Course Notes of ELEC5600
Discrete-Time Linear System Theory
(Internal use only. Do not circulate)
Li Qiu
2018
ii
Contents
1 Basic Concepts
1.1 Signals and Systems . . . . . . . . . . . . . . . .
1.2 Linearity . . . . . . . . . . . . . . . . . . . . . . .
1.3 Causality . . . . . . . . . . . . . . . . . . . . . .
1.4 Time-invariance . . . . . . . . . . . . . . . . . . .
1.5 Stability . . . . . . . . . . . . . . . . . . . . . . .
1.6 Z-Transform and Transfer Functions . . . . . . .
1.7 Elementary System Connections and Operations
1.8 Linear Fractional Transformation . . . . . . . . .
1.9 State Space Systems . . . . . . . . . . . . . . . .
2 LTI
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
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State Space System Analysis
Solution of State Space Equations . . . . . . . . . . . . .
Controllability . . . . . . . . . . . . . . . . . . . . . . . .
Observability . . . . . . . . . . . . . . . . . . . . . . . . .
Kalman Decomposition . . . . . . . . . . . . . . . . . . .
Poles and Zeros . . . . . . . . . . . . . . . . . . . . . . . .
Internal Stability . . . . . . . . . . . . . . . . . . . . . . .
Simple State Space System Connections and Duality . . .
Linear Fractional Transformation of State Space Systems
Controllability indices and Observability indices . . . . . .
3 Realization
3.1 Elementary Realizations .
3.2 Minimal Realizations . . .
3.3 Minimal Realizations from
3.4 Balanced Realizations . .
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Partial Fractional Expansions .
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4 Elementary Feedback Control
4.1 State Feedback . . . . . . . . . . . .
4.2 State Observer . . . . . . . . . . . .
4.3 Observer-Based Controller . . . . . .
4.4 The Set of All Stabilizing Controllers
4.5 General Feedback Systems . . . . . .
4.6 Eigenstructure Placement . . . . . .
iii
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1
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. 19
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23
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iv
CONTENTS
5 System Performance
5.1 H2 Norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 H∞ norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3 Optimal Control Problems . . . . . . . . . . . . . . . . . . . . .
6 An
6.1
6.2
6.3
101
. 101
. 107
. 109
Input-Output Stabilization Theory
113
Coprime Factorization . . . . . . . . . . . . . . . . . . . . . . . . 113
All Stabilizing Controllers Revisited . . . . . . . . . . . . . . . . 119
Graph of Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
7 Riccati Equations and Inequalities
7.1 Generalized Eigenproblem . . . . . .
7.2 Algebraic Riccati Equations . . . . .
7.3 The Stabilizing Solution to the ARE
7.4 Definite AREs . . . . . . . . . . . .
7.5 Riccati inequalities . . . . . . . . . .
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125
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135
138
8 Applications of Riccati Equation
8.1 Introduction . . . . . . . . . . . . . . . . . . . .
8.2 Inner and Weighted All-pass Transfer Functions
8.3 Normalized Coprime Factorization . . . . . . .
8.4 Spectral Factorization . . . . . . . . . . . . . .
8.5 Inner-Outer Factorization . . . . . . . . . . . .
8.6 Bounded Real Lemma . . . . . . . . . . . . . .
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139
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152
153
9 H2 Optimization
9.1 Simple H2 Optimization . . . . . . . . . .
9.2 General H2 Optimization . . . . . . . . .
9.3 Linear Quadratic Regulator Problem . . .
9.4 Full Information H2 Optimization . . . .
9.5 H2 Filtering Problem . . . . . . . . . . . .
9.6 General H2 Optimization Revisited — the
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Separation Principle
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10 H∞ Optimization
10.1 Simple H∞ Optimization . .
10.2 Full Information H∞ Control
10.3 H∞ Filtering Problem . . . .
10.4 General H∞ Optimization . .
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175
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A Mathematical Preliminaries
A.1 Algebraic Structures . . . . . . . . . . . . . . . .
A.1.1 Groups, Rings, and Fields . . . . . . . . .
A.1.2 Equivalence Relations . . . . . . . . . . .
A.1.3 Partial Order . . . . . . . . . . . . . . . .
A.1.4 Linear Spaces and Linear Transformation
A.1.5 Normed Linear Space . . . . . . . . . . .
A.1.6 Inner Product Space . . . . . . . . . . . .
A.2 Matrix Analysis . . . . . . . . . . . . . . . . . . .
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185
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CONTENTS
A.2.1
A.2.2
A.2.3
A.2.4
v
Matrix Operations . . . . . . . . . . . .
Matrix Eigenvalue Problem . . . . . . .
Matrix Factorizations . . . . . . . . . .
Real Symmetric and Hermitian Matrices
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196
197
203
205
vi
CONTENTS
Chapter 1
Basic Concepts
1.1
Signals and Systems
The readers are assumed to be familiar with the notion of a linear (or vector)
space over a field F. For us F is either the real field R or the complex field C.
The most common examples of linear spaces are Fn and Fm×n .
A signal of n-dimension is a bilateral sequence
x = {. . . , x(−2), x(−1), |x(0), x(1), x(2), . . .}
where x(k) ∈ Fn . The underlying time axis is the set of integers Z. The vertical
line is used to indicate the position of time zero. It is often convenient to write
x as an infinite column vector
..
.
x(−1)
[x] =
x(0) .
x(1)
..
.
The set of all signals is denoted by ℓn (Z). Clearly ℓn (Z) is a linear space
(with infinite dimension).
Very often we will deal with unilateral signals
x = {x(0), x(1), . . .}
where x(k) ∈ Fn . The set of such signals is denoted by ℓn (Z+ ). Clearly, ℓn (Z+ )
is also a linear space. It can be considered as a subspace of ℓn (Z) if we identify
{x(0), x(1), . . .}
with
{. . . , 0, 0, |x(0), x(1), . . .}.
Similarly, the set of negative unilateral signal
x = {. . . , x(−2), x(−1)}
1
2
CHAPTER 1. BASIC CONCEPTS
is denoted by ℓn (Z− ), which can also be considered as a subspace of ℓn (Z) by
identifying
{. . . , x(−2), x(−1)}
with
{. . . , x(−2), x(−1), |0, 0, . . .}.
It can be shown that
ℓn (Z− ) ∩ ℓn (Z+ ) = {0}
and
ℓn (Z− ) + ℓn (Z+ ) = ℓn (Z).
We say that ℓn (Z− ) and ℓn (Z+ ) are complementary to each other and write
ℓn (Z− ) ⊕ ℓn (Z+ ) = ℓn (Z).
A system is a map from a subset of ℓm (Z) to ℓp (Z). If G is such a system
we write
G : ℓm (Z) → ℓp (Z).
We often represent a system using a block diagram shown in Figure 1.1. The set
u-
y-
G
Figure 1.1: The block diagram of a system
of u ∈ ℓm (Z) such that Gu is defined is called the domain of G and is denoted
by D(G). The set GD(G) is called the range of G and is denoted by R(G).
The set of pairs (u, Gu) for u ∈ D(G) is called the graph of G and is denoted
by G(G).
Example 1.1
A multiplier M is a system from ℓ2 (Z) to ℓ(Z) defined by
u(−1)
u(0)
u(1)
M ...,
,
,
,...
v(−1)
v(0)
v(1)
= {. . . , u(−1)v(−1), |u(0)v(0), u(1)v(1), . . .}.
Clearly, D(M ) = ℓ2 (Z) and R(M ) = ℓ(Z).
Example 1.2
A moving average system G : ℓm (Z) → ℓp (Z) is defined as
y(k) =
∞
X
G(k, l)u(l)
l=−∞
where G(k, l) ∈ Fp×m . For this system,
m
D(G) = {u ∈ ℓ (Z) :
∞
X
l=−∞
G(k, l)u(l) converges for all k ∈ Z}.
1.2. LINEARITY
3
Here G : Z × Z → Fp×m is called the kernel function of the system. An impulse
is a signal δk of the form
δk = {. . . 0, 0, 1, 0, 0, . . .}
where the only 1 is at time k. Let ei , 1 ≤ i ≤ m, be vectors in Fm whose
elements are all zeros except the i-th one which is 1. Let the input u be an
impulse ei δl . Then the output y = Gu at time k is equal to the i-th column of
G(k, l). Therefore G is also called the impulse response function. If for each k,
there are only a finite number of nonzero G(k, l), then G is said to be FIR (finite
impulse response), otherwise it is said to be IIR (infinite impulse response).
Example 1.3
Let cn+ be the set of all signals x such that limk→∞ x(k) exists. A system G
can be define on cn+ by
Gx = {. . . , 0, | lim x(k), 0 . . .}.
k→∞
For this system D(G) = cn+ ⊂ ℓ(Z), R(G) = {x ∈ ℓ(Z) : x(k) = 0 for all k ̸= 0}.
1.2
Linearity
A system G is said to be linear if for all u, v ∈ D(G) and α, β ∈ F, we have
αu + βv ∈ D(G) and G(αu + βv) = αGu + βGv.
Clearly, the multiplier M is not a linear system, whereas a moving average
system is. The system in Example 1.3 is also a linear system.
Almost all practically interesting linear systems have the form of a moving
average system. (Linear systems not of the form of a moving average system do
exist, e.g., Example 1.3.) We will be only interested in moving average systems
in the following. A moving average system G has a matrix representation
..
..
..
.
.
.
· · · G(−1, −1) G(−1, 0) G(−1, 1) · · ·
[G] =
G(0, 0)
G(0, 1) · · ·
· · · G(0, −1)
· · · G(1, −1)
G(1, 0)
G(1, 1) · · ·
..
..
..
.
.
.
where G is the impulse response function. Here a box is used to identify the
(0, 0)-th elements of the matrix. In this case, y = Gu can be obtained as the
usual matrix multiplication [y] = [G][u].
One of the simplest example of linear systems is the unit delay (forward
shift) system S : ℓm (Z) → ℓm (Z) given by
S{. . . , u(−1), |u(0), u(1), . . .} = {. . . , |u(−1), u(0), u(1), . . .}.
4
CHAPTER 1. BASIC CONCEPTS
The matrix representation of S is
..
.
..
. 0
[S] =
I
0
I
0
..
.
..
.
.
It is easy to see that S is bijective from ℓm (Z) to ℓm (Z). Hence it must have
an inverse. The inverse S −1 is the unit advance (backward shift) system
S −1 {. . . , u(−1), |u(0), u(1), . . .} = {. . . , u(−1), u(0), |u(1), . . .}.
The matrix representation of S −1 is
.. ..
.
.
0
−1
[S ] =
I
0
I
.
..
.
0
..
.
Another simple example of linear systems is the truncation P k , defined by
P k {. . . , u(k − 1), u(k), u(k + 1) . . .} = {. . . , u(k − 1), u(k), 0 . . .}.
The matrix representation of P k is
..
.
I
I
[P k ] =
0
..
.
where the last I is at the (k, k)-th position. We can easily see that P k satisfies
P 2k = P k . Such a map is called an idempotent.
A different truncation Qk can be defined as
Qk {. . . , u(k − 1), u(k), u(k + 1), . . .} = {. . . , 0, u(k), u(k + 1), . . .}.
The matrix representation of Qk is
..
.
0
I
[Qk ] =
I
..
.
1.3. CAUSALITY
5
where the first I is at the (k, k)-th position. We also have Q2k = Qk , i.e., Qk is
an idempotent.
It is easy to see that P k−1 + Qk = I. In particular, P −1 will be denoted by
P − and Q0 will be denoted by P + . We have P − ℓn (Z) = ℓn (Z− ), P + ℓn (Z) =
ℓn (Z+ ) and
[P − ] =
1.3
..
.
,
I
0
0
..
[P + ] =
..
.
.
0
I
.
I
..
.
Causality
A system G : ℓm (Z) → ℓp (Z) is said to be causal if
P ku = P kv
implies
P k Gu = P k Gv
for all k ∈ Z and u, v ∈ D(G). G is said to be strictly causal if
P ku = P kv
implies
P k+1 Gu = P k+1 Gv
for all k ∈ Z and u, v ∈ D(G).
From the definition, we can see that the causality of a linear system can be
easily seen from the impulse response: G is causal iff G(k, l) = 0 for k < l; G is
strictly causal iff G(k, l) = 0 for k ≤ l. This immediately leads to the following
theorem.
Theorem 1.1 A linear system G is causal iff [G] is block lower triangular. A
linear system G is strictly causal iff [G] is strictly block lower triangular.
Examples of causal systems are M , S, P k , and Qk . An example of strictly
causal systems is S. An example of noncausal systems is S −1 .
The opposite of causality is anti-causality. A system G : ℓm (Z) → ℓp (Z) is
said to be anti-causal if
Qk u = Qk v
implies
Qk Gu = Qk Gv
for all k ∈ Z and u, v ∈ D(G). G is said to be strictly anti-causal if
Qk u = Qk v
6
CHAPTER 1. BASIC CONCEPTS
implies
Qk−1 Gu = Qk−1 Gv
for all k ∈ Z and u, v ∈ D(G).
It is easy to see that a linear system G is anticausal iff [G] is block upper
triangular and it is strictly anticausal iff [G] is strictly block upper triangular.
Note the difference between anticausal systems and noncausal systems. Noncausal systems are those which are not causal. Anticausal systems are those
which are “causal in the reverse time”. Examples of anticausal systems are M ,
S −1 , P k , and Qk . An example of strictly anticausal systems is S −1 .
A system which is simultaneously causal and anticausal is called memoryless,
see Exercise 1.6.
1.4
Time-invariance
A system G : ℓm (Z) → ℓp (Z) is said to be time-(shift-)invariant if
GS = SG,
i.e., if G commutes with the unit shift.
A linear time-invariant (LTI) system, in terms of the matrix representation,
has
[G][S] = [S][G].
Let G(k, l) be the (k, l)-th block of [G]. Then it is the (k, l − 1)-th block of
[G][S]. On the other hand, the (k, l − 1)-th block of [S][G] is G(k − 1, l − 1).
This shows that G is time-invariant iff [G] satisfies
G(k, l) = G(k − 1, l − 1)
for all k, l ∈ Z, i.e., G(k, l) depends on only k − l. Therefore we can write the
impulse response as G(k, l) = G(k − l). The matrix representation becomes
..
..
..
.
.
.
..
. G(0) G(−1) G(−2)
..
.
[G] = . G(1) G(0) G(−1) . . ,
.
.
.
G(2) G(1)
G(0)
..
..
..
.
.
.
i.e., the p × m blocks along each diagonal of [G] are the same. Such matrices
are called (p × m block) Toeplitz matrices.
Theorem 1.2 Linear system G is time-invariant iff [G] is p×m block Toeplitz.
The i-th column of G(k), k ∈ Z, is the response of the system to impulse ei δ0 .
For impulse occurring at another time, the response has the same wave form
but is shifted accordingly. Therefore, for LTI systems we call G : Z → Fp×m
the impulse response function.
1.5. STABILITY
7
The input-output relationship of an LTI system becomes
y(k) =
∞
X
G(k − l)u(l).
l=−∞
This relationship is called convolution and is denoted by
y = G ∗ u.
Examples of time-invariant systems are M , S, and S −1 . The truncations
P k and Qk are time-varying.
Two other trivial LTI systems will often be used. The first one is the zero
system, denoted by O, whose output is identically zero no matter what the
input is. The second one is the identity system, denoted by I, whose output is
identically equal to the input. Clearly, the impulse response function of O is
the zero function, i.e., O(k) = 0 for all k ∈ Z, and that of I is given by
(
I if k = 0
I(k) =
0 otherwise.
An LTI system said to have finite impulse response or to be FIR if G(k) is
a sequence with finite many nonzero terms, i.e., there exists K > 0 such that
G(k) = 0 for all |k| > K.
1.5
Stability
There are many ways to define stability. We will present one of the simplest
ways here. A signal x ∈ ℓn (Z) is said to be bounded if
sup |xi (k)| < ∞.
1≤i≤n
k∈Z
The left hand side is also called the amplitude of signal x. The set of all bounded
signals in ℓn (Z) is denoted by ℓn∞ (Z). The space ℓn∞ (Z) becomes a normed linear
space if we simply define the norm of a signal in ℓn∞ (Z) by its magnitude:
∥x∥∞ = sup |xi (k)|.
1≤i≤n
k∈Z
A system G is said to be bounded-input-bounded-output (BIBO) stable if
p
m
ℓm
∞ (Z) ⊂ D(G) and Gℓ∞ (Z) ⊂ ℓ∞ (Z).
Theorem 1.3 Consider a linear system G : ℓm (Z) → ℓp (Z) with impulse response function G
1. G is BIBO stable if
sup
∞ X
m
X
1≤i≤p
l=−∞ j=1
k∈Z
|Gij (k, l)| < ∞.
(1.1)
8
CHAPTER 1. BASIC CONCEPTS
2. G is BIBO stable only if
∞ X
m
X
|Gij (k, l)| < ∞.
(1.2)
l=−∞ j=1
for every 1 ≤ i ≤ p, k ∈ Z.
Proof
Let u be bounded, then
∞ X
m
X
|yi (k)| =
Gij (k, l)uj (l)
l=−∞ j=1
≤ ∥u∥∞
∞ X
m
X
|Gij (k, l)|
l=−∞ j=1
∞
X
≤ ∥u∥∞ sup
m
X
|Gij (k, l)|.
1≤i≤p
l=−∞ j=1
k∈Z
If condition (1.1) holds, then y is also bounded. This shows the sufficiency of
condition (1.1).
Now assume that condition (1.2) does not hold. Then there exist i =
1, 2, . . . , p and k ∈ Z such that
∞ X
m
X
|Gij (k, l)| = ∞.
l=−∞ j=1
Let a particular input u be chosen in the following way:
uj (l) =
Gij (k, l)
.
|Gij (k, l)|
Then u ∈ ℓ∞ and ∥u∥∞ = 1 but
yi (k) =
∞ X
m
X
|Gij (k, l)| = ∞.
l=−∞ j=1
This shows that a bounded input is not even in the domain of the system,
which implies that the system is not BIBO stable. This shows the necessity of
condition (1.2).
2
Notice that the left hand side of (1.1) is the supremum of the absolute
row sums of [G]. The matrix representation [G] has infinitely many different
rows. Theorem 1.3 says that a sufficient condition for the BIBO stability of
G is that all such infinitely many row sums are uniformed bounded and a
necessary condition is that all such infinitely many row sums are finite. These
two conditions are different.
For a BIBO stable system, if the input has a finite magnitude, then the
output also has a finite magnitude. If the stronger condition in Theorem (1.3.1)
is satisfied, the gain in the magnitude, i.e., the largest ratio of the output
maginitude over the input magnitude, is finite.
1.6. Z-TRANSFORM AND TRANSFER FUNCTIONS
9
Theorem 1.4 Let u be the input of a BIBO stable linear system satisfying
condition (1.1) and y be the corresponding output. Then
∞ X
m
X
∥y∥∞
sup
= sup
|Gij (k, l)|.
1≤i≤p
u∈ℓm (Z) ∥u∥∞
k∈Z
u̸=0
l=−∞ j=1
In the LTI case, [G] essentially only has p different row sums. In this case,
the necessary and sufficient conditions coincide.
Corollary 1.1 An LTI system with impulse response G is BIBO stable iff
∞ X
m
X
|Gij (k)| < ∞
k=−∞ j=1
for all i = 1, 2, . . . p.
Example 1.4
Consider SISO LTI system with impulse response
G(k) =
(
0
1
k
if k ≤ 0
if k > 0.
This system is unstable since its impulse response is not absolutely summable,
though the impulse response itself is bounded and converge to zero.
An immediate consequence of Corollary 1.1 is that an LTI FIR system is
always BIBO stable.
Theorem 1.4 also has a simplified version for LTI systems.
Corollary 1.2 A BIBO stable LTI system G has
∞ X
m
X
∥y∥∞
sup
= max
|Gij (k)|.
1≤i≤p
u∈ℓm (Z) ∥u∥∞
k=−∞ j=1
u̸=0
1.6
Z-Transform and Transfer Functions
An impulse response function G : Z → Fp×m of an LTI system G can be
considered as a matrix valued signal. The set of all such functions is denoted
by ℓp×m (Z). Hence ℓn (Z) can be considered as ℓn×1 (Z). Similarly, we can define
ℓp×m (Z+ ) and ℓp×m (Z− ). We can also see
ℓp×m (Z) = ℓp×m (Z+ ) ⊕ ℓp×m (Z− ).
Let G ∈ ℓp×m (Z), then the Z-transform of G is defined as
Ĝ(z) =
∞
X
k=−∞
G(k)z −k .
(1.3)
10
CHAPTER 1. BASIC CONCEPTS
Associated with the Z-transform is a region of convergence (ROC). A region
in the complex plane is a connected open set. The region of convergence of
Ĝ is the region in which the series on the right hand side of (1.3) converges.
Actually, the series on the right hand side of (1.3) converge absolutely in its
ROC. The function Ĝ is analytic in its ROC. We say that Ĝ exists if the region
of convergence is not empty. From the knowledge of complex analysis, we
know that the series is a Laurent series and its ROC has three possibility: if
G ∈ ℓp×m (Z+ ), then the ROC of Ĝ is a punched plane of the form
{z ∈ C : |z| > r1 };
if G ∈ ℓp×m (Z− ), then that of Ĝ is a disk of the form
{z ∈ C : |z| < r2 }.
Here r1 , r2 are nonnegative numbers. Hence Ĝ exists if r1 < ∞ or r2 > 0
respectively. For general G ∈ ℓp×m (Z), we know that
G = G+ + G −
where
G+ = P + G = {. . . , 0, 0, |G(0), G(1), . . .} ∈ ℓp×m (Z+ )
G− = P − G = {. . . , G(−2), G(−1), |0, 0, . . .} ∈ ℓp×m (Z− )
If the ROC of Ĝ+ is {z ∈ C : |z| > r1 } and that of Ĝ− is {z ∈ C : |z| < r2 },
then that of Ĝ is an annulus of the form
{z ∈ C : r1 < |z| < r2 }.
Hence Ĝ exists if r1 < r2 . If r2 ≤ r1 , then G does not have a Z-transform.
Example 1.5
Let G = {. . . , 1, 1, |1, 1, . . .}. Then
∞
X
k=−∞
G(k)z −k =
∞
X
z −k
k=−∞
diverges everywhere, i.e., converges nowhere. Hence G does not have a Ztransform.
The ROC is often left unmentioned when the Z-transform is used. In some
cases, in particular the case when only signals in ℓp×m (Z+ ) are of concern, this
is not harmful since one can identify the ROC from the Z-transform of a signal
in ℓp×m (Z+ ) (it is the region of the form {z ∈ C : |z| > r1 } in which the
Z-transform is analytic). However, in other cases it may lead to ambiguity.
Example 1.6
z
with ROC {z ∈ C : |z| < 1}.
Let G = {. . . , −1, −1, |0, 0, . . .}. Then Ĝ(z) = z−1
z
However if we let G = {. . . , 0, 0, |1, 1, . . .}. Then we also have Ĝ(z) = z−1
, but
the ROC is now {z ∈ C : |z| > 1}.
1.7. ELEMENTARY SYSTEM CONNECTIONS AND OPERATIONS
11
If G is a BIBO stable system, then it is easy to see that the ROC of Ĝ(z)
contains the unit circle T = {z ∈ C : |z| = 1}.
Proposition 1.1 Let G, H ∈ ℓp×m (Z) have Z-transforms Ĝ, Ĥ with ROCs RG
and RH respectively. Let α, β ∈ F. If RG ∩ RH ̸= ∅, then αG + βH has
Z-transform αĜ + β Ĥ with an ROC containing RG ∩ RH .
Proposition 1.2 Let G ∈ ℓp×m (Z) and H ∈ ℓm×q (Z) have Z-transforms Ĝ, Ĥ
with ROCs RG and RH respectively. If RG ∩RH ̸= ∅, then G∗H has Z-transform
ĜĤ with an ROC containing RG ∩ RH .
Proof
Let F = G ∗ H. Then
F̂ (z) =
∞
X
(G ∗ H)(k)z −k =
k=−∞
∞
X
∞
X
G(l)H(k − l)z −k .
k=−∞ l=−∞
P∞
P
−k converge
−l and
Since for z ∈ RG ∩ RH , both ∞
k=−∞ H(k)z
l=−∞ G(l)z
absolutely, it follows that for z ∈ RG ∩ RH ,
F̂ (z) =
∞
X
l=−∞
G(l)z −l
∞
X
H(k − l)z −k+l = Ĝ(z)Ĥ(z)
k=−∞
2
For a system with impulse response G(k), the Z-transform Ĝ of G is called
the transfer function of the system. It is so-called since if û is the Z-transform
of the input u and ŷ be the Z-transform of the output y, then
ŷ = Ĝû
where the ROC of ŷ contains the intersection of the ROC of G and that of u.
The inverse Z-transform returns a signal from its Z-transform. To carry out
this inversion, the ROC is essential. Let a complex function Ĝ, together with an
ROC of the three possible forms above, be given. Then the inverse Z-transform
is given by
I
1
G(k) =
Ĝ(z)z k−1 dz
j2π
where the integral is taken along a circle in the complex plane centered at the
origin that is entirely contained in the ROC. The inverse Z-transform can also
be written as
Z
rk π
Ĝ(rejθ )ejkθ dθ
G(k) =
2π −π
where r > 0 is chosen so that rejθ , θ ∈ [−π, π], is contained in the ROC.
1.7
Elementary System Connections and Operations
A large system is often formed by interconnections of small systems. In this
section, we study some basic interconnections of systems.
12
CHAPTER 1. BASIC CONCEPTS
Let G1 and G2 are two systems from ℓm (Z) to ℓp (Z). If the domains of G1
and G2 have nonempty intersections, the parallel connection or the sum of G1
and G2 , denoted by G1 + G2 , can be naturally defined as
(G1 + G2 )u = G1 u + G2 u
for u ∈ D(G1 ) ∩ D(G2 ). The domain of G1 + G2 is obviously D(G1 ) ∩ D(G2 ).
Graphically, the parallel connection or sum can be represented by the block
diagram in Figure 1.2.
- G1
?y
j -
u
6
- G2
Figure 1.2: Parallel connection or sum
Let G1 : ℓm (Z) → ℓp (Z) and G2 : ℓq (Z) → ℓm (Z) be two systems. If
the domain of G1 and the range of G2 have nonempty intersection, the cascade
connection or product of G1 and G2 , denoted by G1 G2 , can be naturally defined
as
(G1 G2 )u = G1 (G2 u)
for u ∈ {u ∈ D(G2 ) : G2 u ∈ D(G1 )}. Let us denote the set {u ∈ D(G) : Gu ∈
S} by G−1 (S). (Note that here we by no means imply G has an inverse.) Then
the domain of G1 G2 is G−1
2 [D(G1 )]. Graphically, the cascade connection or
product can be represented by the block diagram in Figure 1.3.
u-
G2
- G1
y-
Figure 1.3: Cascade connection or product
It is easy to see that the addition and multiplication of systems preserve
linearity, time-invariance, causality, and BIBO stability.
For linear systems, it can be easily seen that
[G1 + G2 ] = [G1 ] + [G2 ]
and
[G1 G2 ] = [G1 ][G2 ].
Assume that G1 and G2 are LTI with impulse responses G1 and G2 respectively.
Then the impulse response of G1 + G2 is simply G1 + G2 . The impulse response
of G1 G2 , however, is not G1 G2 . Since the response of G1 G2 to impulse δ0 ei is
G1 ∗ (G2 ei ) = (G1 ∗ G2 )ei ,
1.7. ELEMENTARY SYSTEM CONNECTIONS AND OPERATIONS
13
It follows that the impulse response of G1 G2 is G1 ∗ G2 , the convolution of G1
and G2 . Assume that G1 and G2 have transfer functions Ĝ1 and Ĝ2 respectively. Then it follows easily that those of G1 + G2 and G1 G2 are Ĝ1 + Ĝ2 and
Ĝ1 Ĝ2 respectively.
If G : ℓm (Z) → ℓm (Z) is a bijection from D(G) to R(G), then G is said to
be invertible and the inverse of G, denoted by G−1 , is defined by
G−1 y = u ⇐⇒ Gu = y.
for u ∈ D(G) and y ∈ R(G). The domain of G−1 is clearly R(G).
For invertible systems, the inversion preserves linearity and time-invariance.
We leave it as an exercise to verify this statement. However, it does not preserve
causality or stability in general. For example, the unit delay S is causal, but
its inverse, the unit advance S −1 , is not. Suppose that a time-varying system
G has a matrix representation
..
.
1
−
2
2
1
−2
.
[G] =
1
1
2
1
22
..
.
Then G is stable but G−1 is unstable since
..
.
−22
−2
−1
[G ] =
1
.
2
22
..
.
It then is of interest to know when the inverse of a causal system is causal and
when the inverse of a stable system is stable. A clean answer to the latter seems
to be hard, while the following proposition answers the former question.
Proposition 1.3 Assume that G is a linear causal system with impulse response G. Then G is invertible and G−1 is causal iff G(k, k), k ∈ Z, are
invertible.
A linear causal invertible system is said to be causally invertible if its inverse
is causal. Clearly, S is not causally invertible.
For linear invertible systems, we have
[G−1 ] = [G]−1 .
14
CHAPTER 1. BASIC CONCEPTS
For LTI invertible system G, the transfer function of G−1 is the reciprocal of
that of G, i.e.,
d
−1 = Ĝ−1 .
G
For linear system G : ℓm (Z) → ℓp (Z), the conjugate of G, denoted by G∼ ,
is defined as
[G∼ ] = [G]∗
or equivalently
G∼ (k, l) = G(l, k)∗ .
The dual or transpose of G, denoted by G′ , is defined by
G′ (k, l) = G(k, l)′ .
Note the difference between G∼ and G′ . Both conjugation and transposition
preserves time-invariance. The transposition preserves causality and stability,
but the conjugation does not in general. Now let G be LTI with impulse
response G and transfer function Ĝ, i.e.
∞
X
Ĝ(z) =
G(k)z −k .
k=−∞
Then the transfer function of G∼ is
c∼ (z) =
G
∞
X
∞
X
"
G(−k)∗ z −k =
#∗
G(−k)z −k
= G(z −1 )∗
k=−∞
k=−∞
and the transfer function of G′ is
c′ (z) =
G
∞
X
"
′ −k
G(k) z
k=−∞
=
∞
X
#′
G(k)z
−k
= Ĝ′ (z).
k=−∞
Because of this, we define Ĝ∼ as
Ĝ∼ (z) = Ĝ(z̄ −1 )∗ .
1.8
Linear Fractional Transformation
An effective tool in studying feedback connections of systems is the so-called
linear fractional transformation (LFT). First notice that the different input signals and output signals of a system may have quite different physical meanings.
According to their meanings, they can be partitioned into two (or more) groups
as shown in Figure 1.4. For example, some input signals represent noises and
disturbances which cannot be manipulated and others are control inputs which
can be changed by human or devices. Similarly, some output signals of P may
represent those variable in the system that we are interested in and others may
represent the signals measurable by sensors. Sometimes, however, the partition
1.8. LINEAR FRACTIONAL TRANSFORMATION
z
15
w
P
y
u
Figure 1.4: A generalized system
has no physical meaning what so ever and is just for mathematical representation. A system with the input and output signals partitioned is sometimes
called a generalized system.
A reader may notice that the signal flow directions in Figure 1.4 is opposite to what we have used to. This follows a new trend which believes that
the direction in Figure 1.4 is more consistent with the matrix multiplication
convention in linear algebra.
Consider the systems shown in Figures 1.5 and 1.6, where P : ℓm1 +m2 (Z) →
p
+p
ℓ 1 2 (Z) is a system whose input signals and output signals are partitioned
into two groups respectively and Q : ℓp2 (Z) → ℓm2 (Z) is another system which
connects one group of the output signals of P with one group of input signals
of Q.
z
w
P
y
u
- Q
Figure 1.5: Lower linear fractional transformation
- Q
z
w
y
P
u
Figure 1.6: Upper linear fractional transformation
Now assume that we partition P according to its input and output partition.
P 11 P 12
P =
.
P 21 P 22
For Figure 1.5, if I − P 22 Q is invertible, then the system with input w and
16
CHAPTER 1. BASIC CONCEPTS
output z is
P 11 + P 12 Q(I − P 22 Q)−1 P 21 : ℓm1 (Z) → ℓp1 (Z).
This expression is called a lower linear fractional transformation of Q and is
denoted by F l (P , Q), i.e.,
F l (P , Q) = P 11 + P 12 Q(I − P 22 Q)−1 P 21 .
LFT F l (P , Q) is said to be well-posed if I − P 22 Q is invertible. If P and Q
are LTI with transfer functions
P̂11 P̂12
P̂ =
and Q̂
P̂21 P̂22
respectively. Then the transfer function of F l (P , Q) is clearly
P̂11 + P̂12 Q̂(I − P̂22 Q̂)−1 P̂21
which is also written as F l (P̂ , Q̂).
For Figure 1.6, if I − QP 11 is invertible, then system with input u and
output y is
P 22 + P 21 (I − QP 11 )−1 QP 12 : ℓm1 (Z) → ℓp1 (Z).
This expression is called an upper linear fractional transformation of Q and is
denoted by F u (P , Q), i.e.,
F u (P , Q) = P 22 + P 21 (I − QP 11 )−1 QP 12 .
LFT F u (P , Q) is said to be well-posed if I − QP 11 is invertible. The transfer
function of F u (P , Q) is clearly
P̂22 + P̂21 (I − Q̂P̂11 )−1 Q̂P̂12
which is also written as F u (P̂ , Q̂).
The lower and upper LFTs are related in the following way:
P 11 P 12
F l (P , Q) = F l
,Q
P 21 P 22
P 22 P 21
= Fu
,Q
P 12 P 11
O I
O I
= Fu
P
,Q .
I O
I O
Example 1.7
Sums and products are special cases of LFTs:
P1
P 1 + Q = Fl
I
O
P 2 QP 3 = F l
P3
I
O
P2
O
,Q
,Q .
1.8. LINEAR FRACTIONAL TRANSFORMATION
z
17
w
P
y
z̃
HH
HH H
HH
HH
Q
ỹ
u
w̃
ũ
Figure 1.7: Star product
Consider the system connection shown in Figure 1.7, where
P 11 P 12
P =
: ℓm1 +m2 (Z) → ℓp1 +p2 (Z)
P 21 P 22
and
Q=
Q11 Q12
Q21 Q22
: ℓp2 +p3 (Z) → ℓm2 +m3 (Z).
The system with input w, ũ and output z, ỹ is
F l (P , Q11 )
P 12 (I − Q11 P 22 )−1 Q12
: ℓm1 +p3 (Z) → ℓp1 +m3 (Z).
F u (Q, P 22 )
Q21 (I − P 22 Q11 )−1 P 21
This expression is defined as the Redheffer’s star product of P and Q, i.e.,
F l (P , Q11 )
P 12 (I − Q11 P 22 )−1 Q12
P ⋆Q=
.
F u (Q, P 22 )
Q21 (I − P 22 Q11 )−1 P 21
Directly from the physical definition of the star product, we see
F l (P , F l (Q, R)) = F l (P ⋆ Q, R)
F u (Q, F u (P , R)) = F u (P ⋆ Q, R).
Example 1.8
By using the star product, we can easily get the LFT for the composition of
sum and product:
P1 I
O P2
P 1 + P 2 QP 3 = F l
, Fl
,Q
I 0
P3 O
P1 I
O P2
= Fl
⋆
,Q
P3 O
I O
P1 P2
= Fl
,Q .
P3 O
Sometimes we need to deal with the inverse map of an LFT.
18
CHAPTER 1. BASIC CONCEPTS
P 11 P 12
Theorem 1.5 Let P =
: ℓm1 +m2 (Z) → ℓm2 +m1 (Z). If P 12 , P 21 ,
P 21 P 22
and P are invertible, then the LFT map Q 7→ F l (P , Q) is a bijection from
{Q : I − P 22 Q is invertible}
to
−1
−1 is invertible}
{R : I − RP −1
21 P 22 (P 12 + P 11 P 21 P 22 )
and the inverse map is given by R 7→ F u (P −1 , R).
Proof If P 12 , P 21 are invertible, then
P 11 P 12
I −P 11 P −1
O P 12 − P 11 P −1
P 22
21
21
=
P 21 P 22
O
I
P 21
P 22
and
I
−P 22 P −1
12
O
I
P 11 P 12
P 21 P 22
=
P 11
P 12
−1
P 21 − P 22 P 12 P 11 O
.
This shows that if P is also invertible, then P 12 − P 11 P −1
21 P 22 and P 21 −
−1
P
and S 21 = P 21 −
P
are
invertible.
Denote
S
=
P
−
P
P
P 22 P −1
22
11
12
12
11
21
12
P
,
which
are
called
the
Schur
complements
with
respect
to P 21 and
P 22 P −1
11
12
P 12 respectively. Then a straightforward computation (cf., Exercise 1.15) shows
that
S −1
P 22 S −1
−P −1
−1
21
12
21
.
P =
−S 12 P 11 P −1
S −1
21
12
Now assume that
R = F l (P , Q) = P 11 + P 12 Q(I − P 22 Q)−1 P 21 .
Then
−1
−1
−1 −1
Q = [I + P −1
12 (R − P 11 )P 21 P 22 ] P 12 (R − P 11 )P 21
−1
−1
−1
= (S 12 + RP −1
21 P 22 ) (RP 21 − P 11 P 21 )
−1
−1 −1
= S −1
12 (I + RP 21 P 22 S 12 )
−1
−1
−1
−1
× −(I + RP −1
21 P 22 S 12 )P 11 + R(I + P 21 P 22 S 12 P 11 ) P 21
−1
= −S −1
12 P 11 P 21
−1
−1 −1
−1
−1
−1
−1
+ S −1
12 (I + RP 21 P 22 S 12 ) R(P 21 + P 21 P 22 S 12 P 11 P 21 )
−1
−1
−1
−1
−1
= −S −1
12 P 11 P 21 + S 12 (I − RP 21 P 22 S 12 )RS 21
= F u (P −1 , R).
Here we used the fact that
−1
−1
−1
−1
P −1
21 − P 21 P 22 S 12 P 11 P 21 = S 21 ,
see Exercise 1.15.
2
1.9. STATE SPACE SYSTEMS
1.9
19
State Space Systems
A physical dynamic system is not always modelled as an input-output map. It
is often described by difference equations governing their motion. A particular
form of such difference equations is the so-called state space equation, which is
of the form
x(k + 1) = f [x(k), u(k), k]
(1.4)
y(k) = g[x(k), u(k), k]
(1.5)
where x(k) is called the state variable and in general belongs to a subset of a
linear space. If the linear space which x(k) belongs to is finite dimensional, say
Fn , then the state space system is said to be finite dimensional (FD).
A state space system is automatically causal.
A finite-dimensional state space system is said to be linear if f and g are
linear functions of x(k) and u(k). In this case, the state space equation can be
written as
x(k + 1) = A(k)x(k) + B(k)u(k)
y(k) = C(k)x(k) + D(k)u(k)
where A(k), B(k), C(k), D(k) are matrices of appropriate sizes. The impulse
response of a finite-dimensional linear state space system satisfies
k<l
0
D(k)
k=l
G(k, l) =
C(k)A(k − 1) · · · A(l + 1)B(l) k > l.
A finite-dimensional linear state space system is said to be time-invariant
if A(k), B(k), C(k), and D(k) are constant matrix functions. In this case, the
state space equation becomes
x(k + 1) = Ax(k) + Bu(k)
y(k) = Cx(k) + Du(k)
where A, B, C, D are constant matrices. The impulse response of a finitedimensional linear time-invariant (FDLTI) state space system satisfies
k<0
0
D
k=0
G(k) =
CAk−1 B k > 0.
In this course, we will be mainly interested in FDLTI state space systems.
The state space system can be considered as an internal description of a system. In contrast, the input-output relation is an external description of a system. The properties we described in the previous sections are all external properties. An internal description may provides more information about the system. A system with deceivingly good external behavior may have bad internal
behavior so it may not work well. As an extreme example, if g[x(k), u(k), k] = 0
in (1.5), then the system is externally LTI and stable but internally, depending
on the function f , it may be nonlinear, time-varying, or unstable.
20
CHAPTER 1. BASIC CONCEPTS
Exercises
1.1 Show that a system G : ℓm (Z) → ℓp (Z) is linear iff G(G) is a linear
subspace of ℓm+p (Z).
1.2 A subspace U of ℓn (Z) is said to be shift invariant if
SU ⊂ U
where S is the unit shift. Is ℓn (Z+ ) shift invariant? How about ℓn (Z− )?
Show that a linear (moving average) system G : ℓm (Z) → ℓp (Z) is time
invariant iff G(G) is a shift invariant subspace of ℓm+p (Z).
1.3 Show that an FIR moving average system G : ℓm (Z) → ℓp (Z) satisfies
D(G) = ℓm (Z).
1.4 Show that if a moving average system G : ℓm (Z) → ℓp (Z) is causal, then
(I − P k )ℓm (Z) ⊂ D(G) for all k ∈ Z and, in particular, ℓm (Z+ ) ⊂ D(G).
1.5 Show that a linear time-invariant moving average system G is causal if
and only if P − GP + = O.
1.6 A system G is said to be memoryless if u, v ∈ D(G) and u(k) = v(k)
implies (Gu)(k) = (Gv)(k). Show that a linear (moving average) system
is memoryless iff its matrix representation is (block) diagonal.
1.7 Show that the system inversion preserves linearity and time-invariance.
1.8 Prove Proposition 1.3.
1.9 Consider the linear time varying state-space system where the matrices
A(k), B(k), C(k), D(k) are all periodic of period N . Then it is not true
that GS = SG, but what is true? What can be said about [G]?
1.10 In some literature, a system is defined as a map from ℓ(Z+ ) to ℓ(Z+ ).
The unit forward shift system S is defined as
S{x(0), x(1), . . .} = {0, x(0), x(1), . . .}.
