Page 56 - Circuit Analysis II with MATLAB Applications

Chapter 1 Second Order Circuits

s=[1 0.8 9.16]; roots(s)

ans =
-0.4000 + 3.0000i
-0.4000 - 3.0000i

we find that s = – 0.4 + j3 and s = – 0.4 – j3 . Therefore, the total solution is
1
2
– D t
S
v t = v + v Cn = 100 + ke cos Z nS t + M
C
Cf
where
D = R2L = 0.4
e
S
and
2
2
2
2
–
Z = Z – D = 1LC – R e 4L = 9.16 0.16 = 3
e
nS 0 S
Thus,
v t = 100 + ke – 0.4t cos 3t + M (1)
C

and with the initial condition v 0 C = 0 we get 0 = 100 + kcos 0 + M or
kcos M = – 100 (2)
To evaluate and we differentiate (1) with MATLAB and evaluate it at t = . 0
k
M
syms t k phi; v0=100+k*exp( 0.4*t)*cos(3*t+phi); v1=diff(v0)
v1 =

-2/5*k*exp(-2/5*t)*cos(3*t+phi)-3*k*exp(-2/5*t)*sin(3*t+phi)
Thus,
dv – 0.4t – 0.4t
C
-------- = – 0.4k e cos 3t + M – 3ke sin 3t + M

dt
dv
C
-------- = – 0.4kcos – 3ksin M
M
dt t = 0
and with (2)
dv
-------- C = 40 3ksin M (3)
–
dt
t = 0
dv i i
C
C
Also, -------- = ---- = ---- L and at t = 0
dt C C

dv i 0
L
C
-------- = --------------- = 0 (4)
dt C
t = 0
1-44 Circuit Analysis II with MATLAB Applications
Orchard Publications