1. Extra Credit ECE-3163 Signals and Systems Exam II By: Joseph Cunningham

2. Problem 5.45(a)‏ y[n+1] + 0.9 y[n] = 1.9x[n+1]
h[n+1] + 0.9h[n] = 1.9x[n+1]
(1.9)(-0.9)^(n+1)u[n+1] + (1.9)(-0.9)^(n)(.9)u[n] = 1.9(-0.9)^(n+1)(u[n+1] – u[n])‏
(1.9)d(n+1)‏

3. Problem 5.45(c)‏ x[n] = 1 + sin(pi(n)/4) + sin(pi(n)/2)‏
y[n] = h[n] + h[n-1] + h[n-2] = 1.9((-0.9)^(n)u[n] +(-0.9)^(n-1)u[n+1]+ (-0.9)^(n-2)u[n-2])‏

4. Explanation
I still have not grasp this subject effective enough to theoretically break down this problem, but I will explain what I can follow.
An output response y[n] in a discrete-time system is equal to the convolution of the input with the impulse response h[n].
y[n] = x[n] * h[n]
Y(W) = X(W)H(W)‏