Is S injective? Is it surjective? Define the unit backward shift system so
that it is the left inverse of S.
1
1.11 Let a system have a transfer function z−2
with ROC {z ∈ C : |z| < 2}.
Is this system BIBO stable? Is it causal. How about a system with the
same transfer function but a different ROC {z ∈ C : |z| > 2}?
1.12 Consider system with transfer function
1
.
(z − 0.5)(z − 2)
Assume that its ROC has three possibilities: {z : |z| < 0.5}, {z : 0.5 <
|z| < 2}, and {z : |z| > 2}. In each possibility, what are the causality and
stability of the system?
1.9. STATE SPACE SYSTEMS
21
1.13 An anticausal system has transfer function Ĝ(z) =
stable?
1
z−0.5 .
Is this system
1.14 Show that if P 11 is invertible, then
P −1
−P −1
P 12
−1
11
11
F l (P , Q) = F l
,Q .
P 21 P −1
P 22 − P 21 P −1
11
11 P 12
1.15 Show that
P 11 O P 12 O
O O O
I
Q11 Q12
P ⋆ Q = Fl
P 21 O P 22 O , Q21 Q22 .
O
I
O O
1.16 For a system
T 11 T 12
T 21 T 22
: ℓm+p (Z) → ℓm+q (Z),
define a function
T (Q) = (T 21 + T 22 Q)(T 11 + T 12 Q)−1
Show that if T 11 is invertible, then
T (Q) = Fl (P , Q)
for some P . Find this P .
1.17 Derive the impulse response of a finite dimensional linear state space
system.
22
CHAPTER 1. BASIC CONCEPTS
Chapter 2
LTI State Space System
Analysis
2.1
Solution of State Space Equations
Consider an FDLTI state space system described by equation
x(k + 1) = Ax(k) + Bu(k)
(2.1)
y(k) = Cx(k) + Du(k).
This equation is often denoted by (A, B, C, D) or
"
A B
C D
#
.
Graphically, we may represent it using a block diagram shown in Figure 2.1 or
a more detailed diagram shown in Figure 2.2.
"
u
-
A B
C D
#
y
-
Figure 2.1: The block diagram of an LTI state space system
We can see that if x(k0 ) and Qk0 u are known for some k0 ∈ Z, then the
state x(k) and the output y(k) for k > k0 can be obtained iteratively. In other
words, x(k0 ) captures all information of the system before time k0 . It is often
the case that we need to find x(k) or y(k) knowing x(k0 ) and Qk0 u. Simple
23
24
CHAPTER 2. LTI STATE SPACE SYSTEM ANALYSIS
-D
u
- B
- j
-
x
S
- C
? y
- j -
6
A Figure 2.2: Linear state space system structure
algebra shows that
x(k) = Ak−k0 x(k0 ) +
k−1
X
Ak−l−1 Bu(l)
l=k0
y(k) = CAk−k0 x(k0 ) +
k−1
X
CAk−l−1 Bu(l) + Du(k).
l=k0
It is seen that both x(k) and y(k) consist of two parts, the part due to the
initial condition x(k0 ), which is called the zero input response, and the part
due to the input Qk0 u only, which is called the zero state response.
It is seen that if the response starting at time k0 with initial condition
x(k0 ) = x0 and input u is x, then the response starting at time 0 with the same
initial condition x(0) = x0 and a shifted input S −k0 u is S −k0 x. Therefore, in
the following study, we assume that k0 = 0 without loss of generality.
It is now clear that the impulse response of the system is
k<0
0
D
k=0
G(k) =
(2.2)
k−1
CA B k > 0.
The transfer function is given by
Ĝ(z) = D + C(zI − A)−1 B.
A pole of Ĝ must be an eigenvalue of A. Therefore the ROC of Ĝ contains
{z ∈ C : |z| > ρ(A)}. Alternative expressions for the transfer function are
A − zI B
A B
−1
Ĝ(z) =
/11 = F u
,z I .
C
D
C D
The upper LFT expression means that an LTI system has a block diagram
representation as shown in Figure 2.3.
It is easy to see from (2.2) that if A is a nilpotent matrix, i.e., if An = 0,
then the system is FIR. In this case, we also say that the system is deadbeat.
For an FIR or deadbeat system, the zero input response vanishes after certain
period of time. The total response will also vanish after certain period of time
2.1. SOLUTION OF STATE SPACE EQUATIONS
25
- z −1 I
z x̂(z)
x̂(z)
ŷ(z)
A B C D û(z)
Figure 2.3: Another block diagram of an LTI state space system
if the input only lasts for a finite duration of time, as what an impulse input
does. The poles of an FIR or deadbeat system are all equal to zero.
If we carry out state variable transformation x = P z, then the system
becomes
z(k + 1) = P −1 AP z(k) + P −1 Bu(k)
y(k) = CP z(k) + Du(k).
In the abbreviated notation, it becomes (P −1 AP, P −1 B, CP, D) or
"
#
P −1 AP P −1 B
.
CP
D
The solution becomes
z(k) = P
−1
k
A P z(0) +
k−1
X
P −1 Ak−l−1 P P −1 Bu(l)
l=0
y(k) = CAk z(0) +
k−1
X
CAk−l−1 Bu(l) + Du(k).
l=0
It is seen that the state response now is different, but the zero state output
response is still the same. This shows that state variable transformation does
not change the input output relationship but it may change the behavior of the
state response. In particular, the impulse response
k<0
k<0
0
0
D
k=0 =
D
k=0
G(k) =
CP (P −1 AP )k−1 P −1 B k > 0
CAk−1 B k > 0
and the transfer function
Ĝ(z) = D + CP (zI − P −1 AP )−1 P −1 B = D + C(zI − A)−1 B
are not changed by the transformation. Therefore it is a common practice to
use some state variable transformation to simplify the state response but to
keep the input output relation unchanged.
26
CHAPTER 2. LTI STATE SPACE SYSTEM ANALYSIS
Assume that A is diagonalizable, i.e., there exists nonsingular matrix P such
that
λ1
λ2
P −1 AP =
.
.
.
.
λn
If we carry out a state variable transformation x = P z, then under the new
state variable z, the system (with zero input) becomes
λ1
λ2
z(k + 1) =
z(k).
..
.
λn
The solution with initial condition z(0) = z0 becomes
k
λ1
λk2
z(k) =
z0 .
..
.
λkn
The state response of the original system with initial condition x(0) = x0 is
then
k
λ1
λk2
−1
x(k) = P
P x0
.
.
.
k
λn
i.e., x(k) has the form
x(k) = c1 λk1 + c2 λk2 + · · · + cn λkn
where ci are constant vectors depending on x0 . Because of this, the terms
λk1 , . . . , λkn are called the modes of the response x(k). For simplicity, we also call
the eigenvalues λ1 , λ2 , . . . , λn the modes of the system.
If A is not diagonalizable, the response is more complicated.
Example 2.1
If
A=
then
Ak =
λ 1
,
0 λ
λk kλk−1
0
λk
.
Hence
x(k) = c1 λk + c2 kλk−1
where c1 and c2 depend on the the initial condition.
2.2. CONTROLLABILITY
27
In general, if A is not diagonalizable, there exists nonsingular matrix P such
that
J1
J2
P −1 AP =
,
.
.
.
Jl
where
λi
Ji =
1
..
.
..
∈ Cni ×ni
..
. 1
λi
.
and
Jik
where
λi
=
k
j
1
..
.
k
..
k
1
λki
=
..
. 1
λi
.
k
k−ni +1
λk−1
···
i
..
..
.
.
..
.
k
1
..
.
i +1
λk−n
i
,
λik−1
λki
= 0 if j > k, it follows that x(k) is of the form
x(k) =
ni
l X
X
pij (k)λk−j+1
i
i=1 j=1
where pij are polynomial vectors depending on x0 .
2.2
Controllability
Consider system
x(k + 1) = Ax(k) + Bu(k)
(2.3)
whose
output
is not of concern. This system is often referred to as (A, B) or
A B . First let us assume that x(0) = 0. A state x1 ∈ Fn is said to be
reachable if there exists k1 > 0 and u such that x(k1 ) = x1 . Denote the set of
all reachable state by R0 .
Theorem 2.1
R0 = R
B AB A2 B · · · An−1 B
=R
n−1
X
!
k
∗
A BB A
∗k
.
k=0
Proof If x1 ∈ R0 , then there exist a time instance k1 > 0 and an input
sequence u such that
x1 = x(k1 ) =
kX
1 −1
l=0
Ak1 −l−1 Bu(l).
28
CHAPTER 2. LTI STATE SPACE SYSTEM ANALYSIS
By the Cayley-Hamilton theorem, Ak , k ≥ 0, can be expressed as a linear
combination of I, A, . . . , An−1 :
Ak = c1 (k)I + c2 (k)A + c3 (k)A2 + · · · + cn (k)An−1 .
Hence
x1 = B
kX
1 −1
c1 (k1 − l − 1)u(l) + AB
l=0
kX
1 −1
c2 (k1 − l − 1)u(l)
l=0
+ A2 B
kX
1 −1
c3 (k1 − l − 1)u(l) + · · · + An−1 B
l=0
∈R
kX
1 −1
cn (k1 − l − 1)u(l)
l=0
B AB A2 B · · · An−1 B
.
This shows
R0 ⊂ R
B AB A2 B · · · An−1 B
.
The fact that
R
B AB
A2 B
···
An−1 B
=R
n−1
X
!
k
∗
A BB A
∗k
k=0
follows from Exercise A.10.
P
k
∗ ∗k
n
∗ ∗(n−k−1) z.
If x1 = n−1
k=1 A BB A z for some z ∈ F , setting u(k) = B A
Then
n−1
n−1
X
X
x(n) =
An−l−1 BB ∗ A∗(n−l−1) z =
Ak BB ∗ A∗k z = x1 .
l=0
k=0
This shows that
R
B AB A2 B · · · An−1 B
⊂ R0 .
2
R0 is called the reachable subspace of (A, B). We use notation
Mc = B AB A2 B · · · An−1 B
and
Wc (n) =
n−1
X
Ak BB ∗ A∗k ,
k=0
and call them the controllability matrix and the n-step controllability Gramian
of (A, B) respectively. The proof of Theorem 2.1 also shows that if x1 is reachable, it can be reached in n steps.
Proposition 2.1 R0 is smallest A-invariant subspace containing R(B).
Proof
By the Cayley-Harmilton Theorem,
An = −a1 An−1 − a2 An−2 − · · · − an I.
2.2. CONTROLLABILITY
29
Hence
AR0 = R
AB A2 B A3 B · · · An B
AB A2 B · · · An−1 B −a1 An−1 B − a2 An−2 B − · · · − an B
⊂ R B AB A2 B · · · An−1 B
=R
= R0 .
This shows that R0 is A-invariant. Now let us assume that S is another Ainvariant subspace containing R(B). Then
R(B) ⊂ S
R(AB) = AR(B) ⊂ AS ⊂ S
R(A2 B) = AR(AB) ⊂ AS ⊂ S
..
.
Therefore
R0 = R(B) + R(AB) + · · · + R(An−1 B) ⊂ S.
2
Definition 2.1 The system A B is said to be controllable if for all x0 , x1 ∈
Rn , there exist k1 > 0 and u such that x(0) = x0 and x(k1 ) = x1 .
Theorem 2.2 The following statements are equivalent:
1. A B is controllable.
2. R0 = Fn .
3. rankMc = n
4. rankWc (n) = n.
Proof The equivalence of 2,3,4 is given by Theorem 2.1. If A B is
controllable, then every x1 ∈ Fn can be reached from 0 and hence R0 = Fn . If
rankWc (n) = n, let
u(k) = B ∗ A∗(n−k−1)
n−1
X
!−1
Ai BB ∗ A∗i
(x1 − An x0 ).
i=0
Then
n
n−1
X
x(n) = A x0 +
l=0
A
(n−l−1)
∗
BB A
∗(n−l−1)
n−1
X
!−1
i
∗
A BB A
i=0
∗i
(x1 −An x0 ) = x1 .
2
Let {x1 , x2 , . . . , xr } be a basis of R0 . Augment this basis to form a basis of
Fn :
{x1 , · · · , xr , xr+1 , · · · , xn }.
30
CHAPTER 2. LTI STATE SPACE SYSTEM ANALYSIS
From a computational point of view, we should and can choose this basis to be
an orthonormal basis. In this case, following Exercise A.10, {xr+1 , xr+2 , . . . , xn }
forms an orthonormal
basis of N (Wc (n)).
Let P = x1 x2 · · · xn . Since R0 is A-invariant and R(B) ⊂ R0 , we
have
Ã11 Ã12
B̃1
−1
−1
.
P AP =
P B=
0
0 Ã22
If we let x(k) = P z(k), then system (2.9) becomes
z1 (k)
Ã11 Ã12
z1 (k + 1)
B̃1
+
u(k).
=
z2 (k)
z2 (k + 1)
0
0 Ã22
(2.4)
The subsystem
z1 (k + 1) = Ã11 z1 (k) +
Ã12
z2 (k)
B̃1
u(k)
is controllable but the subsystem
z2 (k + 1) = Ã22 z2 (k)
is completely uncontrollable. The decomposition given by (2.4) is called the
controllable and completely uncontrollable decomposition.
The eigenvalues of Ã11 are called the controllable modes of the system
A B and
of Ã22 are called the uncontrollable modes of
the eigenvalues
the system A B .
Theorem 2.3 (Popov-Belevitch-Hautus) A B is controllable iff
rank A − λI B = n
for all λ ∈ C.
Proof If rank A − λI B < n for some λ ∈ C, then there exists x ∈ Cn
such that
x∗ A − λI B = 0,
i.e., x∗ A = λx∗ and x∗ B = 0. Then
x∗ B AB A2 B · · · An−1 B = 0
This shows
the “only if” part.
If A B is not controllable, then it has a controllable and completely
uncontrollable decomposition. In this case,
Ã11 − λI
Ã12
B̃1
rank A − λI B = rank
.
0
Ã22 − λI 0
Clearly, if λ ∈ σ(Ã22 ), then rank A − λI B ≤ n. This shows the “if” part.
2
Since any λ which can make rank A − λI B < n must belong to σ(A),
we can replace λ ∈ C by λ ∈ σ(A) in Theorem 2.3.
2.3. OBSERVABILITY
2.3
31
Observability
Consider system
x(k + 1) = Ax(k)
(2.5)
y(k) = Cx(k)
whose
is assumed to be zero. This system is often referred to as (C, A)
#
" input
A
. An initial state x0 ∈ Fn is said to be unobservable if y(k) = 0 for all
or
C
k ≥ 0 when x(0) = x0 . Denote the set of all unobservable state by N 0 .
Theorem 2.4
N 0 = N
Proof
C
CA
CA2
..
.
CAn−1
= N
If x0 ∈ N 0 , then CAk x0 = y(k) = 0
C
CA
2
x0 ∈ N CA
..
.
CAn−1
n−1
X
!
A∗k C ∗ CAk
.
k=0
for k ≥ 0 and hence
.
(2.6)
On the other hand, if (2.6) holds, then y(k) = CAk x0 = 0 for 0 ≤ k ≤ n − 1. By
the Cayley-Hamilton theorem, we have CAk x0 = 0 for all k ≥ 0. This shows
the first equality. The second equality follow from Exercise A.10.
2
N 0 is called the unobservable subspace of (C, A). We use notation
C
CA
CA2
Mo =
..
.
CAn−1
and
Wo (n) =
n−1
X
A∗k C ∗ CAk
k=0
and call them the observability matrix and the n-step observability Gramian of
(C, A) respectively.
Proposition 2.2 N 0 is the largest A-invariant subspace contained in N (C).
32
CHAPTER 2. LTI STATE SPACE SYSTEM ANALYSIS
Proof If x ∈ AN 0 , then there exists x0 ∈ N 0 such that x = Ax0 . Since
CAk x0 = 0 for all k ≥ 0, it follows that CAk x = CAk+1 x0 = 0 for all k ≥ 0,
i.e., x ∈ N 0 .
Now let us assume that S is another A-invariant subspace contained in
N (C) and let x ∈ S. Then x, Ax, . . . , An−1 x ∈ N (C) which is equivalent to
that Cx = CAx = · · · CAn−1 x = 0, i.e., x ∈ N 0 . Therefore S ⊂ N 0 .
2
"
#
A
Definition 2.2 The system
is said to be observable if every unknown
C
x0 ∈ Fn can be computed from y.
Theorem 2.5 The following statements are equivalent:
"
#
A
1.
is observable.
C
2. N 0 = {0}.
3. rankMo = n.
4. rankWo (n) = n.
Proof The equivalence of 2,3,4 is given by Theorem 2.4. If N 0 ̸= {0}, then
when y" = 0,# x0 can be anything in N 0 and cannot be uniquely determined.
A
Hence
is not observable.
C
Assume that rankWo (n) = n. Since
y(0) = Cx0
y(1) = CAx0
..
.
y(n − 1) = CAn−1 x0 ,
we have
n−1
X
k=0
A∗k C ∗ y(k) =
n−1
X
A∗k C ∗ CAk x0 = Wo (n)x0 .
k=0
This shows that x0 can be solved as
x0 = [Wo (n)]−1
n−1
X
A∗k C ∗ y(k).
2
k=0
Let {xr , · · · , xn } be a basis of N 0 and augment this basis to form a basis of
Fn :
{x1 , . . . , xr−1 , xr , . . . , xn }.
From a computational point of view, we should and can choose this basis to be
an orthonormal basis. In this case, following from Exercise A.10, {x1 , . . . , xr−1 }
forms an orthonormal basis of R(Wo (n)).
2.3. OBSERVABILITY
Let P =
have
33
x1 x2 · · · xn . Since N 0 is A-invariant and N 0 ⊂ N (C), we
Ã11 0
−1
, CP = C̃1 0 .
P AP =
Ã21 Ã22
If we let x(k) = P z(k), then system (2.5), when v = 0 and w = 0, becomes
z1 (k + 1)
Ã11 0
z1 (k)
=
(2.7)
z2 (k + 1)
z2 (k)
Ã21 Ã22
z1 (k)
.
y(k) = C̃1 0
z2 (k)
The subsystem
z1 (k + 1) = Ã11 z1 (k)
y1 (k) = C̃1 z1 (k)
is observable but the subsystem
z2 (k + 1) = Ã22 z2 (k) + Ã21 z1 (k)
y2 (k) = 0
is completely unobservable.
"
# The eigenvalues of Ã11 are called the observable
A
modes of the system
and the eigenvalues of Ã22 are called the unobC
"
#
A
servable modes of the system
. The decomposition (2.7) is called the
C
"
#
A
observable and completely unobservable decomposition of
.
C
"
#
A
Theorem 2.6 (Popov-Belevitch-Hautus)
is observable iff
C
A − λI
rank
=n
C
for all λ ∈ C.
Proof
that
A − λI
If rank
< n for some λ ∈ C, then there exists x ∈ Cn such
C
A − λI
x = 0,
C
i.e., Ax = λx and Cx = 0. Then
C
CA
CA2
..
.
CAn−1
x = 0.
34
CHAPTER 2. LTI STATE SPACE SYSTEM ANALYSIS
This shows
"
#the “only if” part.
A
If
is not observable, then it has an observable/completely unobservC
able decomposition. In this case,
Ã11 − λI
0
A − λI
rank
= rank
Ã21
Ã22 − λI .
C
C̃1
0
A − λI
Clearly, if λ ∈ σ(Ã22 ), then rank
≤ n. This shows the “if” part. 2
C
A − λI
Since any λ which can make rank
< n must belong to σ(A), we
C
can replace λ ∈ C by λ ∈ σ(A) in Theorem 2.6.
2.4
Kalman Decomposition
Consider system
x(k + 1) = Ax(k) + Bu(k)
y(k) = Cx(k) + Du(k).
Define
X 2 = R0 ∩ N 0
X 1 such that X 1 ⊕ X 2 = R0
X 4 such that X 2 ⊕ X 4 = N 0
X 3 such that X 1 ⊕ X 2 ⊕ X 3 ⊕ X 4 = Fn .
Note that there is quite a lot of freedom in choosing X 1 , X 3 , X 4 . We can always
choose them so that
X 1 , X 2 , X 3 , X 4 are orthogonal.
Define P = P1 P2 P3 P4 such that the columns of Pi is a basis of
X i . Then
Ã11 0 Ã13 0
B̃1
Ã21 Ã22 Ã23 Ã24
P −1 B = B̃2
P −1 AP =
0
0
0 Ã33 0
0
0
0 Ã43 Ã44
CP = C̃1 0 C̃3 0 .
Let x = P z. Then the original system becomes
Ã11 0 Ã13 0
Ã21 Ã22 Ã23 Ã24
z(k) +
z(k + 1) =
0
0 Ã33 0
0
0 Ã43 Ã44
y(k) = C̃1 0 C̃3 0 z(k) + Du(k).
B̃1
B̃2
u(k)
0
0
2.4. KALMAN DECOMPOSITION
"
35
#
Ã11 B̃1
-
C̃1
D̃
6
?
u
Ã21
?
"
-
+
#
Ã22 B̃2
0
0
?
l
Ã13
y-
+6
6
6
6
Ã23
6
"
Ã24
Ã33 0
C̃3
#
0
6
?
Ã43
?
"
Ã44 0
0 0
#
Figure 2.4: Kalman decomposition
This is called the Kalman decomposition. Under this decomposition, the system
has the block diagram shown in Figure 2.4.
Consider the impulse response of the system:
F (k) =
D
if k = 0
CAk−1 B if k > 0.
Since
CAk−1 B = CP (P −1 AP )k−1 P −1 B = C̃1 Ãk−1
11 B̃1 ,
36
CHAPTER 2. LTI STATE SPACE SYSTEM ANALYSIS
it follows that
F (k) =
D
if k = 0
k−1
C̃1 Ã11 B̃1 if k > 0,
i.e., the impulse response is only determined by the controllable and observable
part of the system. Similarly, the transfer function of the system satisfies
F̂ (z) = D + C(zI − A)−1 B = D + C̃1 (zI − Ã11 )−1 B̃1 ,
i.e., the transfer function is also only determined by the controllable and observable part of the system. Because of this, the eigenvalues of Ã22 , Ã33 , Ã44
are called hidden modes. Those of Ã33 and Ã44 are called uncontrollable hidden
modes and those of Ã22 and Ã44 are called unobservable hidden modes.
2.5
Poles and Zeros
We first need some preliminary materials on polynomial matrices. A matrix
whose entries are polynomials of a single variable is called a polynomial matrix.
A polynomial matrix can be written as
A(λ) = A0 λl + A1 λl−1 + · · · + Al
where Ai ∈ Fm×n .
Definition 2.3 The following three operations are called elementary row (column) operations on a polynomial matrix:
1. interchanging two rows (columns),
2. multiplying a row (column) by a nonzero constant,
3. multiplying a row (column) by a polynomial and adding it to another row
(column).
It is easy to check that the elementary row (column) operations correspond
to pre(post)-multiplying the polynomial matrix by the following polynomial
matrices, which are obtained by carrying out the same operation to the identity
matrix, respectively:
1
..
1
1
.
.
..
..
.
1
1
·
·
·
b(λ)
0
·
·
·
1
1
.
.. . . ..
..
,
.
c
,
.
.
.
.
.
1
1
1
·
·
·
0
.
..
..
1
.
..
1
.
1
1
2.5. POLES AND ZEROS
37
These matrices are called elementary matrices. Notice that the inverse of an
elementary matrix is still an elementary matrix.
Two polynomial matrices A(λ) and B(λ) of the same size are said to be
equivalent, denoted by A(λ) ∼ B(λ), if there exist elementary matrices
U1 (λ), U2 (λ), . . . , Up (λ)
and
V1 (λ), V2 (λ), . . . , Vq (λ)
such that
B(λ) = Up (λ) · · · U1 (λ)A(λ)V1 (λ) · · · Vq (λ).
It is easy to check that “∼” is an equivalence relation.
Theorem 2.7 Each polynomial matrix A(λ) is equivalent to
γ1 (λ)
γ2 (λ)
.
.
.
γ
(λ)
Γ(λ) =
r
0
..
.
0
(2.8)
where γi (λ), i = 1, . . . , r, are monic polynomials and γi |γi+1 .
The polynomial matrix Γ(λ) is called the Smith canonical form of A. r is
called the normal rank of A and γi (λ) are called the invariant polynomials of
A(λ). The Smith form of a polynomial matrix is unique.
Definition 2.4 A square polynomial matrix U (λ) is said to be unimodular if
det U (λ) is a nonzero constant.
An elementary matrix is apparently unimodular. Furthermore, the product
of a number of elementary matrices is also unimodular. Let W (λ) be a n × n
unimodular matrices. Then there exist elementary matrices U1 (λ), U2 (λ), . . .,
Up (λ) and V1 (λ), V2 (λ), . . . , Vq (λ) such that
γ1 (λ)
γ2 (λ)
W (λ) = Up (λ) · · · U1 (λ)
V1 (λ) · · · Vq (λ).
..
.
γn (λ)
Since the determinants
of W (λ), Uj (λ), Vk (λ) are all nonzero constants, it folQn
lows that i=1 γi (λ) is a nonzero constant and hence γi (λ) = 1 for all i =
1, 2, . . . , n. This shows that W (λ) is the product of a number of elementary
matrices. Therefore, a unimodular matrix can be alternatively defined as the
product of a number of elementary matrices and two polynomial matrices A(λ)
38
CHAPTER 2. LTI STATE SPACE SYSTEM ANALYSIS
and B(λ) can be defined to be equivalent if, alternatively, there exist unimodular matrices U (λ) and V (λ) such that
B(λ) = U (λ)A(λ)V (λ).
Given a polynomial matrix A(λ) of normal rank r, let δi (λ) be the monic
greatest common divisor of all i × i minors of A(λ). Observe that δi (λ) = 0
for i > r and define δ0 (λ) = 1. Then δi (λ)|δi+1 (λ). The polynomials δi (λ), i =
1, 2, . . . , r, are called the determinantal divisors of A(λ). It can be easily shown
that if A(λ) and B(λ) are equivalent, then they have the same determinantal
divisors and hence the same invariant polynomials. Now let A(λ) have Smith
form
γ1 (λ)
γ2 (λ)
.
.
.
.
γr (λ)
0
..
.
0
Then it has
i
Y
δi (λ) =
γj (λ)
j=1
and
δi (λ)
.
δi−1 (λ)
Let A ∈ Fn×n be a square matrix. Then λI − A is a polynomial matrix.
Note that the normal rank of λI − A is always n. Let the invariant polynomials
of λI − A be γi (λ), i = 1, 2, . . . , n. It is easy to see that
γi (λ) =
cA (λ) = δn (λ) = γ1 (λ)γ2 (λ) · · · γn (λ).
This shows that the eigenvalues of A are the roots of the invariant polynomials
of λI − A.
"
#
A B
Definition 2.5 The zeros of the system
are the roots of the invariC D
ant polynomials of
A − λI B
.
C
D
Example 2.2
Consider the system
1
0
"
#
0
A B
= 0
C D
1
0
0
1
0
0
0
1
0
0
0
0
1
0
0
0
0
0
0
1
1
2
0
0
0
0
0
1
1
0
0
0
.
2.5. POLES AND ZEROS
39
This system has two poles at 1 and two poles at 0. Let us carry out the following
elementary operations:
A − λI B
C
D
1−λ
0
0
1−λ
0
0
=
0
0
1
0
0
1
1−λ
0
−λ 1 − λ
0
0
∼
0
λ
0
0
0
0
0
0
0
1
0
0
∼
−λ(λ − 2) 0
0
0
0
0
1−λ
0
0 1 0
0
0 2 1 0
−λ 0 0 1
∼ λ
0 −λ 0 0
0
0
1
0 0 0
0
0
1 0 0
0
0
0 0 1 0
0 0 2 0 λ − 2 1
0
0 0 0 1
∼ 0
λ
0 0 0 0 0
0
1 0 0 0 0
0
0
0 1 0 0
1 0 0
0 0 1 0
0 1 0
0 0 0 0
0 0 0 1
∼ 0 0 1
0 0 0 0
0 0 0
1 0 0 0 0 0 0
0 0 0
0 1 0 0
0
1−λ
0
λ
0
0
0
0
0
0
1
0
1
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
1
0
0
0
0
0
0
1
0
0
1
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
.
0
0
1
0
0 λ(λ − 2)
Hence the system has two zeros which are at 0 and 2
Now consider transfer function
Ĝ(z) =
1
2
0
0
0
0
B(z)
a(z)
where
a(z) = z r + a1 z r−1 + · · · + ar
B(z) = B0 z r + B1 z r−1 + . . . Br .
Let U (z) and V (z) be unimodular matrices such that
γ1 (z)
γ2 (z)
..
.
γr (z)
U (z)B(z)V (z) =
0
..
.
0
0
1
1
0
0
0
40
CHAPTER 2. LTI STATE SPACE SYSTEM ANALYSIS
is in Smith canonical form. Then
β1 (z)
α (z)
1
U (z)Ĝ(z)V (z) =
β2 (z)
α2 (z)
..
.
βr (z)
αr (z)
0
..
.
0
where αi (z) and βi (z) are coprime, βi (z) divides βi+1 (z), αi+1 (z) divides αi (z).
The right hand side is called the Smith-McMillan form of Ĝ(z).
Definition 2.6 The roots of αi are called the poles of Ĝ. The roots of βi are
called the zeros of Ĝ.
Example 2.3
Consider the system in Example 2.1. Its transfer function is
z z−1
1
1
2z
z
1
z
=
.
Ĝ(z) = z −
2
1
z(z − 1)
z−1 z−1
Carry out the following elementary operations
1
0
z
1
z z−1
.
∼
∼
0 z(z − 2)
0 −z + 2
2z
z
Hence the Smith-McMillan form of Ĝ is
1
0
z(z − 1)
z − 2 .
0
z−1
The poles of Ĝ are 0, 1, 1 and the zeros of Ĝ are 2.
We see that the poles and zeros of a state space system and those of its
transfer function are different. However, the poles of the transfer function are
always the poles of the system and the same for the zeros. We explained in the
last section that the missing poles of the system are the hidden modes of the
system. Where do the hidden modes go in the process of computing the transfer
function? They are canceled by some zeros. Those zeros of the system that do
not appear as zeros of the transfer function are called decoupling zeros and they
are exactly equal to the hidden modes. Since they, similar to the hidden modes,
are caused by the uncontrollable and unobservable parts of the system, they
can be classified as uncontrollable decoupling zeros and unobservable decoupling
zeros. It might be convenient to call the zeros of a state space system as its
internal zeros and those of its transfer function as its external zeros.
2.6. INTERNAL STABILITY
2.6
41
Internal Stability
Consider system (2.1). Its zero input state response is described by equation
x(k + 1) = Ax(k)
and its solution is given by
x(k) = Ak x(0).
Definition 2.7 System (2.1) is said to be internally stable if its zero input
state response goes to zero as k → ∞ for all initial conditions.
Apparently, the internal stability of A depends only on A matrix. We often
use the (Schur) stability of square matrix A to mean the internal stability of
system (2.1).
Theorem 2.8 A is stable iff ρ(A) < 1.
Proof Assume that A has an eigenvalue λ with |λ| ≥ 1. Let x0 be an
eigenvector of A corresponding to λ. Then
x(k) = Ak x0 = λk x0 ,
which does not converge to zero as k → ∞. This proves the necessity.
If all eigenvalues of A have absolute values less than 1, by the material in
Section 2.1 x(k) has the form
x(k) =
ni
l X
X
pij (k)λk−j+1
i
i=1 j=1
where λi are the eigenvalues of A and pij are polynomial vectors of k. Hence
limk→∞ x(k) = 0. This shows the sufficiency.
2
An important tool in studying stability (and other issues) is the Lyapunov
analysis.
Define a quadratic function L(x) = x∗ Xx, where X ∈ Hn (F) and X > 0.
The function L(x) can also be written as ∥X −1/2 x∥2 , i.e., it is the weighted
2-norm of x. It has the meaning of the amount of generalized energy stored in
the system when x(k) = x. If
A∗ XA − X < 0,
then
L[x(k + 1)] − L[x(k)] = x∗ (k)(A∗ XA − X)x(k) < 0
for all x(k) ̸= 0, i.e., L[x(k)] is a decreasing function of k as k progresses.
Since L[x(k)] is bounded from below by 0, it has to converge. It can be further
argued that it has to converge to 0. Hence x(k) also converges to zero, i.e., A
is stable. Such a quadratic function L(x) is called a Lyapunov function of A.
Consequently, if A has a Lyapunov function, then it is stable. Notice that not
every positive definite quadratic function is a Lyapunov function. If we take a
42
CHAPTER 2. LTI STATE SPACE SYSTEM ANALYSIS
candidate X > 0, but A∗ XA−X does not turn out to be negative definite, then
L(x) = x∗ Xx is not a Lyapunov function of A and nothing can be said about
the stability of A. The next natural question is then whether a stable matrix
A always has a Lyapunov function. The following theorem gives an affirmative
answer to this question.
Theorem 2.9 A is stable iff there exists X > 0 such that
A∗ XA − X < 0.
Proof It remains to show the necessity. If A is stable, then there exists
P ∈ GL(n, F) such that ∥P −1 AP ∥2 < 1 (Cf. Exercise A.39). This means
(P −1 AP )∗ (P −1 AP ) < I.
Define X = (P P ∗ )−1 . Then immediately A∗ XA − X < 0.
2
∗
One can combine the inequalities X > 0 and A XA − X < 0 together to
state that A is stable iff the LMI
X
0
>0
0 X − A∗ XA
is feasible.
How to determine the feasibility of this LMI is a big issue. Failing to find
such an X by any available means does not rule out its existence. Even if
we know that the LMI is feasible, how to find a solution X, i.e., how to find
a Lyapunov function, is another issue. There are indeed effective numerical
methods to determine the feasibility of an LMI and, if feasible, to find a solution
of it. These methods can of course been used. However, it turns out a better
way to solve the particular LMI on hand exists by starting from an arbitrary
Q > 0 and then solving for a suitable X satisfying equation A∗ XA − X = −Q.
A matrix equation of the form
A∗ XA − X = −Q
(2.9)
is called a Lyapunov equation. This equation is a linear equation in matrix X.
The tool to convert it to the standard form of linear equation is the Kronecker
product.
Let us forget for a while the particular meanings of matrices A, B, C, D in
our context. Let A ∈ Fm×n and B ∈ Fp×q be arbitrary matrices. Then define
the Kronecker product of A and B as
a11 B · · · a1n B
..
..
mp×nq
A⊗B =
.
∈F
.
.
am1 B · · · amn B
Lemma 2.1 Let A ∈ Fm×n , B ∈ Fp×q , C ∈ Fn×l , and D ∈ Fq×r .
1. (A ⊗ B)(C ⊗ D) = AC ⊗ BD.
2.6. INTERNAL STABILITY
43
2. If A, B are invertible, then (A ⊗ B)−1 = A−1 ⊗ B −1 .
3. If A, B are square, then σ(A ⊗ B) = σ(A)σ(B).
We also need the so-called “vec” operator. Let
X=
x1 x2 · · · xn
∈ Fm×n .
Then define
vecX =
x1
x2
..
.
∈ Fmn .
xn
Lemma 2.2 vec(AXB) = (B ′ ⊗ A)vecX.
Taking “vec” in both sides of (2.9), we obtain
(A′ ⊗ A∗ − I)vec(X) = −vecQ.
This is linear equation in a familiar form with unknown vecX.
Proposition 2.3 Lyapunov equation (2.9) has unique solution for all Q iff
1 ̸∈ σ(A)σ(A).
Proof Since σ(A′ ⊗ A∗ − I) = σ(A′ )σ(A∗ ) − 1 = σ(A)σ(A) − 1, it follows
that A′ ⊗ A∗ − I is nonsingular iff 1 ̸∈ σ(A)σ(A).
2
In particular, if A is (Schur) stable, then the Lyapunov equation (2.9) has
a unique solution for each Q.
Proposition 2.4 If the Lyapunov equation has unique solution X for each Q,
then X is Hermitian iff Q is Hermitian.
Proof
If X is Hermitian, then
Q∗ = (−A∗ XA + X)∗ = −A∗ XA + X = Q,
i.e., Q must be Hermitian. If X is not Hermitian, let
1
X1 = (X + X ∗ )
2
1
and X2 = (X − X ∗ ).
2
Then X1 is Hermitian, X2 , which is nonzero, is Skew-Hermitian, i.e., X2∗ =
−X2 , and X = X1 + X2 . Define
Q1 = (−A∗ X1 A + X1 )∗
and Q2 = (−A∗ X2 A + X2 )∗ .
Then Q1 is Hermitian, Q2 is skew-Hermitian, and Q = Q1 + Q2 . Notice that
Q2 is nonzero because of the uniqueness of the solution. Therefore Q is not
Hermitian.
2
44
CHAPTER 2. LTI STATE SPACE SYSTEM ANALYSIS
Theorem 2.10 If A is stable, then for each Q > 0 the solution X of Lyapunov
equation (2.9) is given by
∞
X
X=
A∗k QAk .
k=0
Conversely, if there exist Q > 0 and X > 0 such that the Lyapunov equation
(2.9) is satisfied, then A is stable.
Proof If A is stable, then Proposition 2.3 implies that the Lyapunov equation
(2.9) has a unique solution for each Q. Since
X=
∞
X
A∗k QAk
k=0
converges and satisfies the Lyapunov equation, it must be the unique solution.
Since Q > 0, it follows that X > 0. This shows the first statement.
To prove the second statement, let λ be any eigenvalue of A and x be a
corresponding eigenvector. Then
x∗ A∗ XAx − x∗ Xx = −x∗ Qx.
This gives
(|λ|2 − 1)x∗ Xx = −x∗ Qx.
Since x∗ Qx > 0 and x∗ Xx > 0, it follows that |λ|2 − 1 < 0.
2
In many applications, the Q matrix in Lyapunov equation (2.9) is not positive definite but rather positive semi-definite. From the proof of Theorem 2.10,
it is easy to see that if A is stable, then Q ≥ 0 implies X ≥ 0. However, if
we have Q ≥ 0 and X ≥ 0 (even X > 0) satisfying (2.9), we cannot conclude
that A is stable. For example, if A = 1 and Q = 0, then X = 1 satisfying
(2.9) but A is unstable. In the following, we will see that if we impose some
extra condition on Q when it is positive semi-definite, then we may be able to
conclude the stability of A.
#
"
A
to be
If A is stable, define then the observability Gramian of
C
Wo :=
∞
X
A∗k C ∗ CAk .
k=0
Then X = Wo is the unique solution of the Lyapunov equation
A∗ XA − X = −C ∗ C.
"
Notice that
A
C
(2.10)
#
is observable iff Wo > 0.
"
#
A
Theorem 2.11 Assume that
is observable and Hermitian matrix X
C
satisfies (2.10). Then A is stable if X > 0.
2.6. INTERNAL STABILITY
Proof
Then
45
Let λ be any eigenvalue of A and x be a corresponding eigenvector.
x∗ A∗ XAx − x∗ Xx = −x∗ C ∗ Cx.
This gives
(|λ|2 − 1)x∗ Xx = −x∗ C ∗ Cx.
Since x∗ Xx > 0 and by the PBH observability test x∗ C ∗ Cx > 0, it follows that
|λ|2 − 1 < 0.
2
Similarly, If A is stable, define the controllability Gramian of A B to
be
∞
X
Wc :=
Ak BB ∗ A∗k .
k=0
then Y = Wc is the unique solution of the Lyapunov equation
AY A∗ − Y = −BB ∗ .
Notice that
A B
(2.11)
is controllable iff Wc > 0.
Theorem 2.12 Assume that A B is controllable and Hermitian matrix Y
satisfies (2.11). Then A is stable if Y > 0.
Finally, we relate the internal stability with the BIBO stability.
"
A B
C D
#
Theorem 2.13 If A is stable, then the system
is BIBO stable. If
#
"
A B
is BIBO stable, controllable and observable, then it is internally
C D
stable.
Proof
Each Gij (k) is of the form
if k < 0
0
d
if k = 0
Gij (k) =
k−1
cA b if k > 0,
for some b, c, d. Since cAk−1 b is of the form
q1 (k)λk−1
+ q2 (k)λk−1
+ · · · + ql (k)λlk−1 .
1
2
where λ1 , λ2 , . . . , λl arePthe distinct eigenvalues of A and q1 , q2 , . .P
. , ql are poly∞
k−1
nomials of k, and since k=1 |pi (k)λ
it follows that ∞
k=−∞ |Gij (k)|
P∞1 | converges,
Pm
converges for all i, j. Therefore k=−∞ j=1 |Gij (k)| converges for all i. This
shows the first statement. The proof of the second statement is left as an
exercise.
2
46
2.7
CHAPTER 2. LTI STATE SPACE SYSTEM ANALYSIS
Simple State Space System Connections and Duality
We have seen the elementary system operations such as addition, multiplication,
inversion, conjugation, and transposition from an input output point of view.
If systems are given in terms of state space equations, we are interested in
finding state space equations of the systems resulted from the operations. The
following theorem is stated without proof.
Theorem 2.14 Assume the compatibility of the operations. Then
"
A1 B 1
C1 D 1
"
A1
C1
A1 0
B1
B2
+
= 0 A2
C1 C2 D 1 + D 2
#"
#
A1 B 1 C 2
B1 D2
A2 B2
B1
A2
B2
= 0
D1
C2 D 2
C1 D 1 C2
D1 D2
#−1 "
#
"
A − BD−1 C BD−1
A B
=
C D
−D−1 C
D−1
"
#′ "
#
A B
A′ C ′
.
=
C D
B ′ D′
#
"
A2 B2
C2 D 2
#
It is seen that the sum and product of state space systems are state space
systems. However, the inverse of a state space system is not necessarily a state
space system. This follows from the fact that the system inversion does not
preserve causality. Let us use the unit delay system S as an example to further
demonstrate this point. It is easy to see that
"
#
0 I
S=
,
I 0
but S −1 has no state space representations. Theorem 2.14 implicitly states that
"
A B
C D
#−1
exists and is a state space system iff D is invertible.
