GATE 2020 EE Q.4: solution and answer

Official answer: DSource: Official answer key, GATE 2020 EE (IIT Delhi)

1. Write down the delayed signal. With $y[n]=x[n-k]$, $$y[n]={\left(\frac{1}{2}\right)}^{n-k}1[n-k],$$ which is zero for $n<k$ and equal to ${\left(\frac{1}{2}\right)}^{n-k}$ for $n\ge k$. Because $k>0$, the sequence is right-sided and causal.

2. Take the z-transform directly. $$Y(z)=\sum _{n=k}^{\infty }{\left(\frac{1}{2}\right)}^{n-k}{z}^{-n}.$$ Put $m=n-k$: $$Y(z)=z^{-k}\sum _{m=0}^{\infty }{\left(\frac{1}{2}{z}^{-1}\right)}^m.$$

3. Convergence condition. The geometric series converges only when $\left|\frac{1}{2}{z}^{-1}\right|<1$, that is, when $|z|>\frac{1}{2}$. Its sum there is $\frac{1}{1-\frac{1}{2}{z}^{-1}}$, so $$Y(z)=\frac{z^{-k}}{1-\frac{1}{2}{z}^{-1}},$$ which is the expression given in the question. The factor $z^{-k}$ is finite at every $z\ne 0$, and $z=0$ is already outside $|z|>\frac{1}{2}$, so the delay does not shrink the region.

4. Check with the poles. Written in positive powers, $$Y(z)=\frac{z^{1-k}}{z-\frac{1}{2}}.$$ It has a pole at $z=\frac{1}{2}$. For $k\ge 2$ it also has a pole of order $k-1$ at $z=0$; for $k=1$ there is no pole at the origin. The outermost pole is at $|z|=\frac{1}{2}$. A right-sided sequence has its ROC outside the outermost pole, and because this one is causal the ROC also includes $z=\infty$. So the ROC is $|z|>\frac{1}{2}$.

5. Rule out the other options. (A) $|z|<2$ and (C) $|z|<\frac{1}{2}$ are regions inside a circle, which belong to left-sided sequences. (B) $|z|>2$ lies inside the true ROC but is not all of it, because the series also converges for $\frac{1}{2}<|z|\le 2$. $$x[n-k]⟷\frac{z^{-k}}{1-\frac{1}{2}{z}^{-1}},\text{ROC: }|z|>\frac{1}{2}.$$ The answer is (D).