Extra Credit ECE Signals and Systems Exam II By: Joseph Cunningham

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Presentation transcript:

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

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)‏

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])‏

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)‏