Since a state space system is automatically causal, its conjugate in general
cannot be represented by a state space system unless we define the anti-causal
state space system.
The dual system is very useful in the study of many different dualities in
system theorem. One example of duality is that between controllability and
observability.
Theorem 2.15
2.8. LINEAR FRACTIONAL TRANSFORMATION OF STATE SPACE SYSTEMS47
#
"
#′
A B
A B
1.
is controllable (or observable) iff
is observable (or
C D
C D
controllable).
#
"
A B
are the same as the
2. The controllable (or observable) modes of
C D
"
#′
A B
observable (or controllable) modes of
.
C D
"
The proof of Theorem 2.15 is trivial.
2.8
Linear Fractional Transformation of State Space
Systems
Assume that P has state space equation
x(k + 1) = Ax(k) + B1 w(k) + B2 u(k)
z(k) = C1 x(k) + D11 w(k) + D12 u(k)
y(k) = C2 x(k) + D21 w(k) + D22 u(k)
and Q has state space equation
x̃(k + 1) = Ãx̃(k) + B̃y(k)
u(k) = C̃ x̃(k) + D̃y(k).
Compactly, we write
A B1 B2
P = C1 D11 D12
C2 D21 D22
and
"
Q=
à B̃
C̃
D̃
(2.12)
#
.
The following theorem concerns the state space representation of F l (P , Q) and
F u (P , Q).
Theorem 2.16
"
#
A B1 B2
à B̃
F l C1 D11 D12 ,
C̃ D̃
C2 D21 D22
A + B2 D̃(I − D22 D̃)−1 C2
B2 (I − D̃D22 )−1 C̃
−1
B̃(I − D22 D̃) C2
à + B̃(I − D22 D̃)−1 D22 C̃
=
C1 + D12 (I − D̃D22 )−1 D̃C2
D12 (I − D̃D22 )−1 C̃
B1 + B2 D̃(I − D22 D̃)−1 D21
B̃(I − D22 D̃)−1 D21
D11 + D12 D̃(I − D22 D̃)−1 D21
48
CHAPTER 2. LTI STATE SPACE SYSTEM ANALYSIS
"
#
A B1 B2
à B̃
F u C1 D11 D12 ,
C̃ D̃
C2 D21 D22
A + B1 (I − D̃D11 )−1 D̃C1
B1 (I − D̃D11 )−1 C̃
−1
à + B̃D11 (I − D̃D11 )−1 C̃
B̃(I − D11 D̃) C1
=
C2 + D21 (I − D̃D11 )−1 D̃C1
D21 (I − D̃D11 )−1 C̃
B1 + B2 D̃(I − D11 D̃)−1 D12
B̃(I − D11 D̃)−1 D12
−1
D22 + D21 (I − D̃D11 ) D̃D12
Implicitly, Theorem 2.16 implies that F l (P , Q) is a state space system iff
I − D22 D̃ is invertible and F u (P , Q) is a state space system iff I − D̃D11 is
invertible.
The formulas look complicated. However, in most of the applications, we
have D22 = 0 for the lower LFT or D11 = 0 for the upper LFT. In this case,
the formulas are much simpler and looks more attractive, which are state in the
following for easy reference and better appreciation.
Corollary 2.1
"
#
A B1 B2
à B̃
F l C1 D11 D12 ,
C̃ D̃
C2 D21
0
A + B2 D̃C2
B2 C̃
B1 + B2 D̃D21
B̃D21
B̃C2
Ã
=
C1 + D12 D̃C2 D12 C̃ D11 + D12 D̃D21
"
#
A B1 B2
à B̃
F u C1
0 D12 ,
C̃ D̃
C2 D21 D22
A + B1 D̃C1
B1 C̃
B1 + B2 D̃D12
B̃C1
Ã
B̃D12
=
.
C2 + D21 D̃C1 D21 C̃ D22 + D21 D̃D12
Now if Q is a generalized state space system given by
à B̃1 B̃2
Q = C̃1 D̃11 D̃12 .
C̃2 D̃21 D̃22
Then the state space representation of P ⋆ Q is given by the following theorem.
Theorem 2.17
"
#
à B̃1 B̃2
A B1 B2
A B
C1 D11 D12 ⋆ C̃ D̃
1
11 D̃12 =
C D
C2 D21 D22
C̃2 D̃21 D̃22
2.8. LINEAR FRACTIONAL TRANSFORMATION OF STATE SPACE SYSTEMS49
where
A + B2 D̃11 (I − D22 D̃11 )−1 C2
B2 (I − D̃11 D22 )−1 C̃1
A=
B̃1 (I − D22 D̃11 )−1 C2
à + B̃1 (I − D22 D̃11 )−1 D22 C̃1
B2 (I − D̃11 D22 )−1 D̃12
B1 + B2 D̃11 (I − D22 D̃11 )−1 D21
B=
B̃2 + B̃1 (I − D22 D̃11 )−1 D22 D̃12
B̃1 (I − D22 D̃11 )−1 D21
D12 (I − D̃11 D̃22 )−1 C̃1
C1 + D12 D̃11 (I − D22 D̃11 )−1 C2
C=
D̃21 (I − D22 D̃11 )−1 C2
C̃2 + D̃21 (I − D22 D̃11 )−1 D22 C̃1
D11 + D12 D̃11 (I − D22 D̃11 )−1 D21
D12 (I − D̃11 D22 )−1 D̃12
D=
.
D̃21 (I − D22 D̃11 )−1 D21
D̃22 + D̃21 (I − D22 D̃11 )−1 D22 D̃12
Again if D22 = 0 or D̃11 = 0, then the formula is much simpler and more
elegant.
Corollary 2.2
à B̃1 B̃2
A B1 B2
C1 D11 D12 ⋆
C̃1 D̃11 D̃12
C2 D21
0
C̃2 D̃21 D̃22
B1 + B2 D̃11 D21
B2 D̃12
A + B2 D̃11 C2
B2 C̃1
B̃1 C2
Ã
B̃1 D21
B̃2
=
C1 + D12 D̃11 C2 D12 C̃1 D11 + D12 D̃11 D21 D12 D̃12
D̃21 C2
C̃2
D̃21 D21
D̃22
à B̃1 B̃2
A B1 B2
C1 D11 D12 ⋆
C̃1 D̃11 D̃12
C2 D21 D22
C̃2 D̃21
0
A
B2 C̃1
B1
B2 D̃12
B̃ C
à + B̃1 D22 C̃1
B̃1 D21
B̃2 + B̃1 D22 D̃12
1 2
=
C1
D12 C̃1
D11
D12 D̃12
D̃21 C2 C̃2 + D̃21 D22 C̃1 D̃21 D21 D̃22 + D̃21 D22 D̃12
.
The following corollary follows from Theorem 1.5 and Theorem 2.16.
A B1 B2
Corollary 2.3 Let P = C1 D11 D12 : ℓm1 +m2 (Z) → ℓm2 +m1 (Z). If
C2 D21 D22
D12 , D21 and D are invertible, then Q 7→ F l (P , Q) is bijection from
(
"
#
)
à B̃
Q=
: I − DD̃ is nonsingular
C̃ D̃
to
(
R=
"
Ā B̄
C̄ D̄
#
)
−1
−1
: I − D̄D21
D22 (D12 + D11 D21
D22 )−1 is nonsingular .
50
2.9
CHAPTER 2. LTI STATE SPACE SYSTEM ANALYSIS
Controllability indices and Observability indices
Consider the system
x(k + 1) = Ax(k) + Bu(k),
n×m . Assume
A
B
where
A
∈
and
B
∈
F
is controllable and B =
b1 b2 · · · bm . Then the controllability matrix
Mc = B AB · · · An−1 B
Fn×n
has n linearly independent column vectors, which can form a basis for the ndimensional space. The selection of these n linearly independent vectors from
nm vectors is not unique unless m = 1. Let us arrange the columns of Mc
in the Young diagram shown in the following table to illustrate the selection
procedure.
I
A
A2
A3
A4
A5
A6
A7
b1
1
1
1
0
0
0
0
0
b2
1
0
0
0
0
0
0
0
b3
1
1
1
1
0
0
0
0
b4
0
0
0
0
0
0
0
0
Table 2.1: An example of the controllability matrix searched by rows.
The diagram has m columns representing m inputs {b1 , . . . , bm }, and n
rows representing n powers {I, A, . . . , An−1 }. The (i, j)th cell represents a
corresponding vector Ai−1 bj . There are different ways to choose the n linearly
independent vectors. One way that comes naturally is to search the diagram
by rows, which means the vectors are picked from left to right row by row, and
stop after n vectors are selected.
Fact 2.1 The selected vectors always appear on the top of each column.
Put 1 in the picked cells, and put 0 otherwise. Let the column sum of the
jth be kj . The bag
of numbers k = {k1 , . . . , km } is called the controllability
indices of A B . The
maximal
number k̄ = max{k1 , . . . , km } is called the
A
B
controllability index
of
. For example, the selected vectors of a con
trollable system A B with n = 8 and m = 4 are illustrated in Table 2.1.
Then the controllability indices are {3, 1, 4, 0} and controllability index is 4.
Another scheme that is often discussed in the literature is to search the
diagram by columns. The bag of column sums of the corresponding table is
called the Hermite indices. If we recall the proof
4.1, then the bag
of Theorem
{n1 , . . . , nm } is exactly the Hermite indices of A B .
Lemma 2.3 The controllability indices of A + BF BQ are the same as
those of A B for all F ∈ Fm×n and nonsingular Q ∈ Fm×m .
2.9. CONTROLLABILITY INDICES AND OBSERVABILITY INDICES 51
Proof Let {k1∗ , k2∗ , . . . , kn∗ } represent the row sums of the diagram, which
are the conjugate of the column sums {k1 , k2 , . . . , km }. The conjugate here
means the conjugate sequence of an integer sequence. If {kj } ∈ N m , we define
the conjugate sequence {ki }∗ as
ki∗ = card{i|kj ≥ j}, j = 1, . . . , n.
Therefore, {ki }∗∗ = {ki }, i = 1, . . . , m. Since the vectors are selected by rows,
it follows that the row sums of the diagram can be represented as
k1∗ = dimR(B),
k2∗ = dimR( B AB ) − dimR(B),
..
.
∗
kn = dimR( B AB · · · An−1 B ) − dimR( B AB · · · An−2 B ).
Since Q is a nonsingular matrix,
R([ BQ (A + BF )BQ · · · (A + BF )j BQ ) = R( B AB · · · Aj B ).
Therefore the bag of row sums {k1∗ , k2∗ , . . . , kn∗ } of A + BF BQ is the same
as those of A B . As the column sums are the conjugate of the row
sums, from the definition
of the conjugate
sequence, the bag of column sums
{k1 , k2 , . . . , km } of A + BF BQ is also the same as those of A B .
Similarly, consider the system
x(k + 1) = Ax(k)y(k) = Cx(k),
"
where A ∈
Fn×n
and C ∈
Fp×n .
Assume
A
C
#
is observable and C =
c1
c2
..
.
.
cp
Then the observability matrix Mo =
C
CA
..
.
has n linearly independent
CAn−1
row vectors. The selection of these n linearly independent vectors from pn
vectors is not unique unless p = 1. Let us arrange the columns of Mc in the
Young diagram shown in the following table to illustrate the selection procedure.
c1
c2
c3
I
1
1
1
A
1
0
1
A2
0
0
0
A3
0
0
0
A4
0
0
0
A5
0
0
0
Table 2.2: An example of the observability matrix searched by columns.
52
CHAPTER 2. LTI STATE SPACE SYSTEM ANALYSIS
The diagram has p rows representing p inputs {c1 , . . . , cp }, and n columns
representing n powers {I, A, . . . , An−1 }. The (i, j)th cell represents a corresponding vector Ai−1 bj .Let us search the diagram by columns, which means
the vectors are picked from top to bottom column by column, and stop after n
vectors are selected.
Fact 2.2 The selected vectors always appear on the very left few of each row.
Put 1 in the picked cells, and put 0 otherwise. Let the row sum of ith
" be l#i .
A
The bag of numbers l = {l1 , . . . , lp } is called the observability indices of
.
C
The
maximal
number ¯l = max{l1 , . . . , lp } is called the observability
"
#
" index
# of
A
A
. For example, the selected vectors of a controllable system
with
C
C
n = 8 and p = 4 are illustrated in Table 2.2. Then the observability indices are
{2, 1, 2} and observability index is 2.
"
#
A + LC
Lemma 2.4 The observability indices of
are the same as those of
QC
"
#
A
for all L ∈ Fn×p and nonsingular Q ∈ Fp×p .
C
Proof Let {l1∗ , l2∗ , . . . , ln∗ } represent the column sums of the diagram,
which are the conjugate of the row sums {l1 , l2 , . . . , lp }. Since the vectors are
selected by columns, it follows that the column sums of the diagram can be
represented as
l1∗ = n − dimN (C),
l2∗
C
),
= dimN (C) − dimN (
CA
..
.
ln∗ = dimN (
C
CA
..
.
CAn−2
) − dimN (
Since Q is a nonsingular matrix,
QC
QC(A + LC)
N (
..
.
QC(A + LC)j
)
=
N
(
C
CA
..
.
CAn−1
C
CA
..
.
).
).
CAj
"
#
A
+
LC
Therefore the bag of row sums {l1∗ , l2∗ , . . . , ln∗ } of
is the same as
QC
2.9. CONTROLLABILITY INDICES AND OBSERVABILITY INDICES 53
#
A
. As the column sums are the conjugate of the row sums, from
those of
C
the"definition #
of the conjugate sequence, the"bag of
# column sums {l1 , l2 , . . . , lp }
A + LQ
A
of
is also the same as those of
.
QC
C
"
Exercises
2.1 Compute the step response of the
0 1
x(k + 1) = 0 0
0 0
y(k) = 3 2
following system.
0
1
0 x(k) + 2 u(k).
1
3
2
1 x(k).
2.2 For system
x(k + 1) = Ax(k),
x(0) = x0 ,
show that for each initial condition x0 , there exists a constant α such that
det x(k) Ax(k) · · · An−1 x(k) = α[det(A)]k .
2.3 Consider a second order system
x(k + 1) = Ax(k)
4
when x(0) =
We know that x(1) =
−2
1
1
. Find x(3) when x(0) =
when x(0) =
2
0
2.4 For an internally stable system
"
A B
C D
1
1
and x(1) =
5
−2
.
#
,
prove that when u is a step function of the form u(k) = u0 , k ≥ 0, it holds
lim y(k) = [D + C(I − A)−1 B]u0 .
k→∞
2.5 A mechanical system can often be described by a difference equation
M [p(k + 2) − 2p(k + 1) + p(k)] + D[p(k + 1) − p(k)] + Kp(k) = Bf (k)
where M, D, K ∈ Rn×n , B ∈ Rn×m , and M is nonsingular. Such an
equation is called a second order equation.
54
CHAPTER 2. LTI STATE SPACE SYSTEM ANALYSIS
1. Choose appropriate state variables to obtain a state space realization
of this system with input f (k) and output p(k).
2. Show that this system is controllable if and only if
rank M (z − 1)2 + D(z − 1) + K B = n
for all z ∈ C.
3. Show that this system is always observable.
2.6 Find a control sequence u to drive the state of the system
2 0 0
2
x(k + 1) = 1 1 1 x(k) + 3 u(k)
0 0 0
0
0
0 to the origin. Can you drive the state from the origin to
from
1
0
0 ? Why?
1
2.7 For what values of the parameter α is the system
1
1 α 1
x(k + 1) = 0 1 0 x(k) + 1 u(k)
1
0 0 0
0 1 0
x(k)
y(k) =
1 2 0
controllable? Observable? For those values of α corresponding to an
uncontrollable or unobservable system, find the Kalman decomposition of
the system.
2.8 Show the following two statements:
1. If A B is controllable, then A + BF B is controllable for
all"F . #
"
#
A
A + LC
is observable, then
is observable for all L.
2. If
C
C
2.9 Consider a single input system A B , i.e., B only has one column.
Assume it is controllable. Show that there is a nonsingular matrix P such
that
0 · · · 0 −an
1
1 · · · 0 −an−1
0
P −1 AP = . .
P −1 B = ..
..
..
.
.
.
.
. .
.
0 · · · 1 −a1
0
where ai are the coefficients of the characteristic polynomial of A, i.e.,
det(zI − A) = z n + a1 z n−1 + · · · + an .
2.9. CONTROLLABILITY INDICES AND OBSERVABILITY INDICES 55
#
A
, i.e., C only has one row. Assume
2.10 Consider a single output system
C
it is observable. Show that there is a nonsingular matrix P such that
0
1
···
0
..
..
..
..
.
.
.
P −1 AP = .
0
0
···
1
−an −an−1 · · · −a1
CP = 1 0 · · · 0
"
where ai are the coefficients of the characteristic polynomial of A, i.e.,
det(zI − A) = z n + a1 z n−1 + · · · + an .
2.11 Consider an n-th order system
x(k + 1) = Ax(k) + Bu(k).
A state x0 is said to be K-step null controllable if there exists input u such
that x(0) = x0 and x(K) = 0. A state x0 is said to be null controllable if
it is K-step null controllable for some K ≥ 0. Denote the set of all null
controllable states by C 0 .
1. Show that C 0 = R0 + N (An ). Is C 0 an invariant subspace of A?
2. System A B is said to be null controllable if all possible initial
states are null controllable, i.e., C 0 = Fn . Show that A B is null
controllable iff
rank A − λI B = n
for all λ ∈ C \ {0}.
2.12 Consider an n-th order system
x(k + 1) = Ax(k),
x(0) = x0
y(k) = Cx(k).
A state x1 is said to be K-step unreconstructible if x1 is the solution of
the system at time K for some x0 , i.e., x(K) = x1 for some x0 , but for
all such x0 , it holds y(k) = 0 for all k ≥ 0. A state x1 is said to be
unreconstructible if it is K-step unreconstructible for all K ≥ 0. Denote
the set of all unreconstructible states by N ∞ .
1. Show that N ∞ = N 0 ∩ R(An ). Is N ∞ an invariant subspace of A?
"
#
A
2. System
is said to be reconstructible if the only unreconC
structible state is the origin, i.e., N ∞ = {0}. Show a system is
reconstructible iff
A − λI
rank
=n
C
for all λ ∈ C \ {0}.
56
CHAPTER 2. LTI STATE SPACE SYSTEM ANALYSIS
2.13 Consider a single input system A B , i.e., B has only one column.
Show that if A is diagonalizable but has repeated eigenvalues, then A B
is not controllable no matter what B is.
2.14 Assume that a second order single input system
x(k + 1) = Ax(k)
y(k) = Cx(k)
is observable. If we have observed y(0) and y(1), what will be y(k) for
k > 1?
2.15 Let A B be an n state and m input system over R. A subspace
V ⊂ Rn is said to be A B -invariant if
AV ⊂ V + R(B).
Show that V is (A, B)-invariant if and only if there exists F ∈ Rm×n ,
called a (right) friend of V, such that V is (A + BF )-invariant.
"
#
A
2.16 Let
be an n state and p output system over R. A subspace V ⊂ Rn
C
"
#
A
is said to be
-invariant if
C
A(V ∩ N (C)) ⊂ V.
Show that V is (C, A)-invariant if and only if there exists L ∈ Rn×p , called
a left friend of V, such that V is (A + LC)-invariant.
2.17 Find the controllable and observable part of
1
0
0
2
0
1
0
1
0
0
2
0
1
1
2
0
and its transfer function.
"
A B
C D
#
, then λ ∈ σ(A + BKC) for
"
#
A B
all K. For this reason, λ is also called a fixed mode of
.
C D
2.18 Show that if λ is a hidden mode of
"
2.19 For system
A B
C 0
#
, show that C(zI − A)−1 B = 0 iff R0 ⊂ N0 .
2.9. CONTROLLABILITY INDICES AND OBSERVABILITY INDICES 57
"
2.20 Show that the zeros of
"
A B
C D
#
and those of
A + BF + LC + LDF
C + DF
B + LD
D
#
are the same.
2.21 Assume that D is invertible. Show that the zeros of
"
#
A B
C D
are the eigenvalues of A − BD−1 C.
"
2.22 Show that all hidden modes of a system
A B
C D
#
are internal zeros of
the system.
2.23 Consider a square system
x(k + 1) = Ax(k) + Bu(k)
y(k) = Cx(k) + Du(k)
where x(k) ∈ Fn and u(k), y(k) ∈ Fm . Assume λ is a zero of this system.
Show that there exist u0 ∈ Cm and x0 ∈ Cn , not simultaneously zero, such
that the output response with u(k) = u0 λk and x(0) = x0 is constantly
zero.
2.24 Let A ∈ Fn×n have l different eigenvalues λ1 , λ2 , . . . λl . Among them
λ1 , . . . , λs are inside the unit circle and λs+1 , · · · , λl are outside or on the
unit circle. Let Ei be the eigenspace corresponding to λi . Show that the
trajectory of system
x(k + 1) = Ax(k),
x(0) = x0
satisfies limk→∞ x(k) = 0 for all x0 ∈ E1 ⊕ · · · ⊕ Es .
2.25 Prove Lemmas 2.1 and 2.2.
2.26 An equation of the form
AXB − X = C
where A ∈ Fm×m , B ∈ Fn×n , C ∈ Fm×n are given and X ∈ Fm×n is
unknown, is called a Sylvester equation. Show that such an equation has
a unique solution for all C iff 1 ̸∈ σ(A)σ(B).
58
CHAPTER 2. LTI STATE SPACE SYSTEM ANALYSIS
2.27 State variable transformation is sometimes used to transform the solution
of a Lyapunov equation in to a simple form. Assume that the Lyapunov
equation
A∗ XA − X = −Q
has unique solution X. Consider Lyapunov equation
(P −1 AP )∗ Y (P −1 AP ) − Y = −P ∗ QP
where P is nonsingular. Does it have unique solution? If yes, what is it.
Now assume that Q is Hermitian. Show that there exists P such that Y
is diagonal. Show that in this case if Q > 0 and X > 0, then all principal
submatrices of P −1 AP are stable.
2.28 Show that matrix
0
..
.
A=
is stable if
a1
a2
∥ .
..
∥1 < 1,
1
..
.
0
0
−an −an−1
∥
an
a1
a2
..
.
···
0
..
..
.
.
···
1
· · · −a1
1
∥2 < √ ,
n
or ∥
a1
a2
..
.
1
∥∞ < .
n
an
an
Also
show
that all these conditions are tight in the sense that there exist
a1
a2
.. with
.
an
∥
a1
a2
..
.
∥1 = 1,
an
∥
a1
a2
..
.
1
∥2 = √ ,
n
or ∥
an
a1
a2
..
.
1
∥∞ =
n
an
such that A is unstable.
2.29 Show that a necessary and sufficient condition for the polynomial
a(z) = z 2 + a1 z + a2
to have both roots inside the unit circle is −1 < a2 < 1 and −(a2 + 1) <
a1 < a2 + 1.
2.30 For a square matrix A, show that ∥A∥p < 1, where ∥ · ∥p is the induced
matrix p-norm, implies that A is stable.
2.9. CONTROLLABILITY INDICES AND OBSERVABILITY INDICES 59
2.31 For a given matrix A, show that if there exist X > 0 and Q > 0 such that
A∗ XA − r2 X = −Q,
then ρ(A) < r.
2.32 Let A be stable. Show that the solution X of Lyapunov equation
A∗ XA − X = −Q,
is a monotonically increasing function of Q ∈ Hn (C).
2.33 Verify the formulas in Theorem 2.14.
2.34 Prove
and
"
A B1
C D1
#
"
+
A B2
C D2
#
"
=
A B1 + B2
C D1 + D2
#
"
#
A 0 B1
A B1 + B2
0 A B2 =
.
C
D
C C D
2.35 Show that
à B̃1 B̃2
A B1 B2
C1 D11 D12 ⋆
C̃1 D̃11 D̃12
C2 D21 D22
C̃2 D̃21 D̃22
B2 B2
A B2
D̃11 D̃12
D̃11 C̃1
⋆
C2 D22 ⋆ B̃1 Ã
D21 D22
B̃1 B̃2
= .
C1 D12
D11 D12
D̃11 C̃1
D̃11 D̃12
⋆
⋆
C2 D22
D21 D22
D̃21 C̃2
D̃21 D̃22
60
CHAPTER 2. LTI STATE SPACE SYSTEM ANALYSIS
Chapter 3
Realization
3.1
Elementary Realizations
The purpose of realization is to find a state space representation for a given
transfer function. It is important in controller and filter implementation and
system simulation.
Definition 3.1
#
"
A B
is said to be a realization of Ĝ if
1.
C D
Ĝ(z) = D + C(zI − A)−1 B.
2. The dimension (order) of a realization is the size of the A matrix.
3. A realization of Ĝ is said to be minimal if its dimension is minimal among
all realizations of Ĝ.
It follows from the Kalman decomposition that a minimal realization has
to be controllable and observable and if a transfer function has a realization,
then it has a controllable and observable realization. In this chapter, we will
see whether or not any controllable and observable realization is minimal and
the relationship between various minimal realizations.
A matrix valued function of z is said to be a rational function if all of its
elements are rational functions of z. A rational function Ĝ is said to be proper
if limz→∞ Ĝ(z) exists.
It is clear that the transfer function of an FDLTI state space system is
a proper rational function. Therefore, a nonrational transfer function or a
nonproper rational transfer function cannot have an FDLTI realization. In the
following, we assume that the given transfer function whose realization is to be
found is proper rational and has size p × m. Such a function Ĝ can always be
written as
B1 z r−1 + · · · + Br
+D
(3.1)
Ĝ(z) = r
z + a1 z r−1 + · · · + ar
There are four canonical realizations:
61
62
CHAPTER 3. REALIZATION
1. Controller canonical realization
−a1 I
# I
"
Ac Bc
= ...
Cc D c
0
B1
The dimension of this realization
Toeplitz matrix
I a1 I
0 I
Tc = ... . . .
..
0
.
0
0
· · · −ar−1 I −ar I
···
0
0
..
..
..
.
.
.
···
I
0
· · · Br−1
Br
I
0
..
.
.
0
D
is mr. Define a block upper triangular
· · · ar−2 I ar−1 I
..
..
.
.
ar−2 I
..
..
..
.
.
.
..
.
I
a1 I
···
0
I
Then it is easy to verify that
Tc Bc Ac Bc · · · Ar−1
=I
c Bc
This shows that the matrix formed by the first mr columns of the controllability matrix is always nonsigular. Hence the controller canonical
realization is always controllable. In the special case when the system has
only one input, i.e., m = 1, it holds
1 a1 · · · ar−2 ar−1
0 1 . . . . . . ar−2
.. . . . .
.
−1
.
.
.
.
Mc = Tc = .
.
.
.
.
.. ..
0
.
.
1
a1
0 0 ···
0
1
2. Observer canonical realization
"
Ao
Co
−a1 I
..
#
.
Bo
= −ar−1
Do
−ar I
I
I ···
.. . .
.
.
0 ···
0 ···
0 ···
The dimension of this realization is pr.
Toeplitz matrix
I
0
a1 I
I
.
.
..
..
To =
..
ar−2 I
.
ar−1 I ar−2 I
B1
..
.
.
I Br−1
0 Br
0
D
0
..
.
Define a block lower triangular
··· 0 0
..
..
.
. 0
.
..
..
.
. ..
..
. I 0
· · · a1 I I
3.1. ELEMENTARY REALIZATIONS
Then it is easy to verify that
63
Co
Co Ac
..
.
Co Ar−1
c
To = I.
This shows that the matrix formed by the first mr columns of the controllability matrix is always nonsingular. Hence the observer canonical
realization is always observable. In the special case when the system has
only one output, i.e., p = 1, it holds
1
0
··· 0 0
.. ..
a1
.
. 0
1
. . . . ..
..
.
Mo−1 = To = ...
. .
.
.
..
..
ar−2
. 1 0
.
ar−1 ar−2 · · · 0 1
3. Controllability canonical realization
Recall Toeplitz matrix
Tc =
I a1 I · · · ar−2 I ar−1 I
..
..
.
.
ar−2 I
0 I
..
.. . .
..
..
.
.
.
.
.
..
..
.
.
I
a1 I
0
0 0 ···
0
I
.
Apply similarity transformation to the controller canonical realization using Tc−1 . Then we obtain the mr-dimensional controllability canonical
realization:
"
#
"
#
Aco Bco
Tc Ac Tc−1 Tc Bc
=
Cco Dco
Cc Tc−1
Dc
I
0 ···
0
−ar I
I ···
0
−ar−1 I 0
..
.
..
..
.
..
..
= .
.
.
0 ···
I
−a1 I
0
Bc1 · · · Bc(r−1)
Bcr
D
where
Bc1 · · · Bc(r−1) Bcr
=
B1 · · · Br−1 Br Tc−1 .
This realization is called the controllability canonical realization because
the first mr columns of its controllability matrix form an identity matrix,
i.e.,
Bco Aco Bco · · · Ar−1
= I.
co Bco
64
CHAPTER 3. REALIZATION
In the special case when the system has only one input, i.e., m = 1, it
holds Mc = I.
4. Observability canonical realization
Recall Toeplitz matrix
To =
··· 0 0
..
..
.
. 0
a1 I
I
..
.
..
..
..
.
.
. ..
.
..
..
.
. I 0
ar−2 I
ar−1 I ar−2 I · · · a1 I I
I
0
.
Apply similarity transformation to the observer canonical realization using To . Then we obtain the pr-dimensional observability canonical realization:
#
"
#
"
To−1 Ao To To−1 Bo
Aob Bob
=
Cob Dob
Co To
Do
Bo1
0
I
···
0
..
..
..
..
..
.
.
.
.
.
= 0
0
···
I
Bo(r−1)
−ar I −ar−1 I · · · −a1 I
Bor
I
0
···
0
D
where
Bo1
..
.
Bo(r−1)
Bor
B1
..
= To−1 . .
Br−1
Br
This realization is called the observability canonical realization because
the first pr rows of its observability matrix form an identity matrix, i.e.,
Cob
Cob Aob
= I.
..
.
Cob Ar−1
In the special case when the system has only one input, i.e., p = 1, it
holds Mo = I.
Example 3.1
Consider
1 2
1 2
.
Ĝ(z) =
z(z − 1)
3.2. MINIMAL REALIZATIONS
65
We can easily obtain its four canonical realizations:
"
Ac B c
Cc D c
#
=
"
Ao Bo
Co D o
#
=
"
Aco Bco
Cco Dco
#
=
"
Aob Bob
Cob Dob
#
=
1
0
1
0
0
0
0
1
0
1
0
0
0
0
0
0
1
1
0
0
0
0
2
2
1
0
0
0
0
0
0
1
0
0
0
0
1
0
0
0
1
0
0
1
0
0
0
1
1
0
0
0
0
0
0
1
0
0
0
0
0
0
1
1
0
0
0
0
2
2
0
0
0
0
1
0
0
0
0
0
0
1
0
0
0
0
1
0
1
1
0
0
0
1
2
2
1
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
1
1
0
1
0
0
0
0
1
0
1
0
0
0
0
1
1
0
0
0
0
2
2
0
0
.
From the above discussion, we obtain the following observations:
1. Every proper rational transfer matrix has a realization.
2. Controller and controllability canonical realizations are always controllable.
3. Observer and observability canonical realizations are always observable.
4. Because of 1, every proper rational transfer matrix has a minimal realization and this minimal realization is controllable and observable.
3.2
Minimal Realizations
Let Ĝ be a given p × m proper rational transfer function.
Definition 3.2 The order of a minimal realization of Ĝ is called the McMillan
degree of Ĝ, denoted by δ(Ĝ).
66
CHAPTER 3. REALIZATION
The impulse response G corresponding to Ĝ can be obtained by the inverse
Z-transform. Define (block) Hankel matrices
G(1)
G(2)
···
G(k)
.
G(2)
..
G(k + 1)
G(3)
Γk = .
∈ Fkp×km
..
.
.
..
..
..
.
G(k) G(k + 1) · · · G(2k − 1)
for k = 1, 2, . . . . Such matrices are called block Hankel matrices since the blocks
in each opposite diagonal are the same. Immediately, we have
rankΓk ≤ rankΓk+1
since Γk is a submatrix
" of Γk+1#. Hence rankΓk is an increasing function of k.
A B
Now assume that
is a minimum realization of Ĝ with order δ(Ĝ).
C D
Then
D
k=0
G(k) =
CAk−1 B k > 0.
and
Γk =
=
CAB · · · CAk−1 B
.
CA2 B . .
CAk B
..
.
.
.
.
.
.
.
CAk−1 B CAk B · · · CA2k−2 B
C
CA
B AB · · · Ak−1 B .
..
.
CB
CAB
..
.
(3.2)
CAk−1
"
#
A B
is controllable and observable, it follows that for k ≥ δ(Ĝ),
Since
C D
the left factor has full column rank and the right factor has full row rank. Hence
for k ≥ δ(Ĝ),
rankΓk = rankΓδ(Ĝ) = δ(Ĝ).
i.e., rankΓk saturates when k reaches δ(Ĝ). Therefore we have proved the
following lemma.
Lemma 3.1
1. rankΓk ≤ rankΓk+1 .
2. rankΓk = δ(Ĝ) for all k ≥ δ(Ĝ).
This lemma can be used to find δ(Ĝ). Let n be the order of any realization.
For example, if Ĝ is given by (3.1), we can take n = rm or n = rp. Then
δ(Ĝ) = rankΓn .
3.2. MINIMAL REALIZATIONS
67
Theorem 3.1 A realization is minimal iff it is controllable and observable.
#
"
A B
be a controllable
Proof We only need to prove the sufficiency. Let
C D
and observable realization with dimension n ≥ δ(Ĝ). Let Mc and Mo be the
corresponding controllability and observability matrices. Then
n = rank(Mo Mc ) = rankΓn = rankΓδ(Ĝ) = δ(Ĝ).
This shows that the realization is minimal
2
This theorem implies that there are two ways to establish the minimality
of a given realization. The first is to check whether its order is equal to δ(Ĝ).
The second is to see if it is controllable and observable. Which way is more
convenient depends on the particular situation.
"
#
#
"
A B
à B̃
be two minimal realizations of a
Theorem 3.2 Let
and
C D
C̃ D̃
transfer matrix Ĝ. Then there exists a unique nonsingular matrix P such that
"
# "
#
à B̃
P −1 AP P −1 B
=
.
(3.3)
CP
D
C̃ D̃
The unique P satisfies
P = (Mo∗ Mo )−1 Mo∗ M̃o
and
P −1 = M̃c Mc∗ (Mc Mc∗ )−1 .
Proof
Observe that
M̃o M̃c = Γn = Mo Mc .
Hence
(Mo∗ Mo )−1 Mo∗ M̃o M̃c Mc∗ (Mc Mc∗ )−1 = (Mo∗ Mo )−1 Mo∗ Mo Mc Mc∗ (Mc Mc∗ )−1 = I.
This shows that if
P = (Mo∗ Mo )−1 Mo∗ M̃o .
Then
P −1 = M̃c Mc∗ (Mc Mc∗ )−1 .
Similarly, we also have
M̃o ÃM̃c = Mo AMc , M̃o B̃ = Mo B, C̃ M̃c = CMc .
Hence
P ÃP −1 = (Mo∗ Mo )−1 Mo∗ M̃o ÃM̃c Mc∗ (Mc Mc∗ )−1 = A
P B̃ = (Mo∗ Mo )−1 Mo∗ M̃o B̃ = B
C̃P −1 = C̃ M̃c Mc∗ (Mc Mc∗ )−1 = C.
68
CHAPTER 3. REALIZATION
So we have proved the existence of P . To prove the uniqueness, let P be a
matrix satisfying (3.3). Then
CP
CAP
M̃o =
= Mo P.
..
.
CAn−1 P
This shows
Mo∗ M̃o = Mo∗ Mo P,
i.e., P = (Mo∗ Mo )−1 Mo∗ M̃o is the only possible P .
2
In general, a realization of the transfer function of a BIBO stable system is
not necessarily internally stable. However, a minimal realization is, as stated
in the following corollary.
Corollary 3.1 A system with proper rational transfer function is BIBO stable
iff any of its minimum realization is internally stable.
An indirect way to get a minimal realization of a transfer function is to
obtain an arbitrary realization first and then use Kalman decomposition to get
rid of the uncontrollable or unobservable part. In the rest of this section, we
will present a method to get a minimal realization directly.
Suppose that we have the impulse response G computed from the transfer
function Ĝ and we have already found δ(Ĝ). Let n = δ(Ĝ) and let
G(1)
G(2)
···
G(n)
.
G(2)
..
G(3)
G(n + 1)
Γn = .
..
.
.
..
..
..
.
G(n) G(n + 1) · · · G(2n − 1)
and
Γ̃n =
· · · G(n + 1)
.
. . G(n + 2)
..
.
..
.
G(n + 1) G(n + 2) · · ·
G(2n)
G(2)
G(3)
..
.
G(3)
G(4)
.
..
.
Here Γ̃n is obtained from Γn+1 by removing the first block column and the
last block row or by removing the first block row and the first block column.
Another way to view Γ̃n is that it is the upper right or lower left np × nm
submatrix of Γn+1 .
Factorize Γn as
Γn = Mo Mc
where Mo ∈ Fnp×n and Mc ∈ Fn×nm . Since rankΓn = n, it follows that Mo has
full column rank and Mc has full row rank. This factorization is not unique and
can be done using the singular value decomposition of Γn . We use the symbols
Mo and Mc to denote the matrices resulted from the factorization, the same
as the observability matrix and the controllability matrix of a system, since it
3.3. MINIMAL REALIZATIONS FROM PARTIAL FRACTIONAL EXPANSIONS69
can be shown (as an exercise) that they turn out to be the observability matrix
and the controllability matrix of the minimal realization that we will obtain.
Let Mo† be a left inverse of Mo , i.e., it satisfies Mo† Mo = I. A left inverse is not
unique and we can take, for example, Mo† = (Mo∗ Mo )−1 Mo∗ . Let Mc† be a right
inverse of Mc , i.e., it satisfies Mc Mc† . Again a right inverse is not unique and
we can take, for example, Mc† = Mc∗ (Mc Mc∗ )−1 . Then a minimum realization
of Ĝ is given by
I
"
#
0
†
†
A B
M
Γ̃
M
M
o n c
.
c
..
=
(3.4)
.
C D
0
I 0 · · · 0 Mo
G(∞)
We leave it as an exercise to prove that this is indeed a realization of Ĝ. The
minimality follows from that its order is equal to δ(Ĝ).
3.3
Minimal Realizations from Partial Fractional Expansions
To find the minimal realization by using the method in the last section, one
needs to compute the impulse response from the given transfer function, which
may not be easy in some cases or may not be practical at all since the impulse
response usually contains infinite amount of data, though only finite number of
terms are useful. In this section, we give another way to compute the minimal
realization from the transfer function.
Let us first look at a couple of special cases. First let us consider the case
when the strictly proper transfer function has a form
Ĝ(z) =
B(z)
(z − p1 )(z − p2 ) · · · (z − pl )
where pi ̸= pj for i ̸= j. In other words, we consider the case when the least
common multiple of the denominators of all elements of Ĝ has only simple roots.
In this case, the p × m transfer function can be decomposed into the form
Ĝ(z) =
l
X
Ni
z − pi
i=1
where pi ̸= pj for i ̸= j. Let rankNi = ni . Then we have factorizations
Ni = Ci Bi
where Bi = Fni ×m and Ci = Fp×ni . Here we must have rankBi = rankCi = ni .
Then a realization of Ĝ is given by
p1 In1
B1
"
#
..
..
A B
.
.
=
(3.5)
.
C D
pl Inl Bl
C1
· · · Cl
0
70
CHAPTER 3. REALIZATION
This realization is called the Gilbert realization. Since (pi Ini , Bi ) are controllable pairs and (Ci , pi Ini ) are observable pairs, it follows that
X
l
A − λI
rank A − λI B = rank
=
ni
C
i=1
for λ = pi . Hence the
Pl realization is controllable and observable. This also
shows that δ(Ĝ) =
i=1 ni . In this case, the minimality follows from the
controllability and observability of the realization.
Next let us consider FIR systems. Consider p × m transfer function
Ĝ(z) =
r
X
Ni z −i
i=1
Let
Γr =
N1 N2 · · · Nr
.
N2 N3 . .
0
.
..
. . . ..
.
.
. .
.
Nr 0 · · · 0
Let rankΓr = n. Since Γk , k > r, are formed from Γr by adding more zero
rows and columns, it follows that rankΓk = rankΓr = n for all k > r. Hence
δ Ĝ = rankΓr = n. Factorize
Γr = No Nc
(3.6)
such that Nc ∈ Fn×rm and No ∈ Frp×n . Here we must have rankNc = rankNo =
n. Let Nc† be a right inverse of Nc and No† be left inverse of No , i.e.,
Nc Nc† = I and No† No = I.
It is easy to show that
0
0 I ··· 0
..
. . . . ..
†
I
.
. .
.
†
Nc .
No
No = Nc . .
.
.
.
.
.
.
.
.
. I
.
0 ··· I 0
0
Then a minimum realization of Ĝ is
0
#
"
†
No
A B
=
C D
I
0
I
Nc
..
=
.
0
I
(3.7)
given by
I ··· 0
I
. . . . ..
0
.
. .
No Nc
..
..
.
. I
0
0
0 · · · 0 No
0
I
..
0
†
.
Nc Nc
..
.. ..
.
.
.
0
··· I 0
0 · · · 0 No
0
(3.8)
.
(3.9)
3.3. MINIMAL REALIZATIONS FROM PARTIAL FRACTIONAL EXPANSIONS71
It is left as an exercise to verify that (3.8) and (3.9) indeed give a realization of
Ĝ (Exercise 3.3). Here the minimality follows from the fact that the order of
realization is equal to the McMillan degree.
