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:LdiL(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
LdiL(t)
dt+1
LCiL(t)=1
LdvTh(t)
dt(3.2.20)Referring to Figure 3.2.3, the KCL equation at nodeaisiL(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 voltageCdvC(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
CRThdvC(t)
dt+1
LCvC(t)=1
CdiEQ(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+adx(t)
dt+bx(t)=f(t) (3.2.24)Note for the seriesLCcase,a=RTh
Land b=1
LC(3.2.25)and for the parallelLCcase,a=1
RThCand 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+adxtr(t)
dt+bxtr(t)= 0 (3.2.28)andxss(t) satisfies the following for a particularf(t),
d^2 xss(t)
dt^2+adxss(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