Page 46 - Circuit Analysis II with MATLAB Applications

Chapter 1 Second Order Circuits

or
dv
out
v = – 5 u 10 – 4 -------------
1
dt
and since v = v 0 = 0 , it follows that
0
1
C1
dv
out
------------- = 0 (1.100)
dt
t = 0
The last step in finding the constant k 2 is to differentiate (1.99), evaluate it at t = 0 , and equate it
with (1.100). This is done with MATLAB as follows:
y0=exp( 1000*t)*( 0.024*cos(1000*t)+k2*sin(1000*t))...
+0.024*cos(6280*t) 0.008*sin(6280*t);
y1=diff(y0)
y1 =
-1000*exp(-1000*t)*(-3/125*cos(1000*t)+k2*sin(1000*t))+exp(-
1000*t)*(24*sin(1000*t)+1000*k2*cos(1000*t))-3768/
25*sin(6280*t)-1256/25*cos(6280*t)

or
dv 1000t – 3
out
------------- = – 1000e – § --------- cos 1000t + k sin 1000t ·
2
dt © 125 ¹
+ e – 1000t 24sin 1000t + 1000k cos 1000t
2
3768 1256
– ------------ sin 6280t – ------------ cos 6280t

25 25
and

dv – 3 1256
out
§
·
------------- = – 1000 --------- + 1000k – ------------ (1.101)
2
dt © 125¹ 25
t = 0
Simplifying and equating (1.100) with (1.101) we get
1000k – 26.24 = 0
2
or
k = 0.026
2
and by substitution into (1.99),

v out t = e – 1000t – 0.024cos 1000t + 0.026sin 1000t (1.102)
+ 0.024cos 6280t 0.008sin 6280t
–

1-34 Circuit Analysis II with MATLAB Applications
Orchard Publications