GATE 2024 EE Q.39: solution and answer

Official answer: CSource: Official answer key, GATE 2024 EE (IISc Bengaluru)

1. Start from the definition of the $Z$-transform: $$Y(z)=\sum _{n}y[n]z^{-n}=\sum _{n}x[2n]z^{-n}$$

2. Substitute $m=2n$. As $n$ runs over all integers, $m$ runs over the even integers only, and $z^{-n}=z^{-m/2}={\left(z^{1/2}\right)}^{-m}$: $$Y(z)=\sum _{m\text{even}}x[m]{\left(z^{1/2}\right)}^{-m}$$

3. Write the sum over all $m$ and keep only the even-indexed samples with the selector $\frac{1+(-1{)}^m}{2}$, which is $1$ for even $m$ and $0$ for odd $m$: $$Y(z)=\sum _m\frac{1+(-1{)}^m}{2}x[m]{\left(z^{1/2}\right)}^{-m}$$

4. Split the sum into two halves. The first half is $\frac{1}{2}{\sum }_{m}x[m]{\left(z^{1/2}\right)}^{-m}=\frac{1}{2}X\left(z^{1/2}\right)$.

5. For the second half, $(-1{)}^m=(-1{)}^{-m}$ for every integer $m$, so $(-1{)}^m{\left(z^{1/2}\right)}^{-m}={\left(-z^{1/2}\right)}^{-m}$. Hence the second half is $\frac{1}{2}{\sum }_{m}x[m]{\left(-z^{1/2}\right)}^{-m}=\frac{1}{2}X\left(-z^{1/2}\right)$.

6. Adding the two halves gives the downsampling-by-2 property: $$Y(z)=\frac{1}{2}\left[X\left(z^{1/2}\right)+X\left(-z^{1/2}\right)\right]$$ Which square root of $z$ is taken does not matter, because swapping $z^{1/2}$ and $-z^{1/2}$ leaves the sum unchanged. This is option (C).

7. Quick check with $x[n]=\delta [n-2]$, so $X(z)=z^{-2}$. Then $y[n]=\delta [2n-2]=\delta [n-1]$, so $Y(z)=z^{-1}$. Option (C) gives $\frac{1}{2}\left[{\left(z^{1/2}\right)}^{-2}+{\left(-z^{1/2}\right)}^{-2}\right]=\frac{1}{2}\left[z^{-1}+z^{-1}\right]=z^{-1}$, which matches. Option (B) gives $z$, and options (A) and (D) both give $z^{-4}$, so none of them matches.