GATE 2026 EE Q.11: solution and answer

Official answer: CSource: Official answer key, GATE 2026 EE (IIT Guwahati)

1. Split the two-sided sequence into a right-sided part and a left-sided part: $x[n]=0.9^{n}u[n]+0.9^{-n}u[-n-1]$.

2. Right-sided part (a geometric series): $\sum _{n=0}^{\infty }0.9^n{z}^{-n}=\sum _{n=0}^{\infty }{\left(0.9z^{-1}\right)}^n$ converges if and only if $|0.9z^{-1}|<1$, i.e. $|z|>0.9$.

3. Left-sided part: substitute $m=-n$ to get $\sum _{n=-\infty }^{-1}0.9^{-n}{z}^{-n}=\sum _{m=1}^{\infty }(0.9z{)}^m$, which converges if and only if $|0.9z|<1$, i.e. $|z|<\frac{1}{0.9}$.

4. $X(z)$ exists only where both series converge, so the ROC is the intersection of the two regions: $$0.9<|z|<\frac{1}{0.9}\approx 1.11$$

5. The ring is non-empty because $0.9<1/0.9$ (true for any base $a$ with $|a|<1$), so $X(z)$ exists. As a check, $X(z)=\frac{1-0.81}{(1-0.9z^{-1})(1-0.9z)}$, with poles at $z=0.9$ and $z=1/0.9$, and these two poles are exactly the edges of the annulus.

6. The ROC of any Z-transform is a single connected ring centred on the origin and bounded by poles. It can never be two disjoint regions, which rules out (D). Options (A) and (B) are each the ROC of only one half of the sequence. Answer: (C).