ECE521
Linear Systems
Fall 2009
Homework 6 Solutions
Problem 1
:
Consider a continuous system defined by
A
and
B
,
A
=
0
1
0
0
0
0
1
0
0
0
0
1
1
1

3
4
,
B
=
1
0
0
0
0
0
0
1
Find two different matrices
K
so the eigenvalues of
A

BK
are

1
±
i
and

2
±
i
.
Let
B
= [
b
1
b
2
], and
k
T
1
and
k
T
2
be two row vectors. Note that
A

b
1
k
T
1
=
A

B
parenleftbigg
k
T
1
0
parenrightbigg
and
A

b
2
k
T
2
=
A

B
parenleftbigg
0
k
T
2
parenrightbigg
Hence, operating on
B
and
K
=
parenleftbigg
k
T
1
0
parenrightbigg
or B and
K
=
parenleftbigg
0
k
T
2
parenrightbigg
is the same as operating on
b
1
and
k
T
1
or
b
2
and
k
T
2
.
As long as (
A, b
i
) is controllable, we can place poles at any desired locations. This can be done
by transforming (
A, b
i
) to the cannonical form. Once we have the cannonical we need to choose
free parameters so they match the coefficients of the characteristic polynomial. In our case, the
desired polynomial is
(
λ

(

1 + i))(
λ

(

1

i))(
λ

(

2 + i))(
λ

(

2

i)) =
λ
4
+ 6
λ
3
+ 15
λ
2
+ 18
λ
+ 10
Note that (
A, b
1
) as well as (
A, b
2
) are controllable pairs. Indeed
Q
B
(
A, b
1
) =
1
0
0
0
0
0
0
1
0
0
1
4
0
1
4
13
is nonsingular and (
A, b
2
) is already in the cannonical form (which is always controllable).
(a) We first work with the pair (
A, b
1
). We need to transform the pair (
A, b
1
) to the cannonical
form, say (
ˆ
A,
ˆ
b
1
) where
ˆ
b
1
= [0 0 0 1]
T
. As
A
is in the cannonical form we must have
ˆ
A
=
A
.
We also know (Lecture 13) that
W
=
Q
R
(
A, b
1
)
Q
R
(
ˆ
A,
ˆ
b
1
)

1
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2
will transform (
A, b
1
) to (
ˆ
A,
ˆ
b
1
) where
W

1
AW
=
ˆ
A
and
W

1
b
1
=
ˆ
b
1
From the latter equations we see that the last column of
W
is equal to
b
1
. In addition as
ˆ
A
=
A
the first equation show that
AW
=
WA
.
Let
A
= [
a
1
a
2
a
3
a
4
] and
W
= [
w
1
w
2
w
3
w
4
].
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 Spring '10
 123
 Eigenvalue algorithm, QR algorithm, qo, cannonical form

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