1
MECHATRONICS LAB
ME 140L
SECONDORDER CIRCUIT ANALYSIS
RECALL ME324: MECHANICAL SYSTEMS—2
nd
ORDER SYSTEM
An ideal massspringdamper system
Constitutive equations:
General
Linear
Energy
Mass:
p=p(V)
p=MV
2
2
MV
2M
2
p
=
Damper:
F=F(V)
F=BV
P=BV
2
Spring:
F=F(x)
F=kx
2
2
kx
2k
2
F
=
Compatibility/Continuity:
m
B
m
B
x
x
x
and
x
x
x
k
k
&
&
&
=
=
=
=
M
F
x
M
k
x
M
B
x
kx
x
B

F
x
M
x
m
F
:
Law)
s
(Newton'
Motion
of
Equation
=
+
+
⇒
−
=
=
∑
&
&
&
&
&
&
&
&
The equation of motion for all secondorder linear systems has the form:
frequency
atural
n
n
ω
ratio,
amping
d
ξ
able
input vari
u
,
interest
of
variable
x
:
where
M
B
and
M
k
n
2
n
2
n
n
ω
2
ξ
ω
u
x
ω
x
ω
2
ξ
x
=
=
=
=
=
→
=
=
+
+
&
&
&
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2
M
F
x
M
k
x
M
B
x
=
+
+
&
&
&
By solving this ODE given F, one can characterize the motion of the system.
•
Can change M, B, k to alter the motion of the system
•
Special cases—consider:
a.
No damping—B=0 (no energy dissipation)
M
F
x
M
k
x
=
+
→
&
&
What is frequency of oscillation?
→
=
M
k
n
ω
M
F
x
2
n
ω
x
=
+
&
&
Note: solution is sinusoidal:
t
ω
sin
A
x(t)
n
=
b.
No spring—k=0
ODE)
order

st
(1
M
F
v
M
B
v
x
B
x
=
+
=
+
→
&
&
&
&
Note: solution is exponential:
)

e

A(1
v(t)
α
t
=
Unit step responses to the above two cases:
Step Response
Time (sec)
Amplitude
0
5
10
15
20
25
30
35
40
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
0
1
2
3
4
5
6
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Step Response
Time (sec)
Amplitude
No damper
No spring