Page 44 - Circuit Analysis II with MATLAB Applications

Chapter 1 Second Order Circuits

s s = – D r jE = – 1000 r j1000 = 1000 – 1 r j1
,

1
2
We cannot classify the given circuit as series or parallel and therefore, we should not use the damping
ratio D S or D P . Instead, for the natural response v t we will use the general expression
n
s t s t – Dt
2
1
v t = Ae + Be = e k cos Et + k sin Et (1.93)
n
1
2
where
s s, 2 = – D r jE = – 1000 r j1000
1
Therefore, the natural response is oscillatory and has the form
v t = e – 1000t k cos 1000t + k sin 1000t (1.94)
2
1
n
Since the right side of (1.92) is a sinusoid, the forced response has the form
v t = k cos 6280t + k sin 6280t (1.95)
f
4
3
Of course, for the derivation of the forced response we could use phasor analysis but we must first
derive an expression for the impedance or admittance because the expressions we’ve used earlier are
valid for series and parallel circuits only.

The coefficients k 3 and k 4 will be found by substitution of (1.95) into (1.92) and then by equating
like terms. Using MATLAB we get:
syms t k3 k4; y0=k3*cos(6280*t)+k4*sin(6280*t); y1=diff(y0)

y1 =
-6280*k3*sin(6280*t)+6280*k4*cos(6280*t)
y2=diff(y0,2)

y2 =
-39438400*k3*cos(6280*t)-39438400*k4*sin(6280*t)

y=y2+2*10^3*y1+2*10^6*y0

y =
-37438400*k3*cos(6280*t)-37438400*k4*sin(6280*t)-
12560000*k3*sin(6280*t)+12560000*k4*cos(6280*t)

Equating like terms with (1.92) we get
6
˜
– 37438400 k + 12560000 k ˜ 4 cos 6280t = – 10 cos 6280t (1.96)
3
– 12560000 k ˜ 3 – 37438400 k ˜ 4 sin 6280t = 0
Simultaneous solution of the equations of (1.96) is done with MATLAB.

1-32 Circuit Analysis II with MATLAB Applications
Orchard Publications