Question

1. A causal LTI system is implemented by the difference equation y(n) = 2r(n) - 0.5 y(n-1). (a) Find the frequency response H/(w) of the system. (b) Plot the pole-zero diagram of the system. Based on the pole zero diagram, roughly sketch the frequency response magnitude |H'(w). (c) Indicate on your sketch of H w , its exact values at w=0, 0.5, and . (d) Find the output signal y(n) produced by the input signal (n) = 3 + cos(0.5 in).

Answer 1

al y(n) = 2x(n) - 0.5 glo-1 Z.T. of above Eqh. Y(w) = 2X(W) - 0.54|w). ejw 4(W) | 1 +0.5 ejw} = 2XlW - H(W) = a 2 = 22 Set H(W) - 2 1 4 0.5 W b) HW) = -2 1 +0.5(Cosw-jsinul Hw) - Hoscow -jo.5sinw

THwl. ſu to scow)² + (0 sanw) Hw) - 2 11+0.25c²w + caso +0.25sintw THIwl - V1 +2 +low Inwl - - 13 + cow TH(W) T I ZA H/2) = -2 1 +0szt polea) 121=0.5 zeroz no zero 1 1221 at = 1 c) w=o 141w-oll = 2 √3+1 WE 0.50 TH/WTO-SA)) . -= 1.154 v3to Wak Hlwan)] = 2 = 2 =1.419

+1.414 1.154 Hat d) alnd = 3 + cos (0.5an) H (W) = -2 1+0.5ejw Hwll 2 V3 tcow [H(W) – tan 1-0.5 sino ) 11+0.5 cow! for cos (0.5an) Wo= 0.50 | HlwZ O. sal) = 1.154 (H (W- O.SN) = - tan" 1-0.5) = + 2 6.56 Cos (0.5 in) Olp 1.154 Cos (0.5an+ 26.56)

box 3eos woso Hlwall = 2 143 = 1 LH(0) = 0 3 / 3X1 = 3 y (n) = 3 + 1.154 CCM (0. San +26:56)