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Massachusetts
Institute
of
Technology
Department
of
Computer
Science
and
Electrical
Engineering
6.801/6.866
Machine
Vision
Handed
out:
2004
Sep.
23
Due
on:
2004
Sep.
30
Problem
1:
Suppose
that
the
image
in
a
patch
D
moves
with
velocity
(u,
v)
. To
find
the
bestfit
image
velocity
we
minimize
1
I
=
(uE
x
+
vE
y
+
E
t
)
2
dx
dy
2
D
by
choosing
u
and
v
.
This
is
a
straightforward
calculus
problem—the
stationary
points
of
I
can
be
found
by
solving
the
pair
of
equations
obtained
by
setting
the
derivatives
of
I
with
respect
to
u
and
v
equal
to
zero.
Let
the
solution
so
obtained
be
(u,
v)
=
(u
0
, v
0
)
.
There
usually
is
a
significant
amount
of
measurement
noise
in
real
images,
and
the
estimate
of
the
velocity
will
be
affected
by
this
noise.
We
need
to
determine
how
quickly
I
changes
as
we
move
away
from
the
minimum
at
(u
0
, v
0
)
in
order
to
determine
how
certain
the
location
of
the
minimum
is.
The
solution
is
dependable
if
I
changes
rapidly
as
we
move
away
from
the
minimum,
since
noise
will
then
have
less
of
an
effect
on
the
estimation
of
the
location
of
the
minimum.
(a)
Let
I
denote
the
second
partial
derivative
of
I
in
a
direction
that
lies
an
angle
θ
clockwise
from
the
u
axis.
Show
that
I (θ)
=
cos
2
θ
E
2
+
2
sin
θ
cos
θ
E
x
E
y
+
sin
2
θ
E
y
2
,
x
D
D
D
or
I (θ)
=
(E
x
cos
θ
+
E
y
sin
θ)
2
dx
dy.
D
Conclude
that
I
≥
0
for
all
θ
,
and
that
the
stationary
point
consequently
must
be
a
minimum
(as
opposed
to
a
maximum
or
saddle).
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 Spring '04
 BertholdHorn
 Critical Point, Trigraph, Stationary point, EY, dx dy

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