Question

3. Consider the following system LTI LTI System 2 h2ln] System 1 x [n] hiln) wIn] yIn] with h(n) (0.2)" un),h(n) is the impulse response of 2y(n)-4y(n-1) 2w(n), and x(n) (0.6"u(n). (a) Determine h2(n) (b) Determine the overall impulse response hn) (c) Determine w(n) e Demine e gu x n ) (a) velw mine hrCn) (b) Peke a jin

Answer 1

a) 2y(n) - 4y(n - 1) = 2 w(n) taking 2-transform 2y[z] -4y[z] 2^-1 = 2w[z] y[z]/w[z] = 2/2 - 4z^-1 = 1/1 - 2z^-1 H_2[z] = 1/1 - 2z^-1 taking inverse 2-transform h_2[n] = (2)^n 4[n] b) overall h(n) = h_1(n) times h_2(n) H(2) = H_1(2) H_2(2) h_1 (n) = (0.2)^n 4(n) H_1(z) = 1/1 - 0.02z^-1 H_2 [z] = 1/1 - 2z^-1 H[z] = 1/(1 - 0.2z^+) (1 - 2z^+) taking partial fraction 1/(1 - 0.22^+) (1 - 2z^+) = A/1 - 0.2z^-1 + B/1 - 2z^-1 A Vertical-bar _z^-1 = 5 = 1/1 - 2 times 5 = -1/9 B Vertical-bar _z^-1 = 0.5 = 1/1 - 0.2 times 0.5 = 1.0.9 = 10/9 H(z) = -1/9(1 - 0.02z^-1) + 10/9(1 - 2z^-1) taking inverse z - transform h(n) = 1/9 (0.2)^n u(-n -1) + 10/5 (2)^n u(n) c) w(z) = H_1(z) times (z) = z/z - 0.2 times z/z - 0.06 w[z]/2 = z/(z - 0.2) (2 - 0.6) = -0.5/(z - 0.2) + 1.5 z - 0.6 w[z] = -0.52/z - 0.2 + 1.52/z - 0.6 w(n) = 0.5(0.2)^n u(-n -1) + 1.5 (0.6)^n u(n)..