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