Page 21 - Circuit Analysis II with MATLAB Applications

Response of Series RLC Circuits with DC Excitation

and since
dv
C
i = i = C--------
C
dt
we rewrite (1.18) as
dv dv 2
C
C
RC--------- + LC--------- + v = 15u t (1.19)
dt dt 2 C 0
We observe that this is a non-homogeneous differential equation whose solution will have both the
natural and the forced response components. Of course, the solution of (1.19) will give us the capaci-
tor voltage v t . This presents no problem since we can obtain the current by differentiation of the
C
expression for v t .
C
Substitution of the given values into (1.19) yields

50 – 3 dv C – 3 100 – 3 dv 2 C
------ u 10 -------- + 1 u 10 u ---------10 -------- + v = 15u t
6 dt 6 dt 2 C 0
or

dv 2 dv 5
C
C
-------- + 500-------- + 60000v = 9 u 10 u t (1.20)
C
0
dt 2 dt
The characteristic equation of (1.20) is the same as of that of (1.11) and thus the natural response is
s t s t – 200t – 300t
1
2
v = k e + k e = k e + k e (1.21)
t
2
1
1
2
Cn
Since the right side of (1.20) is a constant, the forced response will also be a constant and we denote it
as v Cf = k 3 . By substitution into (1.20) we get
0 ++ 60000k = 900000
0
3
or
v Cf = k = 15 (1.22)
3
The total solution then is the summation of (1.21) and (1.22), that is,
t
v t = v + v Cf = k e – 200t + k e – 300t + 15 (1.23)
1
2
Cn
C
As before, the constants k 1 and k 2 will be evaluated from the initial conditions. First, using
v 0 = 2.5 V and evaluating (1.23) at t = 0 , we get
C
0
0
v 0 = k e + k e + 15 = 2.5
1
2
C
or
1-9 Circuit Analysis II with MATLAB Applications
Orchard Publications