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3.2 TRANSIENTS IN CIRCUITS 135

xtr(t)=Aest (3.2.30)

Substituting Equation (3.2.30) in Equation (3.2.28), one gets

s^2 Aest+asAest+bAest=0or

(

s^2 +as+b

)

Aest= 0 (3.2.31)

which can be true only if

s^2 +as+b= 0 (3.2.32)

Equation (3.2.32) is known as the characteristic equation of the differential equation. The two
rootss 1 ands 2 of Equation (3.2.32) are given by

s 1 ,s 2 =−

a

2

±

1

2

√

a^2 − 4 b=−

a

2

(

1 ±

√

1 −

4 b

a^2

)

(3.2.33)

Three possibilities arise:

Case 1:The two roots will be real and unequal (distinct), if

(

4 b

a^2

)

<1.

Case 2:The two roots will be real and equal, if

(

4 b

a^2

)

=1.

Case 3:The two roots will be complex conjugates of each other, if

(

4 b

a^2

)

>1.

Let us now investigate these cases in more detail.

Case 1: Roots Real and Unequal (Distinct)

For 4b/a^2 <1, from Equation (3.2.33) it is clear that both roots will be negative ifais

positive. For realistic circuits with positive values ofRTh,L, andC,awill be positive, and

hence both roots will be negative. Withs 1 =−α 1 ands 2 =−α 2 (whereα 1 andα 2 are

positive numbers), the transient (natural) solution will be of the form

xtr(t)=A 1 e−α^1 t+A 2 e−α^2 t (3.2.34)

whereA 1 andA 2 are constants to be determined later. Note that Equation (3.2.34) contains

the sum of two decaying exponentials.

Case 2: Roots Real and Equal

For 4b=a^2 , the roots will be equal and negative, as one can see from Equation (3.2.33).

Withs 1 =s 2 =−a/ 2 =−α(whereαis positive), let us look at the nature of the

transient solution. For the case of real, repeated roots, not onlyxtr(t)=A 1 e−αtsatisfies the

differential equation, Equation (3.2.28), but alsoxtr(t)=A 2 te−αtcan be seen to satisfy

Equation (3.2.28). By superposition, the form of the transient solution will be seen to be

xtr(t)=A 1 e−αt+A 2 te−αt=(A 1 +A 2 t)e−αt (3.2.35)

which contains two constants (to be determined later), as it should for the solution of a

second-order differential equation.