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MASSACHUSETTS
INSTITUTE
OF
TECHNOLOGY
Department
of
Mechanical
Engineering
2.004
Dynamics
and
Control
II
Fall
2007
Problem
Set
#3
Solution
Posted:
Friday,
Sept.
28,
’07
1.
A
second–order
system
has
the
step
response
shown
below.
1
Determine
its
trans
fer
function.
0
0.5
1
1.5
2
t [sec]
Answer:
This
is
under–damped
2nd
order
system.
Starting
from
the
transfer
function
of
the
second
order
system
ω
2
A
n
,
s
2
+2
ζω
n
s
+
ω
2
we
have
to
decide
the
parameters
of
A
(constant),
ζ
(damping
ratio)
and
ω
n
(natural
frequency).
From
the
ﬁnal
value
theorem,
0
0.5
1
1.5
2
f(t) [a.u.]
1
Aω
2
lim
s
n
=
A
s
→
0
s
s
2
n
s
+
ω
n
2
1
a.u.
denotes
arbitrary
units;
its
use
appropriate
when
we
consider
a
function
that
does
not
correspond
to
any
particular
physical
quantity.
1
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²
³
´
µ
¶
·¸
and
the
steady
state
value
is
1
(from
the
given
ﬁgure).
Therefore,
A
=1
.
The
step
response
of
the
under–damped
second
order
system
is
1
−
ae
−
σ
d
t
cos (
ω
d
t
−
φ
)
u
(
t
)
,
where
σ
d
=
ζω
n
and
ω
d
=
ω
n
1
−
ζ
2
.
ζπ
From
the
lecture
note
7
(pp.
26),
%OS
=
exp
−
1
−
ζ
2
:
72%.
Thus
the
damping
ratio
ζ
≈
0
.
1.
To
get
the
natural
frequency,
we
choose
two
peak
points
at
t
1
=0
.
35
sec
and
t
2
.
95
sec.
The
cosine
term
will
be
1
at
the
peaks,
so
that
we
can
consider
exponential
decay
term
only.
f
(
t
1
)=1
−
ae
−
σ
d
t
1
.
72
f
(
t
2
−
ae
−
σ
d
t
2
.
4
Dividing
the
two
equations,
we
obtain
ae
−
σ
d
t
1
1
−
1
.
72
=
.
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This note was uploaded on 02/23/2012 for the course MECHANICAL 2.004 taught by Professor Derekrowell during the Fall '08 term at MIT.
 Fall '08
 DerekRowell
 Mechanical Engineering

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