Solutions to
Homework Set 2

MAE143B, Summer II, 2009
1.1 First, we multiply denominator and numerator of
G
(
s
)by
1
m
and
k
to reach the desired form
G
(
s
)=
1
k
k
m
s
2
+
d
m
s
+
k
m
We can write
d
m
as 2
q
k
m
1
2
d
√
mk
=
d
m
This way, by de±ning
K
=
1
k
,
ω
n
=
q
k
m
and
β
=
d
2
√
mk
we can obtain the desired form
G
(
s
K
ω
2
n
s
2
+2
βω
n
s
+
ω
2
n
1.2 As we learned in the lecture, impulse response of the
G
(
s
)is
y
(
t
G
(
t
). We will simply
take the inverse Laplace of
G
(
s
) by completing squares,
G
(
s
K
ω
2
n
(
s
+
n
)
2
+
ω
2
n
(1
−
β
2
)
We need to take inverse Laplace of this expression.
If we look at the Table A.2 at the
appendix A of the book, we see the following transformation:
L
−
1
{
b
(
s
+
a
)
2
+
b
2
}
=
e
−
at
sin(
bt
)
By writing
G
(
s
)as
G
(
s
Kω
n
p
1
−
β
2
ω
n
p
1
−
β
2
(
s
+
n
)
2
+
ω
2
n
(1
−
β
2
)
Now consider,
b
=
ω
n
p
1
−
β
2
and
a
=
n
. We can reach
G
(
t
)
G
(
t
n
p
1
−
β
2
e
−
n
t
sin(
ω
n
p
1
−
β
2
t
)
The frequency of the oscillation is
ω
d
=
ω
n
p
1
−
β
2
1.3 If
β
≥
1, then
G
(
s
) will not have imaginary poles anymore. It will have real distinct poles;
s
1
=
−
n
+
ω
n
p
β
2
−
1
,s
2
=
−
n
−
ω
n
p
β
2
−
1
Then, we can write
G
(
s
)a
s
G
(
s
C
1
s
−
s
1
+
C
2
s
−
s
2
,mean
s
G
(
t
) will be only exponential (no
sinusoid),
G
(
t
C
1
e
s
1
t
+
C
2
e
s
2
t
1
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View Full Document1.4 Because poles of T.F are on the left hand plane, we know that
G
(
s
) is stable, then we can
use Final Value Theorem,
lim
t
→∞
y
(
t
) = lim
s
→
0
sy
(
s
) = lim
s
→
0
s
1
s
2
+
ms
+
k

{z
}
G
(
s
)
1
s
{z}
input
=
1
k
1.5 We know;
u
(
s
)=
K
p
[
r
(
s
)
−
y
(
s
)]
and
y
(
s
G
(
s
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