Question

If the input to the system described by the difference equation y(n+1) (1/2)x(n+) -x(n) is a) Does it matter what are the initial conditions for nc0 in order to find y(n) for n20? Explain your b) x(n) -u(n) answer. (3 points). Determine the transfer function H(z) and the Frequency Response (H(est) (10 points). Find the amplitude lH(epT)I and the phase He*') as a function of co. Evaluate both for normalized frequency ω T=z/4. ( 10 points) c) Find the steady state response if the input x(nT) sin(n4) (5 points) d) Find h(n) and express h(n) as a sum of delta functions (5 points). Use convolution to find the zero state response if the input is x(nT) sin(n4) u(n) ( 12 points) Compare the answers of parts c) and d) (S points)

Answer 1

"Given - y(n + 1) = 1/2 [x(n + 1) - x(n)] where input, x(n) = u(n) a) Now, for y(n) = 1/2 [x(n + 1 - 1) - x(n - 1)] = 1/2 [x(n) - x(n - 1)] = 1/2 [u(n) - u(n - 1)] [because x(n) = u(n)] So initial condition for n < 0 doesn't matter. \r\n b) because y(n + 1) = 1/2 [x(n + 1) - x(n)] taking z-transform both side, we get - z y(z) = 1/2 [zx(z) - x(z)] doublerightarrow y(z) /x(z) = 1/2 (z - 1/z). doublerightarrow H(z) = y(z) /x(z) = 1/2(1 - z^1). Z = e^jwT therefore H(e^jwT) = 1/2 (1 - e^-jwT) H(e^jwT) = 1/2 [1 - cos wT + j sin wT] therefore