In general, a proper rational transfer function can be decomposed as
Ĝ(z) =
ri
l X
X
i=1 j=1
Nij
+D
(z − pi )j
(3.10)
where each pi is different from others. This decomposition is called the partial
fractional expansion. In this way, the dynamic part of Ĝ is decomposed into
the sum of l subsystems:
Ĝi (z) =
ri
X
j=1
Nij
, i = 1, . . . , l.
(z − pi )j
Each of these subsystems has a distinct pole pi and looks very similar to an
FIR system except a shift in the pole location. Let
"
#
Ai Bi
Ci 0
be a minimal realizations of
ri
X
Nij
j=1
zj
which can be found using the method as in
Ni1 Ni2
Ni2 Ni3
ni = rank .
.
..
..
Niri
0
Then
"
Ai + pi Ini
Ci
Case 2 and has an order
· · · Niri
.
..
0
.. .
.
..
.
···
0
Bi
0
#
is a minimal realization of Ĝi by virtue of Exercise 3.4.
Now putting the minimal realizations of all subsystems, as well as the constant term D, together, we get a realization of the whole system:
A1 + p1 In1
B1
"
#
..
..
A B
.
.
=
.
C D
Al + pl Inl Bl
C1
···
Cl
D
This realization is a minimal realization by virtue of the disjoint feature of the
poles of all subsystems, see Exercise 3.5.
72
CHAPTER 3. REALIZATION
This also shows that for a rational function Ĝ with a partial fractional
expansion as in (3.10),
δ(Ĝ) =
l
X
i=1
ni =
l
X
i=1
rank
Ni1
Ni2
..
.
Niri
3.4
Ni2 · · · Niri
.
0
Ni3 . .
..
.
.
..
..
.
0 ···
0
.
Balanced Realizations
"
#
A B
Let
be a given controllable and observable system. In many appliC D
cations, only its external property, i.e., transfer function, is important, then we
can use similarity transformation to transform the state space representation
to another form with better property. The so-called balanced realization is one
of the commonly used representations for stable systems.
Definition 3.3 A stable state space representation is said to be balanced if
Wc = Wo = diag(σ1 , σ2 , . . . , σn ).
"
Theorem 3.3 For a minimal realization
"
exists a nonsingular matrix P such that
A B
C D
P −1 AP
CP
#
of a stable system, there
#
P −1 B
is balanced.
D
"
A B
C D
#
Let the controllability and observability gramians of
be
"
#
P −1 AP P −1 B
Wc and Wo respectively. Let those of
be W̃c and W̃o
CP
D
respectively. Then
AWc A∗ − Wc = −BB ∗
Proof
implies
P −1 AP P −1 Wc P ∗−1 P ∗ A∗ P ∗−1 − P −1 Wc P ∗−1 = −P −1 BB ∗ P ∗−1 ,
and
A∗ Wo A − Wo = −C ∗ C
implies
P ∗ A∗ P ∗−1 P ∗ Wo P P −1 AP − P ∗ Wo P = −P ∗ C ∗ CP.
This shows W̃c = P −1 Wc P ∗−1 and W̃o = P ∗ Wo P . Since Wc and Wo are positive
definite, then there exists T such that Wo = T ∗ T (e.g., Cholesky factorization)
3.4. BALANCED REALIZATIONS
73
and there exists unitary matrix U such that
λ1
λ2
T Wc T ∗ = U
..
.
∗
U
λn
(spectral factorization). Finally let
√
4
P =T
Then
−1
U
λ1
√
4
λ2
..
√
λ1 √
λ2
W̃c = W̃o =
.
√
4
.
λn
..
.
√
.
λn
√
2
Finally, let σi = λi .
For a stable system, the positive numbers σi , i = 1, 2, . . . , n, in the controllability and observability gramians of its balanced realization (if ordered
in certain way, say, nonincreasingly) are uniquely determined and are called
the Hankel singular values of the system. The reason why the numbers σi ,
i = 1, 2, . . . , n, are called singular values is because the infinite Hankel matrix
Γ∞ has a singular value decomposition as
σ1
σ2
∗
Γ∞ = U
V
.
.
.
σn
where U and V are two isometries given from a balanced realization (Ã, B̃, C̃, D̃)
by
−1/2
σ1
−1/2
σ2
B̃ ÃB̃ Ã2 B̃ · · ·
U =
..
.
−1/2
σn
−1/2
σ1
C̃
−1/2
C̃ Ã
σ2
= C̃ Ã2
..
.
V
∗
..
.
−1/2
.
σn
We leave it as an exercise to prove U and V are indeed isometries.
74
CHAPTER 3. REALIZATION
Exercises
3.1 Verify that the state space system (3.4) indeed gives a realization of an
impulse response G. Also show that
C
CA
Mc = B AB · · · An−1 B and Mo = . .
..
CAn−1
3.2 Write a MATLAB program to implement the procedure to obtain the
minimum realization (3.4). Test it for
z+2
z
1
(z − 1)2 .
Ĝ(z) = z −
1
3
z + 2 z2 + z − 2
3.3 Consider minimum realizations of an FIR system.
1. Prove equality (3.7).
2. Verify that the state space system (3.8)–(3.9) indeed gives a realization of the FIR system.
3. Show that
C
CA
Nc = B AB · · · Ar−1 B and No = . .
.
.
CAr−1
4. The realization depends on the factorization (3.6) and this factorization is not unique. Choose a particular factorization so that the
realization is balanced.
#
"
A B
3.4 Show that if Ĝ(z) has a minimal realization
, then Ĝ(z − p)
C D
"
#
A + pI B
has a minimal realization
.
C
D
3.5 Assume that
"
A1 B1
C1 D 1
#
"
and
A2 B 2
C2 D 2
#
are minimal realizations of Ĝ1 and Ĝ2 respectively. Also assume that
σ(A1 ) and σ(A2 ) are disjoint. Show that
A1 0
B1
0 A2
B2
C1 C2 D 1 + D 2
is a minimal realization of Ĝ1 + Ĝ2 .
3.4. BALANCED REALIZATIONS
75
3.6 Find minimal realizations of
1.
z+2
z+1
Ĝ(z) =
z
z 2 + 3z + 2
2.
Ĝ(z) =
1
z 2 + 3z + 2 ;
z
z+1
1 1 −1
1 1 −2
z +
z .
2 2
1 1
3.7 Find a minimal realization of the system whose impulse response
G(0), G(1), G(2), · · ·
is the Fibonacci sequence
0, 1, 1, 2, 3, 5, 8, 13, . . .
What is its unit step response?
"
3.8 Write a MATLAB program to convert a stable minimal realization
A B
C D
to a balanced realization.
"
3.9 A state space realization
A B
C D
#
of a stable system is said to be in"
#
A B
put normal if Wc = I. Show that if
is an arbitrary conC D
trollable realization, then there exists a nonsingular matrix P such that
(P −1 AP, P −1 B, CP, D) is input normal.
#
"
A B
of a stable system is said to be
3.10 A state space realization
C D
"
#
A B
output normal if Wo = I. Show that if
is an arbitrary obC D
servable realization, then there exists a nonsingular matrix P such that
(P −1 AP, P −1 B, CP, D) is output normal.
"
#
A B
3.11 Show that if
is an arbitrary realization with stable A, then
C D
"
#
P −1 AP P −1
there exists a nonsingular matrix P such that
has
CP
D
controllability gramian and observability gramian
Σ1
Σ1
I
0
and Wo =
Wc =
0
I
0
0
respectively.
#
76
CHAPTER 3. REALIZATION
3.12 (Extra Credit) This problem is to investigate the relationship between
the McMillan degree and the Smith-McMillan form. Let Ĝ be a proper
transfer matrix.
1. Show that δ(Ĝ) = deg α1 (z)+α2 (z)+· · ·+deg αr (z), where αi (z), i =
1, 2, . . . , r, are the diagonal denominators of the Smith-McMillan
form of Ĝ
2. Give a scheme to obtain a minimal realization of Ĝ from its SmithMcMillan form.
Chapter 4
Elementary Feedback Control
4.1
State Feedback
Consider linear state space system
x(k + 1) = Ax(k) + Bu(k).
(4.1)
Let us give it a name G and call it a plant. The simplest control problem is
to alter the dynamics of the system by using a state feedback of the following
form
u(k) = F x(k).
u-
G
x
F Figure 4.1: State feedback
With the control applied, the system becomes the one shown in Figure 4.1
and has state space equation
x(k + 1) = (A + BF )x(k).
(4.2)
System (4.1) is called the open-loop system and the controlled system (4.2) is
called the closed-loop system.
Since the response of the controlled system depends very much on the eigenvalues of A + BF , we are interested in the possibility to assign, by designing F ,
the eigenvalues of A + BF to the desired locations, such as the open unit disk
for stability. The process of assigning eigenvalues of A + BF by designing F is
called pole placement or pole assignment.
Theorem 4.1
of A + BF can be arbitrarily assigned by de The eigenvalues
signing F iff A B is controllable.
77
78
CHAPTER 4. ELEMENTARY FEEDBACK CONTROL
Proof
The necessity is easy. Without loss of generality, we can assume that
A B is of the form of the controllability decomposition:
A11 A12
B1
A=
, B=
.
0 A22
0
If we write F = F1 F2 , then
A11 + B1 F1 A12 + B1 F2
A + BF =
.
0
A22
Therefore, if A B is not controllable, the eigenvalues of A22 cannot be
moved.
For the sufficiency, we will first prove for the special case when m = 1 and
then extend the proof to the multi-input case based on the proof of the single
input case. In the single input case, the controllability matrix
Mc = B AB · · · An−1 B
is square and nonsingular because of the controllability. It is easy to check that
0 · · · 0 −an
1 · · · 0 −an−1
Mc−1 AMc = . .
.
..
.. . . ..
.
0 ··· 1
−a1
where a1 , . . . , an are the coefficients of the characteristic polynomial of A:
cA (z) = z n + a1 z n−1 + · · · + an .
Recall Toeplitz matrix
Tc =
1 a1 · · · an−2 an−1
..
.
0 1
an−2
.. .. . .
..
..
.
.
. .
.
0
···
1
a1
0 0 ···
0
1
Let P = Mc Tc . Then it can be verified that
P
−1
AP
=
Tc−1 (Mc−1 AMc )Tc
P
−1
B =
Tc−1 Mc−1 B
=
1
0
..
.
0
=
.
.
−a1 · · · −an−1 −an
1
···
0
0
..
..
.. ,
..
.
.
.
.
0
···
1
0
4.1. STATE FEEDBACK
79
The eigenvalues of a matrix is completely determined by its characteristic polynomial, and vice versa. Suppose that the desired closed-loop characteristic
polynomial is given by
cA+BF (z) = z n + α1 z n−1 + · · · + αn .
Let
F =
a1 a2 · · · an
−
α1 α2 · · · αn
P −1 .
Then
P −1 (A + BF )P =
−α1 · · · −αn−1 −αn
1
···
0
0
..
..
.. .
..
.
.
.
.
0
···
1
0
Therefore the characteristic polynomial of A+BF is exactly equal to the desired
one.
In the case when m > 1, denote
B = B1 B2 · · · Bm
where Bi ∈ Fn is the ith column of B. Since A B1 is not necessarily
controllable, by using the controllability decomposition, we see that there exists
P1 ∈ GL(n, F) such that
P1−1 AP1 =
Ã11
?
0 Â22
,
P1−1 B1 =
B̃11
0
.
where Ã11 ∈ Fn1 ×n1 , B̃11 ∈ Fn1 , Ã11 B̃11 is controllable, and 0 ≤ n1 ≤ n.
Here we used “?” to mean “irrelevant” elements or blocks. It is possible
that
n1 = 0 in the case when B1 = 0 or n1 = n in the case when A B1 is
controllable. In both these cases, P1 = I. In the latter case, we need to go no
further, but let us pretend we continue the process as if n1 < n. At this stage,
we have
B̃11
? ···
?
−1
P1 B =
.
0 B̂22 · · · B̂2m
Then there exists P2 ∈ GL(n − n1 , F) such that
P2−1 Â22 P2
=
Ã22
?
0 Â33
,
P2−1 B̂22
=
B̃22
0
where Ã22 ∈ Fn2 ×n2 , B̃22 ∈ Fn2 , Ã22 B̃22 is controllable, and 0 ≤ n2 ≤
n − n1 . Again, if Â22 B̂22 is controllable, we have n1 + n2 = n and we can
stop but again we pretend to go on. Keep this process going. Let
P = P1
In1
0
0
P2
In1 +n2
0
0
P3
···
In1 +···+nm−2
0
0
Pm−1
.
80
CHAPTER 4. ELEMENTARY FEEDBACK CONTROL
Then
Ã11 Ã12 · · · Ã1m
B̃11 B̃12 · · · B̃1m
0 Ã22 · · · Ã2m
0 B̃22 · · · B̃2m
−1
P −1 AP = .
,
P
B
=
..
..
..
..
.. .
..
..
..
.
.
.
.
.
.
.
0
0 · · · Ãmm
0
0 · · · B̃mm
where Ãii B̃ii is either controllable or have zero dimension, i.e., ni = 0.
What we have done so far is to have transformed matrices A and B to block
upper triangular forms by a coordinate transformation. Now partition the n
desired closed-loop eigenvalues as m groups, the ith of which has ni numbers. In
the case when F = R, make sure that each complex conjugate pair in the desired
eigenvalues are in the same group. Design F̃ii ∈ F1×ni so that Ãii + B̃ii F̃ii has
eigenvalues equal to the numbers in the ith group. This is a single input pole
placement problem and we now know how to solve it. Finally, let
F̃11 F̃12 · · · F̃1m
0 F̃22 · · · F̃2m
−1
F = .
..
.. P
..
..
.
.
.
0
0
· · · F̃mm
where the off-diagonal blocks F̃ij , i < j, can be arbitrarily chosen. Then
Ã11 + B̃11 F̃11
?
···
?
0
Ã22 + B̃22 F̃22 · · ·
?
P −1 (A + BF )P =
.
..
..
..
..
.
.
.
.
0
0
· · · Ãmm + B̃mm F̃mm
Now it is clear that A + BF has eigenvalues equal to the desired closed-loop
eigenvalues.
2
The proof of Theorem 4.1 gives a constructive way to carry out the pole
placement for controllable system. For a single input controllable system, the
state feedback matrix F is uniquely determined if the desired closed-loop poles
are given. For a multiple input controllable system, however, the state feedback
matrix F is not uniquely determined by the desired closed-loop poles. First the
off-diagonal blocks of F̃ = F P can be arbitrarily chosen. Secondly, the partition
of closed-loop eigenvalues into m groups can be done in a highly non-unique
way. Thirdly, the sequential order of working through the columns of B is taking
naturally in the proof but actually can be done in any other order. A simple
count of the degree of freedom reinforces the non-uniqueness of the design. The
n desired closed-loop poles have n degrees of freedom, whereas the number of
design parameters in F is mn. So the design can only be unique when m = 1.
In the multiple input case, the extra freedom in designing F after achieving the
pole placement can be utilized to address other design concerns. From the proof of Theorem 4.1, it is seen that in the case
when A B is
not controllable, then the uncontrollable modes of A B will remain to be
the eigenvalues of A + BF no matter what F is and the controllable modes of
4.1. STATE FEEDBACK
81
A B are freely assignable. Therefore, if we are only interested in making
A + BF stable, we only need to move the unstable modes of A.
Definition 4.1 A B is said to be stabilizable if there exists F such that
A + BF is stable.
There are several ways to test whether A B is stabilizable.
Theorem 4.2 The following statement are equivalent:
1. A B is stabilizable.
2. The unstable modes of A are all controllable.
3. All uncontrollable modes are stable.
4. rank A − λI B = n for all λ ∈ C with |λ| ≥ 1.
5. The following LMI over Y ∈ Hn (F) and Z ∈ Fm×n is feasible:
Y
Y A∗ + Z ∗ B ∗
> 0.
AY + BZ
Y
Proof
obvious.
stability
exists F
The equivalence between any two statements from 1 to 4 is quite
Let us now prove the equivalence
between
1 and 5. It follows from the
theory in Section 2.6 that A B is stabilizable if and only if there
such that the following matrix inequality over Y ∈ Hn (F) is feasible
Y
0
0 Y − (A + BF )Y (A + BF )∗
Y
0
=
0 Y − AY A∗ − BF Y A∗ − AY F ∗ B ∗ − BF Y F ∗ B ∗
> 0.
Notice that this is not a joint LMI over Y and F . Let F Y = Z. Then the
existence of F and Y > 0 is equivalent to the existence of Z and Y > 0. Hence
A B is stabilizable if and only if there exist Z and Y > 0 such that the
following inequality holds
Y
0
>0
0 Y − AY A∗ − BZA∗ − AZ ∗ B ∗ − BZY −1 Z ∗ B ∗
which is the same as
Y
Y A∗ + Z ∗ B ∗
AY + BZ
Y
>0
because of congruence.
2
The solution to the LMI in the statement 5 of Theorem 4.2 not only verifies
the stabilizability of A B , it also gives a stabilizing state feedback gain
F = ZY −1 and a corresponding quadratic Lyapunov function in terms of Y .
82
4.2
CHAPTER 4. ELEMENTARY FEEDBACK CONTROL
State Observer
Let us consider system
x(k + 1) = Ax(k) + Bu(k)
y(k) = Cx(k) + Du(k).
(4.3)
(4.4)
In many applications, we wish to know how the state x behaves but we cannot
directly measure x. In this case, we wish to estimate the state x from the
measurement of the external variables u and y. This means that we wish to
build a system O, called an observer, with input u and y and output x̃, as
shown in Figure 4.2, so that e = x − x̃ is as small as possible no matter what
the initial conditions of the system to be observed (4.3-4.4) and that of the
observer are.
u-
G
y
- O
x̃-
Figure 4.2: State observer
One idea of constructing an observer is to build an exact duplicate of the
state equation
x̃(k + 1) = Ax̃(k) + Bu(k).
The error system is then
e(k + 1) = Ae(k).
If the initial error is zero, then the error will remain to be zero and the estimate
is exact. However, if the initial error is nonzero, then the behavior of the error
will depend on A, which cannot be adjusted. If A is unstable, then the error
will potentially blow up. Even if A is stable but the eigenvalues are close to the
unit circle, then the error will take a long time to settle to zero.
To eliminate this problem, we feedback the error in the output to adjust the
observer state. The improved observer is of the following form.
x̃(k + 1) = Ax̃(k) + Bu(k) + L[C x̃(k) + Du(k) − y(k)].
With a little reorganization, it becomes
x̃(k + 1) = (A + LC)x̃(k) + (B + LD)u(k) − Ly(k).
(4.5)
This observer is called a Luenberger observer. Now the error dynamics becomes
e(k + 1) = (A + LC)e(k).
(4.6)
To obtain an accurate estimation of x, we wish that the error approaches
zero as quick as possible. The behavior of the error depends very much on the
eigenvalues of A + LC. Therefore, we are interested in the possibility to assign,
by designing L, the eigenvalues of A + LC.
4.2. STATE OBSERVER
83
-D
- B
- j
-
S
x
? y
- j
- C
6
A −?
u
j
+6
-
- B
- j
6
-
D
S
x̃
?
- j ỹ
- C
A L
Figure 4.3: Luenberger observer
Theorem 4.3
" The
# eigenvalues of A + LC can be arbitrarily assigned by deA
signing L iff
is observable.
C
The eigenvalues
of A + LC are the same as those of A′ + C ′ L′ . By
"
#
A
Theorem 2.15,
is observable iff A′ C ′ is controllable. The theorem
C
then follows from Theorem 4.1.
2
In many applications, we are only interested in asymptotic estimation, i.e.,
we only need e(k) → 0 as k → ∞. In this case, we only need to find L so that
A + LC is stable.
"
#
A
Definition 4.2
is said to be detectable if there exists L such that A+LC
C
is stable.
Proof
Clearly, stabilizability and detectability are dual concepts. The following
result is a dual version of Theorem 4.2, whose proof follows in a dual way to
that of Theorem 4.2.
Theorem 4.4 The following statement are equivalent:
"
#
A
1.
is detectable.
C
2. The unstable modes of A are all observable.
84
CHAPTER 4. ELEMENTARY FEEDBACK CONTROL
3. All unobservable modes are stable.
A − λI
4. rank
= n for all λ ∈ C with |λ| ≥ 1.
C
5. The following LMI over X ∈ Hn (F) and Z ∈ Fn×p is feasible
X
XA + ZC
> 0.
A∗ X + Z ∗ C ∗
X
4.3
Observer-Based Controller
In Section 4.1, we see that the state feedback can be used to change the behavior
of a system, in particular, to stabilize a possibly unstable open-loop system.
However, in many applications, the state variables are not available for direct
measurement. Instead, only the output is available for measurement. In this
case the control structure is of the form shown in Figure 4.4.
- G
y
u
K Figure 4.4: Output feedback
The control structure shown in Figure 4.4 is called output feedback. The
feedback system will be called (G, K) for simplicity. (G, K) will be said to
be internally stable if the feedback system is internally stable. If (G, K) is
internally stable, then K is said to stabilize G. In this section, we will consider
a particular form of output feedback controller: the observer-based controller.
One idea in designing an output feedback controller is to feedback the estimated state instead of the real state, as shown in Figure 4.5.
u-
G
y
F
?
6
x̃
O Figure 4.5: Observer-based controller
Consider system (4.3-4.4) and observer (4.5). The estimated state feedback
is
u = F x̃.
4.3. OBSERVER-BASED CONTROLLER
85
Then the closed-loop system is described by the following state space equations
x(k + 1)
A
BF
x(k)
=
.
x̃(k + 1)
−LC A + LC + BF
x̃(k)
Denote e(k) = x(k) − x̃(k). Since
x(k)
I 0
x(k)
=
,
x̃(k)
I −I
e(k)
the equation can be rewritten in terms of the new state variable
x(k + 1)
A + BF
−BF
x(k)
=
.
e(k + 1)
0
A + LC
e(k)
It is seen that the modes of this system are the"eigenvalues
of A + BF and
#
A
A+LC. Therefore, if A B is controllable and
is observable, then the
C
eigenvalues of the closed-loop system can be assigned arbitrarily. If A B is
"
#
A
is detectable, then the closed-loop system can be made
stabilizable and
C
stable. This shows that in order to design an observer based controller, one
can design the state feedback gain F and the observer gain L separately: First
design F to assign the eigenvalues of A + BF and then design L to assign those
of A + LC.
Separation Principle The design of observer based controller can be carried
out in two separate steps. First design a state feedback, ignoring how the state
can be obtained. Then design an observer to estimate the state, ignoring what
the estimate is used for. The final observer based controller is the estimated
state feedback using the state feedback gain designed in the first step and the
state estimate obtained by an observer designed in the second step.
Separation principle holds in many situations of controller design.
The first glance of Figures 4.4 and 4.5 may reveal that they are different
control structures. Actually they are equivalent since the observer
x̃(k + 1) = (A + LC)x̃(k) + (B + LD)u(k) − Ly(k)
together with the control
u(k) = F x̃(k)
can be rewritten as
x̃(k + 1) = (A + LC + BF + LDF )x̃(k) − Ly(k)
u(k) = F x̃(k).
This is exactly a system with input y and output u. Compactly, we write
"
#
A + LC + BF + LDF −L
K=
.
F
0
86
CHAPTER 4. ELEMENTARY FEEDBACK CONTROL
This shows that an observer based controller is a special form of the output
feedback controller. The order of an observer-based controller is n, the same
as the plant order. If the plant is stabilizable and detectable, then an observer
based controller can be designed to stabilize the plant.
What about other forms of stabilizing controllers? Is it possible to design
a controller stabilizing a plant even when the plant is unstabilizable or undetectable? A general FDLTI controller is of the form
"
#
à B̃
K=
.
C̃ D̃
Assume that the state variable of K is x̃. (Do not interpret it as an estimate
of x.) After connecting the plant and the controller, the closed-loop system
becomes
x(k + 1)
A + B D̃(I − DD̃)−1 C
B(I − D̃D)−1 C̃
x(k)
=
.
x̃(k + 1)
x̃(k)
B̃(I − DD̃)−1 C
à + B̃D(I − D̃D)−1 C̃
Clearly, this makes sense only if I − DD̃ and I − D̃D are invertible. (It can
be shown that the invertibility of I − DD̃ and that of I − D̃D are equivalent.)
In this case, we say that the closed-loop system is well-posed. A closed-loop
system which is not well-posed either does not have solution or have infinite
many solutions, so it does not work. Observe that
A + B(I − D̃D)−1 D̃C
B(I − D̃D)−1 C̃
B̃(I − DD̃)−1 C
à + B̃D(I − D̃D)−1 C̃
à + B̃D(I − D̃D)−1 C̃ B̃(I − DD̃)−1
0 I
0 B
A 0
.
+
=
C 0
I 0
0 0
(I − D̃D)−1 C̃
(I − D̃D)−1 D̃
Then we can see that the uncontrollable modes of A B are the same as
those of
A 0
0 B
,
0 0
I 0
"
#
A
and unobservable modes of
are the same as those of
C
0 I
A 0
,
.
C 0
0 0
Therefore they will remain to be the modes of the closed-loop system no matter
how the controller is designed, cf., Exercise 2.24. This leads to the following
theorem.
"
#
A B
Theorem 4.5 There exists a stabilizing controller to stabilize
inC D
"
#
A
is detectable.
ternally iff A B is stabilizable and
C
4.4. THE SET OF ALL STABILIZING CONTROLLERS
4.4
87
The Set of All Stabilizing Controllers
From the knowledge of the Section 4.3, we know that a plant can be internally
stabilized by a controller iff it is stabilizable and detectable. If this condition
is satisfied, we know how to design a controller to accomplish the stabilization.
The controller given in Section 4.3 is an observer based controller. However,
there are other forms of controllers which might be better in some sense, such
as being simpler or consuming less energy, etc. In this section, we will be
concerned with the set of all internally stabilizing controllers. We also need to
restrict the controllers under consideration to FDLTI controllers, since other
types of controllers are outside of the scope of this courses. If we can give a
simple description to this set, then it will potentially be easy to pick a suitable
one depending on the application on hand.
- G
y
u
J
- Q
Figure 4.6: All stabilizing controllers
"
#
A B
Theorem 4.6 Given a stabilizable and detectable plant G =
, let F
C D
and L be such that A+BF and A+LC are stable. Then the set of all stabilizing
controllers is given by
(
"
#
)
à B̃
F l (J , Q) : Q =
, Ã is stable and I + DD̃ is nonsingular
(4.7)
C̃ D̃
where
A + BF + LC + LDF
J =
F
−C − DF
−L B + LD
0
I
I
−D
as shown in Figure 4.6.
Proof
First notice that all FDLTI controllers are given by the set
(
"
#
)
Ā B̄
K=
: det(1 − DD̄) ̸= 0 .
C̄ D̄
88
CHAPTER 4. ELEMENTARY FEEDBACK CONTROL
The condition det(I − DD̄) ̸= 0 is for the well-posedness of the feedback loop.
By Corollary 2.3, Q 7→ F l (J , Q) is a bijection from
(
"
#
)
à B̃
Q=
: det(I + DD̃) ̸= 0
C̃ D̃
to
(
"
K=
Ā B̄
C̄ D̄
#
)
: det(I − DD̄) ̸= 0 .
It follows that the set of all FDLTI controllers is also given by
#
)
(
"
à B̃
and det(I + DD̃) ̸= 0 .
F l (J , Q) : Q =
C̃ D̃
Now connect F l (J , Q) to G as shown in Figure 4.6. The connection of G and
J is an upper linear fractional transformation.
"
#
A + BF + LC + LDF −L B + LD
A B
,
F u (J , G) = F u
F
0
I
C D
−C − DF
I
−D
A + LC + BF + LDF − LDF −LC B + LD − LD
BF
A
B
=
−C − DF + DF
C
−D + D
A + LC + BF −LC B
BF
A
B
=
−C
A + BF
0
=
0
= 0.
C
LC
A + LC
C
0
B
0
0
x
(think about the reason for this
e
notation) and x̃ respectively. Then the closed-loop system is
x(k + 1)
A + BF
LC
B C̃
x(k)
e(k + 1) =
0
A + LC
0 e(k) .
x̃(k + 1)
x̃(k)
0
B̃C
Ã
Let the states of F u (J , G) and Q be
Since the eigenvalues of this matrix are those of A + BF , A + LC, and Ã, it
follows that the closed-loop system is stable iff Q is internally stable.
2
4.5
General Feedback Systems
In applications, the stability requirement is only one, albeit the most crucial
one, requirement to the overall system. Other requirements are in general
4.5. GENERAL FEEDBACK SYSTEMS
89
termed as performance requirements or performance specifications. When the
performance is a consideration in the design, we often model a plant as a generalized system G. Here the input signals are partitioned into two groups. The
first group, called the exogenous input and denoted by w, consists of those input signals which cannot be manipulated, such as noise and disturbance. The
second group, called control input and denoted by u, consists of those input
signals that can be manipulated. The output signals are also partitioned into
two groups. The first group, called the concerned output and denoted by z, consists of those signals in the system whose behavior are of concern. The second
group, called the measured output and denoted by y, consists those signals in
the system which can be measured physically. When the plant is controlled by
a controller K which takes the measurement y and produces the control input
u, the controlled system becomes an LFT shown in Figure 4.7. Typically (not
always), the performance requirements are to make F l (G, K) small in some
sense under the internal stability requirement. The configuration in Figure 4.7
is called the standard setup.
Assume that the generalized plant has state space representation
G=
G11 G12
G21 G22
A B1 B2
= C1 D11 D12
C2 D21 D22
and the controller K has state space representation
"
K=
à B̃
C̃
#
D̃
.
The internal stability of F l (G, K) of course means that the states of G and
K goes to zero as time k goes to infinity for all initial condition when the
exogenous input w is 0.
z
w
G
y
u
- K
Figure 4.7: The standard setup
Theorem 4.7 There exists controller K such
" that#the LFT F l (G, K) is inter
A
nally stable iff A B2 is stabilizable and
is detectable. In this case,
C2
F l (G, K) is internally stable iff (G22 , K) is internally stable.
90
CHAPTER 4. ELEMENTARY FEEDBACK CONTROL
Proof This is obvious since the internal stability has nothing to do with
input w and output z.
2
By Theorems 4.6 and 4.7, the set of all stabilizing controller K for the
standard setup is given by
(
"
F l (J , Q) : Q =
à B̃
C̃
D̃
#
)
, Ã is stable and I + D22 D̃ is nonsingular
(4.8)
where
A + B2 F + LC2 + LD22 F
J=
F
−C2 − D22 F
−L B2 + LD22
0
I
I
−D22
and F, L are matrices such that A + B2 F and A + LC2 are stable, as shown in
Figure 4.8.
z
w
G
y
u
-
J
Q Figure 4.8: The standard setup with a stabilizing controller
The design problem then becomes to choose a stable Q such that the system
with input w and output z, which now depends on Q, satisfy the performance
specifications. To facilitate the design process, we wish this dependency on Q
is simple. Indeed, as stated in the following theorem, the dependency turns out
to be simple (an affine function of Q).
Theorem 4.8 When a stabilizing controller of the form (4.8) is applied to the
standard setup, the system with input w and output z is given by
T 11 + T 12 QT 21
4.5. GENERAL FEEDBACK SYSTEMS
91
where
T 11
T 12
T 21
A + B2 F
B2 F
B1
0
A + LC2 −B1 − LD21
=
C1 + D12 F
D12 F
D11
"
#
A + B2 F B2
=
C1 + D12 D12
#
"
A + LC2 B1 + LD12
.
=
C2
D21
The system from w to z is
Proof
F l (G, F l (J , Q)) = F l (G ⋆ J , Q).
Let T = G ⋆ J . Then by corollary 2.2,
A B1 B2
A + B2 F + LC2 + LD22 F −L B2 + LD22
T = C1 D11 D12 ⋆
F
0
I
−C2 − D22 F
C2 D21 D22
I
−D22
B1
B2
A
B2 F
−LC2 A + B2 F + LC2 + LD22 F − LD22 F −LD21 B2 + LD22 − LD22
=
C1
D12 F
D11
D12
D21
−D22 + D22
C2
−C2 − D22 F + D22 F
A
B2 F
B1
B2
−LC2 A + B2 F + LC2 −LD21 B2
=
C1
D12 F
D11
D12
C2
A + B2 F
0
=
C1 + D12 F
0
−C2
D21
0
B2 F
B1
B2
A + LC2 −B1 − LD21
0
.
D12 F
D11
D12
D21
0
C2
In the last step, we applied certain similarity transformation to the system.
Clearly,
A + B2 F
B2 F
B1
0
A + LC2 B1 + LD21
T 11 =
C1 + D12 F −D12 F
D11
A + B2 F
B2 F
B2
0
A + LC2
0
T 12 =
C1 + D12 F −D12 F D12
A + B2 F
B2 F
B1
0
A + LC2 B1 + LD21
T 21 =
0
C2
D21
A + B2 F
B2 F
B2
0
A + LC2 0 .
T 22 =
0
C2
0
92
CHAPTER 4. ELEMENTARY FEEDBACK CONTROL
After removing the obvious uncontrollable or unobservable modes of T 12 , T 21 ,
and T 22 , we obtain
"
#
A + B2 F
B2
T 12 =
C1 + D12 F D12
"
#
A + LC2 B1 + LD21
T 21 =
C2
D21
T 22 = 0.
In Figure 4.7, the closed-loop transfer function from w to z, considered
as a function of the design variable K̂, is a complicated function and has a
complicated domain, which is the set of the stabilizing and hence useful K.
Theorem 4.8 says that by using the parametrization of all stabilizing controllers
given in (4.8), the closed-loop transfer function from w to z becomes a simple
affine function of the free parameter Q̂ with a nice domain, which is the set
of stable transfer functions. This will facilitate the selection of a stabilizing
controller to satisfy certain performance specifications, which is usually given
in terms of the transfer function from w to z.
4.6
Eigenstructure Placement
From the Section 2.5, we know that each polynomial matrix is equivalent to
the unique Smith Canonical form with invariant
polynomials.
Let [A|B] be a
nth order controllable system with B = b1 b2 · · · bm , and the invariant
polynomials of (λI − A) are {γ1 (λ), . . . , γr (λ)}. Then m ≥ r, and there exists
nonsingular matrices P and Q such that
A1
| b1 ∗ · · · ∗ ∗
. .
.
−1
A2
|
b2 . . .. ..
−1
,
P AP |P BQ =
.
.
.
.
. ∗ ∗
.
|
Ar |
br ∗
where Ai are cyclic with invariant polynomials γi (r) such that γi+1 (λ)|γi (λ), and
the subsystems [Ai |bi ] are controllable. In terms of the linear transformation
in the input space, it can be inferred that the dimension of A1 is the greatest
dimension of subspace that can be controlled by a single control input, the
dimension of A1 along with A2 is the greatest dimension of subspace that can
be controlled by two control inputs, and so on.
Since γi (λ) is exactly the invariant polynomial of (λI − Ai ), it can be infered that there is a one-to-one correspondence of each block in P −1 AP and
U (λ)A(λ)V (λ). It is important to know that the invariant polynomials can
describe the structure of a linear transformation of vector spaces.
The famous Rosenbrock Theorem gives a sufficient and necessary condition
for the arbitrarily eigenstructure placement by static state feedback. It is given
by Rosenbrock whose original proof is based on the polynomial arguments,
4.6. EIGENSTRUCTURE PLACEMENT
93
employing a sequence of matrix manipulations as a tool. However, it is not easy
to understand due to lack of intuitive interpretation. Flamm and Bonilla give
a geometric way to prove it. The main idea of their proof is intuitive, leading
to the convenience of algorithm. Therefore, the geometric proof is adopted in
this article.
Theorem 4.9 Let [A|B] be controllable with controllability indices {ki } and
{γi (λ)} the monic polynomials that satisfy γi+1 (λ)|γi (λ). Then there exists a
feedback F , such that {γi (λ)} is the
P set of invariant polynomials of A + BF iff
k ≺ α, where αi = degγi (λ) and
αi = n.
Proof Let us start from the necessity part.
Let AF = A + BF with F : X → U. Without loss of generality, we assume
that {ki } and {αi } are in non-increasing order. Since the invariant polynomials
of the matrix AF are {γ1 (λ), . . . , γr (λ)}, we only need to prove that
j
X
i=1
ki ≤
j
X
αi , j = 1, · · · , r − 1.
i=1
From the cyclic decomposition, it can be derived that for all S ⊂ X with
dimS = j ≤ r,
j
X
)
≤
αi ,
dim(R S AF S · · · An−1
S
F
i=1
which gives the upper bound for the dimensions of the controllable input subspace. Furthermore, for any j ≤ r, there exists S ⊂ R(B) with dim (S) = j, so
that
j
X
ki ,
dim(R S AF S · · · AFn−1 S ) ≥
i=1
which is also straightforward up to the definition of the controllability indices.
Therefore, the necessity part can be solved.
The sufficiency part is much more difficult and will be proved in a constructive way, i.e., finding the desired feedback given [A|B] and the invariant
polynomials of A + BF . Since the invariant polynomials are related to the
cyclic subspaces generated by the vectors in the input space, the proof follows
the intention to construct such cyclic subspaces with dimensions that satisfy
the majorization inequality. The procedure consists of three main steps.
1. Get the Brunovsky canonical form by a feedback so that the degrees of the
invariant polynomials are the controllability indices.
2. Find a feedback to change the degrees of the invariant polynomials to the
desired dimensions of the cyclic subspaces.
3. Find a feedback to change the invariant polynomials without changing their
degrees.
94
CHAPTER 4. ELEMENTARY FEEDBACK CONTROL
Step 1: given the controllable system[A|B] with the controllability indices
{k1 , . . . , km }, then {b1 , Ab1 , . . . , Ak1 −1 b1 , . . . , bm , Abm , . . . , Akm −1 bm } form a basis of X . Let
H = [b1 , Ab1 , . . . , Ak1 −1 b1 , . . . , bm , Abm , . . . , Akm −1 bm ]−1
= [g1 , g2 , . . . , gn ]T .
Take
T = [gk1 , gk1 A, . . . , gk1 Ak1 −1 , . . . , gk1 +···+km , . . . , gk1 +···+km Akm −1 ]T ,
then
A1 = T AT −1
A11 · · · A1m
.. ,
..
= ...
.
.
Am1 · · · Amm
where Aij ∈ Cki ×kj .
For i = j,
Aii =
0 1 0 ···
0 0 1 ···
.. .. .. . .
.
. . .
0 0 0 ···
∗ ∗ ∗ ···
0
0
..
.
;
1
∗
for i ̸= j,
0 ··· 0
.. . . ..
. .
Aij = .
0 ··· 0
∗ ··· ∗
B11
..
B1 = T B = .
,
;
B1m
and where B1i ∈ Cki ×m ,
0 ··· 0 0 0
.. . . .. .. ..
. . . .
B1i = .
0 ··· 0 0 0
0 ··· 0 1 ∗
··· 0
. . ..
. . ,
··· 0
··· ∗
where 1 is in the i-th column.
Then by properly choosing the feedback F and transformation matrix G, Bc =
BG, Ac = A1 + Bc F is in the Brunovsky canonical form
A11 · · ·
0
.. ,
..
Ac = ...
.
.
0
· · · Amm
4.6. EIGENSTRUCTURE PLACEMENT
95
where Aii ∈ Cki ×ki ,
Aii =
and
0 1 0 ···
0 0 1 ···
.. .. .. . .
.
. . .
0 0 0 ···
0 0 0 ···
0
0
..
.
,
1
0
Bc1
Bc = ... ,
Bcm
where Bci ∈ Cki ×m ,
0 ··· 0 0 0
.. . . .. .. ..
. . . .
Bci = .
0 ··· 0 0 0
0 ··· 0 1 0
··· 0
. . ..
. . ,
··· 0
··· 0
where 1 is in the i-th column. It can be known that Ac has invariant polynomials
λk1 , λk2 , . . . , λkm . Since the controllability indices are invariant under feedback,
for any F , [A + BF |B] has the same Brunovky form.
Step 2: We then start from the Brunovky form of the system. The degrees
of the invariant polynomials are expected to be λα1 , λα2 , . . . , λαm . The desired
feedback will be obtained by a recursion of feedbacks. In this way, the degrees
of the invariant factors will be changed step by step.
Without loss of generality, we will only demonstrate how to find the feedback
and the correponding cyclic spaces with two controllability indices, namely k1
and k2 with k1 ≥ k2 . The desired degrees of invariant polynomials are α1 and
α2 with α1 ≥ α2 . Let j = α1 − k1 . Take F2 : X → U such that
∗
BF2 R1 = 0;
j−1
BF2 Ac b2 = b1 ;
BF2 Ai−1
c b2 = 0, i ∈ {1, . . . , k2 }\{j}.
Define  = Ac + BF2 and b̂1 = Ajc b2 . Then the cyclic subspaces can be constructed respectively as
X1 : = span{b2 , Âb2 , · · · , Âα1 −1 b2 }
X2 : = span{b̂1 , Âb̂1 , · · · , Âα2 −1 b̂1 }.
X1 and X2 are independent Â-invariant and Â-cyclic with invariant polynomials
λα1 and λα2 .Take the transformation matrix
T = [Âα1 −1 b2 , Âα1 −2 b2 , · · · , Âb2 , b2 , Âα2 −1 b̂1 , Âα2 −2 b̂1 , . . . , Âb̂1 , b̂1 ],
then Ā = T −1 ÂT and B̄ = T −1 B. The following of step 2 is the recursion of
the two controllability indices case. Two subspaces can always be dealt with at
96
CHAPTER 4. ELEMENTARY FEEDBACK CONTROL
each time. Therefore, it is sufficient to apply step 2 successively to deal with
the case with more than two controllable indices.
