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The University of Texas at Austin
Department of Electrical and Computer Engineering
EE362K: Introduction to Automatic Control—Fall 2009
Solutions to Problem Set Eight
C. Caramanis
November 18, 2009.
1. The solution to the given diﬀerential equation is
x
(
t
) =
x
(0)
e
at
+
u
(
t
)
*
e
at
where
*
means convolution. Taking the Laplace transform of both sides gives
X
(
s
) =
x
(0)
s

a
+
U
(
s
)
s

a
One may note that the above relation can be readily found from the Laplace transform of the
original diﬀerential equation. To compute the transfer function, set
x
(0) = 0 to get
G
xu
=
X
(
s
)
U
(
s
)
=
1
s

a
s
=
a
is a pole of this transfer function.
u
(
t
) =
e
at
or equivalently
U
(
s
) =
1
s

a
gives
X
(
s
) =
x
(0)
s

a
+
1
(
s

a
)
2
Taking the inverse Laplace transform of both sides gives
x
(
t
) =
L

1
{
x
(0)
s

a
+
1
(
s

a
)
2
}
=
x
(0)
e
at
+
te
at
2. The transfer function of the system is given by
T
(
s
) =
C
(
sI

A
)

1
B
+
D
=
1
J
t
s
2

mgl
3. (a) Setting
u
(
t
) = 0 gives
y
(
n
)
+
a
1
y
(
n

1)
+
...
+
a
n
y
= 0
Substituting
y
(
t
) =
e
λt
in the above equation gives
(
λ
n
+
a
1
λ
n

1
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This note was uploaded on 02/10/2010 for the course EE 362K taught by Professor Friedrich during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Friedrich

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