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MASSACHUSETTS
INSTITUTE
OF
TECHNOLOGY
Department
of
Electrical
Engineering
and
Computer
Science
6.241:
Dynamic
Systems—Fall
2007
Homework
3
Solutions
Exercise
3.2
i)
We
would
like
to
minimize
the
2norm
of
u
,
i.e.,
u
2
.
Since
y
n
is
given
as
2
n
y
n
=
h
i
u
n
−
1
we
can
rewrite
this
equality
as
³
i
=1
⎤
⎡
y
n
=
h
1
h
2
h
n
···
⎢
⎢
⎢
⎣
u
n
−
1
u
n
−
2
.
.
.
u
0
⎥
⎥
⎥
⎦
want
to
±nd
the
u
with
the
smallest
2norm
such
that
y
¯ =
Au.
where
we
assume
that
A
has
a
full
rank
(i.e.
h
i
=
±
0
for
some
i
, 1
≤
i
≤
n
).
Then,
the
solution
reduces
to
the
familiar
form:
u
ˆ =
A
(
AA
)
−
1
¯
y.
n
h
2
i
By
noting
that
AA
=
,
we
can
obtain
ˆ
u
j
as
follows;
i
=1
u
ˆ
j
=
h
j
y
¯
n
h
2
,
i
=1
i
for
j = 0
,
1
,
,
n
−
1
.
ii)
a)
Let’s
introduce
e
as
an
error
such
that
y
n
=
y
−
e
.
It
can
also
be
written
as
y
−
y
n
=
e
.
Then
now
the
quantity
we
would
like
to
minimize
can
be
written
as
r
(
y
−
y
n
)
2
+
u
0
2
+
+
u
n
2
−
1
where
r
is
a
positive
weighting
parameter.
The
problem
becomes
to
solve
the
following
minimization
problem
:
n
u
ˆ =
arg
min
u
i
2
+
re
2
=
arg
min(
u
2
2
+
r e
2
2
)
,
u
u
i
=1
from
which
we
see
that
r
is
a
weight
that
characterizes
the
tradeoﬀ
between
the
size
of
the
±nal
error,
¯
y
−
y
n
,
and
energy
of
the
input
signal,
u
.
In
order
to
reduce
the
problem
into
the
familiar
form,
i.e,
y
−
Ax
,
let’s
augment
√
re
at
the
bottom
of
u
so
that
a
new
augmented
vector,
˜
u
is
⎤
⎡
u
⎦
⎣
u
˜ =
,
√
·
·
·
re
1
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This
choice
of
˜
u
follows
from
the
observation
that
this
is
the
˜
u
that
would
have
u
˜
2
2
=
u
2
2
+
re
2
,
the
quantity
we
aim
to
minimize
.
Now
we
can
write
y
as
follows
⎡
⎤
·
¸
u
y
=
A
.
.
.
1
⎣
⎦
=
A
˜
u
˜ =
Au
+
e
=
y
n
+
e.
√
r
√
·
·
·
re
Now,
ˆ
u
can
be
obtained
using
the
augmented
A
,
A
˜
,
as
u
ˆ =
A
˜
(
A
˜
A
˜
)
−
1
y
=
A
1
AA
+
1
y.
r
√
r
By
noting
that
n
1
¹
1
AA
+
=
h
i
2
+
,
r
r
i
=1
we
can
obtain
ˆ
u
j
as
follows
u
ˆ
j
=
º
n
h
j
h
y
2
1
for
j = 0
,
···
,
n
−
1
.
i
=1
i
+
r
ii)
b)
When
r
=
0,
it
can
be
interpreted
that
the
error
can
be
anything,
but
we
would
like
to
minimize
the
input
energy.
Thus
we
expect
that
the
solution
will
have
all
the
u
i
s
to
be
zero.
In
fact,
the
expression
obtained
in
ii)
a)
will
be
zero
as
r
0.
On
the
other
hand,
the
other
situation
→
is
an
interesting
case.
We
put
a
weight
of
∞
to
the
Fnal
state
error,
then
the
expression
from
ii)
a)
gives
the
same
expression
as
in
i)
as
r
→ ∞
.
Exercise
3.3
This
problem
is
similar
to
Example
3.4,
except
now
we
require
that
˙
p
(
T
) = 0.
t
can
derive,
from
x
(
t
) =
p
¨(
t
),
that
p
(
t
) =
x
(
t
)
∗
tu
(
t
) =
(
t
−
τ
)
x
(
τ
)d
τ
where
∗
denotes
convolution
0
and
u
(
t
)
is
the
unit
step,
deFned
as
1
when
t >
0
and
0
when
t <
0.
(One
way
to
derive
this
is
to
take
x
(
t
) =
p
t
)
to
the
Laplace
domain,
taking
into
account
initial
conditions
,
to
Fnd
the
transfer
function
H
(
s
) =
P
(
s
)
/X
(
s
),
hence
the
impulse
response,
h
(
t
)
such
that
p
(
t
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 Spring '11
 EmilioFrazzoli
 Computer Science, Electrical Engineering, Linear Algebra, Rank, Invertible matrix, Singular value decomposition

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