Question

Consider the infinite-length, discrete-time sequence \(x[n]=0.9^{|n|}\), where \(n\) is an integer. The region of convergence of its Z-transform \(X(z)\) is given by:
(Note: \(z\) is a complex variable)

• A. \(|z|>0.9\)
• B. \(|z|<0.9\)
• C. \(0.9<|z|<1/0.9\)
• D. \(\{z\ \text{such that}\ |z|<0.9\}\cup\{z\ \text{such that}\ |z|>1/0.9\}\)

Hint:

Split the two-sided exponential sequence into a right-sided and a left-sided part, find the ROC of each, and intersect them.

Answer 1

The Correct Option is C

We can also find the ROC by writing the bilateral Z-transform sum directly and applying the ratio test to its two halves.

The Z-transform is:

\[ X(z)=\sum_{n=-\infty}^{\infty}0.9^{|n|}z^{-n}=\underbrace{\sum_{n=0}^{\infty}0.9^{n}z^{-n}}_{\text{first sum}}+\underbrace{\sum_{n=1}^{\infty}0.9^{n}z^{n}}_{\text{second sum}} \]

(the second sum comes from substituting $n\to -n$ in the $n\leq -1$ part of the original sum).

The first sum is a geometric series with common ratio $r_1=0.9z^{-1}$. A geometric series converges exactly when the ratio's magnitude is less than $1$, so:

\[ |0.9z^{-1}|<1 \implies |z|>0.9 \]

The second sum is a geometric series with common ratio $r_2=0.9z$, which converges when:

\[ |0.9z|<1 \implies |z|<\frac{1}{0.9} \]

Both sums must converge together for $X(z)$ to exist, so we take the overlap of the two conditions:

\[ 0.9<|z|<\frac{1}{0.9} \]

This ring-shaped region is exactly option (C).

\[ \boxed{0.9<|z|<1/0.9} \]