8.4 The Natural Response of a Series/Parallel RLC Circuit

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8.3 Finding Initial Values• Example 2 (cont.)4Ω3u(t)A 2Ω__ 12 F ++ vcv R --0.6H20Vi L t = 0 ++ −∴ i (0 ) = i (0 ) = 0LL+ −v (0 ) = v (0 ) =−20VccC.T. Pan 198.3 Finding Initial Values• Example 2 (cont.)4Ω3u(t)A 2Ω__ 12 F ++ vcv R --0.6H20V3A2Ωt = 0 ++ v o (0 + ) -4Ω+v R (0 + )-i c (0 + )i L+ 4i2Ω(0 ) = 3A× = 2A2+4+vR(0 ) = 2A× 2=4V+ 2ic(0 ) = 3A× = 1A2+4+d vL(0 ) 0i+∴L(0 ) = = = 0dt L L+d + ic(0 ) 1vc(0 ) = = = 2dt c 1 2d v+R(0 ) = ?dtC.T. Pan 20VS
8.3 Finding Initial Values• Example 2 (cont.)4Ω3u(t)A 2Ω__ 12 F ++ vcv R --0.6H20Vi L t > 0 +3A2Ω+ v o -4Ω1__ + i L+ 2 F vcv R-- 0.6H20VFrom KVL: − v + v + v + 20=0R o cTake derivatived d dvR() t = vo() t + vc()tdt dt dtd + d + d +∴ vR(0 ) = vo(0 ) + vc(0 ) LL( A)dt dt dtC.T. Pan 218.3 Finding Initial Values• Example 2 (cont.)4Ω3u(t)A 2Ω__ 12 F ++ vcv R --0.6H20Vi L t > 0 +3A2Ω+ v o -4Ω1__ + i L+ 2 F vcv R-- 0.6H20VvR()t vo()tAlso , from KCL: 3A= +2 4Take derivative1 d 1 d0 = vR() t + vo() t LL(B)2 dt 4 dtC.T. Pan 22
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Page 20 and 21: 8.5 The Step Response of a Series/P
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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