Characteristics Equations,
Overdamped-, Underdamped-, and
Critically Damped Circuits
Kevin D. Donohue, University of Kentucky
1
In previous work, circuits were limited to one energy
storage element, which resulted in first-order
differential equations. Now, a second independent
energy storage element will be added to the circuits to
result in second order differential equations:
d 2x
dx
y (t ) =
+ a1
+ a2 x
dt
dt 2
Kevin D. Donohue, University of Kentucky
2
Find the differential equation for the circuit below in
terms of vc and also terms of iL
iL(t)
R
L
C
vs(t)
Show:
d 2 vc
dv
v s (t ) = LC
+ RC c + vc
dt
dt 2
→
v s (t ) d 2 vc R dvc
1
=
+
+
vc
2
LC
L
dt
LC
dt
t
v s (t ) = L
di L
1
+ Ri L +
dt
C
ò
+
vc(t)
_
t
i L (τ )dτ
−∞
→
v s (t ) di L R
1
=
+ iL +
L
dt
L
LC
Kevin D. Donohue, University of Kentucky
ò i (τ )dτ
L
−∞
3
Find the differential equation for the circuit below in
terms of vc and also terms of iL
iL(t)
R
is(t)
Show:
is (t ) = LC
d 2iL
dt
2
L diL
+
+ iL
R dt
C
+
vc(t)
_
→
is (t ) d 2iL
1 diL
1
= 2 +
+
iL
LC
RC dt LC
dt
L
t
is (t ) = C
dvc 1
1
+ vc +
dt
R
L
ò
t
vc (τ )dτ
−∞
→
is (t ) dvc
1
1
=
+
vc +
C
dt
RC
LC
Kevin D. Donohue, University of Kentucky
ò v (τ )dτ
c
−∞
4
Ø
Ø
The method for determining the forced solution is the
same for both first and second order circuits. The
new aspects in solving a second order circuit are the
possible forms of natural solutions and the
requirement for two independent initial conditions to
resolve the unknown coefficients.
In general the natural response of a second-order
system will be of the form:
x(t ) = K1t m exp(− s1t ) + K 2 exp( − s2t )
Kevin D. Donohue, University of Kentucky
5
Ø
Find characteristic equation from homogeneous
equation:
2
0=
Ø
dt 2
+ a1
dx
+ a2 x
dt
Convert to polynomial by the following substitution:
d nx
n
s =
n
dt
Ø
d x
to obtain 0 = s 2 + a1s + a2
Based on the roots of the characteristic equation, the
natural solution will take on one of three particular
forms. Roots given by:
− a ± a 2 − 4a
s1,2 =
1
Kevin D. Donohue, University of Kentucky
1
2
2
6
If roots are real and distinct ( a12 − 4a2 > 0
solution becomes
xn (t ) = a1 exp( s1t ) + a2 exp( s 2t )
), natural
a 1*e xp(-t)+e xp(-2t) for a 1 =[-3:2]
3
2.5
2
a1=2
1.5
re s pons e
Ø
1
0.5
0
-0.5
-1
a1=-3
-1.5
-2
0
0.5
1
1.5
Kevin D. Donohue, University of Kentucky
2
time
2.5
3
3.5
4
7
Ø
If roots are real and repeated (
natural solution becomes
xn (t ) = a1 exp( s1t ) + a2t exp( s1t )
a12 − 4a 2 = 0
→ s1 = s2
),
a 1*e xp(-t)+t*e xp(-t) for a 1 =[-3:2]
2
a1=2
1.5
1
re s pons e
0.5
0
-0.5
-1
a1=-3
-1.5
-2
-2.5
-3
0
0.5
1
1.5
Kevin D. Donohue, University of Kentucky
2
time
2.5
3
3.5
4
8
Ø
If roots are complex ( a12 − 4a2 < 0
becomes: (s1, 2 = α ± jβ )
), natural solution
xn (t ) = exp(αt )[c1 cos( β t ) + c2 sin( β t )]
or
e xp(-t).*(cos (6*pi*t)+2*s in(6*pi*t))
2.5
xn (t ) = A exp(α t ) cos( β t + θ )
2
1.5
−1 æ c2
ö
ç
θ = − tan ç ÷÷
è c1 ø
c1 = A cos(θ )
c2 = − A sin(θ )
1
re s pons e
A = c12 + c22
0.5
0
-0.5
-1
-1.5
-2
0
0.5
1
Kevin D. Donohue, University of Kentucky
1.5
2
time
2.5
3
3.5
4
9
Ø
Find the unit step response for vc and iL for the circuit
below when:
i (t)
L
R
a) R=16Ω
Ω, L=2H, C=1/24 F
b) R=10Ω
Ω, L=1/4H, C=1/100 F
c) R=2Ω
Ω, L=1/3H, C=1/6 F
Show:
3
æ 1
ö
vc (t ) = ç1 + exp(−6t ) − exp( −2t ) ÷u (t )
2
è 2
ø
a)
b) vc (t ) = (1 − exp(−20t ) − 20t exp(−20t ))u(t )
c) vc (t ) = (1 − exp(−3t )(cos(3t ) + sin(3t )))u(t )
L
vs(t)
+
vc(t)
_
C
1
(exp(−2t ) − exp(−6t ))u (t )
8
iL (t ) = (4 exp(−20t ) )u (t )
iL (t ) =
iL (t ) = (exp(−3t ) sin(3t ) )u (t )
Kevin D. Donohue, University of Kentucky
10