Question

Question 1: (2 marks) Find the zero-input response yz(t) for a linear time-invariant (LTI) system described by the following differential equation: j(t) + 5y(t) + 6y(t) = f(t) + 2x(t) with the initial conditions yz (0) = 0 and jz (0) = 10. Question 2: (4 marks) The impulse response of an LTI system is given by: h(t) = 3e?'u(t) Find the zero-state response yzs (t) of the system for each the following input signals using convolution with direct integration. (a) x;(t) = -28(t) + 2e-fu(t) (b) x2(t) = e-fu(t-1)

Answer 1

| | 9° +54 +64 = + 2x Taking Laplace transform SYS) - SA (0) – (0) + 5[ SYS) - go)| 4 Gy(S) S = SX(s) – 4(0) + 2xUS) given y (0)=0 g (0) = 10 SYU) – 10+ 5SY(s) + 6y(S) = SXS) + 2X65) yu) (8' +55+6) = x(S) (S+2)+10 zero input response is due to initial condition also called natural response Where input is zero yos) (5 +55+) = 10 S²+55+6 (S+3) (542) A 5+2 10 Als=2 = B/s 3 = 1 + 2 = 1 lo - o 3 + 2 = 0

DO 6 Ys) = 10 ( R-Hon Taking inverse Laplace (H = locet - e 3t) 44). hath = 3 e 2t ult) to Xit= = 23(2 é zlato His) = 3 Taking Laplace of hit) XIIS) = -2 + 2 Taking Laplace of x (H) Yz5(S) = X(S) H(S) = (-2+ 3) S+2 S3 (5+2)(3+3) = - 5+2 stat A- 6 (61215131 i 3 = -6 Ys - - t ita 28 st2 St3 Yash) - - 6 e 3 t u to r(t) = è tu (t-1) (6) LT - (SH) 44) ě tu (t-1) e Page No.

Y CS - X(S) His) st (5+2) Q 5+1 Alsat = ( HQ) - =-3e (5+2) Yes) = 3 - 3e stl Yet = 3(et - é e 2t) ult Y = 3(ef - ett +1) ju(t) I Shot on realme C2