Page 40 - Circuit Analysis II with MATLAB Applications

Chapter 1 Second Order Circuits

k
Now, we need to evaluate the constants and .
M
With the initial condition v 0 = 5V (1.76) becomes
C
0
v0 = v 0 = ke cos + 2cos – 89.72q = 5

M
C
or
M
kcos | 5 (1.77)
To make use of the second initial condition, we differentiate (1.76) using MATLAB as follows, and
then we evaluate it at t = 0 .

syms t k phi; y0=k*exp( 6.4*t)*cos(4.8*t+phi)+2*cos(6400*t 1.5688);
y1=diff(y0); % Differentiate v(t) of (1.76)
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)-12800*sin(6400*t-1961/1250)
or
dv – 6.4ke – 6.4t cos 4.8t + M – 4.8ke – 6.4t sin 4.8t + M – 12800sin 6400t – 1.5688
------ =

dt
and
dv
–
------ = – 6.4kcos M 4.8ksin M 12800sin – 1.5688 = – 6.4kcos M 4.8ksin + 12800 (1.78)
–
–

M
dt
t = 0
With (1.77) we get
dv = – 32 4.8ksin + 12800 | – 4.8ksin + 12832 (1.79)
------
–
M
M
dt
t = 0
dv
dv
Also, i = C------ or ------ = i ---- C and
C
dt
C
dt
dv i 0 + i 0 + i 0 – + i 0 – +
C
L
S
R
------ = --------------- = -----------------------------------------------------
dt + C C
t = 0
or
+
dv = i 0 – v 0 R – i 0 20 5 50 – 2 11456 (1.80)
e
–
e
C
S
L
----------------------------------------------------------- =
------------------------------- =
------
dt C 1640
e
t = 0
k
Equating (1.79) with (1.80) and solving for we get
– 4.8ksin + 12832 = 11456
M
1-28 Circuit Analysis II with MATLAB Applications
Orchard Publications