This preview shows page 1. Sign up to view the full content.
90
Chapter 15.
Solutions: Case Study: Blind Deconvolution: A Matter of Norm
∂F
(
b
e
,
f
)
∂f
±
q
=
m
X
i
=1
(
k
i±
+
e
i±
)(
k
iq
+
e
iq
)=((
K
+
E
)
T
(
K
+
E
))
±q
,
where outofrange entries in summations are assumed to be zero and
R
is a matrix
whose nonzero entries are components of
r
.So
g
=
∇
F
(
b
e
,
f
)=
·
D
2
b
e
−
F
T
r
(
K
+
E
)
T
r
¸
,
H
(
b
e
,
f
·
D
2
+
F
T
FR
T
+
F
T
(
K
+
E
)
R
+(
K
+
E
)
T
F
(
K
+
E
)
T
(
K
+
E
)
¸
.
The Newton direction is the solution to
H
(
b
e
,
f
)
p
=
−
g
.
CHALLENGE 15.3.
The least squares problem is of the form
min
x
±
Ax
−
b
±
2
,
where
x
=
·
Δ
b
e
Δ
f
¸
and
A
and
b
are the given matrix and vector. So to minimize
±
Ax
−
b
±
2
=(
Ax
−
b
)
T
(
Ax
−
b
), we set the derivative equal to zero, obtaining
A
T
Ax
−
A
T
b
=
0
.
The solution to this equation is a minimizer if the second derivative
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.
 Fall '11
 Dr.Robin

Click to edit the document details