0195136047.pdf

136 TIME-DEPENDENT CIRCUIT ANALYSIS

Case 3: Complex Conjugate Roots

For

(

4 b/a^2

)

>1, as seen from Equation (3.2.33), the roots will be complex conjugates of

each other. Withs 1 =−α+jβands 2 =−α−jβ(whereαandβare positive numbers),

the transient solution will be of the form

xtr(t)=A 1 e(−α+jβ)t+A 2 e(−α−jβ)t=e−αt

(

A 1 ejβt+A 2 e−jβt

)

(3.2.36)

in whichA 1 andA 2 will turn out to be complex conjugates of each other, since the transient

solution must be real. An alternative, and more convenient, form of Equation (3.2.36) is

given by

xtr(t)=e−αt(C 1 cosβt+C 2 sinβt) (3.2.37)

in whichC 1 andC 2 are real constants to be determined later. Another alternative form of

Equation (3.2.37) can be shown to be

xtr(t)=Ce−αtsin(βt+φ) (3.2.38)

Note that in all three cases [see Equations (3.2.34), (3.2.35), and (3.2.38)] with nonzeroα‘s

the transient solution will eventually decay to zero as time progresses. The transient solution is

said to beoverdamped, critically damped,andunderdamped,corresponding to Cases 1, 2, and

3, respectively. Justification of the statement can be found from the solution of the following

example.

EXAMPLE 3.2.5

Consider the circuit of Figure 3.2.2 fort>0 with zero initial conditions,vTh(t)=1 V (dc), and

RTh= 2 ;L=1 H. Determine the complete response forvC(t) for capacitance values of:

(a) (25/9) F

(b) 1.0 F

(c) 0.5 F

Solution

The KVL equation around the loop is given by

2 i(t)+ 1

(

di(t)

dt

)

+

1

C

∫t

−∞

i(τ)dτ= 1. 0

Since

vC(t)=

1

C

∫t

−∞

i(τ)dτ and i(t)=iC(t)=C

dvC(t)

dt

the differential equation in terms ofvC(t) is obtained as

2 C

dvC(t)

dt

+C

d^2 vC(t)

dt^2

+vC(t)= 1. 0