0195136047.pdf

3.2 TRANSIENTS IN CIRCUITS 137

The steady-state responsevC,ss(t)due to a dc source is determined by replacing inductors with
short circuits and capacitors with open circuits,

vC,ss(t)=V 0 = 1 .0V

[Note thatvC,ss(t) is of the form of the excitation and is a constantV 0 ; substitution ofV 0 for
vC,ss(t) gives 0+ 0 +V 0 = 1 .0.]
The transient response is obtained from the force-free (homogeneous) equation, which is

2 C

dvC,tr

dt

+C

d^2 vC,tr(t)

dt^2

+vC,tr(t)= 0

AssumingvC,tr(t) to be of the formAest, the values ofsare determined from

Aest

[

2 Cs+Cs^2 + 1

]

=0orCs^2 + 2 Cs+ 1 = 0

and the roots are

s 1 ,s 2 =

− 2 C±

√

( 2 C)^2 − 4 C

2 C

=− 1

(

1 ±

√

1 −

1

C

)

(a) ForC= 25 /9 F, the values ofs 1 ands 2 are

s 1 =− 0. 2 and s 2 =− 1. 8

The transient response is given by (see Case 1 with real and unequal roots)

vC,tr(t)=A 1 e−^0.^2 t+A 1 e−^1.^8 tV

and the total (complete) response is

vC(t)=vC,ss(t)+vC,tr(t) or vC(t)= 1. 0 +A 1 e−^0.^2 t+A 2 e−^1.^8 tV

whereA 1 andA 2 are to be evaluated from the initial conditions,

vC( 0 −)= 0 ,iL( 0 −)=i( 0 −)= 0

By the continuity principle,vC( 0 +)=0 andiL( 0 +)=0. Also,iL( 0 +)=iC( 0 +)and

dvC/dt

(

0 +

)

=iC( 0 +)/C=0. ExpressingdvC(t)/dtasdvC(t)/dt=− 0. 2 A 1 e−^0.^2 t−

1. 8 A 2 e−^1.^8 t, evaluatingdvC( 0 +)/dtandvC( 0 +)yields

vC( 0 +)= 0 = 1 +A 1 +A 2

dvC

dt

(

0 +

)

= 0 =− 0. 2 A 1 − 1. 8 A 2

Simultaneous solution yieldsA 1 =− 1 .125 andA 2 = 0 .125. Hence the complete

solution is

vC(t)= 1. 0 − 1. 125 e−^0.^2 t+ 0. 125 e−^1.^8 tV

(b) ForC= 1 .0 F, the values ofs 1 ands 2 are both equal;s 1 =s 2 =−1. Corresponding to

Case 2 with real and equal roots, the transient response is given by Equation (3.2.35),

vC, n(t)=(A 1 +A 2 t)e−αt=(A 1 +A 2 t)e−t