EE263s Summer 200910
Laurent Lessard
EE263s homework 3
1.
Linearizing range measurements.
Consider a single (scalar) measurement
y
of the distance or range of
x
∈
R
n
to a fixed point or beacon at
a
,
i.e.
,
y
=
bardbl
x
−
a
bardbl
.
(a) Show that the linearized model near
x
0
can be expressed as
δy
=
k
T
δx
, where
k
is the unit vector
(
i.e.
, with length one) pointing from
a
to
x
0
. Derive this analytically, and also draw a picture
(for
n
= 2) to demonstrate it.
(b) Consider the error
e
of the linearized approximation,
i.e.
,
e
=
bardbl
x
0
+
δx
−
a
bardbl − bardbl
x
0
−
a
bardbl −
k
T
δx.
The relative error of the approximation is given by
η
=
e/
bardbl
x
0
−
a
bardbl
. We know, of course, that the
absolute value of the relative error is very small provided
δx
is small. In many specific applications,
it is possible and useful to make a stronger statement, for example, to derive a bound on how
large the error can be. You will do that here. In fact you will prove that
0
≤
η
≤
α
2
2
where
α
=
bardbl
δx
bardbl
/
bardbl
x
0
−
a
bardbl
is the relative size of
δx
. For example, for a relative displacement of
α
= 1%, we have
η
≤
0
.
00005,
i.e.
, the linearized model is accurate to about 0
.
005%. To prove
this bound you can proceed as follows:
•
Show that
η
=
−
1 +
radicalbig
1 +
α
2
+ 2
β
−
β
where
β
=
k
T
δx/
bardbl
x
0
−
a
bardbl
.
•
Verify that

β
 ≤
α
.
•
Consider the function
g
(
β
) =
−
1 +
radicalbig
1 +
α
2
+ 2
β
−
β
with

β
 ≤
α
.
By maximizing and
minimizing
g
over the interval
−
α
≤
β
≤
α
show that
0
≤
η
≤
α
2
2
.
Solution.
(a) For the linearized model we have
δy
=
parenleftbigg
∂y
∂x
parenrightbigg
δx
so all we have to do is to compute the matrix
∂y/∂x
. Since
y
=
bardbl
x
−
a
bardbl
we have
y
2
= (
x
−
a
)
T
(
x
−
a
)
and differentiating both sides with respect to
x
gives
2
∂y
∂x
y
= 2(
x
−
a
)
T
and therefore
∂y
∂x
=
(
x
−
a
)
T
y
=
(
x
−
a
)
T
bardbl
x
−
a
bardbl
,
so
δy
=
k
T
δx
with
k
= (
x
−
a
)
/
bardbl
x
−
a
bardbl
. Clearly,
k
points from
a
to
x
and is of length one since
k
T
k
=
(
x
−
a
)
T
(
x
−
a
)
bardbl
x
−
a
bardbl
2
= 1
.
1
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a
k
bardbl
x
0
−
a
bardbl
bardbl
x
−
a
bardbl
x
0
δx
k
T
δx
x
(b)
•
First we show that
η
=
−
1 +
radicalbig
1 +
α
2
+ 2
β
−
β
where
β
=
k
T
δx/
bardbl
x
0
−
a
bardbl
. Note that
e
=
bardbl
x
0
+
δx
−
a
bardbl − bardbl
x
0
−
a
bardbl −
k
T
δx
=
bardbl
x
0
−
a
bardbl
parenleftbigg
bardbl
x
0
−
a
bardbl
x
0
−
a
bardbl
+
δx
bardbl
x
0
−
a
bardbl
bardbl −
1
−
k
T
δx
bardbl
x
0
−
a
bardbl
parenrightbigg
,
and after dividing both sides by
bardbl
x
0
−
a
bardbl
and using
k
= (
x
0
−
a
)
/
bardbl
x
0
−
a
bardbl
,
β
=
k
T
δx/
bardbl
x
0
−
a
bardbl
and
η
=
e/
bardbl
x
0
−
a
bardbl
we get
η
=
bardbl
k
+
δx
bardbl
x
0
−
a
bardbl
bardbl −
1
−
β.
(1)
But
bardbl
k
+
δx
bardbl
x
0
−
a
bardbl
bardbl
=
radicalBigg
parenleftbigg
k
+
δx
bardbl
x
0
−
a
bardbl
parenrightbigg
T
parenleftbigg
k
+
δx
bardbl
x
0
−
a
bardbl
parenrightbigg
=
radicalBigg
bardbl
k
bardbl
2
+ 2
k
T
δx
bardbl
x
0
−
a
bardbl
+
bardbl
δx
bardbl
2
bardbl
x
0
−
a
bardbl
2
.
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 Summer '08
 BOYD,S
 Linear Algebra, Rank

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