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A Solution Manual and Notes for:
Kalman Filtering: Theory and Practice using MATLAB
by Mohinder S. Grewal and Angus P. Andrews.
John L. Weatherwax
∗
April 30, 2012
Introduction
Here you’ll find some notes that I wrote up as I worked through this excellent book. There
is also quite a complete set of solutions to the various end of chapter problems. I’ve worked
hard to make these notes as g ood as I can, but I have no illusions that they are perfect. If
you feel that that there is a better way to accomplish or explain an exercise or derivation
presented in t hese no t es; or that one or more of the explanations is unclear, incomplete,
or misleading, please tell me. If you find an error of any kind – technical, gra mmat ical,
typogr aphical, whatever – please tell me that, too. I’ll gladly add to the acknowledgments
in later printings the name of the first person to bring each problem to my att ention. I
hope you enjoy this book as much as I have and that these notes might help the further
development of your skills in Kalman filtering.
Acknowledgments
Special thanks to (most recent comments are listed fir st): Bobby Motwani and Shantanu
Sultan for finding var ious typos from the text. All comments (no matter how small) are
much appreciated. In fact, if you find these notes useful I would appreciate a contribution
in the form of a solution t o a problem that is not yet worked in these notes. Sort of a “take
a penny, leave a penny” type of approach. Remember: pay it forward.
∗
wax@alum.mit.edu
1
Chapter 2: Linear Dynamic Systems
Notes On The Text
Notes on Example 2.5
We are told that the fundament al solution Φ(t) to the differential equation
d
n
y
dt
= 0 when
written in companion for m as the matrix
dx
dt
= F x or in components
d
dt
x
1
x
2
x
3
.
.
.
x
n−2
x
n−1
x
n
=
0 1 0
0 0 1
0 0 0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 1 0
0 0 1
0 0 0
x
1
x
2
x
3
.
.
.
x
n−2
x
n−1
x
n
,
is
Φ(t) =
1 t
1
2
t
2
1
3!
t
3
···
1
(n−1)!
t
n−1
0 1 t
1
2
t
2
···
1
(n−2)!
t
n−2
0 0 1 t ···
1
(n−3)!
t
n−3
0 0 0 1 ···
1
(n−4)!
t
n−4
.
.
.
.
.
.
0 0 0 0 ··· 1
.
Note here the only nonzero values in the matrix F are the ones on its first superdiagonal.
We can verify this by showing that the given Φ(t) satisfies the differentia l equation and has
the correct initial conditions, t hat is
Φ(t)
dt
= F Φ(t) and Φ(0) = I. That Φ(t) has the correct
initial conditions Φ(0) = I is easy to see. For the t derivative of Φ(t) we find
Φ
′
(t) =
0 1 t
1
2!
t
2
···
1
(n−2)!
t
n−2
0 0 1 t ···
1
(n−3)!
t
n−3
0 0 0 1 ···
1
(n−4)!
t
n−4
0 0 0 0 ···
1
(n−5)!
t
n−5
.
.
.
.
.
.
0 0 0 0 ··· 0
.
From the above expressions for Φ(t) and F by considering the given product F Φ(t) we see
that it is equal to Φ
′
(t) derived above as we wanted to show. As a simple modification of
the above example consider what the fundamental solution would be if we were g iven the
following companion form for a vector of unknowns x
d
dt
ˆx
1
ˆx
2
ˆx
3
.
.
.
ˆx
n−2
ˆx
n−1
ˆx
n
=
0 0 0
1 0 0
0 1 0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
0 0 0
1 0 0
0 1 0
ˆx
1
ˆx
2
ˆx
3
.
.
.
ˆx
n−2
ˆx
n−1
ˆx
n
=
ˆ
F
ˆx
1
ˆx
2
ˆx
3
.
.
.
ˆx
n−2
ˆx
n−1
ˆx
n
.
Note in this example the only nonzero va lues in
ˆ
F are the ones on its first subdiagonal. To
determine Φ(t) we note that since this coefficient matrix
ˆ
F in this case is the transpose of
the first system considered above
ˆ
F = F
T
the system we are asking to solve is
d
dt
ˆ
x = F
T
ˆ
x.
Thus the fundamenta l solution to this new problem is
ˆ
Φ(t) = e
F
T
t
= (e
F t
)
T
= Φ(t)
T
,
and that this later matrix looks like
ˆ
Φ(t) =
1 0 0 0 ··· 0
t 1 0 0 ··· 0
1
2
t
2
t 1 0 ··· 0
1
3!
t
3
1
2
t
2
t 1 ··· 0
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
1
(n−1)!
t
n−1
1
(n−2)!
t
n−2
1
(n−3)!
t
n−3
1
(n−4)!
t
n−4
··· 1
.
