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&iacute;•„&ecirc;&cedil;&deg;&euml;&iexcl; #2

# &iacute;•„&ecirc;&cedil;&deg;&euml;&...

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13. Fr equency r esponse Sunghoon Kwon Sl i de cr edi t : Pr of essor Roger Howe, St anf or d

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A Second Order System v S ( t ) L v C ( t ) + - R C i L i R i C v L + - Wher e does t he i nduct or come f r om? ( ~ 1nH/ mm “ f or any wi r e” ) Do st ep r esponse: v S ( t ) j umps t o V DD at t = 0 t = 0 V DD
“Step Response” of R-L-C circuit v S ( t ) L v C ( t ) + - R C i L i R i C v L + - I ni t i al condi t i ons: v C ( t =0 - ) = 0 V & i L ( t =0 - ) = 0 A C L i i = ( 29 dt dv C t d t v L C t L = 0 1 + - = C C DD L v R dt dv C V v I nduct or vol t age: t = 0 V DD C L R i i i = = f or t 0

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Second-Order Differential Equation for  v C (t) dt dv C t d v R dt dv C V L C t C C DD = + - 0 1 Subst i t ut i ng f or v L : 2 2 1 dt v d C v R dt dv C V L C C C DD = + - Di f f er ent i at i ng bot h si des: DD C C C V v dt dv RC dt v d LC = + + 2 2 Rear r angi ng t er ms:
St eady- st at e sol ut i on: v C, ss = V DD ) ( t Tr ansi ent sol ut i on: v C, t r = ? guess v C, t r = ae st and subst i t ut e: ( 29 ( 29 0 2 = + + st st st ae ae RCs ae LCs Short Review : Solving the 2 nd  Order O.D.E. DD C C C V v dt dv RC dt v d LC = + + 2 2 ) ( t v S 0 1 2 = + + LC s L R s - ± - = LC L R L R s 1 2 2 2 2 , 1 : Char act er i st i c f unct i on Compl i cat ed…… - > Phasor anal ysi s! !

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Thr ee cases f or s1 and s2 Get t wo negat i ve r eal r oot s ( over damped) - ± - = LC L R L R s 1 2 2 2 2 , 1 LC L R 1 2 LC L R 1 2 = LC L R 1 2 < Get a si ngl e negat i ve r oot s ( cr i t i cal damped) I nt er est i ng case! ( Under damped) v C, t r = ae st
Under damped case 2 2 2 , 1 2 1 2 1 2 2 - ± - = - ± - = L R LC j L R LC L R L R s t L R LC j t L R t L R LC j t L R DD C e e a e e a

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