Step 3: This step aims to assign the coefficients of the invariant polynomials
while the degrees remain the same. Without loss of generality, we can still focus
on the two controllability indices case. Assume that after step 2, we obtain that
Ā1 0 | ∗ b2
Ā|B̄ =
,
0 Ā2 | b̂1 0
and the desired invariant factors are
γi (λ) = λα1 + (ai1 + ai2 λ + · · · + aiαi ), i = 1, 2.
In order to assign the coefficients, we apply the third feedback
0
0
···
0
−a21 −a22 · · · −a2α2
F3 =
,
−a11 −a12 · · · −a1α1
0
0
···
0
then we can get the new matrix
h
i à X | ∗ b2 1
Ã|B̄ =
,
0 Ã2 | b̂1 0
where à = Ā+B̄F3 . The basis of à is {e1 , e2 , · · · , eα1 , eα1 +1 , eα1 +2 , · · · , eα1 +α2 },
which can be written in the following form
(
(j)
ej = γ1 (Ã1 )b2 , j = 1, 2, · · · , α1 ,
(j)
eα1 +j = γ2 (Ã2 )b̂1 , j = 1, 2, · · · , α2 ,
(0)
(j)
(j−1)
(λ)−aij ), j = 1, · · · , αi .
where for i = 1, 2, γi (λ) = γi (λ) and γi (λ) = λ1 (γi
We can express the column vectors xj of the matrix X in the form of this basis
as
α1
X
xj =
xk,j ek = (Ā1 ∆j + θj )b2 , j = 1, · · · , α2 ,
k=1
P 1 −1
(k+1)
(Ã1 ) and θj = αk=1
xk,j a1k+1 +xα1 ,j , j = 1, · · · , α2 .
where ∆j = k=1 xk,j γ1
In order to set the X part to be 0 by feedback, a new basis needs to be taken
as
ēj = ej ; j = 1, 2, · · · , α1
ēα1 +j = eα1 +j − ∆j b2 , j = 1, 2, · · · , α2 .
Pα1 −1
Then the matrix is in the form
"
#
h
i
Ø1 Xα1 β | ∗ b2
˜
Ã|B̄ =
,
0
Ø2
| b̂1 0
where XαT1 = [0 · · · 0 1]1×α1 and β = [β1 · · · βα2 ].
Applying the feedback
0 0 ··· 0
0
0
···
0
F4 =
,
0 0 · · · 0 −β1 −β2 · · · −βα2
we can finish our proof.
4.6. EIGENSTRUCTURE PLACEMENT
97
Exercises
4.1 Let
1
1
A=
0
0
9
0
7
0
9
0
0
1
7
0
,
1
0
1
0
B=
2
0
0
0
.
1
0
Find F such that the eigenvalues of A + BF are all equal to 0.
4.2 Show that A B is null controllable (controllable to zero) iff there
exists F such that A + BF is nilpotent.
4.3 Let
A=
1 2
,
2 4
B=
1
.
2
Is A B controllable, null controllable, or stabilizable? If possible, find
F such that A + BF is deadbeat.
4.4 The state feedback discussed in Section 4.1 is called static state feedback.
In general state feedback can be dynamic, i.e., we can use u = Kx where
K is a state space system. Show that a system can be stabilized by a
dynamic state feedback iff it can be stabilized by a static state feedback.
4.5 The general form of an observer is
xo (k + 1) = Ao xo (k) + Bo1 u(k) + Bo2 y(k)
x̃(k) = Co xo (k) + Do1 u(k) + Do1 y(k).
The observer given in the Section 4.2 (called full order observer) has n-th
order. Show that an (n − p)-th order observer (called a minimal order
observer) exists, where p = rankC, if and only if (C, A) is detectable.
Give such an observer.
"
#
A B
4.6 Suppose a system
can be stabilized by a static output feedback,
C 0
i.e., there exist a matrix K such that A + BKC is stable. Show that this
static feedback controller is indeed a member of the set given in Theorem
4.6.
"
#
A B
4.7 For a stabilizable and detectable system
. Characterize the set
C D
of all strictly proper stabilizing controllers.
"
#
A B
4.8 Given system
, find the condition under which there is an
C 0
output-feedback controller so that the closed-loop system is deadbeat.
Assume that this condition is satisfied, characterize the set of all controllers yielding deadbeat closed-loop systems.
98
CHAPTER 4. ELEMENTARY FEEDBACK CONTROL
4.9 Consider the system shown in Figure 4.9. Assume that x is the source signal and C represents a communication channel with known stable transfer
function. Also assume that d is noise and W is a weighting system describing the characteristics of the noise. The design problem is to design
an equalizer E so that the transmission error is as small as possible. Convert the problem into the standard setup.
d
?
W
x-
C
?
- j - E
y
-
Figure 4.9: Equalization
4.10 A tracking system is shown in Figure 4.10. r is the reference signal. d is a
disturbance. The plant P is given. The problem is to design the controller
C so that the closed-loop system is internally stable and the tracking error
e is as small as possible. Convert the problem into a standard setup.
r - jeC
+
6
−
d
u- ?
j - P
y
-
Figure 4.10: Tracking system
4.11 A so-called two degree of freedom controller structure is shown in Figure 1.
The
plant# is assumed to have a stabilizable and detectable realization
"
A B
. Notice that the controller C = C 1 C 2 has two groups
C 0
of inputs, the first group being r and the second z. Assume that matrices
F and L have been found such that A + BF and A + LC are stable.
Characterize the set of all controllers C such that the closed-loop system
is internally stable.
4.6. EIGENSTRUCTURE PLACEMENT
r-
C
u-
99
P
-z
6
Figure 4.11: Two degree of freedom structure
100
CHAPTER 4. ELEMENTARY FEEDBACK CONTROL
Chapter 5
System Performance
5.1
H2 Norm
For A, B ∈ Fp×m , define their inner product to be
∗
⟨A, B⟩ = tr(A B) =
p X
m
X
a∗ij bij .
i=1 j=1
The induced norm of A is then
v
uX
m
p
p
u p X
∗
∥A∥F = ⟨A, A⟩ = tr(A A) = t
|aij |2 .
i=1 j=1
The subscript “F ”, standing for Frobenius, is used to distinguish this norm
from other possible matrix norms.
A signal G ∈ ℓp×m (Z) is said to belong to ℓ2p×m (Z) if
∞
X
∥G(k)∥2F < ∞.
k=−∞
For G ∈ ℓp×m
(Z), its ℓ2 norm is defined as
2
v
u X
u ∞
∥G∥2 = t
∥G(k)∥2 .
F
k=−∞
Often we also call ∥G∥22 the energy of G. We can also define the inner product
of F, G ∈ ℓp×m (Z) as
⟨F, G⟩ =
∞
X
tr [F (k)∗ G(k)] .
−k=∞
The right hand side converges since
∞
X
k=−∞
∞
X
1
1
|trF (k) G(k)| ≤
∥F (k)∥2F + ∥G(k)∥2F < ∞.
2
2
∗
k=−∞
101
102
CHAPTER 5. SYSTEM PERFORMANCE
Clearly
∥G∥22 = ⟨G, G⟩,
i.e., the ℓ2 norm is induced by this inner product.
We also need two subspaces of ℓp×m (Z):
ℓp×m
(Z+ ) = ℓp×m (Z+ ) ∩ ℓp×m
(Z)
2
2
ℓp×m
(Z− ) = ℓp×m (Z− ) ∩ ℓp×m
(Z).
2
2
Recall the truncation system Pk and Qk defined in Chapter 1. We have
Q0 ℓp×m
(Z) = ℓp×m
(Z+ )
2
2
P−1 ℓp×m
(Z) = ℓp×m
(Z− ).
2
2
It is easy to see that
(Z)
(Z− ) = ℓp×m
(Z+ ) + ℓp×m
ℓp×m
2
2
2
and
ℓp×m
(Z+ ) ⊥ ℓp×m
(Z− ),
2
2
i.e., ℓp×m
(Z+ ) and ℓp×m
(Z− ) are orthogonal complement to each other.
2
2
Denote the unit circle in the complex plane by T and the open unit disk by
D. A function Ĝ : T → Cp×m is said to belong to Lp×m
if
2
Z π
∥Ĝ(ejω )∥2F dω < ∞.
−π
For Ĝ ∈
Lp×m
,
2
its L2 norm is defined as
Z π
1/2
1
jω 2
∥Ĝ∥2 =
∥Ĝ(e )∥F dω
.
2π −π
If Ĝ is the transfer function of a system, then Ĝ(ejω ) is called the frequency
response of the system. Hence ∥Ĝ∥2 is the root-mean-square (RMS) value of the
frequency response. Sometimes ∥Ĝ∥22 is also called the energy of the frequency
response.
For F̂ , Ĝ ∈ Lp×m
, we can also define their inner product as
2
Z π
1
⟨F̂ , Ĝ⟩ =
trF̂ (ejω )∗ Ĝ(ejω )dω.
2π −π
Again the integral on the right hand side is bounded since
Z π
Z π
1
1
jω ∗
jω
jω 2
jω 2
|trF̂ (e ) Ĝ(e )|dω ≤
∥F̂ (e )∥F + ∥Ĝ(e )∥F dω < ∞.
2
−π
−π 2
We also need two subspaces of Lp×m
. The space Hp×m
consists of functions
2
2
p×m
in L2
which are proper and analytic outside of D. The space H⊥p×m
consists
2
of functions in Lp×m
which
are
analytic
inside
D
and
vanish
at
0.
At this
2
⊥p×m
moment, let us consider the symbol H2
as a notation; we will later show
⊥p×m
that H2
is indeed the orthogonal complement of Hp×m
.
2
In the following, we will omit the dimension indicators in the notation of
spaces ℓ2 , H2 , etc., if the dimension is not important or can be deduced easily
from the context.
5.1. H2 NORM
103
Example 5.1
Proper rational functions with no poles on or outside of the unit circle, such as
z+1
1
z+2
,
, 2
z − 0.5 z z + 0.2
are in H2 .
A rational function with no poles on or inside of the unit circle, which also
vanishes at 0, such as
z
z 2 (z − 1)
, z,
,
z+3
(z + 4)(z − 5)
are in H⊥
2.
The relationship between the time domain spaces ℓ2 (Z), ℓ2 (Z+ ), and ℓ2 (Z+ )
and the frequency domain spaces L2 , H2 , and H⊥
2 is given in the following
theorem.
Theorem 5.1
1. G ∈ ℓ2 (Z) iff Ĝ ∈ L2 . Furthermore ∥G∥2 = ∥Ĝ∥2 .
2. G ∈ ℓ2 (Z+ ) iff Ĝ ∈ H2 .
3. G ∈ ℓ2 (Z− ) iff Ĝ ∈ H⊥
2.
A rigorous proof of Theorem 5.1 is quite involved. However, the idea (at
least for the only if part) is not hard. P
Let G ∈ ℓ2 (Z). Then G(k) → 0 as
−k converges inside the
k → ∞ and k → −∞. This shows that −1
k=−∞ G(k)z
P∞
unit circle. and k=0 G(k)z −k converges outside of the unit circle. Our current
knowledge does not allow us to say any thing about their convergence on the
unit circle. It can be claimed by using some advanced mathematics that they
do converge (at almost all points) on the unit circle. Let us agree on this point.
Then
Z π
Z π
jω 2
∥Ĝ(e )∥F dω =
trĜ(ejω )∗ Ĝ(ejω )dω
−π
−π
#" ∞
#
Z π " X
∞
X
∗ jkω
−jlω
=
tr
G(k) e
G(l)e
dω.
−π
Observe that
Z
π
k=−∞
l=−∞
ejkω e−jlω dω = 2πδ0 (k − l).
−π
Therefore
∥Ĝ∥22
1
=
2π
Z
π
∥Ĝ(e
−π
jω
)∥2F dω
=
∞
X
k=−∞
trG(k)∗ G(k) = ∥G∥22 .
104
CHAPTER 5. SYSTEM PERFORMANCE
With the first statement, the second statement becomes easy since for G ∈
ℓ2 (Z+ ),
∞
X
Ĝ(z) =
G(k)z −k .
k=0
It is clearly proper and analytic outside the unit circle. Similarly, the third
statement is easy since G ∈ ℓ2 (Z− ),
Ĝ(z) =
−1
X
G(k)z −k .
k=−∞
It is clearly analytic inside the unit circle and vanishes at 0.
Roughly, this theorem says that H2 consists of transfer functions of stable
causal systems and H⊥
2 consists of transfer functions of stable strictly anti-causal
systems.
The sets of rational members of L2 , H2 , and H⊥
2 are denoted by RL2 , RH2 ,
⊥
and RH2 respectively. We will be mainly interested in these sets, since they
results from the study of FDLTI systems. Clearly, RL2 is the set of rational
functions without poles on T, RH2 is the set of rational functions without poles
on T or outside of D, and RH⊥
2 is the set of rational functions without poles
on T or inside of D and vanishes at 0.
Since any signal in ℓ2 (Z) can be uniquely decomposed into the sum of a signal
in ℓ2 (Z+ ) and a signal in ℓ2 (Z− ). Theorem 5.1 implies that a function in L2 can
be uniquely decomposed into the sum of a function in H2 and H⊥
2 . What one
can do is to take the inverse Z-transform of this function, do the decomposition
in ℓ2 (Z), and take the Z-transforms of the two decomposed terms. However, for
a function in RL2 , we can simply do a partial fractional decomposition, group
the term corresponding to poles inside the unit circle and the constant term to
form the term in RH2 , and group the terms corresponding to poles outside of
the unit circle and the terms involving z k for positive k to form the term in
RH⊥
2 . This shows
H2 + H⊥
2 = L2 .
Theorem 5.1 also implies that for F, G ∈ ℓ2 (Z),
⟨F, G⟩ = ⟨F̂ , Ĝ⟩.
This shows that ℓ2 (Z) and L2 are abstractly the same if we identify G ∈ ℓ2 (Z)
with Ĝ. In mathematical language the two spaces ℓ2 (Z) and L2 are said to be
isometrically isomorphic. Now let F̂ ∈ H2 and Ĝ ∈ H⊥
2 and let F, G be their
inverse Z-transforms, then
⟨F̂ , Ĝ⟩ = ⟨F, G⟩ = 0
since F ∈ ℓ2 (Z+ ) and G ∈ ℓ2 (Z− ). This shows that
H2 ⊥ H⊥
2.
Hence H⊥
2 is indeed the orthogonal complement of H2 . We use PH2 and PH⊥
to denote the orthogonal projection onto H2 and H⊥
2.
2
5.1. H2 NORM
105
Z
- H⊥
2
ℓ2 (Z− )
6
6
PH ⊥
P−1
2
Z
- L2
ℓ2 (Z) PH2
Q0
Z
?
?
- H2
ℓ2 (Z+ )
Figure 5.1: A commutative diagram
Recall that the conjugate G∼ of an LTI system G with impulse response G
has impulse response
G∼ (k) = G(−k)∗ .
If the transfer function of G is Ĝ then that of G∼ is given by
Ĝ∼ (z) = Ĝ(z̄ −1 )∗ .
A transfer function Û is said to be allpass or lossless if
Û ∼ Û = I.
A transfer function V̂ is said to be co-allpass or co-lossless if
V̂ V̂ ∼ = I.
Proposition 5.1 If Û is an allpass transfer function and V̂ is a co-allpass
transfer function, then
∥Û ĜV̂ ∥2 = ∥Ĝ∥2
for all Ĝ ∈ L2 .
The following two theorems give the physical meanings of the ℓ2 (L2 ) norm.
Theorem 5.2 Let G be the impulse response of an LTI system. Assume that
G ∈ ℓ2 (Z). Let yi be the response of the system to ei δ0 . Then
∥Ĝ∥22
=
m
X
i=1
∥yi ∥22 .
106
CHAPTER 5. SYSTEM PERFORMANCE
Proof Trivial.
A standard white noise u ∈ ℓm (Z) is a stationary stochastic process satisfying E[u(k)] = 0 and E[u(k)u(l)∗ ] = δk−l I, where E[·] is the expectation
operator..
Theorem 5.3 Let G be the impulse response of an LTI system. Assume that
G ∈ ℓ2 (Z). Let y be the response of the system to a standard while noise. Then
y is a stationary process with variance equal to ∥Ĝ∥22 , i.e.,
∥Ĝ∥22 = E[y ∗ (t)y(t)].
Proof
Since
∞
X
y(k) =
G(k − l)u(l),
l=−∞
it follows that
E[y ∗ (k)y(k)] = trE[y(k)y ∗ (k)]
∞
∞
X
X
= trE
G(k − l1 )u(l1 )
u∗ (l2 )G∗ (k − l2 )
l1 =−∞
= tr
= tr
= tr
=
∞
X
∞
X
l2 =−∞
G(k − l1 )E[u(l1 )u∗ (l2 )]G∗ (k − l2 )
l1 =−∞ l2 =−∞
∞
X
G(k − l)G∗ (k − l)
l=−∞
∞
X
G(l)G∗ (l)
l=−∞
∥G∥22 .
2
Definition 5.1 The L2 norm of a function Ĝ ∈ H2 is called the H2 norm of
Ĝ.
Since the aim in the engineering system design is to produce a causal stable
system, it is the H2 norm that is usually used to measure the performance of a
system.
Let us now see how to compute the H2 norm of a stable state space system.
Recall that the controllability "gramian #Wc and the observability gramian Wo
A B
of a stable state space system
are defined to be the solutions of the
C D
Lyapunov equations
AWc A∗ − Wc = −BB ∗
and
A∗ Wo A − Wo = −C ∗ C.
5.2. H∞ NORM
107
Also
Wc =
∞
X
Ak BB ∗ A∗k
k=0
and
∞
X
Wo =
A∗k C ∗ CAk .
k=0
"
Theorem 5.4 For a stable state space system G =
A B
C D
#
,
∥Ĝ∥22 = tr(DD∗ + CWc C ∗ ) = tr(D∗ D + B ∗ Wo B).
Proof
Since
k<0
0
D
k=0
G(k) =
CAk−1 B k > 0,
it follows that
∥G∥22
∗
= trDD +
∞
X
trCAi BB ∗ A∗i C ∗
i=0
∞
X
∗
= trDD + trC
!
i
∗
∗i
C∗
A BB A
i=0
∗
∗
= tr(DD + CWc C )
and
∥G∥22
∗
= trD D +
∞
X
trB ∗ A∗i C ∗ CAi B
i=0
= trD∗ D + trB ∗
∞
X
!
A∗i C ∗ CA B
i=0
∗
∗
= tr(D D + B Wo B).
2
5.2
H∞ norm
For a matrix A ∈ Fp×m , another norm is defined as
∥A∥s = σ1 (A).
Here we assume that the singular values are ordered in a non-increasing way:
σ1 (A) ≥ σ2 (A) ≥ · · · ≥ σmin{p,m} (A). Hence σ1 (A) is the largest singular value
of A. This norm is called the spectral norm. It can be shown that
∥A∥s =
∥Ax∥2
.
x∈Fm ,x̸=0 ∥x∥2
sup
108
CHAPTER 5. SYSTEM PERFORMANCE
Consequently, the spectral norm is also called the matrix norm induced from
the Hölder 2-norm.
Denote the unit circle in the complex plane by T and the open unit disk by
p×m
D. A function Ĝ : T → Cp×m is said to belong to L∞
if
ess
∥Ĝ(ejω )∥s < ∞.
sup
ω∈[−π,π]
For Ĝ ∈ Lp×m
∞ , its L∞ norm is defined as
∥Ĝ∥∞ = ess
sup
∥Ĝ(ejω )∥s .
ω∈[−π,π]
If Ĝ is the transfer function of a system, then ∥Ĝ∥∞ is the peak of the frequency
response.
p×m
We also need two subspaces of Lp×m
consists of functions
∞ . The space H∞
p×m
in L∞ which are proper and analytic outside of D. The space H⊣p×m
consists
∞
of functions in Lp×m
which
are
analytic
inside
D
and
vanish
at
0.
Here
the
∞
⊣p×m
p×m
symbol H∞
is just as a notation. Since L∞ is not an inner product space,
there is no such concepts as orthogonal complement.
In the following, we will omit the dimension indicators in the notation of
spaces L∞ , H∞ , etc., if the dimension is not important or can be deduced easily
from the context.
Example 5.2
Proper rational functions with no poles on or outside of the unit circle, such as
z+1
1
z+2
,
, 2
z − 0.5 z z + 0.2
are in H∞ .
A rational function with no poles on or inside of the unit circle, such as
z
z 2 (z − 1)
, z,
,
z+3
(z + 4)(z − 5)
are in H⊣∞ .
The sets of rational members of L∞ , H∞ , and H⊣∞ are denoted by RL∞ ,
RH∞ , and RH⊣∞ respectively. We will be mainly interested in these sets, since
they results from the study of FDLTI systems. Clearly, RL∞ is the set of
rational functions without poles on T, RH∞ is the set of rational functions
without poles on T or outside of D, and RH⊣∞ is the set of rational functions
without poles on T or inside of D and vanishes at the origin. We see that the
sets RL∞ , RH∞ , and RH⊣∞ contain the same elements as RL2 , RH2 , and
RH⊥
2 respectively. The difference is that the norm used in RL∞ , RH∞ , and
RH⊣∞ are different from that used in RL2 , RH2 , and RH⊥
2.
The following theorem give the physical meanings of the L∞ norm.
Theorem 5.5 Let Ĝ ∈ L∞ . Then
∥Ĝ∥∞ =
∥Ĝû∥2
∥Ĝû∥2
= sup
.
û∈L2 ,û̸=0 ∥û∥2
û∈H2 ,û̸=0 ∥û∥2
sup
5.3. OPTIMAL CONTROL PROBLEMS
109
Definition 5.2 The L∞ norm of a function Ĝ ∈ H∞ is called the H∞ norm
of Ĝ.
Since the aim in the engineering system design is to produce a causal stable
system, it is the H∞ norm that is usually used to measure the performance of
a system.
5.3
Optimal Control Problems
z
w
G
y
u
- K
Figure 5.2: The standard setup
Consider Figure 5.2. Assume that G is given. The general H2 optimization
problem is to design K so that the overall system is internally stable and the
H2 norm of the transfer function from w to z
∥F l (Ĝ, K̂)∥2
is minimized.
The general H∞ optimization problem is to design K so that the overall
system is internally stable and the H∞ norm of the transfer function from w to
z
∥F l (Ĝ, K̂)∥∞
is minimized.
Let
A B1 B2
G = C1 D11 D12 .
C2 D21 D22
Then it is seen that necessary conditions for the solvability of the above problems are (A, B2 ) is stabilizable and (C2 , A) is detectable. Assume that these
conditions are satisfied. Then the set of all stabilizing controllers can be characterized by
(
"
#
)
à B̃
K = F l (J , Q) : Q =
, Ã is stable and I + D22 D̃ is nonsingular
C̃ D̃
where
A + B2 F + LC2 + LD22 F
J=
F
−C2 − D22 F
−L B2 + LD22
0
I
I
−D22
110
CHAPTER 5. SYSTEM PERFORMANCE
and F, L are matrices such that A + B2 F and A + LC2 are stable, as shown
in Figure 4.8. With this characterization, the closed loop system with input w
and output z becomes
F l (G, K) = T 11 + T 12 QT 21
where
T 11
T 12
T 21
A + B2 F
B2 F
B1
0
A + LC2 −B1 − LD21
=
C1 + D12 F
D12 F
D11
#
"
A + B2 F B2
=
C1 + D12 D12
"
#
A + LC2 B1 + LD12
.
=
C2
D21
Therefore the general H2 optimization problem becomes to design a stable
system Q such that
∥T̂11 + T̂12 Q̂T̂21 ∥2
is minimized and the general H∞ optimization problem becomes to design a
stable system Q such that
∥T̂11 + T̂12 Q̂T̂21 ∥∞
is minimized. These problems can be considered as special cases of the general
H2 and H∞ optimizations, but they have the advantage that the “fraction”
in the in the original linear fractional transformation disappeared. The special
function Q̂ → T̂11 + T̂12 Q̂T̂21 is an affine function on a linear space. Hence
the above minimization problems are convex minimization problems, which is
easier to solve at least in principle. These problems are called H2 and H∞
model matching problems.
In the next few chapters, we will investigate what additional solvability
conditions are needed for the H2 and H∞ optimization problems and how the
problems can be solved when the solvability conditions are satisfied. The solutions follow the above idea of converting to model matching problems and then
solving the model matching problems.
Exercises
5.1 A state space method is given in Theorem 5.4 to compute the H2 norm
of a causal stable system. Assume now that a noncausal stable system is
given by a rational transfer function. Give a procedure to compute its L2
norm using Theorem 5.4.
P
−k ∈ H . Find
5.2 Let T̂ (z) = ∞
2
k=0 T (k)z
min ∥T̂ (z) + z −n Q̂(z)∥2 .
Q̂∈H2
5.3. OPTIMAL CONTROL PROBLEMS
111
5.3 A SISO causal stable system G has transfer function Ĝ and impulse response G. Show that
∞
X
∥Ĝ∥∞ ≤
|G(k)|.
k=0
Show that if G(k) ≥ 0 for all k ≥ 0, then
∥Ĝ∥∞ =
∞
X
k=0
|G(k)|.
112
CHAPTER 5. SYSTEM PERFORMANCE
Chapter 6
An Input-Output Stabilization
Theory
6.1
Coprime Factorization
We’ve seen the state space interpretation of all stabilizing controllers and general feedback systems in the previous sections. However it is not the only interpretation. In particular, the whole theory can be restated by transfer functions
alone. This is in fact what came out first around 1970s-80s and was later put
into the state space form. As a preliminary knowledge we first introduce the
coprime factorization of proper stable real rational transfer function matrices
(hereafter abbreviated as matrices over RH∞ ) in this section. Then armed
with this new weapon, we restate the conclusions on the internal stability, the
Youla-Kučera parametrization and the closed-loop transfer function in the next
section. Moreover, some new results can be obtained, as the reader will find in
Section 6.3.
We begin with the concept of coprimeness. Recall that in the ring of integers
or the ring of polynomials, we say that two of the elements are coprime if
there exist another two elements from the same ring such that the Bézout
Identity holds. In fact this idea can be extended to the matrices over RH∞ .
However since the multiplication between matrices does not commute, we need
two definitions for different cases.
Definition 6.1 Let M̂ and N̂ be two matrices over RH∞ with the same number
of columns. They are said to be right coprime if there exist two matrices X̂ and
Ŷ also over RH∞ such that the following equality holds
X̂ M̂ + Ŷ N̂ = I.
Definition 6.2 Let M̂ and N̂ be two matrices over RH∞ with the same number
of rows. They are said to be left coprime if there exist two matrices X̂ and Ŷ
also over RH∞ such that the following equality holds
M̂ X̂ + N̂ Ŷ = I.
113
114
CHAPTER 6. AN INPUT-OUTPUT STABILIZATION THEORY
By the coprimeness we can introduce the coprime factorization, which turns
out to be a very powerful tool not only in the content stated here, but also in
many of the later topics in this notes. As expected, there are two versions for
this concept.
Definition 6.3 A transfer function matrix Ĝ is said to have a right coprime
factorization if there exists a pair of right coprime matrices M̂ , N̂ over RH∞
such that
Ĝ = N̂ M̂ −1 .
Definition 6.4 A transfer function matrix Ĝ is said to have a left coprime
ˆ Ñ
ˆ over RH
factorization if there exists a pair of left coprime matrices M̃,
∞
such that
ˆ −1 Ñ.
ˆ
Ĝ = M̃
A natural question one may ask about the definitions is whether the factorizations exist for all transfer function matrices. This is indeed true for the
proper real rational ones, which is confirmed by the following theorem.
Theorem 6.1 For each proper real rational transfer function matrix Ĝ with
a stabilizable and detectable realization (A, B, C, D), there exist eight matrices
ˆ Ñ,
ˆ X̃,
ˆ Ỹˆ over RH such that Ĝ = N̂ M̂ −1 = M̃
ˆ −1 Ñ
ˆ are right
M̂ , N̂ , X̂, Ŷ ,M̃,
∞
and left coprime factorizations respectively, and also the following equality holds
"
#
ˆ −Ỹˆ M̂ Ŷ X̃
= I.
ˆ M̃
ˆ
N̂ X̂
−Ñ
Such a factorization is called a doubly coprime factorization of Ĝ. One particular choice of such factorizations is given by
A + BF B −L
M̂ Ŷ
=
F
I
0 ,
N̂ X̂
C + DF D I
"
#
A + LC −B − LD L
ˆ
ˆ
X̃ −Ỹ
=
F
I
0 .
ˆ M̃
ˆ
−Ñ
−D
I
C
where F and L are chosen so that A + BF and A + LC are stable.
Proof Since the special choice is presented and always exists, it only remains
to verify that the choice indeed gives a doubly coprime factorization.
First verify that Ĝ is given by the factorization. Only the right coprime
factorization is verified since the other part is similar. Noticing that A + BF is
stable, introduce a new variable v := u − F x, then it satisfies
x(k + 1) = (A + BF )x(k) + Bv(k)
u(k) = F x(k) + v(k)
y(k) = (C + DF )x(k) + Dv(k)
6.1. COPRIME FACTORIZATION
115
Hence it’s easy to figure out that M̂ is exactly the transfer function from v to
u and N̂ is the one from v to y, i.e. u = M̂ v, y = N̂ v. So Ĝu = y = N̂ M̂ −1 u
for any u, or Ĝ = N̂ M̂ −1 .
One can alternatively verify this by directly multiplying the two systems.
−1
A + BF B
A + BF B
−1
N̂ M̂
=
C + DF D
F
I
A −B
A + BF B
=
C + DF D
F
I
A + BF BF
B
0
A −B
=
C + DF DF D
A + BF 0 0
0
A B
=
C + DF C D
A B
=
C D
= Ĝ.
Then verify the last identity, which also implies coprimeness between M̂ and
ˆ and Ñ
ˆ . Since both the systems are invertible, it suffices to
N̂ , and between M̃
verify that
#
−1 " ˆ
M̂ Ŷ
X̃ −Ỹˆ
=
.
ˆ M̃
ˆ
N̂ X̂
−Ñ
But this is not hard to see since
−1
M̂ Ŷ
N̂ X̂
I 0 −1
I 0 −1
F
− B −L
A + BF − B −L
D I
C + DF
D I
=
−1 −1
I 0
F
I 0
D I
C + DF
D I
A + BF − (BF − LC) −B − LD L
=
F
I
0
C
−D
I
"
#
ˆ −Ỹˆ
X̃
=
.
ˆ M̃
ˆ
−Ñ
However, although coprime factorizations always exist, they are highly nonunique. Define a matrix over RH∞ to be unimodular if its inverse exists and
ˆ if
is also a matrix over RH∞ . Then for any unimodular matrices D̂ and D̃,
"
#
ˆ −Ỹˆ M̂ Ŷ X̃
=I
ˆ M̃
ˆ
N̂ X̂
−Ñ
116
CHAPTER 6. AN INPUT-OUTPUT STABILIZATION THEORY
gives a doubly coprime factorization of Ĝ, then we have
"
#"
#
ˆ −D̂−1 Ỹˆ
ˆ −1
D̂−1 X̃
M̂ D̂ Ŷ D̃
ˆ Ñ
ˆ
ˆ M̃
ˆ
ˆ −1
−D̃
D̃
N̂ D̂ X̂ D̃
"
#"
#
"
#
ˆ −Ỹˆ M̂ Ŷ D̂
D̂−1 0
0
X̃
=
ˆ
ˆ −1
ˆ M̃
ˆ
N̂ X̂
0
D̃
0 D̃
−Ñ
= I
and it also gives a doubly coprime factorization of Ĝ. Moreover, the coprime
factorization is still non-unique even up to unimodular equivalence. As one will
see later in Corollary 6.2, for any Q̂ ∈ RH∞ , if
"
#
ˆ −Ỹˆ M̂ Ŷ X̃
=I
ˆ M̃
ˆ
N̂ X̂
−Ñ
then
"
ˆ + Q̂Ñ
ˆ
X̃
ˆ
−Ñ
ˆ
−Ỹˆ − Q̂M̃
ˆ
M̃
#
M̂
N̂
Ŷ + M̂ Q̂
X̂ + N̂ Q̂
=I
again gives a doubly coprime factorization of Ĝ.
The result is better if we consider the general right (or left) coprime factorizations rather than the doubly coprime factorizations. We are able to show
that, such a general coprime factorization is unique up to a multiplication by a
unimodular equivalence over RH∞ .
Corollary 6.1 The right (or left) coprime factorization of a given system Ĝ is
unique up to right (or left) multiplications by a unimodular matrix over RH∞ .
Proof
Assume that Ĝ = N̂ M̂ −1 is one of the right coprime factorizations
and X̂ M̂ + Ŷ N̂ = I. Then for any unimodular matrix D over RH∞ , N̂ D and
M̂ D are also coprime since
D−1 X̂ M̂ D + D−1 Ŷ N̂ D = I.
Moreover N̂ D(M̂ D)−1 = N̂ DD−1 M̂ −1 = N̂ M̂ −1 = Ĝ, hence it’s also a right
coprime factorization of Ĝ.
It remains to show that all other right coprime factorizations of Ĝ can be
expressed in such a form for some unimodular D. Suppose that Ĝ = T̂ Ŝ −1 is
another arbitrary right coprime factorization. Then since both M̂ and Ŝ are
invertible in RH∞ , they are in fact unimodular. Hence we may choose the
unimodular matrix
D = M̂ −1 Ŝ,
such that N̂ D = N̂ M̂ −1 Ŝ = T̂ Ŝ −1 Ŝ = T̂ and M̂ D = M̂ M̂ −1 Ŝ = Ŝ.
The parallel proof for left coprime factorizations is omitted here.
Nevertheless, lacking uniqueness does not hurt its significance in the control
theory. As mentioned at the beginning, we will finish this section with the
6.1. COPRIME FACTORIZATION
117
G j
6
?
- j - K
Figure 6.1: The system setup
internal stability condition, and introduce more of its consequences in the next
two sections.
We need one more lemma before really begin. Consider the feedback system
(G, K) depicted in Figure 6.1, then
Lemma 6.1 Assume that the closed-loop system (G, K) in Figure 6.1 is wellposed. Then it is internally stable if and only if
I K̂
Ĝ I
−1
is stable.
Proof I − K̂ Ĝ and I − ĜK̂ are invertible since the problem is well-posed.
Hence
−1 I K̂
(I − K̂ Ĝ)−1
−K̂(I − ĜK̂)−1
=
.
Ĝ I
−Ĝ(I − K̂ Ĝ)−1
(I − ĜK̂)−1
By definition (G, K) is internally stable if and only if all of its transfer functions
from inputs to outputs are stable. But they are exactly the entries in the above
matrix on RHS, with only possible differences in the sign. This finishes the
proof.
Here comes the central theorem which has many useful applications. Assume
ˆ −1 Ñ,
ˆ K̂ = V̂ Û −1 = Ũ
ˆ −1 Ṽˆ are (not necessarily doubly)
that Ĝ = N̂ M̂ −1 = M̃
coprime factorizations. Then we have the following theorem.
Theorem 6.2 The following statements are all equivalent.
1. (G, K) is internally stable.
−1
M̂ V̂
2.
is stable.
N̂ Û
"
3.
ˆ
Ũ
ˆ
−Ñ
−Ṽˆ
ˆ
M̃
#−1
is stable.
ˆ Û − Ñ
ˆ V̂ )−1 is stable.
4. (M̃
ˆ M̂ − Ṽˆ N̂ )−1 is stable.
5. (Ũ
118
CHAPTER 6. AN INPUT-OUTPUT STABILIZATION THEORY
Proof First we consider 1⇔2.
Note that
I K̂
Ĝ I
M̂
0
Interestingly,
−1
0
Û
X̂ Ŷ
T̂ Ŝ
−1
I
V̂ Û −1
=
N̂ M̂ −1
I
−1
M̂ V̂
M̂ −1
0
=
N̂ Û
0
Û −1
−1
M̂ 0
M̂ V̂
=
.
N̂ Û
0 Û
M̂
N̂
and
M̂
0
0
Û
V̂
Û
are right coprime since
0
T̂
+
Ŷ
0
M̂
N̂
V̂
Û
= I,
where X̂ M̂ + Ŷ N̂ = Ŝ Û + T̂ V̂ = I. This implies
X̂ Ŷ
T̂ Ŝ
I K̂
Ĝ I
−1
+
0
T̂
Ŷ
0
=
M̂
N̂
V̂
Û
−1
,
−1
−1
I K̂
M̂ V̂
is stable. But by
is stable if and only if
Hence
Ĝ I
N̂ Û
Lemma 6.1, the latter is equivalent to say that (G, K) is stable. Hence 1⇔2.
Similar argument also holds when proving 1⇔3 and hence is omitted.
Next we see that 2 or 3⇒4 and 5. Indeed if 2 or 3 holds then
"
ˆ M̂ − Ṽˆ N̂
Ũ
0
0
ˆ Û − Ñ
ˆ V̂
M̃
#
"
=
ˆ
Ũ
ˆ
−Ñ
−Ṽˆ
ˆ
M̃
#
M̂
N̂
V̂
Û
is invertible, which implies 4 and 5.
It only remains to show that 4 or 5⇒1. Take 5 for example. Since
"
1 holds if
similar.
I
P̂
ˆ M̂
Ũ
N̂
K̂
I
−1
Ṽˆ
I
"
ˆ −1 Ṽˆ
I
Ũ
=
N̂ M̂ −1
I
"
ˆ
M̂ 0
Ũ M̂
=
0 I
N̂
#−1
Ṽˆ
I
#−1 "
ˆ
Ũ
0
0
I
#
,
#−1
is stable, or equivalently 5 holds. Proving 4⇒1 is
6.2. ALL STABILIZING CONTROLLERS REVISITED
6.2
119
All Stabilizing Controllers Revisited
One of the most important applications of the coprime factorization makes
use of the non-uniqueness of coprime factorizations and gives the complete
description on all stabilizing controllers of a transfer function matrix Ĝ, as
stated in the following corollary.
ˆ Ñ,
ˆ X̃,
ˆ Ỹˆ ) be a doubly coprime factorization
Corollary 6.2 Let (M̂ , N̂ , X̂, Ŷ , M̃,
of Ĝ. Then the set of all stabilizing controllers of Ĝ is given by
n
o
ˆ + Q̂Ñ
ˆ )−1 (Ỹˆ + Q̂M̃
ˆ ), Q̂ is stable.
K̂ := (Ŷ + M̂ Q̂)(X̂ + N̂ Q̂)−1 = (X̃
Again we are going to prove only half way and leave the left coprime
−1
M̂ Ŷ
is stable. Then
case to the reader. By Theorem 6.2 we know that
N̂ X̂
for any stable Q̂,
Proof
M̂
N̂
Ŷ + M̂ Q̂
X̂ + N̂ Q̂
−1
=
I −Q̂
0 I
M̂
N̂
Ŷ
−X̂
−1
is also stable, i.e. K̂ = (Ŷ + M̂ Q̂)(X̂ + N̂ Q̂)−1 is an internally stabilizing
controller.
Conversely consider any internally stabilizing controller K̂ with right coˆ Û − Ñ
ˆ V̂ is invertible
prime factorization K̂ = V̂ Û −1 . Then by Theorem 6.2 M̃
in RH∞ . Now we construct Q̂ and show that K̂ can be expressed in the desired
form. Define
ˆ Û − Ñ
ˆ V̂ )−1 − Ŷ ],
Q̂ = M̂ −1 [V̂ (M̃
which is obviously stable, then it’s straight forward to verify that
ˆ Û − Ñ
ˆ V̂ )−1
Ŷ + M̂ Q̂ = V̂ (M̃
ˆ −1
= K̂(I − ĜK̂)−1 M̃
and
ˆ − Ñ
ˆ K̂)−1 − Ŷ ]
X̂ + N̂ Q̂ = X̂ + Ĝ[K̂(M̃
ˆ −1
= X̂ − ĜŶ + ĜK̂(I − ĜK̂)−1 M̃
ˆ −1 + ĜK̂(I − ĜK̂)−1 M̃
ˆ −1
= M̃
ˆ −1
= (I − ĜK̂)−1 M̃
Hence K̂ = (Ŷ + M̂ Q̂)(X̂ + N̂ Q̂)−1 for some stable Q̂.
There is also another more natural proof to the converse direction. Consider
"
#
"
#
ˆ M̂ V̂ ˆ Û
Ỹˆ
−X̃
I Ỹˆ V̂ − X̃
=
ˆ Û − Ñ
ˆ V̂ ,
ˆ M̃
ˆ
N̂ Û
0 M̃
−Ñ
120
CHAPTER 6. AN INPUT-OUTPUT STABILIZATION THEORY
hence
M̂
N̂
V̂
Û
M̂
N̂
"
M̂
=
=
N̂
#
ˆ Û
Ỹˆ V̂ − X̃
ˆ Û − Ñ
ˆ V̂
0 M̃
#
ˆ Û − Ñ
ˆ V̂ ) + M̂ (Ỹˆ V̂ − X̃
ˆ Û )
Ŷ (M̃
ˆ Û − Ñ
ˆ V̂ ) + N̂ (Ỹˆ V̂ − X̃
ˆ Û )
X̂(M̃
Ŷ
X̂
"
I
ˆ Û − Ñ
ˆ V̂ is unimodular, hence we can denote Q̂ = (Ỹˆ V̂ −
By Theorem 6.2, M̃
ˆ Û )(M̃
ˆ Û − Ñ
ˆ V̂ )−1 ∈ RH , then
X̃
∞
or
M̂
N̂
"
V̂
Û
M̂
N̂
=
M̂
N̂
Ŷ + M̂ Q̂
X̂ + N̂ Q̂
ˆ Û − Ñ
ˆ V̂ )−1
V̂ (M̃
ˆ Û − Ñ
ˆ V̂ )−1
Û (M̃
"
#
#
0
,
ˆ Û − Ñ
ˆ V̂
0 M̃
=
I
M̂
N̂
Ŷ + M̂ Q̂
X̂ + N̂ Q̂
.
