Transcript
Chapter 8, Solution 29. (a)
s2 + 4 = 0 which leads to s 1,2 = j2 (an undamped circuit) v(t) = V s + Acos2t + Bsin2t 4V s = 12 or V s = 3 v(0) = 0 = 3 + A or A = -3 dv/dt = -2Asin2t + 2Bcos2t dv(0)/dt = 2 = 2B or B = 1, therefore v(t) = (3 – 3cos2t + sin2t) V
(b)
s2 + 5s + 4 = 0 which leads to s 1,2 = -1, -4 i(t) = (I s + Ae-t + Be-4t) 4I s = 8 or I s = 2 i(0) = -1 = 2 + A + B, or A + B = -3
(1)
di/dt = -Ae-t - 4Be-4t di(0)/dt = 0 = -A – 4B, or B = -A/4 From (1) and (2) we get A = -4 and B = 1 i(t) = (2 – 4e-t + e-4t) A (c)
s2 + 2s + 1 = 0, s 1,2 = -1, -1 v(t) = [V s + (A + Bt)e-t], V s = 3. v(0) = 5 = 3 + A or A = 2 dv/dt = [-(A + Bt)e-t] + [Be-t] dv(0)/dt = -A + B = 1 or B = 2 + 1 = 3 v(t) = [3 + (2 + 3t)e-t] V
(2)
(d)
s2 + 2s +5 = 0, s 1,2 = -1 + j2, -1 – j2 i(t) = [I s + (Acos2t + Bsin2t)e-t], where 5I s = 10 or I s = 2 i(0) = 4 = 2 + A or A = 2 di/dt = [-(Acos2t + Bsin2t)e-t] + [(-2Asin2t + 2Bcos2t)e-t] di(0)/dt = -2 = -A + 2B or B = 0 i(t) = [2 + (2cos2t)e-t] A
