This preview shows pages 1–3. Sign up to view the full content.
MEEN 364
Notes from Session on October 6, 2004
304 Fermier Hall, 6:00~7:00 pm
1
)
Consider the electromechanical system shown below. The input to the system is the voltage,
‘e(t)’ and the output of the system is the displacement of the block of mass, ‘m’. The voltage
drop across the core and the force developed by the core are proportional to the current, ‘i
2
(t)’
and are given by the relations,
)
(
)
(
2
t
i
k
t
e
b
b
=
and
)
(
)
(
2
t
i
k
t
F
t
=
, where k
b
and k
t
are constants.
a)
Derive the governing differential equations of motion for this system.
b)
Represent the derived equations in statespace form.
k
b
i
1
(t)
i
2
(t)
R
F(t)
e(t)
c
e
b
(t)
C
o
r
e
LOOP 1
LOOP 2
L
m
Writing the voltage balance equation for LOOP 1, we get
).
(
)
(
)
(
1
t
e
t
V
t
Ri
c
=
+
(
1
)
Similarly writing the voltage balance equation for LOOP 2, we get
.
0
)
(
)
(
)
(
,
0
)
(
)
(
)
(
2
2
2
=
−
+
⇒
=
−
+
t
V
t
i
k
dt
t
di
L
t
V
t
e
dt
t
di
L
c
b
c
b
(
2
)
Free body diagram of the block of mass, ‘m’
)
(
t
kx
)
(
t
x
b
&
)
(
t
F
Writing the Newton’s second law of motion, we get
),
(
)
(
)
(
)
(
t
F
t
kx
t
x
b
t
x
m
=
+
+
&
&
&
).
(
)
(
)
(
)
(
2
t
i
k
t
kx
t
x
b
t
x
m
t
=
+
+
⇒
&
&
&
(
3
)
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentMEEN 364
Notes from Session on October 6, 2004
304 Fermier Hall, 6:00~7:00 pm
Equations (1), (2) and (3) represent the governing differential equations of motion.
Statespace Representation
Let the states of the system be chosen as
).
(
)
(
),
(
)
(
),
(
)
(
),
(
)
(
4
.
3
2
1
2
t
x
t
x
t
x
t
x
t
x
t
V
t
x
t
i
c
=
=
=
=
Substituting the above relations in the differential equations we get
()
).
(
)
(
)
(
)
(
),
(
)
(
,
)
(
)
(
)
(
1
)
(
)
(
)
(
1
)
(
)
(
1
)
(
),
(
1
)
(
)
(
4
3
1
4
.
4
3
.
1
2
2
2
1
2
.
2
1
1
.
t
x
m
b
t
x
m
k
t
x
m
k
t
x
t
x
t
x
t
x
R
t
x
R
t
e
C
t
i
R
t
V
R
t
e
C
t
i
t
i
C
t
x
t
x
L
t
x
L
k
t
x
t
c
b
−
−
=
=
⎟
⎠
⎞
⎜
⎝
⎛
−
−
=
⎟
⎠
⎞
⎜
⎝
⎛
−
−
=
−
=
+
−
=
Therefore representing the above equations in matrix form, we get
).
(
0
0
1
0
)
(
)
(
)
(
)
(
0
1
0
0
0
0
0
1
1
0
0
1
)
(
)
(
)
(
)
(
4
3
2
1
4
.
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Staff
 Controls

Click to edit the document details