Easy to verify that Ŷ + M̂ Q̂ and X̂ + N̂ Q̂ are coprime (Consider −Ñ and M̃ ).
So for any stabilizing controller K̂ = V̂ Û −1 , there exists a corresponding stable
Q̂ as defined such that one of the coprime factorizations of K̂ is of the form
K̂ = (Ŷ + M̂ Q̂)(X̂ + N̂ Q̂)−1 .
We still need to verify that
ˆ + Q̂Ñ
ˆ )−1 (Ỹˆ + Q̂M̃
ˆ ).
(Ŷ + M̂ Q̂)(X̂ + N̂ Q̂)−1 = (X̃
But noticing that
ˆ + Q̂Ñ
ˆ )(Ŷ + M̂ Q̂) − (Ỹˆ + Q̂M̃
ˆ )(X̂ + N̂ Q̂)
(X̃
ˆ Ŷ + X̃
ˆ M̂ Q̂ + Q̂Ñ
ˆ Ŷ + Q̂Ñ
ˆ M̂ Q̂
= X̃
ˆ X̂ − Q̂M̃
ˆ N̂ Q̂
−Ỹˆ X̂ − Ỹˆ N̂ Q̂ − Q̂M̃
ˆ Ŷ − Ỹˆ X̂) + (X̃
ˆ M̂ − Ỹˆ N̂ )Q̂
= (X̃
ˆ Ŷ − M̃
ˆ X̂) + Q̂(Ñ
ˆ M̂ − M̃
ˆ N̂ )Q̂
+Q̂(Ñ
= Q̂ − Q̂
= 0
the verification is done.
Thus we have shown a parallel result to the one in Section 4.4. Such identification of K̂ by Q̂ is usually called the Youla-Kučera Parametrization. Figure
6.2 shows the block diagrams.
By this corollary we can describe the closed-loop system in terms of transfer function matrices. The description turns out to be analogous to the one
appeared before in the state space form. Consider the standard setup in Figure
4.7 instead of Figure 6.1. Assume that G has a stabilizable and detectable
realization
A B1 B2
G11 G12
G=
= C1 D11 D12 ,
G21 G22
C2 D21 D22
6.2. ALL STABILIZING CONTROLLERS REVISITED
- −Ñ
- j
M̃ - −M̃
121
- j
?
?
Q
- X̃ −1
˜ −N
Q
?
- j
Ỹ
-
ˆ + Q̂Ñ
ˆ )−1 (Ỹˆ + Q̂M̃
ˆ)
a) K̂ = (X̃
?
- j
Y
X −1 b) K̂ = (Ŷ + M̂ Q̂)(X̂ + N̂ Q̂)−1
Figure 6.2: All stabilizing controllers
and (A, B2 ) is stabilizable, (C2 , A) is detectable. Then we have the following
lemma.
Lemma 6.2 A controller K stabilizes G if and only if it stabilizes G22 .
Proof
The necessity is obvious. To prove the sufficiency, note that the
realization of G22 is inherited from G, hence they share the same system matrix
A. Once G22 is stabilized, the closed-loop system matrix becomes stable, then
G is also stabilized.
By this lemma we can extend the conclusions above to the standard system setup, except that all “Ĝ” have to be replaced by “Ĝ22 ”. Again find a
ˆ Ñ,
ˆ X̃,
ˆ Ỹˆ ) of Ĝ . Define systems
doubly coprime factorization (M̂ , N̂ , X̂, Ŷ , M̃,
22
T 1 , T 2 , T 3 whose transfer functions are listed below.
T̂1 := Ĝ11 + Ĝ12 M̂ Ỹˆ Ĝ21
ˆ Ĝ
= Ĝ + Ĝ Ŷ M̃
11
12
21
T̂2 := Ĝ12 M̂
ˆ Ĝ
T̂ := M̃
3
21
And K̂ is given by a Youla-Kučera parametrization. Then
Theorem 6.3 The closed-loop transfer function is given by
Fl (G, K) = T 1 + T 2 QT 3 .
Proof
By the definition of linear fraction transformation,
Fl (G, K) = Ĝ11 + Ĝ12 (I − K̂ Ĝ22 )−1 K̂ Ĝ21 .
Note that
(I − K̂ Ĝ22 )−1
ˆ + Q̂Ñ
ˆ )−1 (Ỹˆ + Q̂M̃
ˆ )N̂ M̂ −1 )−1
= (I − (X̃
ˆ + Q̂Ñ
ˆ )M̂ − (Ỹˆ + Q̂M̃
ˆ )N̂ ]−1 (X̃
ˆ + Q̂Ñ
ˆ)
= M̂ [(X̃
ˆ + Q̂Ñ
ˆ)
= M̂ (X̃
122
CHAPTER 6. AN INPUT-OUTPUT STABILIZATION THEORY
So further we have
ˆ )Ĝ
Fl (G, K) = Ĝ11 + Ĝ12 M̂ (Ỹˆ + Q̂M̃
21
ˆ
ˆ Ĝ
= Ĝ + Ĝ M̂ Ỹ Ĝ + Ĝ M̂ Q̂M̃
11
12
21
12
21
which finishes the proof.
On the other hand consider Fl (G, K) = Ĝ11 + Ĝ12 K̂(I − Ĝ22 K̂)−1 Ĝ21 and
K̂ = (Ŷ + M̂ Q̂)(X̂ + N̂ Q̂)−1 . With similar procedure as above we have
ˆ Ĝ + Ĝ M̂ Q̂M̃
ˆ Ĝ
Fl (G, K) = Ĝ11 + Ĝ12 Ŷ M̃
21
12
21
and the rest follows.
6.3
Graph of Systems
Another application which did not show up in the earlier contents, is the graph
description of systems. It addresses the geometric view of the problem and
reveals the advantage of coprime factorizations. To introduce this corollary, we
need some preliminary knowledge. Consider G ∈ H∞ . Obviously it’s a linear
operator on H2 such that GH2 ⊂ H2 . If G ̸∈ H∞ then GH2 ̸⊂ H2 , but its
image of some subset of H2 will stay in H2 . To be more precise we introduce
the following concepts.
Definition 6.5 Let G be some given system. Then define
D(G) = {u ∈ H2 : Gu ∈ H2 .}
to be the domain of G, and
R(G) = GD(G)
to be the range of G.
With the domain and the range we can define the graph of a system.
Definition 6.6 Given a system G, the graph of G is defined to be
G(G) = {(u, Gu) : u ∈ D(G)}.
Similarly the inverse graph of G is defined to be
G −1 (G) = {(Gu, u) : u ∈ D(G)}.
The graph and inverse graph of a system are linear subspaces of H2 × H2 . It
turns out that they can precisely described by coprime factorizations, which is
shown in the next lemma. This can be considered to be the geometric meaning
of coprime factorizations.
6.3. GRAPH OF SYSTEMS
123
Lemma 6.3 Let the transfer function matrix of G has a right coprime factorization Ĝ = N̂ M̂ −1 . Then
N̂
M̂
−1
H2 .
H2 , G (G) =
G(G) =
M̂
N̂
Proof First we prove that D(G) = M̂ H2 . For any u ∈ H2 , we have M̂ u ∈
H2 and G[M̂ u] = ĜM̂ u = N̂ u ∈ H2 . Hence M̂ u ∈ D(G), which implies
D(G) ⊃ M̂ H2 . Conversely if u ∈ D(G) then u ∈ H2 and Ĝu = N̂ M̂ −1 u ∈ H2 .
The latter implies that M̂ −1 u ∈ H2 since N̂ ∈ RH∞ . But then u ∈ M H2 or
D(G) ⊂ M̂ H2 . Hence the two sets must be equal.
Then not surprisingly, R(G) = GD(G) = ĜM̂ H2 = N̂ H2 , and hence both
G(G) and G −1 (G) can be expressed in the desired form. Now we are ready to
state the corollary. Consider the system setup in Figure 6.1.
Corollary 6.3 (G, K) is stable if and only if
G(G) ⊕ G −1 (K) = H2 × H2 .
Proof
Suppose (G, K) have right coprime factorizations Ĝ = N̂ M̂ −1 and
−1
K̂ = V̂ Û
respectively. Then by Theorem 6.2, it is stable if and only if
P̂ −1 ∈ RH∞ , where
M̂ V̂
.
P̂ =
N̂ Û
This is equivalent to say that P̂ is an automorphism on H2 × H2 . Note that
P̂ is such an automorphism if and only if P̂ H2 × H2 = H2 × H2 . But by
Lemma 6.3, P̂ H2 × H2 = G(G) + G −1 (K). In particular when P is invertible,
it has a trivial kernel, in this case the sum is really a direct sum. Therefore
P̂ H2 ×H2 = H2 ×H2 if and only if G(G)⊕G −1 (K) = H2 ×H2 , which completes
the last part of the proof.
124
CHAPTER 6. AN INPUT-OUTPUT STABILIZATION THEORY
Chapter 7
Riccati Equations and
Inequalities
7.1
Generalized Eigenproblem
The generalized eigenproblem concerns a pair of square matrices A, B ∈ Cn×n .
A matrix pair (A, B) is said to be singular if for all (α, β) ∈ C × C,
det(βA − αB) = 0,
otherwise it is said to be regular. We will not be interested in singular matrix
pairs. Therefore in the following we always assume that the matrix pairs are
regular.
In C2 \ {0}, define equivalent relation
α1
α2
≡
β1
β2
if
α1
β1
=τ
α2
β2
α
for some nonzero τ ∈ C. The equivalent class containing
is denoted by
β
⟨α, β⟩. Clearly the set of such equivalent classes can be identified with C ∪ {∞},
the one-point campactification of C or the Riemann sphere, by identifying ⟨α, β⟩
by α/β.
Definition 7.1
1. ⟨α, β⟩ is said to be a generalized eigenvalue of (A, B) if
det(βA − αB) = 0.
2. If ⟨α, β⟩ is a generalized eigenvalue of (A, B), then a nonzero vector x ∈
Fn is said to be a (right) generalized eigenvector corresponding to ⟨α, β⟩
if
βAx = αBx
125
126
CHAPTER 7. RICCATI EQUATIONS AND INEQUALITIES
and nonzero vector y ∈ Fn is said to be a left generalized eigenvector
corresponding to ⟨α, β⟩ if
βy ∗ A = αy ∗ B.
Clearly, in the case when B is nonsingular, ⟨α, β⟩ is a generalized eigenvalue
of (A, B) if and only if α/β is an eigenvalue of B −1 A or AB −1 . A corresponding
generalized eigenvector of (A, B) is a corresponding eigenvector of B −1 A and
a corresponding left generalized eigenvector of (A, B) is a corresponding left
eigenvector of AB −1 .
Two matrix pencils (A, B) and (C, D) are said to be equivalent if there are
nonsingular matrices P and Q so that C = P AQ and D = P BQ. Clearly,
equivalent matrix pencils have the same generalized eigenvalues. For a given
pencil (A, B), it is possible to use P and Q to transform it into an equivalent
pair with a simple form, such as diagonal form or triangular form so that the
generalized eigenvalues can be easily read out. For numerical considerations,
such P and Q are better to be unitary matrices. The following theorem gives
an answer.
Theorem 7.1 (QZ decomposition) Let ⟨αi , βi ⟩, i = 1, 2, . . . , n, be the generalized eigenvalues of regular pair (A, B) ∈ Cn×n × Cn×n in any given order, there
exist unitary matrices U and V such that
S = U ∗ AV
and
T = U ∗ BV
are upper triangular and ⟨αi , βi ⟩ = ⟨sii , tii ⟩.
Theorem 7.1 is the basis to compute the generalized eigenvalues and the
deflating subspaces, generalized versions of eigenspaces.
A subspace X ⊂ Cn is said to be a deflating subspace if
dim(AX + BX ) ≤ dim(X ).
The one dimensional subspaces spanned by generalized eigenvectors are deflating subspaces. Let X be an m-dimensional deflating subspace. Then there
exists an m-dimensional subspace Y such that AX ⊂ Y and BX ⊂ Y. Let the
columns of X ∈ Cn×m give a basis of X and those of Y ∈ Cn×m give a basis of
Y. Then there exists a regular pair (A1 , B1 ) ∈ Cm×m × Cm×m such that
AX = Y A1
and BX = Y B1 .
If ⟨α, β⟩ is an eigenvalue of (A1 , B1 ), then it is also an eigenvalue of (A, B).
Here (A1 , B1 ) can be considered as the representation of (A, B) on X so we
denote it by (A, B)|X .
In the QZ factorization, if we partition U , V , S, T consistently as
∗ S11 S12
U1 U2 A V1 V2
=
0 S22
∗ T11 T12
U1 U2 B V1 V2
=
0 T22
7.2. ALGEBRAIC RICCATI EQUATIONS
127
so that S11 , T11 ∈ Cm×m , then
AV1 = U1 S11
and BV1 = U1 T11
This shows that X = R(V1 ) is the deflating subspace corresponding to the
generalized eigenvalues ⟨α1 , β1 ⟩, . . . , ⟨αm , βm ⟩.
Let E be a nonsingular matrix. A matrix pair (A, B) ∈ Fn×n × Fn×n is said
to be symplectic with respect to E if
AEA∗ = BEB ∗ .
A set of numbers in C ∪ {∞} are said to be symmetric to the unit circle if λ is a
member, then so is 1/λ, or equivalently if ⟨α, β⟩ is a member, then so is ⟨β̄, ᾱ⟩
Proposition 7.1 The generalized eigenvalues of a regular symplectic pair are
symmetric to the unit circle.
Proof
If ⟨α, β⟩ is a generalized eigenvalue of (A, B), i.e.,
det(βA − αB) = 0
then
det(βA∗ − αB ∗ ) = 0,
i.e., ⟨α, β⟩ is a generalized eigenvalue of (A∗ , B ∗ ). Hence there exists nonzero
x ∈ Cn such that
βA∗ x = αB ∗ x.
Multiplying AE from the left, we get
βAEA∗ x = βBEB ∗ x = αAEB ∗ x.
Multiplying BE from the left, we get
βBEA∗ x = αBEB ∗ x = αAEA∗ x.
Since (A∗ , B ∗ ) is a regular pencil, A∗ x and B ∗ x cannot be zero simultaneously.
Therefore, ⟨β, α⟩ is also a generalized eigenvalue of (A, B) with corresponding
generalized eigenvector EA∗ x or EB ∗ x.
2
7.2
Algebraic Riccati Equations
"
A B
C D
#
Given a p × m state space system
and a p × p signature matrix
Ip1
0
J =
, the associated algebraic Riccati equation (ARE) is of the
0 −Ip2
following form
A∗ XA−X +C ∗ JC −(A∗ XB +C ∗ JD)(B ∗ XB +D∗ JD)−1 (B ∗ XA+D∗ JC) = 0.
(7.1)
128
CHAPTER 7. RICCATI EQUATIONS AND INEQUALITIES
Let us use the following notation to denote the lower Schur complement:
Q11 Q12
Cl
= Q11 − Q12 Q−1
(7.2)
22 Q21 .
Q21 Q22
Then the ARE can be written as
∗ A B
X 0
A B
X 0
Cl
−
=0
C D
0 J
C D
0 0
or
X = Cl
A B
C D
∗ X 0
0 J
A B
C D
.
It is often a practice to write
∗ Q S
C
.
J C D =
S∗ R
D∗
Then the ARE can also be written as
A∗ XA − X + Q − (A∗ XB + S)(B ∗ XB + R)−1 (B ∗ XA + S ∗ ) = 0.
(7.3)
In the special case when R = D∗ JD is invertible and S = 0, the ARE becomes
A∗ XA − X + Q − A∗ XBR−1 B ∗ (I + XBR−1 B ∗ )−1 XA = 0
(7.4)
A∗ (I + XBR−1 B ∗ )−1 XA − X + Q = 0.
(7.5)
or
The forms (7.4)-(7.5) are used in many other studies of ARE. However, the
assumptions that R is invertible and S = 0 may not be valid in general applications. Furthermore, these forms have undesirable asymmetric looks. So we
will take (7.1) or (7.3) as the standard form of ARE.
If J = I or equivalently p2 = 0, then the Riccati equation is said to be
definite, otherwise it is said to be indefinite. A matrix X is said to be a solution
to (7.1) if B ∗ XB + D∗ JD is invertible and the equation is satisfied. Clearly
a necessary condition for the existence of a solution is N (B) ∩ N (D) = {0};
otherwise B ∗ XB + D∗ JD will never be invertible. This will always be assumed
in the following. A solution X to (7.1) is said to be a stabilizing solution if
A − B(B ∗ XB + D∗ JD)−1 (B ∗ XA + D∗ JC)
is stable.
Let X be a stabilizing solution of the Riccati equation (7.1) and
F = −(B ∗ XB + D∗ JD)−1 (B ∗ XA + D∗ JC).
(7.6)
Then (7.1) can be rewritten as
(A + BF )∗ X(A + BF ) − X = −(C + DF )∗ J(C + DF ).
(7.7)
This is a Lyapunov equation. By the discussion after Theorem 2.10, we obtain
the following result:
7.2. ALGEBRAIC RICCATI EQUATIONS
129
Proposition 7.2 A stabilizing solution of algebraic Riccati equation (7.1) is
Hermitian.
The next result says that the stabilizing solution, if exists, is unique.
Proposition 7.3 Riccati equation (7.1) has at most one stabilizing solution.
Suppose that X1 and X2 are stabilizing solutions of (7.1). Define
Proof
F1 = −(B ∗ X1 B + D∗ JD)−1 (B ∗ X1 A + D∗ JC)
F2 = −(B ∗ X2 B + D∗ JD)−1 (B ∗ X2 A + D∗ JC).
Since X1 and X2 are Hermitian matrices, we have
A∗ X1 A − X1 + C ∗ JC + F1∗ (B ∗ X1 A + D∗ JC) = 0
A∗ X2 A − X2 + C ∗ JC + (A∗ X2 B + C ∗ JD)F2 = 0.
Subtracting these two equations, we get
A∗ (X1 − X2 )A − (X1 − X2 ) + F1∗ (B ∗ X1 A + D∗ JC) − (A∗ X2 B + C ∗ JD)F2 = 0.
Notice that
F1∗ (B ∗ X1 A + D∗ JC) − (A∗ X2 B + C ∗ JD)F2
= F1∗ (B ∗ X1 A + D∗ JC) − F1∗ (B ∗ X2 A + D∗ JC) + (A∗ X1 B + C ∗ JD)F2
− (A∗ X2 B + C ∗ JD)F2 + F1∗ (B ∗ X2 A + D∗ JC) − (A∗ X1 B + C ∗ JD)F2
= F1∗ B ∗ (X1 − X2 )A + A∗ (X1 − X2 )BF2
+ F1∗ [−(B ∗ X2 B + D∗ JD) + (B ∗ X1 B + D∗ JD)]F2
= F1∗ B ∗ (X1 − X2 )A + A∗ (X1 − X2 )BF2 + F1∗ B ∗ (X1 − X2 )BF2 .
Therefore
(A + BF1 )∗ (X1 − X2 )(A + BF2 ) − (X1 − X2 ) = 0.
This is a Sylvester equation in X1 − X2 , see Exercise 2.26. Since both A + BF1
and A + BF2 are stable, the only solution is X1 − X2 = 0.
2
Consequently, we use “the stabilizing solution” in contrast to “a solution”
to the ARE.
"
#
A B
Im1
0
Given p × m system
and m × m matrix J =
, we
0 −Im2
C D
will also encounter the dual Riccati equation
AY A∗ − Y + BJB ∗ − (AY C ∗ + BJD∗ )(CY C ∗ + DJD∗ )−1 (CY A∗ + DJB ∗ ) = 0.
(7.8)
In this case, we assume that C D has full row rank. A matrix Y is said
to be a solution to (7.8) if CY C ∗ + DD∗ is invertible and (7.8) is satisfied. A
solution Y to (7.8) is said to be a stabilizing solution if
A − (AY C ∗ + BJD∗ )(CY C ∗ + DJD∗ )−1 C
is stable.
The following results are dual versions of the above results and can be proved
in a dual way.
130
CHAPTER 7. RICCATI EQUATIONS AND INEQUALITIES
Proposition 7.4 A stabilizing solution of algebraic Riccati equation (7.8) is
Hermitian.
Proposition 7.5 Algebraic Riccati equation (7.8) has at most one stabilizing
solution.
For the convenience of discussion, we then call ARE (7.1) the primal ARE.
For the reasons that will become apparent later, we also call primal ARE (7.1)
the control ARE or CARE and the dual ARE (7.8) the filtering ARE or FARE.
7.3
The Stabilizing Solution to the ARE
The stabilizing solution of ARE (7.1), if exists, can be solved via a generalized
eigenvalue problem. First let us consider the special case when D∗ JD is invertible. In this case, it is more convenient and concise to use the notation (7.2).
Then the ARE takes the form (7.3). Consider 2n × 2n matrix pencil
I
BR−1 B ∗
A − BR−1 S ∗ 0
.
(7.9)
,
(M, N ) =
0 A∗ − SR−1 B ∗
−Q + SR−1 S ∗ I
It is easy to verify that
M
0 −I
I 0
M∗ = N
0 −I
I 0
N ∗.
0 −I
. Then
This means that (M, N ) is a symplectic pair with respect to
I 0
the generalized eigenvalues of (M, N ) are symmetric to the unit circle. Assume
that (M, N ) has no generalized eigenvalues on the unit circle. Then it has n
V1
∈ C2n×n span
inside the unit circle and n outside. Let the columns of
V2
the deflating subspace corresponding to the n generalized eigenvalues inside the
unit circle.
Theorem 7.2 In the case when R is nonsingular, the Riccati equation (7.3)
has a stabilizing solution if and only if (M, N ) has no generalized eigenvalues on
the unit circle and V1 is invertible. In this case, the unique stabilizing solution
of (7.3) is given by X = V2 V1−1 .
Proof
If X is a stabilizing solution of (7.3) and
F = −(B ∗ XB + R)−1 (B ∗ XA + S ∗ ),
then there holds
A − BR−1 S ∗ 0
I
I
BR−1 B ∗
I
=
(A + BF ).
−Q + SR−1 S ∗ I
X
0 A∗ − SR−1 B ∗
X
I
This shows that R
is the deflating subspace of (M, N ) with respect
X
the eigenvalues of A + BF which are also the generalized eigenvalues of (M, N ).
7.3. THE STABILIZING SOLUTION TO THE ARE
131
This implies that (M, N ) has n generalized eigenvalues inside the unit circle,
hence by the symmetry it also has n outside and has no on the unit circle. Any
other basis of the deflating subspace will be given by the columns of
I
V1
X
for some nonsingular V1 . This
shows the necessity.
V1
If the columns of
span the deflating subspace corresponding to the
V2
U1
generalized eigenvalues inside the unit circle, then there exist
∈ C2n×n
U2
and (M1 , N1 ) ∈ Cn×n × Cn×n such that
V1
U1
A − BR−1 S ∗ 0
=
M1
−Q + SR−1 S ∗ I
V2
U2
V1
I
BR−1 B ∗
U1
=
N1
0 A∗ − SR−1 B ∗
V2
U2
where the generalized eigenvalues of (M1 , N1 ) are all inside the unit circle.
This implies that N1 is nonsingular and these generalized eigenvalues are the
eigenvalues of N1−1 M1 . If V1 is nonsingular and we let X = V2 V1−1 , then
I
I
BR−1 B ∗
I
A − BR−1 S ∗ 0
V1 N1−1 M1 V1−1 .
=
X
0 A∗ − SR−1 B ∗
X
−Q + SR−1 S ∗ I
This gives
A − BR−1 S ∗ = (I + BR−1 B ∗ X)V1 N1−1 M1 V1−1
and
−Q + SR−1 S ∗ + X = (A∗ − SR−1 B ∗ )XV1 N1−1 M1 V1−1
Hence
−Q + SR−1 S ∗ + X = (A∗ − SR−1 B ∗ )X(I + BR−1 B ∗ X)−1 (A − BR−1 S ∗ ).
Reorganizing gives (7.3), i.e., X is a solution to the ARE. Also
A − B(B ∗ XB + R)−1 (B ∗ XA + S ∗ ) = (I + BR−1 B ∗ X)−1 (A − BR−1 S ∗ )
= V1 N1−1 M1 V1−1
which is stable. Therefore X is a stabilizing solution. This completes the proof
of the sufficiency.
2
∗
In general when R = D JD is not necessarily nonsingular, the stabilizing solution of the ARE can be found via an expanded generalized eigenvalue
problem. Consider (n + p + n + m) × (n + m + n + p) matrix pencil
A B 0 0
I 0 0
0
C D 0 0 0 0 0 J −1
(7.10)
0 0 I 0 , 0 0 A∗ C ∗ .
0 0 0 0
0 0 B ∗ D∗
132
CHAPTER 7. RICCATI EQUATIONS AND INEQUALITIES
Proposition 7.6 The matrix pencil (7.10) has at least p eigenvalues at 0, at
least m at ∞, and the rest 2n eigenvalues are symmetric to the unit circle.
Proof We will show that the pencil (7.10) can be deflated to a 2n × 2n symplectic pencil after removing a p dimensional deflating subspace corresponding
to the eigenvalues at 0 and an m dimensional deflating subspace corresponding
to the eigenvalues at ∞ and that the 2n × 2n pencil has eigenvalues symmetric
to the unit circle. With appropriate row transformations, (7.10) is equivalent
to
I 0 0
0
A
B
0 0
−C ∗ JC −C ∗ JD I 0 0 0 A∗
0
.
,
∗
∗
∗
−D JC −D JD 0 0 0 0 B
0
0 0 0 J −1
C
D
0 0
Use the notation (7.2) to rewrite the pencil as
A
B 0
−Q −S I
−S ∗ −R 0
C
D 0
I 0 0
0
0
∗
0
0 0 0 A
,
0
0 0 0 B∗
0 0 0 J −1
0
.
Clearly,
0
0
R
0
I
is the deflating subspace of this pencil corresponding to eigenvalues at 0. After
removing this subspace, the deflated pencil becomes
A
B 0
I 0 0
−Q −S I , 0 0 A∗ .
−S ∗ −R 0
0 0 B∗
Since N (B) ∩ N (D) = {0}, there exists matrix X0 such that B ∗ X0 B + R is
invertible. Applying row transformations to the above pencil, we obtain
A
B
0
I
0 0
−Q
−S
I ,
0
0 A∗
∗
∗
∗
∗
−B X0 A − S −B X0 B − R 0
−B X0 0 B ∗
and further
A − B(B ∗ X0 B + R)−1 (B ∗ X0 A + S ∗ )
0
0
−Q + S(B ∗ X0 B + R)−1 (B ∗ X0 A + S ∗ )
0
I ,
∗
∗
∗
−B X0 A − S
−B X0 B − R 0
I − B(B ∗ X0 B + R)−1 B ∗ X0 0
B(B ∗ X0 B + R)−1 B ∗
S(B ∗ X0 B + R)−1 B ∗ X0
0 A∗ − S(B ∗ X0 B + R)−1 B ∗ .
∗
−B X0
0
B∗
7.3. THE STABILIZING SOLUTION TO THE ARE
133
0
Clearly, R I is a deflating subspace of this pencil corresponding to the
0
eigenvalues at ∞ and after removing this space, the deflated pencil becomes
A − B(B ∗ X0 B + R)−1 (B ∗ X0 A + S ∗ ) 0
,
−Q + S(B ∗ X0 B + R)−1 (B ∗ X0 A + S ∗ ) I
I − B(B ∗ X0 B + R)−1 B ∗ X0
B(B ∗ X0 B + R)−1 B ∗
.
S(B ∗ X0 B + R)−1 B ∗ X0
A∗ − S(B ∗ X0 B + R)−1 B ∗
This pencil is symplectic with respect to
0 −I
I 0
, so its generalized eigen-
values are symmetric to the unit circle.
2
Assume that (7.10) does not have eigenvalue on the unit circle. Let U be its
deflating subspace corresponding to theeigenvalues
inside the unit circle. Let
U1
U2
(n+p+n+m)×(n+p) . It is easy
a basis of U be given by the columns of
U3 ∈ C
U4
0
0
0
0
to see that R
0 ⊂ U. Let V be any complement of R 0 in U
I
I
V1
V2
(n+p+n+m)×n .
and let a basis of V be given by the columns of
V3 ∈ C
V4
Theorem 7.3 The Riccati equation (7.5) has a stabilizing solution if and only
if the matrix pencil has no generalized eigenvalue on the unit circle and V1
is invertible. In this case, the unique stabilizing solution of (7.5) is given by
X = V3 V1−1 and the corresponding state feedback gain is given by F = V2 V1−1 .
Proof: Let X be the stabilizing solution of (7.5). Let
F = −(B ∗ XB + D∗ JD)−1 (B ∗ XA + D∗ JC).
Then it is straightforward to verify that
A B 0 0
I 0
C D 0 0 F 0
0 0 I 0 X 0
0 0 0 0
0 I
I
0
=
0
0
0 0
0
I 0
−1
0 0 J
F 0
0 A∗ C ∗ X 0
0 B ∗ D∗
0 I
A + BF
0
J(C + DF ) 0 .
I 0
F 0
Hence U = R
X 0 . An arbitrary complement of R
0 I
0
0
in U is
0
I
134
CHAPTER 7. RICCATI EQUATIONS AND INEQUALITIES
I
F
given by V = R
X for arbitrary V4 . This shows the necessity. On the
V
4
V1
V2
other hand, if
V3 is obtained as in the theorem with V1 nonsingular, then
V4
M11 M12
there exists a stable matrix M =
such that
M21 M22
A B 0 0
V1
C D 0 0 V2
0 0 I 0 V3
0 0 0 0
0
0
I 0 0
0
0 0 0 J −1
0
=
0 0 0 A∗ C ∗
I
0 0 B ∗ D∗
V1
V2
V3
0
0 0
M11 M12 .
0 M21 M22
I
This can be rewritten as
AV1 + BV2 0
V1 M11
V1 M12
−1
CV1 + DV2 0
J M21
J −1 M22
=
∗
∗
∗
V3
0
A V3 M11 + C M21 A V3 M12 + C ∗ M22
0
0
B ∗ V3 M11 + D∗ M21 B ∗ V3 M12 + D∗ M22
,
which leads to M12 = 0, M22 = 0, and
M11 = V1−1 (AV1 + BV2 )
M21 = J(CV1 + DV2 ).
Hence
V3 = A∗ V3 V1−1 (AV1 + BV2 ) + C ∗ J(CV1 + DV2 )
0 = B ∗ V3 V1−1 (AV1 + BV2 ) + D∗ J(CV1 + DV2 ).
This gives
V2 V1−1 = −(B ∗ V3 V1−1 B + D∗ JD)−1 (B ∗ V3 V1−1 A + D∗ JC)
and
A∗ V3 V1−1 A − V3 V1−1 + C ∗ JC
−(A∗ V3 V1−1 B + C ∗ JD)(B ∗ V3 V1−1 B + D∗ JD)−1 (B ∗ V3 V1−1 A + D∗ JC) = 0.
Also note that
A − B(B ∗ V3 V1−1 B + D∗ JD)−1 (B ∗ V3 V1−1 A + D∗ JC) = V1−1 M11 V1
is stable. It then follows that V3 V1−1 is the stabilizing solution. This proves the
sufficiency. It also follows that the corresponding state feedback gain is
F = −(B ∗ V3 V1−1 B + D∗ JD)−1 (B ∗ V3 V1−1 A + D∗ JC) = V2 V1−1 .
2
7.4. DEFINITE ARES
135
According to Theorem 7.3, the computation of the solution of the ARE can
start by computing abasis
of
U and then follow by computing a basis of V,
0
0
the complement of R
0 in U. The latter step can actually to avoided.
I
This is because
U2
U3
U1
U4
−1
=
V2 0
V3 0
V1 0
V4 I
−1
=
F 0
.
X 0
Hence we have the following theorem, which gives an alternative way to solve
the ARE.
Theorem 7.4 The Riccati equation (7.5) has a stabilizing solution if and
only
U1
if the matrix pencil has no generalized eigenvalue on the unit circle and
U4
is invertible. In this case, the unique stabilizing solution of (7.5) is given by
−1 U1
I
X = U3
and the corresponding state feedback gain is given by
U4
0
−1 U1
I
F = U2
.
U4
0
Theorems 7.3 and 7.4 give implicit necessary and sufficient conditions on
the existence of a stabilizing solution to an ARE along with a method to find
the stabilizing solution. The condition is implicit in the sense that it does not
depend on the problem data explicitly.
7.4
Definite AREs
The following theorem gives an explicit necessary and sufficient condition for
the existence of a stabilizing solution to a definite ARE.
Theorem 7.5 In the case when J = I, ARE (7.1) has a stabilizing solution iff
the two conditions
1. (A, B) is stabilizable;
2.
A − ejω I B
C
D
has full column rank for all ω ∈ R.
Assume these two conditions are satisfied and X is the stabilizing solution, then
X ≥ 0.
The
of (A,
necessity
B) being stabilizable is obvious. To show the
A − ejω I B
necessity of
having full column rank for all ω ∈ R, we show
C
D
Proof
136
CHAPTER 7. RICCATI EQUATIONS AND INEQUALITIES
A − ejω I B
that if
C
D
an eigenvalue of
has reduced column rank for some ω ∈ R, then ejω is
A − B(B ∗ XB + D∗ D)−1 (B ∗ XA + D∗ C)
for every positive semidefinite solution X of (7.1). Suppose that
for some
x
u
A − ejω I B
C
D
x
u
=0
̸= 0. This can also be written as
B
D
A − ejω I
x.
u=−
C
(7.11)
B
= {0} is assumed, it follows that x ̸= 0. Assume that X is
Since N
D
any Hermitian solution of (7.1). Pre-multiplying (7.11) by B ∗ X D∗ gives
u = −(B ∗ XB + D∗ D)−1 (B ∗ XA + D∗ C)x + (B ∗ XB + D∗ D)−1 ejω B ∗ Xx.
Then
0 = x∗ [A∗ XA − X + C ∗ C − (A∗ XB + C ∗ D)(B ∗ XB + D∗ D)−1 (B ∗ XA + D∗ C)]x
= x∗ A∗ XAx − x∗ Xx + x∗ C ∗ Cx + x∗ (A∗ XB + C ∗ D)[u − (B ∗ XB + D∗ D)−1 ejω B ∗ Xx)]
= x∗ A∗ X(Ax + Bu) − x∗ Xx + x∗ C ∗ (Cx + Du)
− x∗ (A∗ XB + C ∗ D)(B ∗ XB + D∗ D)−1 ejω B ∗ Xx
= x∗ A∗ Xejω x − x∗ Xx + [u − (B ∗ XB + D∗ D)−1 ejω B ∗ Xx)]∗ ejω B ∗ Xx
= (x∗ A∗ + u∗ B ∗ )Xejω x − x∗ Xx − x∗ XBe−jω (B ∗ XB + D∗ D)−1 ejω B ∗ Xx
= x∗ e−jω Xejω x − x∗ Xx − x∗ XB(B ∗ XB + D∗ D)−1 B ∗ Xx
= −x∗ XB(B ∗ XB + D∗ D)−1 B ∗ Xx.
This implies that B ∗ Xx = 0. Therefore
[A − B(B ∗ XB + D∗ D)−1 (B ∗ XA + D∗ C)]x
= Ax + B[u − (B ∗ XB + D∗ D)−1 ejω B ∗ Xx)]
= Ax + Bu = ejω x,
∗
jω
i.e., A − B(B ∗ XB + D∗ D)−1 (B ∗ XA +
D C) has an eigenvalue at e . This
jω
A−e I B
shows the necessity of
having full column rank.
C
D
We next show that if the two conditions are satisfied, then the matrix pencil
(7.10) does not have a generalized eigenvalue on the unit circle and the matrix
7.4. DEFINITE ARES
137
U1
U2
U3 formed by any basis of the deflating subspace corresponding to its
U4
U1
eigenvalues inside the unit circle has a nonsingular submatrix
.
U4
On the contrary, assume that the matrix pencil (7.10) has a generalized
eigenvalue on the unit circle, i.e.,
I 0 0
0
v1
A B 0 0
C D 0 0
0
I
jω 0 0
v2
0 0 I 0 − e 0 0 A∗ C ∗ v3 = 0
v4
0 0 0 0
0 0 B ∗ D∗
∗
for a vector v1∗ v2∗ v3∗ v4∗
̸= 0. This yields
v1
A − ejω I B
0
v4
=
C
D
v2
ejω I
v3
I − ejω A∗ −ejω C ∗
= 0.
v4
−ejω B ∗ −ejω D∗
The second equation above implies
∗ ∗ A − ejω I B
v3 v4
=0
C
D
and further
v3∗
v4∗
A − ejω I B
∗ ∗ v1
0
v4 = ejω v4∗ v4 = 0.
= v3 v4
C
D
v2
ejω I
Hence v4 = 0. This shows
v3∗ A − ejω I B = 0
and
A − ejω I B
C
D
v1
v2
= 0.
Conditions 1 and 2 then force v1 = 0, v2 = 0, and v3 = 0, which together with
v4 = 0 results in a contradiction.
Now let a basis
the deflating subspace of matrix pencil (7.10) be given by
of
U1
U2
U1
(n+p+n+m)×(n+p)
the columns of
∈C
. We need to show N
=
U3
U4
U4
{0}. First notice that there is a stable matrix M ∈ C(n+p)×(n+p) such that
A B 0 0
U1
I 0 0
0
U1
C D 0 0 U2 0 0 0
I
U2
0 0 I 0 U3 = 0 0 A∗ C ∗ U3 M
0 0 0 0
U4
0 0 B ∗ D∗
U4
138
CHAPTER 7. RICCATI EQUATIONS AND INEQUALITIES
which gives
and
A B
C D
U1
U2
=
U1
U4
M
(7.12)
∗
I
A C∗
U3
U3 =
M.
(7.13)
0
B ∗ D∗
U4
U1
U1
Suppose that dim N
≥ 1. For each x ∈ N
, it holds
U4
U4
B
U1
U2 x =
M x.
D
U4
U1
Let us use the fact that N
is M -invariant. (This is yet to be proved!!!)
U4
Then
B
U2 x = 0
D
It then follows from the basic assumption N (B) ∩ N (D) = {0} that
U2 x = 0.
U1
,
Let x be an eigenvector of the restricted linear transformation M to N
U4
U1
i.e., x ∈ N
is nonzero and M x = λx for some λ ∈ D. From (7.13),
U4
we obtain
∗
∗ U3
A C∗
A
I
U3 λx.
U3 x =
Mx =
∗
∗
B D
U4
B∗
0
If λ = 0, then U3 x = 0. If λ ̸= 0, then
∗
A − λ̄−1
U3 x = 0.
B∗
which also implies U3 x = 0 due to the stabilizability of (A, B). This shows that
there exists a nonzero x such that U x = 0 which is in contradiction
with
the
U1
assumption that dim R(U ) = n + p. Therefore, we mush have N
=
U4
{0}.
2
#
"
A B
The dual ARE for data
and J is the primal ARE for data
C D
"
#
A∗ C ∗
and J. So it can be solved using the same method as for the
B ∗ D∗
primal problem.
Theorem 7.6 In the case when J = I, ARE
(7.8) has a stabilizing solution
A − ejω I B
iff (C, A) is detectable and
has full row rank for all ω ∈ R. If
C
D
the condition is satisfied and Y is the stabilizing solution, then Y ≥ 0.
7.5
Riccati inequalities
Chapter 8
Applications of Riccati
Equation
8.1
Introduction
Normalized coprime factorization, spectral factorization and inner-outer factorization have wide applications in control theory. For example, normalized
coprime factorization can be used in controller parametrization problems of linear systems [?] and inner-outer factorization has been employed in the model
matching problem [?, ?]. It can be seen later in this report that spectral factorization is crucial to prove the bounded real lemma.
There is much research available on factorization theory [?, ?, ?, ?, ?]. However, most of current results involve certain unnecessary assumptions which
narrow the application potential of these factorizations. In this project, factorizations are further investigated with those unnecessary assumptions removed.
Especially, the spectral factorization is dealt with in a general unified framework
which includes different formulations in the literature as special cases. In order
to find those factorizations’ realizations for the sake of numerical computability,
corresponding algebraic Riccati equations need to be solved.
Another issue which attracts much attention is the bounded real lemma.
It has played a crucial role in optimal control, stability analysis, etc. [?, ?].
Researchers have put effort in studying this lemma [?, ?], however, the proof
is still not satisfactory. No direct proof from the bounded realness to a Riccati
equation’s stabilizing solution was given. In this project, it is shown that the
bounded real lemma is almost a special case of spectral factorization which
gives a convincing and concise proof.
8.2
Inner and Weighted All-pass Transfer Functions
Inner functions and weighted all-pass functions play important roles in factorization theory. Results of this section will be employed in later discussions.
Definition 8.1 A transfer function G(z) ∈ RH∞ is said to be inner if G∼ (z)G(z) =
I and outer if it has a right inverse G−1 (z) ∈ RH∞ .
139
140
CHAPTER 8. APPLICATIONS OF RICCATI EQUATION
Definition 8.2 A transfer function G(z) is said to be Σ-weighted all-pass if
G∼ (z)ΣG(z) = I for given nonzero Hermitian matrix Σ.
The following theorem gives the sufficient condition for a transfer function
to be inner by a state space approach. It also provides the necessary condition
given the realization is controllable.
A
C
Theorem 8.1 Let G(z) =
B
D
, where A is stable.
1. If the following equations are satisfied:
D∗ C + B ∗ Wo A = 0,
∗
(8.1)
∗
D D + B Wo B = I,
(8.2)
where Wo is the observability Gramian of (C, A), then G(z) is inner.
