ISRO EC 2023 Q.20: solution and answer

Official answer: BSource: ISRO's revised and final answer key of 16 February 2024

1. The sequence $x[n]$ is real, so its $N$-point DFT has conjugate symmetry: $X(N-k)=X^{\ast }(k)$. This follows from $X(N-k)={\sum }_{n=0}^{N-1}x[n]e^{-j2\pi (N-k)n/N}={\sum }_{n=0}^{N-1}x[n]e^{j2\pi kn/N}=X^{\ast }(k)$ for real $x[n]$.

2. Here $N=8$, so $X(6)=X(8-2)=X^{\ast }(2)$.

3. Given $X(2)=3+j$, its conjugate is $X^{\ast }(2)=3-j$.

4. Hence $X(6)=3-j$. The other given values $X(0),X(1),X(3)$ are not needed: $X(3)$ pairs with $X(5)$, and $X(1)$ with $X(7)$.

5. The other options are the wrong partners: $2-3j=X^{\ast }(3)=X(5)$, $1-j=X^{\ast }(1)=X(7)$, and $1+j=X(1)$ itself. None of them is $X(6)$.