Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8EE201 Lecture 22 P. 1Introduction to Second Order CircuitsLC CircuitsCombining an inductor and capacitor.L+_CvLiLiC+_vCKVL: vC(t) = vL(t)KCL: ic(t) = -iL(t)-iL(t) = ic(t) = Cdvc(t)dtRecall:diL(t)dt=1LvL(t)Take derivative of Eqn. (22.1)-diL(t)dt= -1Lvc(t) = Cd2vc(t)dt2(22.1)EE201 Lecture 22 P. 2d2vc(t)dt2=-1LCvc(t) (22.2) Similarly,LdiL(t)dt= vL (t) = vC (t)Differentiate w.r.t time,d2iL(t)dt2= 1Ldvc(t)dt= 1LCic (t)d2iL(t)dt2=-1LCiL (t) (22.3)EE201 Lecture 22 P. 3Eqns. 22.2 and 22.3 are often written as:d2vc(t)dt2+1LC vc(t) = 0 (22.2a)d2iL(t)dt2+1LCiL(t) = 0 (22.3a)General differential equation for LC circuitsd2x(t)dt2+1LCx(t) = 0 (22.4)Possible solutions for Eqn.(22.4)x(t) = A cos (t)x(t) = B sin (t)x(t) = A cos (t) + B sin (t)All valid for specific values of A, B, and .Eqn. 22.4 describes the behavior of the capacitor voltage or inductor current in an LC circuit.Strategy for finding A, B, Solve for second derivative of trial solution. EE201 Lecture 22 P. 4 d2dt2[ A cos (t) + B sin (t)] +2 [ A cos (t) + B sin (t)] = 0Comparing this to Eqn.(22.4),2 = 1LC = 1LCFrom trigonometry, if initial condition x(t0) is known, constants A, B can be found.Suppose t0 = 0 x(0) = A cos (0) + B sin (0)x(0) = AEE201 Lecture 22 P. 5If dx(0)dtis known,dx(0)dt= -A sin (0) + B cos (0)dx(0)dt= x'(0) = B Example: Find vc(t) and iL (t). The switch opens at t = 0 s.vc(t)+_100.2A0.1Ft = 00.9HiL(t)10Note: if to ≠ 0, then solve simultaneous equations for x(to) and x'(to).EE201 Lecture 22 P. 6Step1: Establish initial conditionsCircuit configuration at t = 0- sec.iL (0-) = 1/101/10+1/100.2A = 0.1A = iL (0+)Step2: Solve for constants A, B, for capacitor voltage when t ≥ 0 sec.vc(t) = A cos (t) + B sin (t)A = 0 (t0 = 0)vc (0-) = vc (0+) = 0 Vvc(t)+_100.2A0.1F0.9HiL(t)10From inspection,EE201 Lecture 22 P. 7iL(0+) = -CBB = -iL(0+)C= -0.1(0.1)() = 1LC= 10.09 = 3.33 s-1B = -0.3 Vdvc(t)dt= -A sin (t) + B cos (t)dvc(t)dt= -iL(t)CiL(t) = CA sin (t) - CB cos (t)EE201 Lecture 22 P. 8diL(0+)dt=1LvL(0+) = 1Lvc(0+) = 0diL(0+)dt= 0 = -A sin (t) + B cos (t)B = 0iL (t) = 0.1 cos (3.33t) AWrite vc(t) expressionvc(t) = A cos (t) + B sin (t)vc(t) = -0.3sin (3.33t) VStep 3: Repeat steps for inductor current when t ≥ 0 sec.iL(t) = A cos (t) + B sin (t)iL(0+) = A = 0.1 ANote: A and B are different for vc(t) and iL(t)
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