The correct option is A -0.4
Y(z)=0.2X(z)−0.5z−2X(z)+0.4z−3X(z)
x(n)=[−1,1,0,−1]
=−1δ(n)+1δ(n−1)+0δ(n−2)−1δ(n−3)
X(z)=−1+z−1−z−3
Y(z)=X(z).H(z)=(0.2−0.5z−2+0.4z−3)(−1+z−1−z−3)
=−0.2+0.2z−1+0.5z−2−1.1z−3+0.4z−4+0.5z−5−0.4z−6
y(n)=[−0.2,0.2,0.5,−1.1,0.4,0.5,−0.4]
y(n)=−0.4