Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9+ + = F (2) + + = F (1)_EE201 Lecture 25 P. 12nd Order Circuits with Constant InputsiL as circuit variablevc(t)Differential equations for RLC series circuits: d2 iL (t) R d iL (t) 1 iL (t) dt2 L dt LCvc as circuit variable d2 vc (t) R d vc (t) 1 vc (t) dt2 L dt LCLRCiL(t)++_vs(t)EE201 Lecture 25 P. 2Parallel RLC circuits:RCiLLvc(t)+_ + + = F (4) + + = F (3)iL as circuit variable d2 iL (t) 1 d iL (t) 1 iL (t) dt2 RC dt LCvc as circuit variable d2 vc (t) 1 d vc (t) 1 vc (t) dt2 RC dt LCEE201 Lecture 25 P. 3General form of differential equation:General solution formx(t) = xn(t) + xFEqn. (5) has the characteristic equations2 + bs +c = 0xn(t) – solution satisfying homogenous differential equation (i.e. F = 0)xF -- constant accounting for non-zero forcing functions1 , s2 = -b ± b2 - 4c 2Therefore, 3 cases will be considered + b + c x (t) = F (5) d2 x (t) d x (t) dt2 dtEE201 Lecture 25 P. 4Case 1: s1 , s2 are real and distinctSolution:x (t) = K1 e s1(t-to) + K2 e s2(t-to) + xF xF = x()Solve for K1 , K2 using x (0+) = K1 + K2 + xFx' (0+)= s1 K1 + s2 K2 Case 2: s1 , s2 are complex and distincts1 = - + jds2 = - - jdSolutionx (t) = e -(t-to) [A cos (dt) + B sin (dt) ] + xFEE201 Lecture 25 P. 5xF = x()x (0+)= A + xFx' (0+)= -A + dBCase 3 : s1 = s2 and are real Solution: x (t) = (K1 + K2t )e s1(t-to) + xF xF = x() where x (0+) = K1+ xF x '(0+) = s1 K1 + K2 evaluated at to = 0EE201 Lecture 25 P. 6Example: Find vc(t) for t ≥0 assuming vc(0-) =1V and iL(0-) =3/2 A. 10u(t) V_+2/3 0.5 F1 HiL(t)vc(t)Step 1: Set up initial conditions vc(0+) = vc(0-) = 1ViL (0+) = iL(0-) = 3/2 A+_ 10V+_2/3 iL(0+) = 3/2 Avc(0+) = 1Vic(t)- vL(0+) +iR(0+)Circuit at t = 0+_+EE201 Lecture 25 P. 7From KVL, vL(0+) = 10 - vc(0+) = 9 VSolving for iR(0+) iR(0+) = vc(0+) /R= 3/2 AFrom KCL,ic(0+) = iL(0+) - iR(0+) = 0 AStep 2: Set up solution form for vc(t) withsource zero parallel RCLEE201 Lecture 25 P. 8s2 + [1/(RC)] s + 1/(LC) = s2 + 3s + 2 = 0s1= -1 ; s2= -2 (Case 1 solution)vc(t) = K1 e s1t + K2 e s2t + vF vc(t) = K1 e -t + K2 e -2t + vFStep 3: Solve for unknown constantsAt t = , 10V+-2/3 vF = vc() = 10VFrom initial conditions, K1 and K2 are found.vc+_EE201 Lecture 25 P. 9vc(0+) = K1 + K2 + vF K1 +K2 = -9 v'c(t) = (1/C) ic(0+) = 2 ic(0+) = 0v'c(0+) = - K1 - 2 K2 = 0K1 = -18K2 = 9Step 4: Obtain final solutionvc (t) = K1 e s1t + K2 e s2t + vFvc (t) = (-18e -t + 9 e -2t +10) V t
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