Page 33 - Circuit Analysis II with MATLAB Applications

Response of Parallel GLC Circuits with DC Excitation

i R i L i C

`
vt
10u t A 10 H 1 640 F
e
0

Figure 1.13. Circuit for Example 1.4

Solution:
2
Since the natural response is to be critically damped, we must have Z = 64 because the L and C val-
0
ues are the same as in the previous example. Please refer to (1.43). We must also have

G
1
1
D = ------- = ----------- = Z = ------- = 8
0
P
2C 2RC LC
or
2
1
1
--- = 8 u --------- = ------
R 640 40
or R = 40 : and thus s = s = – D = – 8 . The natural response will have the form
P
1
2
– D t – 8t
vt = v t = e P k + k t or v t = v t = e k + k t (1.60)
2
n
1
2
n
1
Using the initial condition v 0 = 5V and evaluating (1.60) at t = 0 , we get
C
v0 = e k + 0 1 k 0 = 5

2
or
k = 5 (1.61)
1
and (1.60) simplifies to
vt = e – 8t 5 + k t (1.62)
2
As before, we need to compute the derivative dv dte in order to apply the second initial condition and
find the value of the constant k 2 .
We obtain the derivative using MATLAB as follows:

syms t k2; v0=exp( 8*t)*(5+k2*t); v1=diff(v0); % v1 is 1st derivative of v0

v1 =
-8*exp(-8*t)*(5+k2*t)+exp(-8*t)*k2

1-21 Circuit Analysis II with MATLAB Applications
Orchard Publications