. For convergence, $|b^{-1}z|<1⟹|z|<|b|$.

Combining them:

ROC: The overall ROC is the intersection of the two regions: $|a|<|z|<|b|$. For the ROC to exist, we must have $|a|<|b|$. The ROC visually is a ring (annulus) in the z-plane, strictly bounded between the circles of radius $|a|$ and $|b|$.