Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12EE201 Lecture 24 P. 1Undriven RLC Circuits (continued)General form of second-order differential equation representing behavior of RLC circuits.d2x(t)dt2+ bdx(t)dt+ c x(t) = 0For nontrivial solutions, characteristic equation must be satisfied.s2 + bs + c = 0Solutions to this equation are separated into 3 categories.Case 1: s1 s2, but s1, s2 are real waveform is overdamped and general solution is of the form x(t) = K1es1(t-to) + K2es2(t-to) , t t0EE201 Lecture 24 P. 2Case 2: s1 s2, s1 and s2 are complex conjugates.Waveform is underdamped and general solution is of form x(t) = e-(t-to) [Acos(d(t-t0)) + Bsin (d(t-t0)) ]t t0Case 3: s1= s2 Waveform is critically damped and general solution is of formx(t) = [K1 + K2(t-t0) ]es1(t-to) t t0EE201 Lecture 24 P. 3Example: Find vc(t) for t 0.31u(-t)A_+8u(-t)V(1/18) F+_vL(t)iL(t)2HStep1: Find initial conditions For t = 0-, inductor is shortened , capacitor is open iL(0-)= 1A= iL(0+) vC(0-)= -8V = vC(0+)}Continuity principles for L and C_+vC(t)EE201 Lecture 24 P. 4 iC(0-) = 0A vL(0-) = 0VAt t = 0+, independent sources are deactivated.Because of source deactivation, circuit appears as:3+_vC(0+)iL(0+)=1A2HiR(0+)1/18FiC(0+)Since elements are in parallel;vR(0+) = vL(0+) = vC(0+) = -8VEE201 Lecture 24 P. 5Calculate iR(0+) from vR(0+) using Ohm’s Law iR(0+) = vR(0+)R= -8V3=-83AUse KCL to find iC(0+) iC(0+) = -iL(0+) - iR(0+) = -1-( -8/3) iC(0+) = (5/3) AInitial conditions; iC(0+) = (5/3)A ; vC(0+) = -8V iL(0+) = 1A ; vL(0+) = -8VStep2: Analyze RLC parallel circuit.For parallel RLC circuits, general solution isKest (s2 + bs + c) = 0EE201 Lecture 24 P. 6Where b = 1RCand c = 1LCs2 + 1RCs + 1LC= 0s2 + 6s +9 = 0 = (s + 3)2s = -3 s-1Therefore, s1 = s2 = -3 s-1 (case 3) vC(t) = (K1 + K2t) e-3tt 0Step3: Solve for unknown constants and obtain expression for vC(t). vC(0+) = (K1 + K2 t) e-3t = K1 = -8 VEE201 Lecture 24 P. 7dvC(0+)dt=1CiC (0+) = 18 x (5/3) = 30dvC(0+)dt= [ -3(K1 + K2t) e-3t + K2 e-3t ]t = 0+dvC(0+)dt= -3 K1 + K2 = 30K2 = 6 V/sFinal solution:vc(t) = (-8 + 6t)e-3t VEE201 Lecture 24 P. 8Example: Find iL(t) for t 0+_+_vc(t)8mF80V15vR(t)+_+_vL(t)t = 0Step1: Find initial conditionsFor t = 0- , inductor is shorted , capacitor is openiL (0-) = 2A = iL (0+) vC (0-) = 254080V = 50V = vC (0+) 256.25 HEE201 Lecture 24 P. 9For t = 0++_iL(0+) = 2AvR(0+)+_vC(t) = 50ViC (t) = -iL (t) = -2A ; vR(0+) = -50V vL (0+) = -2(25)+50 = 0Step 2: Set up solution for series circuit6.25 H258mFd2iL(t)dt2+RLdiL(t)dt+1LCiL(t) = 0s2 + bs + c = s2 + s + = 0RL1LCEE201 Lecture 24 P. 10s2 + 4s + 20 = 0s = -b b2 – 4c2= -4 16 – 802s = -2 j4 j = -1s1 = -2 - j4 s2 = -2 + j4 Solution is of formx(t) = e-t [A cos(dt) + B sin (dt) ] = e-2t [A cos (4t) + B sin (4t) ]EE201 Lecture 24 P. 11Step3: Find values for unknown constants.iL(t) = e-2t [A cos (4t) + B sin (4t) ]t = 0+= AA = iL(0+) = 2AiL'(0+) = diL(0+)dt=1LvL(0+) = 0 diL(0+)dt= e-2t [d A sin (dt) + d B cos (dt) ] - e-t [A cos(dt) + B sin (dt) ]EE201 Lecture 24 P. 12diL(0+)dt= - A + dB = -2A + 4B = 0B = 1Step 4: Write solutioniL(t) = e-2t [2 cos (4t) + sin (4t) ]
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