Page 43 - Circuit Analysis II with MATLAB Applications

Other Second Order Circuits

and by substitution of given numerical values into (1.85) through (1.87), we get

1
1
1
1
1
·
§ ------------------ + ------------------ + ------------------ v + 25 u 10 – 9 dv 1 ------------------v = ------------------v
-------- –
© 5 4 4 ¹ 1 dt 4 out 5 in
2 u 10 4 u 10 5 u 10 4 u 10 2 u 10
or
– 3 – 9 dv 1 – 4 – 5
0.05 u 10 v + 25 u 10 -------- – 0.25 u 10 v out = 0.5 u 10 v in (1.88)
1
dt
dv
out
–
4
v = – 5 u 10 ------------- (1.89)
1
dt
2
dv d v out
–
4
-------- = – 5 u 10 --------------- (1.90)
1
dt dt 2
Next, substitution of (1.89) and (1.90) into (1.88) yields
2
dv d v out
–
out ·
4
0.05 u 10 – 3 § – 5 u 10 ------------- + 25 u 10 – 9 – 5 u 10 – 4 --------------- (1.91)
© dt ¹ dt 2
– 0.25 u 10 – 4 v out = 0.5 u 10 – 5 v in
or
2
d v out dv out
–
4
–
7
– 125 u 10 – 13 --------------- – 0.25 u 10 ------------- – 0.25 u 10 – 4 v out = 10 v in
dt 2 dt
Division by 125– u 10 – 13 yields
2
d v out 3 dv out 6 5
---------------- + 2 u 10 ------------- + 2 u 10 v out = – 1.6 u 10 v in
dt 2 dt
or
2
d v out 3 dv out 6 6
---------------- + 2 u 10 ------------- + 2 u 10 v out = – 10 cos 6280t (1.92)
dt 2 dt
We use MATLAB to find the roots of the characteristic equation of (1.92).
syms s; y0=solve('s^2+2*10^3*s+2*10^6')

y0 =
[ -1000+1000*i]
[ -1000-1000*i]

that is,

1-31 Circuit Analysis II with MATLAB Applications
Orchard Publications