0195136047.pdf

146 TIME-DEPENDENT CIRCUIT ANALYSIS

F 1 (s)=

N 2 (s)

(s+a+jb)(s+a−jb)D 2 (s)

(3.3.14)

or

F 1 (s)=

K 1

s+(a+jb)

+

K 2

s+(a−jb)

+

N 3 (s)

D 2 (s)

(3.3.15)

The procedure for findingK 1 andK 2 is the same as outlined earlier for unrepeated linear factors

or simple poles. Thus we have

K 1 = lim

s→(−a−jb)

[(s+a+jb)F 1 (s)] (3.3.16)

and

K 2 = lim

s→(−a+jb)

[(s+a−jb)F 1 (s)] (3.3.17)

It can be shown thatK 1 andK 2 are conjugates of each other. The terms in the time function,

L−^1 [F(s)], due to the complex poles ofF 1 (s), are then found easily.

Alternate Representation for Complex Poles

Complex poles can be combined to yield a quadratic term in the partial fraction expansion. The

representation may best be illustrated by considering one real pole and two complex conjugate

poles. Let us then consider

F 1 (s)=

N 2 (s)

(s−p 1 )(s+a+jb)(s+a−jb)

=

N 2 (s)

(s−p 1 )

[

(s+a)^2 +b^2

]

=

N 2 (s)

(s−p 1 )

(

s^2 +As+B

) (3.3.18)

which can be rewritten as

F 1 (s)=

K 1

s−p 1

+

K 2 s+K 3

s^2 +As+B

(3.3.19)

EvaluatingK 1 as before, one has

K 1 =lim

s→p 1

[(s−p 1 )F 1 (s)] (3.3.20)

It follows from Equations (3.3.18) and (3.3.19) that

N 2 (s)=K 1

(

s^2 +As+B

)

+(K 2 s+K 3 )(s−p 1 ) (3.3.21)

Since Equation (3.3.21) must hold for all values ofs, the coefficients of various powers ofson

both sides of the equality must be equal. These equations of equality are then solved to determine

K 2 andK 3.

Even if many pairs of complex conjugate poles occur, this procedure may be used, re-

membering that the partial fraction for each complex conjugate pair will be of the form

discussed.