Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11+ + = 0 (2) + + = 0 (1)EE201 Lecture 23 P. 1Undriven RLC CircuitsiL as circuit variable:+vc(t)_RLC series circuits d2 iL (t) R d iL (t) 1 iL (t) dt2 L dt LCvc as circuit variable: d2vc (t) R dvc (t) 1 vc (t) dt2 L dt LCLRCiL(t)+ + = 0 (4) + + = 0 (3)EE201 Lecture 23 P. 2RLC parallel circuitsRCiLL d2 iL (t) 1 d iL (t) 1 iL (t) dt2 RC dt LCiL as circuit variablevc as circuit variable d2 vc (t) 1 d vc (t) 1 vc (t) dt2 RC dt LC+vc(t)_K + b K + c K e st = 0 + b + c x (t) = 0 (5)EE201 Lecture 23 P. 3General Form of Differential Equations for RLC Circuits: d2 x (t) d x (t) dt2 dt Trial solution for Eqn. (5)x(t) = K e st d2 e st d e st dt2 dt K e st (s2 + bs +c) = 0For non trivial solutions(s2 + bs + c) = 0This is the “characteristic equation” (6)EE201 Lecture 23 P. 4s1 , s2 = -b ± b2 - 4c 2s1 , s2 are natural frequenciesThree solution cases for Eqn. (7)Case 1: Solution forms (b2 - 4c) > 0 s1 , s2 are real and distinct General form is:x (t) = K1 e s1t + K2 e s2t (8)(7)Find K1 and K2 using initial conditions. If t0= 0, x (0+) = K1 e s10 + K2 e s20 = K1 + K2(9)dx (0+) dt Solve Eqns. (9) and (10) to obtain K1 , K2 = x' (0+) = s1 K1 + s2 K2 (10)Solution for Eqn. 6EE201 Lecture 23 P. 5Case 2: Solution forms (b2 - 4c) < 0 s1 , s2 are complex and distinctComplex number: Z = X + jY where j = -1 s1 , s2 = ± j = - ± jd (11)where = b/2 and d = 4c-b2 /2 s1 = - + jds2 = - - jd s1 = s2*s1 and s2 are complex conjugates -b 4c - b2 22General form is:x (t) = K1 e s1t + K2 e s2t Because s1 = s2*, es1t and e s2t are complex conjugates. Therefore, for x(t) to be real, K2 = K1* .EE201 Lecture 23 P. 6 The solution becomes, x (t) = e -t [A cos (dt) + B sin (dt) ] (12)This is equivalent to x (t) = Ke -t cos (dt + ) K= A2+B2 = tan-1 (-B/A)A, B are found from initial conditions x (0+) = e -t [A cos (dt) + B sin (dt) ] t=0+ x (0+) = A x' (0+) = -A + dBNote that, d= (n2 - 2)1/2EE201 Lecture 23 P. 7Case 3 : Solution forms b2 = 4c s1 = s2 and are real The solution form is x (t) = (K1 + K2t) e s1t (13) where x (0+) = K1 x' (0+) = s1 K1 + K2 As before, if to ≠ 0, solve simultaneous equations for x(to+) and x' (to+) to find K1 and K2.EE201 Lecture 23 P. 8Undamped response x(t)tx(t)Typical for LC circuitsUnderdamped responseTypical of case 2; LCR circuits where Re [s1]< 0 (= - )decays as exp(-t)EE201 Lecture 23 P. 9Critically dampedexponential decayx(t)tTypical of case 3; LCR circuits where s1 = s2Over damped responseNon oscillatoryx(t)tTypical of case 1; LCR circuit where s1and s2 are negativeEE201 Lecture 23 P. 10he amplitude of iL(t) decays exponentially with time, but is oscillatory. Therefore, response is underdamped.General solution for case 2 isiL (t) = e -t [A cos (dt) + B sin (dt) ]+vc(t)_Example: Find the value of the capacitance C if iL(t) = 50 e –10t sin (10 3 t) A for t 0.L10CiL(t)EE201 Lecture 23 P. 11Comparing both expressions for iL(t), A=0 d=10 3 B= 50Comparing equations (1), (5) and (12) = b/2 = R/(2L) = 0.5 R/L = 10 L = 0.5 HBut d = (n2 - 2)1/2 = 10 3 n2 = 1/LC = 300 + 100 = 400 n = 20 LC = 1/ 400 C = 1/(400L) = 1/(400)(0.5) C = 0.005
View Full Document