Verification of the Solution to the Continuous Linear System
We are told that a solution to the continuous linear system with a time dependent companion
matrix F (t) is given by
x(t) = Φ(t)Φ(t
0
)
−1
x(t
0
) + Φ(t)
Z
t
t
0
Φ
−1
(τ)C(τ)u(τ)dτ . (1)
To verify this take the derivative of x(t) with respect to time. We find
x
′
(t) = Φ
′
(t)Φ
−1
(t
0
) + Φ
′
(t)
Z
t
t
0
Φ
−1
(τ)C(τ)u(τ)dτ + Φ(t)Φ
−1
(t)C(t)u(t)
= Φ
′
(t)Φ
−1
(t)x(t) + C(t)u(t)
= F (t)Φ(t)Φ
−1
(t)x(t) + C(t)u(t)
= F (t)x(t) + C(t)u(t) .
showing that the expression given in Equation 1 is indeed a solution. Note that in the above
we have used the fact that for a fundamental solution Φ(t) we have Φ
′
(t) = F (t)Φ(t).
Problem S olutions
Problem 2.2 (the companion matrix for
d
n
y
dt
n
= 0)
We begin by defining the following functions x
i
(t)
x
1
(t) = y(t)
x
2
(t) = ˙x
1
(t) = ˙y(t)
x
3
(t) = ˙x
2
(t) = ¨x
1
(t) = ¨y(t)
.
.
.
x
n
(t) = ˙x
n−1
(t) = ··· =
d
n−1
y(t)
dt
n−1
,
as the components of a state vector x. Then the companion form for this system is given by
d
dt
x(t) =
d
dt
x
1
(t)
x
2
(t)
.
.
.
x
n−1
(t)
x
n
(t)
=
x
2
(t)
x
3
(t)
.
.
.
x
n
(t)
d
n
y (t)
dt
n
=
0 1 0 0 ··· 0
0 0 1 0 ··· 0
0 0 0 1 ··· 0
.
.
.
1
0 . . . 0 0
x
1
(t)
x
2
(t)
.
.
.
x
n−1
(t)
x
n
(t)
= F x(t)
With F the companion matrix given by
F =
0 1 0 0 ··· 0
0 0 1 0 ··· 0
0 0 0 1 ··· 0
.
.
.
1
0 . . . 0 0
.
Which is of dimensions of n ×n.
Problem 2.3 (the companion matrix for
dy
dt
= 0 and
d
2
y
dt
2
= 0 )
If n = 1 the above specifies to the differential equation
dy
dt
= 0 and the companion matrix F
is the zero matrix i.e. F = [0]. When n = 2 we are solving the differential equation given
by
d
2
y
dt
2
= 0, and a companion matrix F given by
F =
0 1
0 0
.
Problem 2.4 (the fundamental solution matrix for
dy
dt
= 0 and
d
2
y
dt
2
= 0)
The fundamental solution matrix Φ(t) satisfies
dΦ
dt
= F (t)Φ(t) ,
with an initial condition Φ(0) = I. When n = 1, we have F = [0], so
dΦ
dt
= 0 giving that
Φ(t) is a constant, say C. To have the initial condition hold Φ(0) = 1, we must have C = 1,
so that
Φ(t) = 1 . (2)
When n = 2, we have F =
0 1
0 0
, so that the equation satisfied by Φ is
dΦ
dt
=
0 1
0 0
Φ(t) .
If we denote the matr ix Φ(t) into its components Φ
ij
(t) we have that
0 1
0 0
Φ(t) =
0 1
0 0
Φ
11
Φ
12
Φ
21
Φ
22
=
Φ
21
Φ
22
0 0
,
so the differential equations for the components of Φ
ij
satisfy
dΦ
11
dt
dΦ
12
dt
dΦ
21
dt
dΦ
22
dt
=
Φ
21
Φ
22
0 0
.
Solving the scalar differential equations above for Φ
21
and Φ
22
using the known initial con-
ditions for them we have Φ
21
= 0 and Φ
22
= 1. With these results the differential equations
for Φ
11
and Φ
12
become
dΦ
11
dt
= 0 and
dΦ
12
dt
= 1 ,
so that
Φ
11
= 1 and Φ
21
(t) = t .
Thus the fundamenta l solution matrix Φ(t) in the case when n = 2 is
Φ(t) =
1 t
0 1
. (3)
Problem 2.5 (the state transition matrix for
dy
dt
= 0 and
d
2
y
dt
2
= 0 )
Given the fundamental solution matrix Φ(t) for a linear system
dx
dt
= F (t)x the state transi-
tion matrix Φ(τ, t) is given by Φ(τ)Φ(t)
−1
. When n = 1 since Φ(t) = 1 the state tr ansition
matrix in this case is Φ(τ, t) = 1 also. When n = 2 since Φ(t) =
1 t
0 1
we have
Φ(t)
−1
=
1 −t
0 1
,
so that
Φ(τ)Φ(t)
−1
=
1 τ
0 1
1 −t
0 1
=
1 −t + τ
0 1
.
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