8.4 The Natural Response of a Series/Parallel RLC Circuit

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8.5 The Step Response of a Series/Parallel RLCCircuitCase (ii) Let2di di 1 di 1 I s2dt dt RC dt LC LCv= L ⇒ + + i = , t > 0Step 2. Complete solution = i h (t)+i p (t)Step 3.ph( )i t = IsSt 1 St 2⎧ Ae1+ Ae2overdamped⎪i t A At e critically damped−αt() = ⎨( 1+2 )⎪ −αt( cos + sin )⎩e A1 wtdA2wtdundererdampedInitial Condition+ −( 0 ) ( 0 )i = i = I0di d vL ( 0)(0)+Vdt dt L L+0C.T. Pan L = vL⇒ i = =438.5 The Step Response of a Series/Parallel RLCCircuit• Example1 F2- -t < 0 , i(0 )=0 , v(0 ) =12Vt>01 F2C.T. Pan44
8.5 The Step Response of a Series/Parallel RLCCircuitC.T. PanMethod (a)v 1 dvFrom KCL ⇒ i = +2 2 dt(A)diFrom KVL ⇒ 4i+ 1 + v = 12dt(B)Substitute (A) into (B)2dv 1 dv 1 d v(2v+ 2 )+( + )+ v= 122dt 2 dt 2 dt2⇒ d v dv5 6v24V2dt+ dt+ =Characteristic equation( s+ 2)( s+ 3) = 0s=− 2, s =−31 2458.5 The Step Response of a Series/Parallel RLCCircuitv () t = A e + A eh−2t−3t1 224vp() t = = 4V6∴ vt () = 4+ A e + A e−2t−3t1 2+ -Initial Condition v(0 )= v(0 )=12V+t=01 F2C.T. PandvC = icdt++dv(0 ) ic(0 ) −646∴ = = =−12 V/Sdt C 1/2
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Page 5: 8.3 Finding Initial ValuesUnder tra
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Page 28 and 29: 8.6 State EquationStep1. Pick a tre
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8.4 The Natural Response of a Series/Parallel RLC Circuit

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