Page 34 - Circuit Analysis II with MATLAB Applications

Chapter 1 Second Order Circuits

Then,
dv – 8t – 8t
------= – 8e 5 + k t + k e

dt 2 2
and

dv
------ = – 40 + k (1.63)
dt 2
t = 0
dv dv i
Also, i = C------ or ------ = ---- C and
C
dt
C
dt
+ + +
dv = i 0 I i 0 – R i 0 – (1.64)
S
C
L
--------------- =
------
-------------------------------------------
dt C C
t = 0 +
or
7.875
dv = I – v 0 R – i 0 10 5 40 – 2 ---------------- = 5040 (1.65)
–
e
e
L
S
C
------------------------------------------------- =
------------------------------- =
------
dt C 1640 1640
e
e
t = 0
k
Equating (1.63) with (1.65) and solving for we get
2
– 40 + k = 5040
2
or
k = 5080 (1.66)
2
and by substitution into (1.62), we obtain the total solution as
vt = e – 8t 5 + 5080t V (1.67)

Check with MATLAB:
syms t; y0=exp( 8*t)*(5+5080*t); y1=diff(y0)% Compute 1st derivative
y1 =
-8*exp(-8*t)*(5+5080*t)+5080*exp(-8*t)
y2=diff(y0,2) % Compute 2nd derivative
y2 =
64*exp(-8*t)*(5+5080*t)-81280*exp(-8*t)
y=y2/640+y1/40+y0/10 % Verify differential equation, see (1.40)
y =
0

The plot is shown in Figure 1.14 where we have used the following MATLAB code:

1-22 Circuit Analysis II with MATLAB Applications
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