The correct choice is (d) {9,7,9,11}
The explanation: The four point DFT of h(n) is H(k)=1+2e^-jkπ/2+3 e^-jkπ (k=0,1,2,3)
Hence H(0)=6, H(1)=-2-j2, H(3)=2, H(4)=-2+j2
The four point DFT of x(n) is X(k)= 1+2e^-jkπ/2+2 e^-jkπ+3e^-3jkπ/2(k=0,1,2,3)
Hence X(0)=6, X(1)=-1-j, X(2)=0, X(3)=-1+j
The product of these two four point DFTs is
Ŷ(0)=36, Ŷ(1)=j4, Ŷ(2)=0, Ŷ(3)=-j4
The four point IDFT yields ŷ(n)={9,7,9,11}
We can verify as follows
We know that from the previous question y(n)={1,4,9,11,8,3}
ŷ(0)=y(0)+y(4)=9
ŷ(1)=y(1)+y(5)=7
ŷ(2)=y(2)=9
ŷ(3)=y(3)=11.