2. If (A, B) is controllable and G(z) is inner, then (8.1) and (8.2) are satisfied.
Proof
Denote ρ as the spectral radius of A. Since G(z) is a causal
system, it follows that
G(z) =
∼
G(z) =
∞
X
G(k)z −k ,
k=0
0
X
∗ −k
G(−k) z
k=−∞
=
∞
X
G(k)∗ z k ,
|z| > ρ(A),
(8.3)
1
,
ρ(A)
(8.4)
|z| <
k=0
Define U (z) = G∼ (z)G(z). In view of (8.3) and (8.4), we have U (z) =
1
|z| < ρ(A)
, where
P∞
k=−∞ U (k)z
(P
∞
G∗ (k + i)G(i), k ≥ 0,
U (k) = Pi=0
∞
∗
i=0 G (i)G(−k + i), k < 0.
(
D
Substitute G(i) =
CAi−1 B
(8.5)
i=0
into (8.5) yields
i>0
P∞
∗ (A∗ )k+i−1 C ∗ CAi−1 B + B ∗ (A∗ )k−1 C ∗ D, k > 0,
i=1 B P
∗
∗ i−1 C ∗ CAi−1 B,
U (k) = D∗ D + ∞
k = 0,
i=1 B (A )
P
∞
∗
∗ i−1 C ∗ CA−k+i−1 B + D ∗ CA−k−1 B,
k < 0.
i=1 B (A )
Further computation leads to the following:
P∞
∗ i−1 C ∗ CAi−1 B
U (0) = D∗ D + B ∗
i=1 (A )
= D∗ D + B ∗ Wo B,
(8.6)
k , ρ(A)
<
8.2. INNER AND WEIGHTED ALL-PASS TRANSFER FUNCTIONS
141
and when k > 0,
P
∗
∗ k+i−1 C ∗ CAi−1 B
U (k) = B ∗ (A∗ )k−1 C ∗ D + ∞
i=1 B (A
P)∞
∗
∗
k−1
∗
∗
∗
k
∗ i−1 C ∗ CAi−1 B
= B (A ) C D + B (A )
i=1 (A )
= B ∗ (A∗ )k−1 (C ∗ D + A∗ Wo B),
(8.7)
when k < 0,
P
U (k) = ∞
B ∗ (A∗ )i−1 C ∗ CA−k+i−1
B + D∗ CA−k−1 B
i=1
P
∞
∗ i−1 C ∗ CAi−1 A−k B + D ∗ CA−k−1 B
= B∗
i=1 (A )
∗
= (B Wo A + D∗ C)A−k−1 B.
(8.8)
1. In view of (8.6), (8.7) and (8.8), it follows that U (0) = I and U (k) = 0
when k ̸= 0 which implies G(z) is inner.
2. Since G(z) is inner, it follows that
D∗ D + B ∗ Wo B = 0,
B ∗ (A∗ )k−1 (C ∗ D + A∗ Wo B) = 0, k > 0
(8.9)
In the presence of the condition that (A, B) is controllable, it can be
indicated from (8.9) that C ∗ D + A∗ Wo B = 0. This completes the proof.
Corollary 8.1 Let G(z) ∈ RH∞ , then G(z) is inner if and only if it has an
isometric realization.
Ai Bi
be an isometric realProof We prove the if part first. Let
Ci D i
ization, then
∗
Ai Ci∗ Ai Bi
=I
Bi∗ Di∗ Ci Di
from which it follows:
A∗i Ai + Ci∗ Ci = I,
(8.10)
Di∗ Ci
Di∗ Di
= 0,
(8.11)
= I.
(8.12)
+ Bi∗ Ai
+ Bi∗ Bi
It can be inferred from (8.10) that I is the observability Gramian of (Ci , Ai ).
In view of (8.11) and (8.12), Theorem 8.1 (i) yields that G(z) is inner.
A B
Now consider the only if part. Let
be a minimal realization of
C D
G(z). According to Theorem 8.1 (ii), since (A, B) is controllable and G(z) is
inner, we know that the following equations are satisfied:
D∗ C + B ∗ Wo A = 0,
∗
∗
D D + B Wo B = I.
(8.13)
(8.14)
142
CHAPTER 8. APPLICATIONS OF RICCATI EQUATION
Since A is stable, it follows that the observability Gramian W
a
o > 0. Make
Ai B i
− 21
and
state transformation via Wo , denote the new realization by
Ci Di
apply (8.13) and (8.14), we get
−1
1
1
− 21
A∗i Ai + Ci∗ Ci = Wo 2 A∗ Wo2 Wo2 AWo
−1
−1
− 12
+ Wo 2 C ∗ CWo
− 12
= Wo 2 (A∗ Wo A + C ∗ C)Wo
= I,
− 12
Di∗ Ci + Bi∗ Ai = DCWo
1
1
− 12
+ B ∗ Wo2 Wo2 AWo
− 12
= (DC + B ∗ Wo A)Wo
= 0,
1
1
Di∗ Di + Bi∗ Bi = D∗ D + B ∗ Wo2 Wo2 B
= D ∗ D + B ∗ Wo B
= I.
Ai Bi
It follows that
is an isometric realization.
Ci D i
As for the weighted all-pass transfer functions, we present the following
theorem which can be proved by the same approach as that of Theorem 8.1.
A B
and Σ be a nonzero Hermitian matrix.
Theorem 8.2 Let G(z) =
C D
1. If there exists a matrix X = X ∗ such that the following equations are
satisfied:
A∗ XA − X + C ∗ ΣC = 0,
(8.15)
∗
∗
(8.16)
∗
∗
(8.17)
D ΣC + B XA = 0,
D ΣD + B XB = I,
then G(z) is Σ-weighted all-pass.
2. If (A, B) is controllable and G(z) is Σ-weighted all-pass, then (8.15),
(8.16) and (8.17) are satisfied.
8.3
Normalized Coprime Factorization
Definition 8.3 A right coprime factorization G(z) = N (z)M −1 (z) with N (z), M (z) ∈
RH∞ is called a right normalized coprime factorization if
M ∼ (z)M (z) + N ∼ (z)N (z) = I
M (z)
i.e.,
is inner. Similarly, a left coprime factorization G(z) = M̃ −1 (z)Ñ (z)
N (z)
with M̃ (z), Ñ (z) ∈ RH∞ is called a left normalized coprime factorization if
[M̃ (z) Ñ (z)] is co-inner.
8.3. NORMALIZED COPRIME FACTORIZATION
143
In order to find all the right normalized coprime factorizations of a given
transfer function, the following lemma is given for preparation.
Lemma 8.1 The right normalized coprime factorization of a given system is
unique up to a right multiplication by a unitary matrix.
Proof By Corollary 4.1 in [?], the right coprime factorization is unique
up to a right multiplication by a unimodular matrix over RH∞ . Specifically, let
G(z) = N (z)M −1 (z) and G(z) = T (z)S −1 (z) be two arbitrary right coprime
factorizations of G(z), then a certain unimodular matrix D(z) can be found
such that T (z) = N (z)D(z) and S(z) = M (z)D(z).
We proceed to prove D(z) is actually unitary when the right coprime factorization is normalized. The fact that D(z) ∈ RH∞ yields the following:
D(z) =
D∼ (z) =
∞
X
D(k)z −k ,
k=0
0
X
D(−k)∗ z −k .
k=−∞
S(z)
M (z)
are inner, it is easy to see that D(z) is also
and
Since both
T (z)
N (z)
inner, i.e., D∼ (z)D(z) = I. This indicates that D∼ (z) = D−1 (z) ∈ RH∞ and
further implies that D(−k)∗ = 0 when k < 0 which is equivalent to D(k) = 0
when k > 0. Therefore, D(z) = D(0) is a constant matrix. Moreover, since
D∗ D = I and D is invertible, it follows that D is unitary. This completes the
proof.
Lemma 8.1 in fact gives a way to parameterize all the right normalized
coprime factorizations of a given system. Specifically, let G(z) = N (z)M −1 (z)
be one particular choice, then all the right normalized coprime factorizations
can be parameterized as follows:
T (z) = N (z)D,
S(z) = M (z)D,
where D is unitary and G(z) = T (z)S −1 (z).
The next theorem is to find realizations of all the right normalized coprime
factorizations of a given system.
A B
Theorem 8.3 Let G(z) =
. Assume that (A, B) is stabilizable and
C D
(C, A) is detectable, then all the right normalized coprime factorizations G(z) =
N (z)M −1 (z) are given as follows:
A + BF
M (z)
=
F
N (z)
C + DF
BP
P ∈ RH∞ ,
DP
(8.18)
144
CHAPTER 8. APPLICATIONS OF RICCATI EQUATION
where
F = −(I + D∗ D + B ∗ XB)−1 (D∗ C + B ∗ XA),
1
P = (I + D∗ D + B ∗ XB)− 2 U for an arbitrary unitary matrix U
and X is the stabilizing solution to the Riccati equation
A∗ XA − X + C ∗ C − (A∗ XB + C ∗ D)(I + D∗ D + B ∗ XB)−1 (B ∗ XA + D∗ C) = 0.
(8.19)
Proof
According to Lemma 8.1, it suffices to show that (8.18) gives a
right normalized coprime factorization for one specific unitary matrix U . For
simplicity, choose U = I. The Riccati equation (8.19) can be rewritten in the
standard form:
∗ ∗ D
C
C
C
∗
∗
− A XB +
A XA − X +
I
0
0
0
(8.20)
−1 ∗ ∗ C
D
D
D
∗
∗
= 0.
+ B XB
B XA +
·
0
I
I
I
Since (C, A) is detectable, the matrix
A − ejω I B
C
D
0
I
has full column rank for all ω ∈ [0, 2π). In view of the assumption that (A, B) is
stabilizable, it can be indicated that (8.20) has a stabilizing solution X. Hence,
A + BF is stable and M (z), N (z) ∈ RH∞ . Furthermore, we have X ≥ 0 and
1
I + D∗ D + B ∗ XB > 0 which implies that (I + D∗ D + B ∗ XB)− 2 is well defined.
By Theorem 4.1 and Corollary 4.1 in [?], we
know (8.18) gives a
already
M (z)
is inner. Denote
right coprime factorization. It suffices to show
N (z)
A B
M (z)
=
,
N (z)
C D
F
P
where A = A + BF , B = BP , C =
, D =
. Since X is
C + DF
DP
the stabilizing solution to (8.20), it follows that X = W o , where W o is the
observability Gramian of (C, A). Then we have
∗
∗
∗
∗
D C + B W o A =D C + B XA
=P ∗ F + P ∗ D∗ C + P ∗ D∗ DF + P ∗ B ∗ XA + P ∗ B ∗ XBF
=P ∗ ((I + D∗ D + B ∗ XB)F + D∗ C + B ∗ XA)
=0,
∗
∗
∗
∗
D D + B W o B =D D + B XB
=P ∗ (I + D∗ D + B ∗ XB)P
1
1
=(I + D∗ D + B ∗ XB)− 2 (I + D∗ D + B ∗ XB)(I + D∗ D + B ∗ XB)− 2
=I.
8.3. NORMALIZED COPRIME FACTORIZATION
145
M (z)
Application of Theorem 8.1 yileds that
is inner. This completes the
N (z)
proof.
The same approach can be applied to the left normalized coprime factorization and similar results can be obtained. The following two theorems are
given with proofs omitted since they are just analogous to those of the right
normalized coprime factorization.
Theorem 8.4 The left normalized coprime factorization of a given system is
unique up to a left multiplication by a unitary matrix.
A B
Theorem 8.5 Let G(z) =
. Assume that (A, B) is stabilizable and
C D
(C, A) is detectable, then all the left normalized coprime factorizations G(z) =
M̃ −1 (z)Ñ (z) are given as follows:
M̃ (z) Ñ (z) =
A + LC
QC
L
Q
B + LD
QD
∈ RH∞ ,
where
L = −(AY C ∗ + BD∗ )(I + DD∗ + CY C ∗ )−1 ,
1
Q = U (I + DD∗ + CY C ∗ )− 2 for an arbitrary unitary matrix U
and Y is the stabilizing solution to the dual Riccati equation
AY A∗ − Y + BB ∗ − (AY C ∗ + BD∗ )(I + DD∗ + CY C ∗ )−1 (CY A∗ + DB ∗ ) = 0.
So far, the right and left normalized coprime factorization have been handled respectively. One natural extension is to consider the doubly normalized
coprime factorization.
Definition 8.4 G(z) is said to have a doubly normalized coprime factorization
if there exists M (z), N (z), M̃ (z), Ñ (z), X(z), Y (z), X̃(z), Ỹ (z) ∈ RH∞ such
that the following conditions are satisfied:
1. G(z) = N (z)M −1 (z) = M̃ −1 (z)Ñ (z).
M (z)
is inner and [M̃ (z) Ñ (z)] is co-inner.
N (z)
X̃(z) −Ỹ (z) M (z) Y (z)
= I.
N (z) X(z)
−Ñ (z) M̃ (z)
2.
3.
146
CHAPTER 8. APPLICATIONS OF RICCATI EQUATION
A B
. Assume that (A, B) is stabilizable and
Theorem 8.6 Let G(z) =
C D
(C, A) is detectable, then there exists a doubly normalized coprime factorization
G(z) = N (z)M −1 (z) = M̃ −1 (z)Ñ (z). One particular realization is given as
follows:
A + BF BP −LQ−1
M (z) Y (z)
,
=
F
P
0
N (z) X(z)
−1
C + DF DP
Q
A + LC −(B + LD) L
X̃(z) −Ỹ (z)
= P −1 F
P −1
0 ,
−Ñ (z) M̃ (z)
QC
−QD
Q
where
F = −(I + D∗ D + B ∗ XB)−1 (D∗ C + B ∗ XA),
1
P = (I + D∗ D + B ∗ XB)− 2 ,
L = −(AY C ∗ + BD∗ )(I + DD∗ + CY C ∗ )−1 ,
1
Q = (I + DD∗ + CY C ∗ )− 2
and X is the stabilizing solution to the Riccati equation
A∗ XA − X + C ∗ C − (A∗ XB + C ∗ D)(I + D∗ D + B ∗ XB)−1 (B ∗ XA + D∗ C) = 0,
Y is the stabilizing solution to the dual Riccati equation
AY A∗ − Y + BB ∗ − (AY C ∗ + BD∗ )(I + DD∗ + CY C ∗ )−1 (CY A∗ + DB ∗ ) = 0.
Proof
The first two conditions in Definition 8.4 follow directly from
Theorem 8.3 and Theorem 8.5. We only need to verify the third condition is
also satisfied. Direct computation yields
X̃(z) −Ỹ (z) M (z) Y (z)
N (z) X(z)
−Ñ (z) M̃ (z)
LQ−1
A + LC −BF + LC −BP
0
A + BF
BP
−LQ−1
.
=
−1
−1
P F
P F
I
0
QC
QC
0
I
I −I
After a state variable transformation via P =
, we get
0 I
X̃(z) −Ỹ (z) M (z) Y (z)
N (z) X(z)
−Ñ (z) M̃ (z)
A + LC
0
0
0
0
A + BF BP −LQ−1
=
−1
P F
0
I
0
QC
0
0
I
=I.
8.4. SPECTRAL FACTORIZATION
147
Therefore, all the three conditions in Definition 8.4 are satisfied and the proof
is done.
It is worth mentioning that unlike the right or left normalized coprime
factorization, the doubly normalized coprime factorization is not unique up to
unimodular equivalence.
8.4
Spectral Factorization
Different formulations of spectral factorizations are frequently encountered in
the H2 /H∞ optimization problem. For example, spectral factorization of G∼ (z)G(z)
can be used to obtain the inner-outer factorization of G(z). On proving the famous bounded real lemma, spectral factorization of γ 2 I − G∼ (z)G(z) needs to
be conducted. To treat these different formulations in a unified framework, we
define the spectral factorization in the following general sense:
Definition 8.5 Let G(z) ∈ RL∞ have no zeros on the unit circle where G(z)
is not necessarily square. Given signature matrix J, then G∼ (z)JG(z) is said
to have a spectral factorization if there exists an outer transfer function Φ(z)
such that G∼ (z)JG(z) = Φ∼ (z)Φ(z). Similarly, G(z)JG∼ (z) is said to have
a co-spectral factorization if there exists an outer transfer function Φc (z) such
that G(z)JG∼ (z) = Φc (z)Φ∼
c (z).
It is known [?] that a stabilizing state feedback gain F can yield a right
coprime factorization G(z) = N (z)M −1 (z) over RH∞ where
A + BF
M (z)
=
F
N (z)
C + DF
BH
H ∈ RH∞
DH
and H is an arbitrary invertible matrix. It follows that
∼
G∼ (z)JG(z) = M −1 (z)N ∼ (z)JN (z)M −1 (z).
If we can find a special coprime factorization such that N (z) is J-weighted
∼
all-pass, then G∼ (z)JG(z) = M −1 (z)M −1 (z). This will lead to a spectral
factorization when G(z) ∈ RH∞ if we let Φ(z) = M −1 (z). For the general
case when G(z) is not necessarily stable, an additional co-spectral factorization
of M (z)M ∼ (z) will be needed. Using the above approach, we begin with the
following lemma:
A B
where A has no eigenvalues on the unit
Lemma 8.2 Let G(z) =
C D
circle. Assume that (A, B) is stabilizable and (C, A) is observable. Given signature matrix J, then the following statements are equivalent:
1. G∼ (ejω )JG(ejω ) > 0 for all ω ∈ [0, 2π).
148
CHAPTER 8. APPLICATIONS OF RICCATI EQUATION
2. The Riccati equation
A∗ XA − X + C ∗ JC − (A∗ XB + C ∗ JD)(D∗ JD + B ∗ XB)−1 (B ∗ XA + D∗ JC) = 0
(8.21)
has a stabilizing solution X such that D∗ JD + B ∗ XB > 0.
Moreover, if either one of the above statements is satisfied, then all the factorizations G∼ (z)JG(z) = Ψ∼ (z)Ψ(z) with Ψ−1 (z) ∈ RH∞ are given as follows:
"
#
B
A
Ψ(z) =
,
1
1
−U (D∗ JD + B ∗ XB) 2 F U (D∗ JD + B ∗ XB) 2
where F = −(D∗ JD+B ∗ XB)−1 (B ∗ XA+D∗ JC) and U is an arbitrary unitary
matrix.
Proof
1. (a) =⇒ (b).
Without loss of generality it can be assumed that A and B are in the
canonical form:
B1
A1 A2
,
,B =
A=
0
0 A3
with (A1 , B1 ) controllable and A3 stable. Also, let X and C be partitioned
with compatible dimensions:
X1 X2
, C = [C1 C2 ].
X=
X2∗ X3
Then, G(z) = D + C1 (zI − A1 )−1 B1 and (8.21) is equivalent to the following three equations:
A∗1 X1 A1 − X1 + C1∗ JC1 − (A∗1 X1 B1 + C1∗ JD)(D∗ JD + B1∗ X1 B1 )−1
(B1∗ X1 A1 + D∗ JC1 ) = 0,
(8.22)
∗
∗
−1
∗
∗
A1 − B1 (D JD + B1 X1 B1 ) (B1 X1 A1 + D JC1 ) X2 A3 − X2 + A∗1 X1 A2
+C1∗ JC2 − (A∗1 X1 B1 + C1∗ JD)(D∗ JD + B1∗ X1 B1 )−1 (B1∗ X1 A2 + D∗ JC2 ) = 0,
(8.23)
A∗3 X3 A3 − X3 + A∗2 X1 A2 + A∗3 X2∗ A2 + A∗2 X2 A3 + C2∗ JC2 − (A∗2 X1 B1 + A∗3 X2∗ B1
+C2∗ JD)(D∗ JD + B1∗ X1 B1 )−1 (B1∗ X1 A2 + B1∗ X2 A3 + D∗ JC2 ) = 0.
(8.24)
Since G∼ (ejω )JG(ejω ) > 0, it follows that all the zeros of G∼ (z)JG(z) are
symmetric to the unit circle and none of them are on the unit circle. This
makes sure of the existence of the factorization G∼ (z)JG(z) = Φ∼ (z)Φ(z)
such that Φ(z) has all its zeros inside the unit circle. Therefore, according
8.4. SPECTRAL FACTORIZATION
149
to our previous discussion, a special coprime factorization can be found
such that N (z) is J-weighted all-pass. Partition F compatibly as [F1 F2 ],
then
A1 + B1 F1 A2 + B1 F2 B1 H
A1 + B1 F1 B1 H
0
A3
0
=
N (z) =
.
C1 + DF1
DH
C1 + DF1
C2 + DF2
DH
Since (A1 , B1 ) is controllable, by Theorem 8.2 (ii), there exists X1 = X1∗
such that
(A1 + B1 F1 )∗ X1 (A1 + B1 F1 ) − X1 + (C1 + DF1 )∗ J(C1 + DF1 ) = 0,
(8.25)
(DH)∗ J(C1 + DF1 ) + (B1 H)∗ X1 (A1 + B1 F1 ) = 0,
∗
∗
(DH) J(DH) + (B1 H) X1 B1 H = I.
(8.26)
(8.27)
From (8.27), we have HH ∗ = (D∗ JD + B1∗ X1 B1 )−1 which implies that
D∗ JD + B1∗ X1 B1 > 0. Solving for H yields
1
H = (D∗ JD + B1∗ X1 B1 )− 2 U
where U is an arbitrary unitary matrix. Substitute H into (8.26), we get
H ∗ ((D∗ JD + B1∗ X1 B1 )F1 + (B1∗ X1 A1 + D∗ JC1 )) = 0.
Solving for F1 yields F1 = −(D∗ JD + B1∗ X1 B1 )−1 (B1∗ X1 A1 + D∗ JC1 ).
Then substitute F1 into (8.25), we get
A∗1 X1 A1 − X1 + C1∗ JC1 − (A∗1 X1 B1 + C1∗ JD)(D∗ JD + B1∗ X1 B1 )−1
(B1∗ X1 A1 + D∗ JC1 ) = 0.
This shows that X1 is indeed a solution to (8.22). In addition, it is a
stabilizing solution since F1 is designed to be a stabilizing state feedback gain of (A1 , B1 ). With this and the stableness of A3 , equation
(8.23) can be uniquely solved for X2 . It then follows that (8.24)
can be
X1 X2
uniquely solved for X3 . This finishes the construction of X =
X2∗ X3
to be a solution to the Riccati equation (8.21). Let F2 = −(D∗ JD +
B1∗ X1 B1 )−1 (B1∗ X1 A2 + B1∗ X2 A3 + D∗ JC2 ), then
F = [F1 F2 ] = −(D∗ JD + B ∗ XB)−1 (B ∗ XA + D∗ JC).
Since A + BF is stable due to the requirements of the coprime factorization, it follows that X is the stabilizing solution to (8.21). Finally, the
positive definiteness of D∗ JD + B ∗ XB follows from D∗ JD + B ∗ XB =
D∗ JD + B1∗ X1 B1 > 0.
2. (b) =⇒ (a).
150
CHAPTER 8. APPLICATIONS OF RICCATI EQUATION
It is easy to verify that G∼ (z)JG(z) = Ψ∼ (z)Ψ(z) by some tedious computation. Since X is the stabilizing solution to (8.21), we have A + BF
is stable. Since
"
#
1
A + BF B(D∗ JD + B ∗ XB)− 2 U −1
−1
Ψ (z) =
,
1
F
(D∗ JD + B ∗ XB)− 2 U −1
it follows that Ψ−1 (z) ∈ RH∞ . Therefore, Ψ(z) has no zeros on the unit
circle and G∼ (ejω )JG(ejω ) = Ψ∼ (ejω )Ψ(ejω ) > 0 for all ω ∈ [0, 2π).
It follows directly that Lemma 8.2 would result in a spectral factorization
when G(z) ∈ RH∞ . See the following theorem:
A B
. Assume that A is stable. Given signaTheorem 8.7 Let G(z) =
C D
ture matrix J, then the following statements are equivalent:
1. G∼ (ejω )JG(ejω ) > 0 for all ω ∈ [0, 2π).
2. The Riccati equation
A∗ XA − X + C ∗ JC − (A∗ XB + C ∗ JD)(D∗ JD + B ∗ XB)−1 (B ∗ XA + D∗ JC) = 0
has a stabilizing solution X such that D∗ JD + B ∗ XB > 0.
Moreover, if either one of the above statements is satisfied, then all the spectral
factorizations G∼ (z)JG(z) = Φ∼ (z)Φ(z) are given as follows:
"
Φ(z) =
A
1
∗
−U (D JD + B ∗ XB) 2 F
B
1
∗
U (D JD + B ∗ XB) 2
#
,
where F = −(D∗ JD+B ∗ XB)−1 (B ∗ XA+D∗ JC) and U is an arbitrary unitary
matrix.
A parallel theorem on co-spectral factorization when G(z) ∈ RH∞ is given
as follows:
A B
. Assume that A is stable. Given signaC D
ture matrix J, then the following statements are equivalent:
Theorem 8.8 Let G(z) =
1. G(ejω )JG∼ (ejω ) > 0 for all ω ∈ [0, 2π).
2. The dual Riccati equation
AY A∗ − Y + BJB ∗ − (AY C ∗ + BJD∗ )(DJD∗ + CY C ∗ )−1 (CY A∗ + DJB ∗ ) = 0
has a stabilizing solution Y such that DJD∗ + CY C ∗ > 0.
8.4. SPECTRAL FACTORIZATION
151
Moreover, if either one of the above statements is satisfied, then all the cospectral factorizations G(z)JG∼ (z) = Φc (z)Φ∼
c (z) are given as follows:
#
"
1
A −L(DJD∗ + CY C ∗ ) 2 V
,
Φc (z) =
1
C
(DJD∗ + CY C ∗ ) 2 V
where L = −(AY C ∗ + BJD∗ )(DJD∗ + CY C ∗ )−1 and V is an arbitrary unitary
matrix.
We are now in a position to consider the spectral factorization under the
general situation when G(z) is not necessarily stable.
A B
Theorem 8.9 Let G(z) =
where A has no eigenvalues on the unit
C D
circle. Assume that (A, B) is stabilizable and (C, A) is detectable. Given signature matrix J, if G∼ (ejω )JG(ejω ) > 0 for all ω ∈ [0, 2π), then all the spectral
factorizations G∼ (z)JG(z) = Φ∼ (z)Φ(z) are given as follows:
Φ(z)
"
=
−Z (D∗ JD + B ∗ XB)−1 + F Y F ∗
#
−L
A + BF + LF
− 1
2
F
Z (D∗ JD + B ∗ XB)−1 + F Y F ∗
− 1
2
,
where X is the stabilizing solution to Riccati equation
A∗ XA − X + C ∗ JC − (A∗ XB + C ∗ JD)(D∗ JD + B ∗ XB)−1 (B ∗ XA + D∗ JC) = 0,
Y is the stabilizing solution to the dual Riccati equation
(A + BF )Y (A + BF )∗ − Y + B(D∗ JD + B ∗ XB)−1 B ∗ − ((A + BF )Y F ∗ + B(D∗ JD
−1
+B ∗ XB)−1 ) F Y F ∗ + (D∗ JD + B ∗ XB)−1
F Y (A + BF )∗ + (D∗ JD + B ∗ XB)−1 B ∗ = 0
and
F = −(D∗ JD + B ∗ XB)−1 (B ∗ XA + D∗ JC),
L = −((A + BF )Y F ∗ + B(D∗ JD + B ∗ XB)−1 ) F Y F ∗ + (D∗ JD + B ∗ XB)−1
−1
According to Theorem 8.2, there exist factorizations G∼ (z)JG(z) =
where
"
#
A
B
Ψ(z) =
,
1
1
−U (D∗ JD + B ∗ XB) 2 F U (D∗ JD + B ∗ XB) 2
Proof
Ψ∼ (z)Ψ(z)
and U is an arbitrary unitary matrix. Since
"
#
1
A + BF B(D∗ JD + B ∗ XB)− 2 U −1
−1
Ψ (z) =
∈ RH∞ ,
1
F
(D∗ JD + B ∗ XB)− 2 U −1
it follows that
−1
∼
G∼ (z)JG(z) = Ψ∼ (z)Ψ(z) = Ψ−1 (z)Ψ−1 (z)
.
.
152
CHAPTER 8. APPLICATIONS OF RICCATI EQUATION
Let M (z) = Ψ−1 (z), then M (z) ∈ RH∞ and
G∼ (z)JG(z) = (M (z)M ∼ (z))−1 .
Since M −1 (z) = Ψ(z) and Ψ(z) has no poles on the unit circle, it follows that
M (z) has no zeros on the unit circle, i.e., M (ejω )M ∼ (ejω ) > 0 for all ω ∈ [0, 2π).
Then by Theorem 8.8, let J = I, co-spectral factorizations M (z)M ∼ (z) =
Π(z)Π∼ (z) can be conducted where
"
Π(z) =
A + BF
F
#
1
−L (D∗ JD + B ∗ XB)−1 + F Y F ∗ 2 V
1
(D∗ JD + B ∗ XB)−1 + F Y F ∗ 2 V
and V is an arbitrary unitary matrix. Let
Φ(z)
=Π−1 (z)
"
=
−Z (D∗ JD + B ∗ XB)−1 + F Y F ∗
− 1
2
F
Z (D∗ JD + B ∗ XB)−1 + F Y F ∗
where Z = V −1 , then Φ(z), Φ−1 (z) ∈ RH∞ and
G∼ (z)JG(z) = (M (z)M ∼ (z))−1 = (Π(z)Π∼ (z))−1 = Φ∼ (z)Φ(z).
This completes the proof.
8.5
#
−L
A + BF + LF
Inner-Outer Factorization
Definition 8.6 Let G(z) ∈ RH∞ . G(z) is said to have an inner-outer factorization if there exists Wi (z) and Wo (z) such that G(z) = Wi (z)Wo (z) where
Wi (z) is inner and Wo (z) is outer.
A B
Theorem 8.10 Let G(z) =
have no zeros on the unit circle. AsC D
sume that A is stable. Then all the inner-outer factorizations G(z) = Wi (z)Wo (z)
are given as follows:
"
#
1
A + BF B(D∗ D + B ∗ XB)− 2 U −1
Wi (z) =
,
1
C + DF D(D∗ D + B ∗ XB)− 2 U −1
"
#
A
B
Wo (z) =
,
1
1
−U (D∗ D + B ∗ XB) 2 F U (D∗ D + B ∗ XB) 2
where X is the stabilizing solution to Riccati equation
A∗ XA − X + C ∗ C − (A∗ XB + C ∗ D)(B ∗ XB + D∗ D)−1 (B ∗ XA + D∗ C) = 0
and F = −(B ∗ XB + D∗ D)−1 (B ∗ XA + D∗ C), U is an arbitrary unitary matrix.
− 1
2
8.6. BOUNDED REAL LEMMA
153
Proof By Theorem 8.7, let J = I, then spectral factorizations G∼ (z)G(z) =
Φ∼ (z)Φ(z) can be conducted where
"
#
B
A
Φ(z) =
.
1
1
−U (D∗ D + B ∗ XB) 2 F U (D∗ D + B ∗ XB) 2
"
1
A + BF B(D∗ D + B ∗ XB)− 2 U −1
Let Wi (z) = G(z)Φ−1 (z) =
1
C + DF D(D∗ D + B ∗ XB)− 2 U −1
and Wo (z) = Φ(z). Apparently, Wo (z) is outer. Since
∼
Wi∼ (z)Wi (z) = G(z)Φ−1 (z)
G(z)Φ−1 (z)
#
∈ RH∞
∼
= Φ−1 (z)G∼ (z)G(z)Φ−1 (z)
∼
= Φ−1 (z)Φ∼ (z)Φ(z)Φ−1 (z)
= I,
it follows that Wi (z) is inner. This completes the proof.
8.6
Bounded Real Lemma
A B
. Assume that A is stable, γ > 0. Then
C D
the following statements are equivalent:
Theorem 8.11 Let G(z) =
1. ∥G(z)∥∞ < γ.
2. The Riccati equation
A∗ XA − X + C ∗ C − (A∗ XB + C ∗ D)(−γ 2 I + D∗ D + B ∗ XB)−1 (B ∗ XA + D∗ C) = 0
(8.28)
has a stabilizing solution X such that −γ 2 I + D∗ D + B ∗ XB < 0.
A B
−I 0
Proof By Theorem 8.7, let G̃(z) = C D
and J =
, then
0 I
0 γI
the following two statements are equivalent:
1. G̃∼ (ejω )J G̃(ejω ) > 0 for all ω ∈ [0, 2π).
2. The Riccati equation
A∗ XA − X − C ∗ C − (A∗ XB − C ∗ D)(γ 2 I − D∗ D + B ∗ XB)−1 (B ∗ XA − D∗ C) = 0
(8.29)
has a stabilizing solution X such that γ 2 I − D∗ D + B ∗ XB > 0.
It is straightforward to see that (a) is equivalent to (a∗ ). It follows that the
theorem can be proved by way of showing the equivalence between (b) and (b∗ ).
Given X0 to be the stabilizing solution to (8.29) such that γ 2 I − D∗ D +
∗
B X0 B > 0, then A − B(γ 2 I − D∗ D + B ∗ X0 B)−1 (B ∗ X0 A − D∗ C) is stable.
154
CHAPTER 8. APPLICATIONS OF RICCATI EQUATION
Let X1 = −X0 , then it can be verified that X1 is a solution to (8.28) such that
−γ 2 I + D∗ D + B ∗ X1 B < 0. Let F = −(−γ 2 I + D∗ D + B ∗ X1 B)−1 (B ∗ X1 A +
D∗ C). Since
A + BF = A − B(−γ 2 I + D∗ D + B ∗ X1 B)−1 (B ∗ X1 A + D∗ C)
= A − B(γ 2 I − D∗ D + B ∗ X0 B)−1 (B ∗ X0 A − D∗ C),
it follows that A + BF is stable which indicates that X1 is indeed a stabilizing
solution to (8.28). This shows that (b∗ ) implies (b). Using the same method
we can also show that (b) implies (b∗ ) and the proof is done.
Chapter 9
H2 Optimization
9.1
Simple H2 Optimization
In this section, we consider a very simple H2 optimization problem: SISO model
matching problem. Consider the system shown in Figure 5.2.
- T 11
?
j 6
- T 21
- Q
- T 12
Figure 9.1: Model matching problem
Let T 11 , T 12 , and T 21 be SISO casual stable systems with transfer functions
T̂11 , T̂12 , T̂13 ∈ H2 . The problem is to find a causal stable system Q with transfer
function Q̂ ∈ H2 such that
∥T̂11 + T̂12 Q̂T̂21 ∥2
is minimized. Since the systems are all SISO, the transfer functions commute.
Rename T̂12 T̂21 = T̂2 and T̂11 = T̂1 . Then the problem becomes
min ∥T̂1 + T̂2 Q̂∥2 .
Q̂∈RH2
The idea in solving the problem is as follows. First we factorize T̂2 into
T̂2 = T̂2i T̂2o
such that T̂2i is allpass in the sense that
∼
T̂2i
T̂2i = 1
155
CHAPTER 9. H2 OPTIMIZATION
156
−1
and T̂2o is minimum phase in the sense that both T̂2o and T̂2o
are in H2 . Notices
∼
that multiplication of T̂2i does not change the H2 norm and the map Q̂ 7→ T̂2o Q̂
is a one-one correspondence in H2 . Then
min ∥T̂1 + T̂2 Q̂∥2 . =
∼
(T̂1 + T̂2i T̂2o )Q̂)∥2
min ∥T̂2i
Q̂∈RH2
Q̂∈RH2
=
∼
T̂1 + T̂2o Q̂)∥2
min ∥T̂2i
Q̂∈RH2
=
∼
T̂1 + Q̂1 )∥2 .
min ∥T̂2i
Q̂1 ∈RH2
∼ T̂ is in general not in H . It is rather in L . Therefore the minimizing Q̂
T̂2i
1
2
2
1
∼ T̂ onto H . The minimizing Q̂
is then chosen to cancel the projection of T̂2i
1
2
−1
.
can then be obtained from the minimizing Q̂1 by multiplication of T̂20
The details of the design is as follows. Assume that T̂2 (z) has no zeros on
the unit circle. Then it can be written as
T̂2 (z) =
(z − z1 ) · · · (z − zb )(z − zb+1 ) · · · (z − zd )
(z − p1 )(z − p2 ) · · · (z − pn )
where z1 , . . . , zb are outside of the unit circle and zb+1 , . . . , zd are inside of the
unit circle. Rewrite T̂2 as
T̂2 (z) =
z −(n−d) (z − z1 ) · · · (z − zb ) z n−d (z1∗ z − 1) · · · (zb∗ z − 1)(z − zb+1 ) · · · (z − zd )
.
(z1∗ z − 1) · · · (zb∗ z − 1)
(z − p1 )(z − p2 ) · · · (z − pn )
Let us denote the first factor by T̂2i and the second factor by T̂2o , i.e.,
T̂2i (z) =
z −(n−d) (z − z1 ) · · · (z − zb )
(z1∗ z − 1) · · · (zb∗ z − 1)
T̂2o (z) =
z n−d (z1∗ z − 1) · · · (zb∗ z − 1)(z − zb+1 ) · · · (z − zd )
.
(z − p1 )(z − p2 ) · · · (z − pn )
∼ T̂ = 1, it follows that
Then T̂2 = T̂2i T̂2o is the desired factorization. Since T̂2i
2i
∼
∥T̂1 + T̂2 Q̂∥2 = ∥T̂2i
T̂1 + T̂2o Q̂∥2 .
Since T̂2o has all poles and zeros inside the unit circle and its numerator and
denominator degrees are the same, T̂2o Q̂ ∈ RH2 iff Q̂ ∈ RH2 .
Then the minimization problem becomes
∼
min ∥T̂2i
T̂1 + Q̂1 ∥2
Q̂1 ∈RH2
and the minimum solution of this minimization problem is related to that of
−1
the original problem by Q̂ = T̂2o
Q̂1 .
∼
∼
Now T̂2i T̂1 ∈ RL2 . Hence T̂2i T̂1 can be decomposed into
∼
T̂2i
T̂1 = T̂c + T̂ac
9.2. GENERAL H2 OPTIMIZATION
157
where T̂c ∈ RH2 is causal and T̂ac ∈ RH⊥
2 is anticausal. Therefore
∼
T̂1 + Q̂1 ∥22 =
min ∥T̂2i
Q̂1 ∈RH2
min ∥T̂c + T̂ac + Q̂1 ∥22 =
Q̂1 ∈RH2
min ∥T̂c + Q̂1 ∥22 + ∥T̂ac ∥22
Q̂1 ∈RH2
and the optimal solution is Q̂1 = −T̂c .
The optimal solution of the original optimization problem is then given by
−1
Q̂ = −T̂2o
T̂c .
Example 9.1
Let
z+3
z + 0.5
T̂1 (z) =
and
Then
T̂2 (z) =
z+2
.
z − 0.5
T̂2 (z) =
z + 2 −2z − 1
−2z − 1 z − 0.5
i.e.,
T̂2i (z) = −
z+2
2z + 1
Then
∼
T̂2i
(z)T̂1 (z) = −
T̂2o (z) = −
and
1
z +2
2 z1 + 1
2z + 1
.
z − 0.5
z+3
z+3
z
= −2
=
− 3.
z + 0.5
z+2
z+2
2z+1
Therefore, letting Q̂1 = − z−0.5
Q̂ leads to
min ∥T̂1 + T̂2 Q̂∥2 =
∼
min ∥T̂2i
T̂1 + Q̂1 ∥2
Q̂1 ∈RH2
Q̂∈RH2
and the minimization problem on the right has solution Q̂1 (z) = 3. Finally, the
minimization problem on the left has solution Q̂(z) = −3 z−0.5
2z+1 and
min ∥T̂1 + T̂2 Q̂∥2 =
Q̂1 ∈RH2
Q̂∈RH2
9.2
∼
min ∥T̂2i
T̂1 + Q̂1 ∥2 = ∥
General H2 Optimization
z
w
G
y
u
- K
Figure 9.2: The standard setup
1
z
∥2 = √ .
z+2
3
CHAPTER 9. H2 OPTIMIZATION
158
Consider Figure 9.2. Assume that G is given. The general H2 optimization
problem is to design K so that the overall system is internally stable and the
H2 norm of the transfer function from w to z
∥F l (Ĝ, K̂)∥2
is minimized. Let
A B1 B2
G = C1 D11 D12 .
C2 D21 D22
(9.1)
The following assumption is needed for the existence and uniqueness of the
H2 optimal controller.
Assumption 9.1
1. (A, B2 ) is stabilizable and (C2 , A) is detectable.
A − ejω I B2
2.
has full column rank for all ω ∈ R.
C1
D12
A − ejω I B1
3.
has full row rank for all ω ∈ R.
C2
D21
The solution of H2 optimization depends on the following Riccati equations:
A∗ XA − X + C1∗ C1
∗
∗
− (A∗ XB2 + C1∗ D12 )(B2∗ XB2 + D12
D12 )−1 (B2∗ XA + D12
C1 ) = 0(9.2)
AY A∗ − Y + B1 B1∗
∗
∗ −1
)(C2 Y C2∗ + D21 D21
) (C2 Y A∗ + D21 B1∗ ) = 0.(9.3)
− (AY C2∗ + B1 D21
Under Assumption 9.1, Riccati equations (9.2)-(9.3) have unique stabilizing
∗ D
∗
solutions X and Y respectively. Then B2∗ XB2 + D12
12 > 0 and C2 Y C2 +
∗ > 0. Define
D21 D21
F
∗
∗
= −(B2∗ XB2 + D12
D12 )−1 (B2∗ XA + D12
C1 )
∗
∗ −1
L = −(AY C2∗ + B1 D21
)(C2 Y C2∗ + D21 D21
)
A B1
Y C2∗
∗
∗ −1
∗
H = −(B2∗ XB2 + D12
D12 )−1 B2∗ X D12
(C2 Y C2∗ + D21 D21
) .
∗
C1 D11
D21
We first consider a special case when D22 = 0. This special case occurs in
practice very often and it leads to simpler formula for the optimal controller.
Theorem 9.1 Assume the plant G satisfies Assumption 9.1 and D22 = 0.
