082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

19.6 Constant coefficient equations 573
Equating coefficients of cosωt gives
−ω 2 LE + E C =V 0 ω
CV
sothatF = 0andE = 0
ω
. It follows that the particular integral is
1−ω 2 LC
i =
CV 0 ω cos ωt. Finally, the general solution is
1−ω 2 LC
t t
i=Acos√ +Bsin√ + CV 0ωcos ωt
LC LC 1−ω 2 LC
The terms zero-input response and zero-state response were introduced in Section 19.4
inconnectionwithfirst-orderequations.Thisterminologyisidenticalwhenwedealwith
second-order equationsasthe following example illustrates.
Engineeringapplication19.8
ParallelRLC circuit
Figure19.7showsaparallelRLCcircuitwhichhasacurrentsourcei S
(t).Theinductor
current,i, can befound bysolvingthe second-order differential equation
LC d2 i
dt + L di
2 Rdt +i=i S (t) fort0
It would be a useful exercise for you to derive this equation. Note that this equation
issecond order, and inhomogeneous due tothe source termi S
(t).
i S (t)
R L C
i
Figure19.7
AparallelRLC circuit.
SupposeL = 10 H,R = 10 ,C = 0.1 F andi S
(t) = e −2t and that the initial
conditions arei = 1 and di
dt =2whent=0.
(a) Obtain the solution of this equation subject to the given initial conditions and
hence statethe inductor currenti.
(b) Obtain the zero-input response. This isthe solution wheni S
(t) = 0.
(c) Obtain the zero-state response. This is the solution of the inhomogeneous equation
subject to the conditionsi = 0 and di = 0 att = 0. It corresponds to there
dt
being no initial energy inthe circuit.
(d) Show that the solution in (a) is the sum of the zero-input response and the zerostateresponse.
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