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134 TIME-DEPENDENT CIRCUIT ANALYSIS


WithiL(t)=iC(t)=i(t),vL(t)=LdiL(t)/dt, andvC(t)=C^1

∫t
−∞iC(τ)dτ, one obtains
the following equation in terms of the inductor current:

L

diL(t)
dt

+

1
C

∫τ

−∞

iL(τ)dτ+RThiL(t)=vTh(t) (3.2.19)

Differentiating Equation (3.2.19) and dividing both sides byL, one gets
d^2 iL(t)
dt^2

+

RTh
L

diL(t)
dt

+

1
LC

iL(t)=

1
L

dvTh(t)
dt

(3.2.20)

Referring to Figure 3.2.3, the KCL equation at nodeais

iL(t)+iC(t)+

vC(t)
RTh

=iEQ(t) (3.2.21)

WithvL(t)=vC(t)=v(t),iC(t)=CdvdtC(t), andiL(t)=L^1

∫t
−∞vC(τ)dτ, one obtains the
following equation in terms of the capacitor voltage

C

dvC(t)
dt

+

1
L

∫t

−∞

vC(τ)dτ+

vC(t)
RTh

=iEQ(t) (3.2.22)

Differentiating Equation (3.2.22) and dividing both sides byC, one gets
d^2 vC(t)
dt^2

+

1
CRTh

dvC(t)
dt

+

1
LC

vC(t)=

1
C

diEQ(t)
dt

(3.2.23)

Equations (3.2.20) and (3.2.23) can be identified to be second-order, constant-coefficient,
linear ordinary differential equations of the form
d^2 x(t)
dt^2

+a

dx(t)
dt

+bx(t)=f(t) (3.2.24)

Note for the seriesLCcase,

a=

RTh
L

and b=

1
LC

(3.2.25)

and for the parallelLCcase,

a=

1
RThC

and b=

1
LC

(3.2.26)

Because of the linearity of Equation (3.2.24), the solution will consist of the sum of a transient
solutionxtr(t) and a steady-state solutionxss(t),

x(t)=xtr(t)+xss(t) (3.2.27)
wherextr(t) satisfies the homogeneous differential equation withf(t)=0,
d^2 xtr(t)
dt^2

+a

dxtr(t)
dt

+bxtr(t)= 0 (3.2.28)

andxss(t) satisfies the following for a particularf(t),
d^2 xss(t)
dt^2

+a

dxss(t)
dt

+bxss(t)=f(t) (3.2.29)

For obtaining the steady-state solution due to a dc source, recall replacing inductors with
short circuits and capacitors with open circuits. For the steady-state solution due to a sinusoidal
source, simple methods have been developed in Section 3.1.
Now for the transient solution, a possible form forxtr(t) in order to satisfy Equation (3.2.28)
is the exponential functionest. Thus let us try a solution of the form
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