Page 36 - Circuit Analysis II with MATLAB Applications

Chapter 1 Second Order Circuits

1
1
G
D = ------- = ----------- = ------------------------------------- = 6.4
P
2C 2RC 2 u 50 u 1640
e
or
2
D = 40.96
P
and as before,
1
1
2
Z = ------- = ---------------------------- = 64
0
LC 10 u 1 640
e
2
Also, Z ! D 2 P . Therefore, the natural response is underdamped with natural frequency
0
2
2
–
Z nP = Z – D = 64 40.96 = 23.04 = 4.8
P
0
Since v = 0 , the total response is just the natural response. Then, from (1.48),
f
– D t
vt = v t = ke P cos Z t + M = ke – 6.4t cos 4.8t + M (1.69)

nP
n
and the constants and will be evaluated from the initial conditions.
k
M
From the initial condition v 0 = v0 = 5V and (1.69) we get
C
0
v0 = ke cos 0 + M = 5

or
kcos M = 5 (1.70)
k
To evaluate the constants and we differentiate (1.69), we evaluate it at t = 0 , we write the equa-
M
tion which describes the circuit at t = 0 + , and we equate these two expressions. Using MATLAB we
get:
syms t k phi; y0=k*exp( 6.4*t)*cos(4.8*t+phi); y1=diff(y0)
y1 =
-32/5*k*exp(-32/5*t)*cos(24/5*t+phi)-24/5*k*exp(-32/5*t)*sin(24/
5*t+phi)

pretty(y1)
- 32/5 k exp(- 32/5 t) cos(24/5 t + phi)
- 24/5 k exp(- 32/5 t) sin(24/5 t + phi)

Thus,

dv – 6.4ke – 6.4t cos 4.8t + M – 4.8ke – 6.4t sin 4.8t + M (1.71)
------ =

dt
and

1-24 Circuit Analysis II with MATLAB Applications
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