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MASSACHUSETTS
INSTITUTE
OF
TECHNOLOGY
Department
of
Mechanical
Engineering
2.004
Dynamics
and
Control
II
Fall
2007
Problem
Set
#6
Solution
Posted:
Friday,
Oct.
26,
’07
1.
Starting
with
the
Supplement
to
Lecture
12
(“Supplement”
for
short,)
modify
the
time–domain
torque
balance
equation
(3)
on
the
motor
shaft
and
force
balance
equation
(5)
on
the
mass
to
include
the
compliances.
Answer:
Since
we
have
compliance
terms
in
the
system,
the
spring
torque
T
c
and
force
F
c
are
T
C
(
t
)
=
C
r
θ
(
t
)
,
F
C
(
t
)
=
C
t
x
(
t
)
.
Note
that
the
spring
torque
and
force
are
proportional
to
displacement
θ
(
t
)
and
x
(
t
)
respectively,
which
are
time
integrals
of
ω
(
t
)
and
v
(
t
).
Thus
equations
(3)
and
(5)
will
be
t
K
m
i
(
t
)
=
Jω
˙ (
t
) +
Dω
(
t
) +
C
r
ω
(
τ
)
dτ
+
T
p
(
t
)
,
0
t
F
p
(
t
)
=
Mv
˙(
t
) +
f
v
v
(
t
) +
C
t
v
(
τ
)
dτ.
0
2.
Laplace
transform
your
new
torque/force–balance
equations
and
repeat
the
steps
leading
to
equation
(13)
of
the
Supplement.
The
KVL
equation
should
be
identi
cal
to
(1–2)
since
we
are
neglecting
the
inductance,
as
we
did
in
the
Supplement.
Obtain
the
transfer
function
V
(
s
)
/V
s
(
s
)
of
the
compliant
system
without
feed
back.
After
substituting
C
r
= 2N
·
m
/
rad,
C
t
=
10N
/
m
and
the
remaining
numerical
values
from
the
Supplement,
your
transfer
function
should
be
0
.
3162
s
.
s
2
+ 2
s
+ 3
You
should
use
this
transfer
function
in
the
rest
of
the
problem
set,
even
if
you
can’t
get
your
algebra
from
question
1
to
match
it.
Answer:
Applying
the
Laplace
transform
to
the
previous
results,
we
obtain
Ω(
s
)
K
m
I
(
s
)
=
Js
Ω(
s
) +
D
Ω(
s
) +
C
r
+
T
p
(
s
)
,
s
V
(
s
)
F
p
(
s
)
=
MsV
(
s
) +
f
v
V
(
s
) +
C
t
.
s
1
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Using
the
above
two
relations,
we
obtain
the
Laplace
transform
of
the
current
i
(
t
) as
J
+
r
2
M
D
+
r
2
f
v
C
r
+
r
2
C
t
1
I
(
s
) =
s
+
+
Ω(
s
)
.
rK
m
rK
m
rK
m
s
Combining
this
result
with
KVL,
J
+
r
2
M
D
+
r
2
f
v
C
r
+
r
2
C
t
1
V
(
s
)
V
s
(
s
) =
s
+
+
RV
(
s
) +
K
v
,
rK
m
rK
m
rK
m
s
r
we
end
up
with
the
transfer
function
given
by
rK
m
s
V
(
s
)
R
(
J
+
r
2
M
)
=
.
V
s
(
s
)
2
D
+
r
2
f
v
+
K
m
K
v
/R
C
r
+
r
2
C
t
s
+
s
+
J
+
r
2
M
J
+
r
2
M
Substituting
the
numerical
values,
we
find
0
.
3162
s
G
(
s
) =
.
s
2
+ 2
s
+ 3
3.
Now
consider
the
complete
feedback
system,
which
includes
the
differential
ampli
fier
with
reference
voltage
input
and
velocity
feedback.
Write
expressions
for
the
open
loop
transfer
function
and
the
closed–loop
transfer
function
of
this
system
with
the
feedback
gain
K
as
a
variable.
Answer:
Open
loop
transfer
function
G
(
s
) is
0
.
3162
s
.
s
2
+ 2
s
+ 3
Closed
loop
transfer
function
is
0
.
3162
Ks
.
s
2
+ (2 + 0
.
3162
K
)
s
+ 3
4.
What
is
the
order,
type,
and
nature
(
i.e.
,
overdamped,
undamped,
etc.)
of
this
system?
Back
up
your
answer
with
a
sketch
of
the
open–loop
zeros
and
poles.
Answer:
Obviously
this
system
has
two
poles,
so
it
is
2nd
order
system.
The
number
of
pure
integrators
is
zero,
thus
it
is
Type
0.
(The
steady–state
error
is
non–zero
for
step
input
and
zero
for
ramp
input.)
The
open
loop
system
is
underdamped,
but
the
closed
loop
system
transitions
from
underdamped
to
overdamped
as
the
gain
K
increases
because
the
poles
are
p
1
,
2
=
−
(2 + 0
.
3162
K
)
±
(2 + 0
.
3162
K
)
2
−
12)
/
2
.
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 Fall '08
 DerekRowell
 Mechanical Engineering

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