GATE 2023 EE Q.19: solution and answer

Official answer: BSource: Official answer key, GATE 2023 EE (IIT Kanpur)

1. Poles of $X(z)$: $z=\frac{1}{5},\frac{2}{3},3$, so the pole radii are $|z|=\frac{1}{5},\frac{2}{3},3$. A valid ROC is an annulus bounded by poles that contains no pole.

2. Rule: the DTFT $X(e^{j\omega })=X(z){|}_{z=e^{j\omega }}$ converges (i.e. $x[n]$ is absolutely summable) only if the ROC contains the unit circle $|z|=1$.

3. The four possible ROCs are $|z|<\frac{1}{5}$, $\frac{1}{5}<|z|<\frac{2}{3}$, $\frac{2}{3}<|z|<3$ and $|z|>3$. Only $\frac{2}{3}<|z|<3$ contains $|z|=1$, because $\frac{2}{3}<1<3$.

4. (A) $|z|>3$: this is the right-sided (causal) ROC. It lies entirely outside the unit circle, so the DTFT does not converge. Physically, the pole at $z=3$ then contributes a term proportional to $3^{n}u[n]$, which grows without bound.

5. (C) Left-sided sequence: the ROC must lie inside the innermost pole, i.e. $|z|<\frac{1}{5}$. This is inside the unit circle, so the DTFT does not converge.

6. (D) Right-sided sequence: the ROC must lie outside the outermost pole, i.e. $|z|>3$, the same as (A). The DTFT does not converge.

7. (B) $\frac{2}{3}<|z|<3$ contains the unit circle, so the DTFT converges. Check by partial fractions: $$X(z)=\frac{150}{49}\frac{z}{z-\frac{1}{5}}-\frac{180}{49}\frac{z}{z-\frac{2}{3}}+\frac{30}{49}\frac{z}{z-3}$$ With this ROC the poles at $\frac{1}{5}$ and $\frac{2}{3}$ are inside (right-sided terms) and the pole at $3$ is outside (left-sided term): $$x[n]=\frac{150}{49}{\left(\frac{1}{5}\right)}^{n}u[n]-\frac{180}{49}{\left(\frac{2}{3}\right)}^{n}u[n]-\frac{30}{49}{3}^{n}u[-n-1]$$ Every term decays away from $n=0$, so $x[n]$ is a two-sided, absolutely summable sequence. Answer: (B).