Page 22 - Circuit Analysis II with MATLAB Applications

Chapter 1 Second Order Circuits

k + k = – 12.5 (1.24)
1
2
Also,
dv dv i dv i 0 5
C
C
L
C
L
i = i = C-------- -------- = ---- and -------- = ------------ = ------------------------- = 300 (1.25)
C
L
dt
dt
C
dt
t = 0 C 100 10 – 3
--------- u
6
Next, we differentiate (1.23), we evaluate it at t = 0 and equate it with (1.25). Then,
dv – 200t – 300t dv
C
-------- = – 200k e – 300k e and -------- C = – 200k 300k 2 (1.26)
–
1
2
1
dt dt
t = 0
Equating the right sides of (1.25) and (1.26) we get
– 200k 300k = 300
–
1
2
or
k – – 1.5k = 1.5 (1.27)
2
1
From (1.24) and (1.27), we get k = – 34.5 and k = 22 . By substitution into (1.23), we obtain the
1
2
total solution as
v t = 22e – 300t – 34.5e – 200t + 15 u t (1.28)

0
C
Check with MATLAB:
syms t % Define symbolic variable t
y0=22*exp( 300*t) 34.5*exp( 200*t)+15; % The total solution y(t)
y1=diff(y0) % The first derivative of y(t)
y1 =
-6600*exp(-300*t)+6900*exp(-200*t)
y2=diff(y0,2) % The second derivative of y(t)
y2 =
1980000*exp(-300*t)-1380000*exp(-200*t)
y=y2+500*y1+60000*y0 % Summation of y and its derivatives
y =
900000

Using the expression for v t we can find the current as
C
dv 100 – 3 – 200t – 300t – 200t – 300t
C
i= i = i C C-------- = --------- u 10 = 6900e – 6600e = 115e – 110e A (1.29)
L
6
dt
1-10 Circuit Analysis II with MATLAB Applications
Orchard Publications