Question

2. (11 pts) a) and b) Use the tables of Laplace transform pairs and properties to find the Laplace transforms for the following funetions: a) x(t)-(t-2) u(t) Electeie Caits 1 2s Ci) e2(1-2s) 2e2s Cii) Cin 1-2s 2 Civ) b) x(t)- d (3sin(3t)) u(t) dt -18s Ci) (s+9)2 9s CH) s+9 (ii) -3 (iv) +9 +9 e) and d) Find the inverse transform x(t) for each transform X(s) below. Assume all x(t) are multiplied by u(t) (left out for brevity). You are given the MATLAB output of the"residue" command with the k- line omitted. 10s c) X(s)- (s+1)(s 2+3s +4) >> [r.p,k]-residue([10 0).conv((1 1].(1 3 4))) 2.5000+4.7246 -1.5000- 1.3229 r2.5000-4.7246 -5.0000 -1.0000 p--1.5000+ 1.3229 abs(r)- 5.3453 angle(r)* 180/pi--62.1147 5.3453 5.0000 62.1147 180.0000 (ii)-5et+10.7e1cos(1.3t-62.10) (i)-5et+ 10.7e5tcos(1.3t+ 62.10) (iv) -Set+ e1s(2.5cos(1.3t) -4.7sin(1.3t) (iii) -5et+5-3e15tcos(1.3t-62.10) 2s +3 d) X(s)- 2) 2 (s+3)(s + [r.p,k]-residue([2 3].conv(conv([1 2).[1 2] ),.[1 3))) -1.0000 3.0000 r--3.0000 -2.0000 -3.0000 -2.0000 P (iv)-3e-3t 4+3te2t-e2t (iii)-3e-3t+ 2ea (ii) -3e-3t- 2est 2et (i)-3e-3t+ 3e-2t-te at e) Which of the following this question, circle all the answers that are true. are reasons for using Laplace circuit analysis compared to other methods? FOw (i) it is easier than solving differential equations in the time-domain for higher-order circuits (ii) it can only handle circuits with one independent source and no initial conditions (ii) it can handle more kinds of time-dependent sources than phasors (iv) it includes both natural and forced response (v) all of the above are true

Answer 1

X(t) = 9 t - 2) 4(t) x(t0 = t4 (t) - 24 (t) X(s) = 1/ s^2 - 2/ s = 1 - 2s/s^2 (iii) is correct x(t) = d /dt 9 3 sin (3t)) 4 (t) X(S) = 3sx 3/s^2 - 9 (iv) is correct X(S) = 2.5 - 4.7246 I/s - [1.5 + 1.3229 i] + 2.5 + 4.7264 i/ s - [- 1.5 plusminus 1.3229 i] - s/s - 9 - 1) \r\n x(t) = (2.5 - 4.7246 i) e^(- 1.5 + 1.327 9 I) t + 9 2.5 + 4.7246 i) e^(- 1.5 - 1.3229 i) t - 5 e^-t = e^- 1.5 (t0 [cos (1.3229 t) + I sin (1.3229)t] [2.5 - 1.7246 i] + e^- 1.5 t [cos (1.3229 t) - I sin (1.3229 t)] [2.5 + 4.7246 i] - 5 e^t = -5 e^-t