Then the unique H2 optimal controller is given by
#
"
A + B2 F + LC2 − B2 HC2 −L + B2 H
K opt =
.
F − HC2
H
9.2. GENERAL H2 OPTIMIZATION
159
The optimal H2 norm is
∗
∥F(Ĝ, K̂opt )∥22 = tr(D11
D11 ) + tr(A∗ XAY + XB1 B1∗ + C1∗ C1 Y − XY )
∗
∗ 1/2 2
−∥(B2∗ XB2 + D12
D12 )1/2 H(C2 Y C2∗ + D21 D21
) ∥F .
Proof Let X and Y be the stabilizing solutions of (9.2)-(9.3). Then all
stabilizing controllers are characterized by a linear fractional transformation
K = Fl (J , Q),
(9.4)
where
A + B2 F + LC2 −L B2
J =
F
0
I .
−C2
I
0
Now it follows from Theorem 4.8 that under the controllers characterized in
(9.4), the closed loop transfer function is
Fl (Ĝ, K̂) = T̂11 + T̂12 Q̂T̂21
where
T 11
T 12
T 21
A + B2 F
B2 F
B1
0
A + LC2 B1 + LD21
=
C1 + D12 F
D12 F
D11
B2
A + B2 F
=
C1 + D12 F D12
A + LC2 B1 + LD21
=
.
C2
D21
We need two claims.
Claim 9.1
∼
∗
T̂12
(z)T̂12 (z) = B2∗ XB2 + D12
D12 ,
∼
∗
T̂21 (z)T̂21
(z) = C2 Y C2∗ + D21 D21
.
In the following, we use temporary notation
∗
M = B2∗ XB2 + D12
D12 ,
∗
N = C2 Y C2∗ + D21 D21
.
Claim 9.2
∼
∼
T̂12
T̂11 T̂21
∈
H⊥
2
+
B2∗ X
∗
D12
A B1
Y C2∗
.
∗
C1 D11
D21
Define
Û =
∼
M −1/2 T̂12
−1
∼
I − T̂12 M T̂12
,
V̂ =
∼ N −1/2 I − T̂ ∼ N −1 T̂
T̂21
21 .
21
CHAPTER 9. H2 OPTIMIZATION
160
Then Claim 9.1 implies Û ∼ Û = I and V̂ V̂ ∼ = I, i.e., Û is allpass and V̂ is
co-allpass. Hence
∥Fl (Ĝ, K̂)∥22
= ∥T̂11 + T̂12 ĜT̂21 ∥22
= ∥Û (T̂11 + T̂12 Q̂T̂21 )V̂ ∥22
∼ T̂ T̂ ∼ N −1/2 + M 1/2 Q̂N 1/2
∼ T̂ (I − T̂ ∼ N −1 T̂ )
M −1/2 T̂12
M −1/2 T̂12
11 21
11
21
2
21
= ∥
∼ )T̂ T̂ ∼ N −1/2
−1 T̂ ∼ )T̂ (I − T̂ ∼ N −1 T̂ ) ∥2 .
(I − T̂12 M −1 T̂12
(I
−
T̂
M
11 21
12
21
12 11
21
So minimizing ∥Fl (Ĝ, K̂)∥22 is equivalent to minimizing
∼ −1/2
∼
N
+ M 1/2 Q̂N 1/2 ∥22
T̂11 T̂21
∥M −1/2 T̂12
for stable Q̂. From Claim 9.2, we know that
∼
∼ −1/2
M −1/2 T̂12
T̂11 T̂21
N
A B1
Y C2∗
∗
∈ M −1/2 B2∗ X D12
N −1/2 + H⊥
2.
∗
C1 D11
D21
Since M 1/2 Q̂N 1/2 is causal, it can only be used to cancel the constant term of
∼
∼ −1/2
T̂11 T̂21
N
.
M −1/2 T̂12
The optimal Q̂ then is given by
M
1/2
Q̂opt N
1/2
= −M
−1/2
B2∗ X
∗
D12
A B1
C1 D11
Y C2∗
∗
D21
N −1/2 .
Hence Q̂opt = H. This shows that
"
K opt = F l (J , Qopt ) = F l (J , H) =
A + B2 F + LC2 − B2 HC2 −L + B2 H
F − HC2
H
The optimal H2 norm is given by
∥Fl (Ĝ, K̂)∥22 = ∥T̂11 ∥22 − ∥M 1/2 Q̂opt N 1/2 ∥22 = ∥T̂11 ∥22 − ∥M 1/2 HN 1/2 ∥22 .
It is easy to verify that
T 11 =
A + B2 F
C1 + D12 F
B1
D11
+ T 12
A + LC2 B1 + LD21
F
0
.
Denote
TF =
A + B2 F
C1 + D12 F
B1
D11
,
TL =
A + LC2 B1 + LD21
F
0
.
#
.
9.2. GENERAL H2 OPTIMIZATION
161
Straightforward computation shows that T̂F and T̂12 T̂L are orthogonal to each
other and Claim 5.1 implies ∥T̂12 T̂L ∥2 = ∥M 1/2 T̂L ∥2 . Hence,
∥T̂11 ∥22 = ∥T̂F ∥22 + ∥M 1/2 T̂L ∥22
∗
= tr(D11
D11 + B1∗ XB1 ) + trM 1/2 F Y F ∗ M 1/2
∗
= tr(D11
D11 + B1∗ XB1 )
∗
∗
+tr(A∗ XB2 + C1∗ D12 )(B2∗ XB2 + D12
D12 )−1 (B2∗ XA + D12
C1 )Y
∗
= tr(D11
D11 + B1∗ XB1 ) + tr(A∗ XA + C1∗ C1 − X)Y.
This proves the expression for the optimal H2 norm.
It remains to show the two claims. We leave it as exercises.
2
G̃
z
w
j
G
−
6
D22
ỹ
ũ
D22
?
- j - K
K̃
Figure 9.3: Loop transformation
What if D22 ̸= 0? Let us consider the loop transformation shown in Figure
9.3. From an external point of view, the system in Figure 9.3 and that in
Figure 9.2 are exactly the same. Now call the system in the upper dash box
G̃ and the system in lower dash box K̃. Then K internally stabilizes G iff K̃
ˆ minimizes
internally stabilizes G̃, and K̂ minimizes ∥F l (Ĝ, K̂)∥2 if and only if K̃
ˆ K̃)∥
ˆ . Notice that if
∥F (G̃,
l
2
A B1 B2
G = C1 D11 D12 ,
C2 D21 D22
then
A B1 B2
G̃ = C1 D11 D12 .
C2 D21
0
ˆ K̃)∥
ˆ .
Hence K̃ opt can be obtained to internally stabilize G̃ and minimize ∥F l (G̃,
2
The optimal K which internally stabilizes G and minimizes ∥F l (Ĝ, K̂)∥2 is then
given by
0 I
−1
K opt = (I + K̃ opt D22 ) K̃ opt = F l
, K̃ opt .
I D22
CHAPTER 9. H2 OPTIMIZATION
162
Theorem 9.2 Assume the plant G satisfies Assumption 9.1. Then the unique
H2 optimal controller is given by
#!
"
A + B2 F + LC2 − B2 HC2 −L + B2 H
0 I
K opt = F l
,
I D22
F − HC2
H
"
#
A + B2 F̃ + L̃C2 − B2 H̃C2 − Ẽ −L̃ + B2 H̃
=
F̃ − H̃C2
H̃
where
H̃ F̃
L̃ Ẽ
=
0 I
I D22
H F
⋆
.
L 0
The optimal H2 norm is
∗
∥Fl (Ĝ, K̂opt )∥22 = tr(D11
D11 ) + tr(A∗ XAY + XB1 B1∗ + C1∗ C1 Y − XY )
∗
∗ 1/2 2
−∥(B2∗ XB2 + D12
D12 )1/2 H(C2 Y C2∗ + D21 D21
) ∥F .
9.3
Linear Quadratic Regulator Problem
Consider system
x(k + 1) = Ax(k) + Bu(k)
x(0) = x0
y(k) = Cx(k) + Du(k).
Assume that the state x(k) is available for measurement. The linear quadratic
regulator problem is to find control signal u ∈ ℓ2 (Z+ ) so that x ∈ ℓ2 (Z+ ) and
the ℓ2 norm of the output y
∥y∥2 = ∥Cx + Du∥2 =
(∞
X
)1/2
[Cx(k) + Du(k)]∗ [Cx(k) + Du(k)]
k=0
is minimized.
Assumption 9.2
1. (A, B) is stabilizable.
A − ejω I B
2.
has full column rank for all ω ∈ R.
C
D
Let X be the stabilizing solution of Riccati equation
A∗ XA−X +C ∗ C −(A∗ XB+C ∗ D)(B ∗ XB+D∗ D)−1 (B ∗ XA+D∗ C) = 0 (9.5)
and let
F = −(B ∗ XB + D∗ D)−1 (B ∗ XA + D∗ C).
Theorem 9.3 The unique optimal control signal is given by u(k) = F x(k) and
the optimal value of ∥y∥22 is given by x∗0 Xx0 .
9.4. FULL INFORMATION H2 OPTIMIZATION
Proof
becomes
163
Let u(k) = F x(k) + v(k) for some v ∈ ℓ2 (Z+ ). Then the system
x(k + 1) = (A + BF )x(k) + Bv(k)
x(0) = x0
y(k) = (C + DF )x(k) + Dv(k).
Denote AF = A + BF and CF = C + DF . Hence
ŷ(z) = CF (zI − AF )−1 zx0 + [CF (zI − AF )−1 B + D]v̂(z).
Let
T̂1 (z) = CF (zI − AF )−1 z
and
T̂2 (z) = CF (zI − AF )−1 B + D.
Then
ŷ = T̂1 x0 + T̂2 v̂.
∼
Claim 9.3 Tˆ2 T̂1 ∈ H⊥
2.
This claim implies that
∼
⟨T̂1 x0 , T̂2 v̂⟩ = ⟨Tˆ2 T̂1 x0 , v̂⟩ = 0
for all v̂ ∈ H2 . Therefore
min ∥y∥22 = min (∥T̂1 x0 ∥22 + ∥T̂2 v̂∥22 ) = ∥T̂1 x0 ∥22
v̂∈H2
v̂∈H2
and the minimum is achieved at v̂ = 0. Since X is the observability gramian of
(CF , AF ),
∥T̂1 x0 ∥22 = ∥CF (zI − AF )−1 zx0 ∥22 = ∥CF (zI − AF )−1 x0 ∥22 = x∗0 Xx0 .
It remains to prove Claim 9.3. By using (7.7), we have
∼
Tˆ2 (z)T̂1 (z)
= [B ∗ z(I − zA∗F )−1 CF∗ + D∗ ]CF (zI − AF )−1 z
= B ∗ z(I − zA∗F )−1 CF∗ CF (zI − AF )−1 z + D∗ CF (zI − AF )−1 z
= B ∗ z(I − zA∗F )−1 (X − A∗F XAF )(zI − AF )−1 z + D∗ CF (zI − AF )−1 z
= B ∗ z(I − zA∗F )−1 (X − zA∗F X + zA∗F X − A∗F XAF )(zI − AF )−1 z + D∗ CF (zI − AF )−1 z
= B ∗ zX(zI − AF )−1 z + B ∗ z(I − zA∗F )−1 A∗F Xz + D∗ CF (zI − AF )−1 z
∈ (B ∗ zX + D∗ CF )(zI − AF )−1 z + H⊥
2
= (B ∗ zX − B ∗ XAF + B ∗ XA + B ∗ XBF + D∗ C + D∗ DF ](zI − AF )−1 z + H⊥
2
= (B ∗ X(zI − AF ) + B ∗ XA + D∗ C + (B ∗ XB + D∗ D)F ](zI − AF )−1 z + H⊥
2
= B ∗ Xz + H⊥
2
= H⊥
2.
This completes the proof.
2
CHAPTER 9. H2 OPTIMIZATION
164
z
x
w
G
w
u
@
@
- K
Figure 9.4: The full information setup
9.4
Full Information H2 Optimization
The full information H2 optimization problem concerns the following system
x(k + 1) = Ax(k) + B1 w(k) + B2 u(k)
z(k) = C1 x(k) + D11 w(k) + D12 u(k)
x(k)
.
y(k) =
w(k)
In this system, the measurement y contains the complete information of the
state x and the external disturbance w. The problem is to design a feedback
controller in the form of û = K̂ ŷ such that the H2 norm of the system from
w to z is minimized. Using the LFT notation, we represent the plant by a
generalized plant
A
B1
B2
C1
D11
D12
G=
I 0 0 .
0
I
0
Then the problem is to design K̂ to minimize
∥F l (Ĝ, K̂)∥2 .
This problem looks like a special problem of the general H2 optimization problem studied in Section 6.3, but actually it is not since Assumption 6.1.3 is not
satisfied. Nevertheless, we make the following assumption.
Assumption 9.3
1. (A, B2 ) is stabilizable.
A − ejω I B2
2.
has full column rank for all ω ∈ R.
C1
D12
Let X be the stabilizing solution of Riccati equation
∗
∗
A∗ XA−X+C1∗ C1 −(A∗ XB2 +C1∗ D12 )(B2∗ XB2 +D12
D12 )−1 (B2∗ XA+D12
C1 ) = 0
and define
∗
A B1
∗
∗
−1
∗
F F0 = −(B2 XB2 + D12 D12 )
B2 X D12
.
C1 D11
9.4. FULL INFORMATION H2 OPTIMIZATION
165
Theorem 9.4 The unique optimal controller K opt is given by K̂opt = F F0 ,
i.e., the optimal control signal is given by uopt (k) = F x(k) + F0 w(k). The optimal closed loop system is
"
#
A + B2 F
B1 + B2 F0
Gc = Fl (G, K opt ) =
,
C1 + D12 F D11 + D12 F0
and
∗
∗
∗
∗
∥Ĝc ∥22 = tr B1∗ XB1 + D11
D11 − (B1∗ XB2 + D11
D12 )(B2∗ XB2 + D12
D12 )−1 (B2∗ XB1 + D12
D11 ) .
Proof
becomes
Let u(k) = F x(k) + v(k) for some v ∈ ℓ2 (Z+ ). Then the system
x(k + 1) = (A + B2 F )x(k) + B1 w(k) + B2 v(k)
y(k) = (C1 + D12 F )x(k) + D11 w(k) + D12 v(k).
Denote AF = A + BF and CF = C1 + D12 F . Hence
ŷ(z) = [CF (zI − AF )−1 B1 + D11 ]ŵ(z) + [CF (zI − AF )−1 B1 + D12 ]v̂(z).
Let
T̂0 (z) = CF (zI − AF )−1 B1 + D11
and
T̂12 (z) = CF (zI − AF )−1 B2 + D12 .
Then
ŷ = T̂0 ŵ + T̂12 v̂.
Denote
∗
M = B2∗ XB2 + D12
D12 .
Notice that Claim 6.1 implies that
∼
M −1/2 T̂12
Û =
∼
I − T̂12 M −1 T̂12
satisfies Û ∼ Û = I. Hence
∼ T̂ ŵ + M −1/2 v̂
M −1/2 T̂12
0
∥ŷ∥2 = ∥Û ŷ∥2 = ∥
∥2 .
∼ T̂ )ŵ
(T̂0 − T̂12 M −1 T̂12
0
∼ T̂ ∈ H⊥ + B ∗ XB + D ∗ D .
Claim 9.4 T̂12
0
1
2
2
12 11
This shows that the optimal v̂ ∈ H2 is given by
∗
v̂opt = −M −1 (B2∗ XB1 + D12
D11 )ŵ = F0 ŵ.
In the time domain, this means
vopt (k) = F0 w(k),
CHAPTER 9. H2 OPTIMIZATION
166
i.e.,
uopt (k) = F x(k) + F0 w(k).
The optimal closed loop system follows immediately. The optimal closed loop
cost follows from the straightforward computation.
2
This theorem shows that the unique optimal full information control is a
static state feedback plus a static disturbance feedforward, even though dynamic controllers are allowed in the problem formulation. The readers are also
suggested to compare similarities and differences between the full information
H2 optimization problem and the LQR problem.
9.5
H2 Filtering Problem
Let us first consider the standard filtering problem. The schematic diagram is
shown in Figure 9.5. Here G is a plant driven by an exogenous input w. It
has two groups of output: the measurement output y and the concerned output
z which is to be estimated. The filter K tries to produce an estimate z̃ of z
so that the estimation error e is small. The H2 filtering problem is to design
a filter K for a given plant G so that the H2 norm between w and error e is
minimized.
z
w-
+?
j e−
G
y
- K
z̃
6
Figure 9.5: The standard filtering problem
The plant G is assumed to have state space description
x(k + 1) = Ax(k) + B1 w(k)
z(k) = C1 x(k) + D11 w(k)
y(k) = C2 x(k) + D21 w(k).
In the compact notation, we have
A B1
G = C1 D11
C2 D21
Let us make the following assumption.
Assumption 9.4
1. (C2 , A) is detectable.
A − ejω I B1
2.
has full row rank for all ω ∈ R.
C2
D21
9.5. H2 FILTERING PROBLEM
167
Let Y be the unique stabilizing solution of FARE
∗
∗ −1
AY A∗ −Y +B1 B1∗ −(AY C2∗ +B1 D21
)(C2 Y C2∗ +D21 D21
) (C2 Y A∗ +D21 B1∗ ) = 0.
(9.6)
Define
L
A B1
Y C2∗
∗ −1
=−
(C2 Y C2∗ + D21 D21
) .
(9.7)
L0
C1 D11
D21
Theorem 9.5 The unique optimal filter is given by
"
#
A + LC2
−L
K opt =
.
C1 + L0 C2 −L0
The optimal transfer function from w to e is
"
#
A + LC2
B1 + LD21
Gf =
C1 + L0 C2 D11 + L0 D21
and
∗
∗
∗ −1
∗
∥Ĝf ∥22 = tr C1 Y C1∗ + D11 D11
− (C1 Y C2∗ + D11 D21
)(C2 Y C2∗ + D21 D21
) (C2 Y C1∗ + D21 D11
) .
Proof
Denote
"
G1 =
A B1
C1 D11
#
"
,
G2 =
A B1
C2 D21
#
Then the transfer matrix from w to e is
Ĝ1 − K̂ Ĝ2
The problem becomes to minimize
∥Ĝ1 − K̂ Ĝ2 ∥2
subject to K̂ ∈ H∞ and Ĝ1 − K̂ Ĝ2 ∈ H2 . An observer which estimates C1 x
from y is
"
#
A + LC2 −L
O=
C1
0
Let K = O + F . Then the system from the noise w to the estimation error e is
G1 − KG2 = G1 − OG2 − F G2
"
# "
#"
#
A B1
A + LC2 −L
A B1
=
−
− F G2
C1 D11
C1
0
C2 D21
"
#
A + LC2 B1 + LD21
=
− F G2 .
C1
D11
CHAPTER 9. H2 OPTIMIZATION
168
Denote
A + LC2 B1 + LD21
C1
D11
#
A + LC2 B1 + LD21
=
C2
D21
#
"
A + LC2 L
.
=
C2
I
#
"
T0 =
"
T 21
T tmp
−1
T̂21 gives a left coprime factorization. Hence
It is easy to check that Ĝ2 = T̂tmp
−1
K̂ ∈ H∞ and Ĝ1 − K̂ Ĝ2 ∈ H2 if and only if F̂tmp = F̂ T̂tmp
∈ H∞ . The transfer
matrix from w to e now becomes
T̂0 − F̂tmp T̂21
The optimal filtering becomes to design F̂tmp ∈ H2 such that
∥T̂0 − F̂tmp T̂21 ∥2
is minimized.
∼ = N where
From Claim 7.1, we have T̂21 T̂21
∗
N = C2 Y C2∗ + D21 D21
.
This shows that
V̂ =
∼ N −1/2 I − T̂ ∼ N −1 T̂
T̂21
21
21
satisfies V̂ V̂ ∼ = I. Hence
∥T̂0 − F̂tmp T̂21 ∥2 = ∥(T̂0 − F̂tmp T̂21 )V̂ ∥2
∼ N −1/2 − F̂
1/2 T̂ − T̂ T̂ ∼ N −1 T̂
= ∥ T̂0 T̂21
tmp N
0
0 21
21 ∥2 .
∼ ∈ H⊥ + C Y C ∗ + D D ∗
Claim 9.5 T̂0 T̂21
1
11 21
2
2
This gives that the optimal F̂tmp is
∗
F̂tmp = (C1 Y C2∗ + D11 D21
)N −1 = −L0 .
This implies that the optimal F is
"
F = −L0 T tmp =
A + LC2 −L
L0 C2
−L0
#
A + LC2
−L
C1 + L0 C2 −L0
#
.
The optimal filter K is
"
K =O+F =
.
9.6. GENERAL H2 OPTIMIZATION REVISITED — THE SEPARATION PRINCIPLE169
The optimal transfer function from w to e is given by
"
#
#
"
A + LC2 B1 + LD21
A + LC2 B1 + LD21
+ L0
T 0 − F tmp T 21 =
C1
D11
C2
D21
"
#
A + LC2
B1 + LD21
=
.
C1 + L0 C2 D11 + L0 D21
The optimal cost follows from straightforward computation.
2
Next we consider a more general filtering problem: filtering with known
input. The schematic diagram is in Figure 9.6. Here u is the known input
to the system which can be used in the estimation. The observer tries to
produce an estimate z̃ of z so that the estimation error e is small and possibly
independent of the known input u. The optimal H2 observation problem is to
design a causal and stable observer O so that the error e is independent of u
and the H2 norm between w and e is minimized. It turns out that this problem
can be easily solved by using the result of the standard filtering problem. Let
A B1 B2
G = C1 D11 D12 .
C2 D21
0
Consider the filtering ARE (9.6) and let Y be the stabilizing solution. Also
define L and L0 as in (9.7). Then the optimal observer is given by
A + LC2
−L B2
O opt =
C1 + L0 C2 −L0 D12
and the optimal system from w to e is still the same as Gf , where
"
#
A + LC2
B1 + LD21
Gf =
.
C1 + L0 C2 D11 + L0 D21
It is easy to check that the error e is independent from u.
+
e
z
?
−
j
6 z̃
w
y
O G
u
Figure 9.6: The filtering problem with a known input.
9.6
General H2 Optimization Revisited — the Separation Principle
In the general H2 optimization problem, we see that if we can measure the
information needed for the full information control, namely, if the measurement
CHAPTER 9. H2 OPTIMIZATION
170
y(k) contains x(k) and w(k), or even only the optimal full information control
signal uf i (k) = F x(k) + F0 w(k), then we can exactly construct the optimal full
information control uf i (k) from the measurement y(k). The controller obtained
this way is clearly the optimal one. In the case when we cannot exactly construct
uf i (t) from y(t), an heuristic idea is to estimate the optimal control signal uf i (t)
in an optimal way and then feedback the estimated optimal control signal. It
turns out the optimal controller for general output feedback case obtained in
Section 7.3 indeed has such an interpretation.
Let the optimal full information controller has been designed as in Section
6.5. Let us then consider the optimal H2 estimation problem defined by
A B1 B2
Gest = F
F0
0
C2 D21 0
i.e., the signal to be estimated is uf i (k). According to the development in the
last section, we obtain the optimal filter as
A + LC2
−L B2
O opt =
C1 + L0 C2 −L0 0
where L and L0 are obtained from the stabilizing solution Y of Riccati equation
∗
∗ −1
AY A∗ −Y +B1 B1∗ −(AY C2∗ +B1 D21
)(C2 Y C2∗ +D21 D21
) (C2 Y A∗ +D21 B1∗ ) = 0.
as
L
L0
=
∗ )
−(AY C2∗ + B1 D21
∗
∗ −1
∗ ) (C2 Y C2 + D21 D21 ) .
−(F Y C2∗ + F0 D21
The optimal estimation error is given by the H2 norm of the system
"
#
A + LC2 B1 + LD21
Gf =
.
F + L0 C2 F0 + L0 D21
Notice that the Riccati equation, as well as L, is independent of F and F0 ,
hence of C1 , D11 , D12 . Whereas L0 , does depend on F and F0 , hence indirectly depends on the control Riccati equation. Substituting F and F0 to the
expression of L0 , we obtain
∗
∗
∗
∗ −1
L0 = (B2∗ XB2 + D12
D12 )−1 [(B2∗ XA + D12
C1 )Y C2∗ + (B2∗ XB1 + D12
D11 )](C2 Y C2∗ + D21 D21
)
A B1
Y C2∗
∗
∗ −1
∗
= (B2∗ XB2 + D12
D12 )−1 B2∗ X D12
(C2 Y C2∗ + D21 D21
)
C1 D11
D21
= −H.
Now consider a controller shown in Figure 9.7, where u(k) = ũf i . Then the
controller can be expressed in the standard form as
#
"
A + B2 F + LC2 + B2 L0 C2 −L − B2 L0
K=
F + L0 C2
−L0
"
#
A + B2 F + LC2 − B2 HC2 −L + B2 H
=
.
F − HC2
H
9.6. GENERAL H2 OPTIMIZATION REVISITED — THE SEPARATION PRINCIPLE171
z
w
G
y
ũf i
?
- O
Figure 9.7: The feedback of the optimally estimated optimal full information
control
This is exactly the optimal controller for the general output feedback H2 optimization problem. This shows that the optimal controller can be designed
in two steps. The first step is to design an optimal full information controller,
assuming the information needed is available. The second step is to design an
optimal estimator to estimate the optimal full information control signal. There
is an asymmetry here: the first step does not depend on the second step but
the second step does depend on the first step (in rather a minor way).
It can also be shown that the optimal cost satisfies
∗
∥Fl (Ĝ, K̂opt )∥22 = ∥Ĝc ∥22 + ∥(B2∗ XB2 + D12
D12 )1/2 Ĝf ∥22 .
(HW2: Prove or disprove this.) Again we have an asymmetry here: the second
term is only a slightly modified optimal estimation error.
Now we can justify the following separation principle:
Separation Principle: The optimal H2 controller consists of the feedback of
the optimally estimated optimal full information control signal. The optimal
cost function is the superposition of the optimal cost of the full information
control and a modified optimal estimation error.
Exercises
9.1 A state space method is given in Theorem 5.4 to compute the H2 norm
of a causal stable system. Assume now that a noncausal stable system is
given by a rational transfer function. Give a procedure to compute its L2
norm using Theorem 5.4.
9.2 Consider the model matching problem with
T̂11 (z) =
1
,
4z + 1
T̂12 (z) =
1
,
z
T̂21 (z) =
z+4
.
z
Find Q̂ such that
∥T̂11 + T̂12 Q̂T̂21 ∥2
is minimized and find the minimum value of the above H2 norm.
P
−k ∈ H . Find
9.3 Let T̂ (z) = ∞
2
k=0 T (k)z
min ∥T̂ (z) + z −n Q̂(z)∥2 .
Q̂∈H2
CHAPTER 9. H2 OPTIMIZATION
172
9.4 In communications, we often hope a transmission system to be a pure
delay so that there is no distortion in the output from the input. However,
physical limitations often prevent us from designing an exact pure delay
system. In this case, we strive for a system as close to a pure delay as
possible. This leads to the following optimization problem:
J(d) = min ∥z −d − F̂ (z)Q̂(z)∥2 .
Q̂∈H2
Solve this problem for F̂ (z) =
z−2
z .
Show that limd→∞ J(d) = 0.
9.5 Let F̂ be a stable transfer function without zero on the unit circle and
with one real zero λ outside of unit circle. Show that
min ∥1 − F̂ (z)Q̂(z)∥22 = 1 −
Q̂∈H2
1
.
λ2
9.6 A noise cancellation problem can be formulated as in Figure 9.8. Here n
1
is a standard white noise, P̂ (z) = 2z+1
, and Ŵ (z) = z−2
z .
1. Characterize all system K such that the closed loop system is stable.
2. Find K such that the variance of the residue r is minimized.
n
?
W
- P
? r
- j -
K Figure 9.8: Noise cancellation
9.7 A double integrator, with input u and output y, has a transfer function
1
. Let us choose y(k) and y(k + 1) − y(k) as state variables. Design a
(z−1)2
state feedback controller such that ∥y∥22 + ∥u∥22 is minimized for all initial
condition.
9.8 Consider a feedback system shown in the Figure 9.9. Suppose that we
wish to
so that the H2 norm of the transfer function
design
the
controller
w1
z1
from
to
is minimized.
w2
z2
1. Formulate this problem as a standard H2 optimization problem.
9.6. GENERAL H2 OPTIMIZATION REVISITED — THE SEPARATION PRINCIPLE173
w1
- j - P
6
z2
C z1
?
j
w2
Figure 9.9: Feedback system for stabilization
2. For P (z) =
9.9 Prove Claim 9.1.
9.10 Prove Claim 9.2.
1
z−1 ,
solve the problem.
174
CHAPTER 9. H2 OPTIMIZATION
Chapter 10
H∞ Optimization
10.1
Simple H∞ Optimization
In this section, we consider a very simple H∞ optimization problem: SISO
model matching problem. Consider the system shown in Figure 10.1.
- T 11
?
j 6
- T 21
- Q
- T 12
Figure 10.1: Model matching problem
Let T 11 , T 12 , and T 21 be SISO casual stable systems with transfer functions
T̂11 , T̂12 , T̂21 ∈ H∞ . The problem is to find a causal stable system Q with
transfer function Q̂ ∈ H∞ such that
∥T̂11 + T̂12 Q̂T̂21 ∥∞
is minimized. Since the systems are all SISO, the transfer functions commute.
Rename T̂12 T̂21 = T̂2 and T̂11 = T̂1 . Then the problem becomes
min ∥T̂1 + T̂2 Q̂∥∞ .
Q̂∈RH∞
Before presenting the solution to the simple H∞ optimization problem. Let
us consider a related problem called Nehari problem: Given R̂ ∈ RH⊥
∞ , find
X̂ ∈ RH∞ to minimize
∥R̂ − X̂∥∞ .
Note that the naive “solution” X̂ = 0 is not a minimizing solution in gen∼
∼
eral. Since R̂ ∈ RH⊥
∞ , it follows that R̂ ∈ RH∞ . Let R̂ have state space
realization
"
#
A B
.
C 0
175
CHAPTER 10. H∞ OPTIMIZATION
176
Here D is zero since D = R̂∼ (∞) = R̂(0) = 0. Let Wc and Wo be the controllability and observability gramians of (C, A, B). Let λ be the squre
Theorem 10.1 Let α2 be the square root of the largest eigenvalue of Wc Wo
and let w be a corresponding eigenvector. Then
min ∥R̂ − X̂∥∞ = α.
X̂∈RH∞
and the unique minimizing X̂ is given by
X̂ = R̂ − α......
The idea in solving the simple H∞ optimization problem is as follows. First
we factorize T̂2 into
T̂2 = T̂2i T̂2o
such that T̂2i is inner in the sense that
∼
T̂2i
T̂2i = 1
−1
are in H∞ . Notices that
and T̂2o is outer in the sense that both T̂2o and T̂2o
∼
multiplication of T̂2i does not change the H∞ norm and the map Q̂ 7→ T̂2o Q̂ is
a one-one correspondence in RH∞ . Then
min ∥T̂1 + T̂2 Q̂∥∞ . =
Q̂∈RH∞
∼
min ∥T̂2i
(T̂1 + T̂2i T̂2o )Q̂)∥∞
Q̂∈RH∞
=
∼
min ∥T̂2i
T̂1 + T̂2o Q̂)∥∞
Q̂∈RH∞
=
min
Q̂1 ∈RH∞
∼
∥T̂2i
T̂1 + Q̂1 )∥∞ .
∼ T̂ is in general not in RH . It is rather in RL
T̂2i
1
∞
∞ and hence can be decomposed in a unique way as the sum of a function in RH∞ and a function in
RH⊥
∞:
∼
∼
∼
T̂2i
T̂1 = ΠH∞ T̂2i
T̂1 + ΠH∞
⊥ T̂2i T̂1
Therefore
min
Q̂1 ∈RH∞
∼
∥T̂2i
T̂1 + Q̂1 )∥∞ =
min
Q̂2 ∈RH∞
∼
∥ΠH∞
⊥ T̂2i T̂1 − Q̂2 )∥∞
(10.1)
The right hand side of (10.1) is simply a Nehari problem and the minimizing
Q̂2 can be obtained from Theorem 10.1. The minimizing Q̂1 for the right hand
side of (10.1) is then
∼
Q̂1 = −ΠH∞ T̂2i
T̂1 − Q̂2 .
The minimizing Q̂ can then be obtained from the minimizing Q̂1 by
−1
Q̂ = T̂20
Q̂1 .
10.2. FULL INFORMATION H∞ CONTROL
177
The details of the design is as follows. Assume that T̂2 (z) has no zeros on
the unit circle. Then it can be written as
T̂2 (z) =
(z − z1 ) · · · (z − zb )(z − zb+1 ) · · · (z − zd )
(z − p1 )(z − p2 ) · · · (z − pn )
where z1 , . . . , zb are outside of the unit circle and zb+1 , . . . , zd are inside of the
unit circle. Rewrite T̂2 as
T̂2 (z) =
z −(n−d) (z − z1 ) · · · (z − zb ) z n−d (z1∗ z − 1) · · · (zb∗ z − 1)(z − zb+1 ) · · · (z − zd )
.
(z1∗ z − 1) · · · (zb∗ z − 1)
(z − p1 )(z − p2 ) · · · (z − pn )
Let us denote the first factor by T̂2i and the second factor by T̂2o , i.e.,
T̂2i (z) =
z −(n−d) (z − z1 ) · · · (z − zb )
(z1∗ z − 1) · · · (zb∗ z − 1)
T̂2o (z) =
z n−d (z1∗ z − 1) · · · (zb∗ z − 1)(z − zb+1 ) · · · (z − zd )
.
(z − p1 )(z − p2 ) · · · (z − pn )
∼ T̂ = 1, it follows that
Then T̂2 = T̂2i T̂2o is the desired factorization. Since T̂2i
2i
∼
∥T̂1 + T̂2 Q̂∥∞ = ∥T̂2i
T̂1 + T̂2o Q̂∥∞ .
Since T̂2o has all poles and zeros inside the unit circle and its numerator and
denominator degrees are the same, T̂2o Q̂ ∈ RH∞ iff Q̂ ∈ RH∞ .
Then the minimization problem becomes
min
Q̂1 ∈RH∞
∼
∥T̂2i
T̂1 + Q̂1 ∥∞
and the minimum solution of this minimization problem is related to that of
−1
the original problem by Q̂ = T̂2o
Q̂1 .
∼
∼
Now T̂2i T̂1 ∈ RL∞ . Hence T̂2i T̂1 can be decomposed into
∼
T̂2i
T̂1 = T̂c + T̂ac
where T̂c ∈ RH∞ is causal and T̂ac ∈ RH⊥
∞ is anticausal. Therefore
min
Q̂1 ∈RH∞
∼
∥T̂2i
T̂1 + Q̂1 ∥∞ =
min
∥T̂c + T̂ac + Q̂1 ∥∞ =
Q̂1 ∈RH∞
min
∥T̂ac − Q̂2 ∥∞
Q̂2 ∈RH∞
where the minimizing Q̂2 and the minimizing Q̂1 are related by Q̂1 = T̂c − Q̂2 .
The optimal solution of the original optimization problem is then given by
−1
Q̂ = T̂2o
(T̂c − Q̂2 ).
where Q̂2 is the solution of the Nehari problem
min
Q̂2 ∈RH∞
∥T̂ac − Q̂2 ∥∞
CHAPTER 10. H∞ OPTIMIZATION
178
z
x
w
G
w
u
@
@
- K
Figure 10.2: The full information setup
10.2
Full Information H∞ Control
Similar to the full information H2 control optimization problem, the full information H∞ control problem concerns the following system
x(k + 1) = Ax(k) + B1 w(k) + B2 u(k)
z(k) = C1 x(k) + D11 w(k) + D12 u(k)
x(k)
.
y(k) =
w(k)
In this system, the measurement y contains the complete information of the
state x and the external disturbance w. The problem is to design a feedback
controller in the form of û = K̂ ŷ such that the H∞ norm of the system from
w to z satisfies certain specification. Using the LFT notation, we represent the
plant by a generalized plant
A
B1
B2
C1
D11 D12
.
G=
I
0
0
0
0
I
Then the problem is to design K̂ so that
∥F l (Ĝ, K̂)∥∞ < 1.
(10.2)
We again make the following assumption.
Assumption 10.1
1. (A, B2 ) is stabilizable.
A − ejω I B2
2.
has full column rank for all ω ∈ R.
C1
D12
We will have to use Riccati equation
A∗ XA − X + C1∗ C1 − A∗ X B1
∗ ∗
B1
∗
D12
×
X B1 B2 + D11
∗
B2
∗
B1
×
B2∗
B2 + C1∗ D11 D12
−1
D11
I 0
−
D12
0 0
∗ D11
XA +
C1 = 0.
∗
D12
(10.3)
10.2. FULL INFORMATION H∞ CONTROL
179
This is an indefinite Riccati equation. Introducing temporary compact notation
C1
D11 D12
I 0
, D=
, J=
.
B = B1 B2 , C =
0
I
0
0 −I
Then the Riccati equation can be rewritten in a more familiar form
A∗ XA−X +C ∗ JC −(A∗ XB +C ∗ JD)(B ∗ XB +D∗ JD)−1 (B ∗ XA+D∗ JC) = 0.
In case a stabilizing solution X exists, define
∗
∗ C
∗
∗
F F0 = −(B2∗ XB2 + D12
D12 )−1 B2∗ XA + D12
1 B2 XB1 + D12 D11 .
Notice that X being a stabilizing solution means that
A − B(B ∗ XB + D∗ JD)−1 (B ∗ XA + D∗ JC)
is stable, not directly that A + B2 F is stable.
Theorem 10.2 There exists a controller that satisfies the H∞ norm specification (10.2) if and only if the Riccati equation (10.3) has a stabilizing solution
X satisfying
1. X ≥ 0;
∗ D
2. B2∗ XB2 + D12
12 > 0;
∗
∗ D − I B ∗ XB + D ∗ D
B1 XB1 + D11
11
2
1
11 12
3. Sl
< 0.
∗ D
∗ D
B2∗ XB1 + D12
B2∗ XB2 + D12
11
12
Under these conditions, one of such controllers is given by K̂(z) =
i.e., a feedback control of the form u(k) = F x(k) + F0 w(k).
F
F0 ,
Proof We first show the sufficiency by construction. To simplify the manipulation, let us simplify the notation a bit. Denote
∗
∗
M00 M10
M20
∗
M = M10 M11 M21
M20 M21 M22
∗
∗
C1 A
X 0 0
∗ C
− 0 I 0 .
= B1∗ X A B1 B2 + D11
1 D11 D12
∗
∗
B2
D12
0 0 0
Then the Riccati equation can be written as
−1 ∗
M11 M21
M10
∗
∗
M00 − M10 M20
= 0.
M21 M22
M20
Conditions 2-3 imply that M has a factorization of the form
∗
L00 0
0
0 0 0
L00 0
0
M = L10 L11 0 0 −I 0 L10 L11 0
L20 L21 L22
0 0 I
L20 L21 L22
CHAPTER 10. H∞ OPTIMIZATION
180
where
L00 = I
−1
−1
∗
∗
L10 = (−M11 + M21
M22
M21 )−1/2 (−M10 + M21
M22
M20 )
−1
∗
L11 = (−M11 + M21
M22
M21 )1/2
−1/2
L20 = M22
L21 =
L22 =
M20
−1/2
M22 M21
1/2
M22
Now the controller can be rewritten as
K̂(z) =
−1
−1
M20 −M22
M21
−M22
=
−1
−L−1
22 L20 −L22 L21 .
The closed-loop system is
# "
#
"
Ac Bc
B1 − B2 L−1
A − B2 L−1
22 L21
22 L20
=
.
Fl (G, K) =
−1
Cc D c
C1 − D12 L−1
22 L20 D11 − D12 L22 L21
We need to show that Fl (Ĝ, K̂) ∈ H∞ and ∥Fl (Ĝ, K̂)∥∞ < 1.
We first prove Fl (Ĝ, K̂) ∈ H∞ , i.e., Ac is stable. This requires the following
claim.
Claim 10.1 A∗c XAc − X +
Proof
A∗c XAc
Cc∗ L∗10
Cc
L10
= 0.
(of Claim 8.1):
−X +
Cc∗
L∗10
Cc
L10
−1 ∗
−1
−1 ∗
∗
∗
= (A − B2 L−1
22 L20 ) X(A − B2 L22 L20 ) − X + (C1 − D12 L22 L20 )(C1 − D12 L22 L20 ) + L10 L10
−1
∗
∗
∗
= A∗ XA − X + C1∗ C1 − (A∗ XB2 + C1∗ D12 )L−1
22 L20 − L20 L22 (B2 XA + D12 C1 )
−1
∗
∗
∗
+L∗20 L−1
22 (B2 XB2 + D12 D12 )L22 L20 + L10 L10
−1 ∗
M11 M21
M10
∗
∗
= M00 − M10 M20
M21 M22
M20
= 0.
Since
−1 ∗
0
M11 M21
M10
Cc
L−1
11
A−B
= Ac − B1 B2
M21 M22
M20
L10
0 L22 L21
Cc
is stable, hence
, Ac is detectable. Consequently by a modified verL10
sion of Theorem 2.11 we conclude Ac is stable.
We then prove ∥Fl (Ĝ, K̂)∥∞ < 1 by showing the following claim.
10.3. H∞ FILTERING PROBLEM
Claim 10.2 Let
181
Ac B c
P = Cc D c
L10 L11
Then P ∼ P = I.
Since Fl (Ĝ, K̂) is a submatrix of P̂ which is allpass, we immediately get
∥Fl (Ĝ, K̂)∥∞ ≤ 1. (HW3: Prove Claim 8.2 and complete the argument to show
that ∥Fl (Ĝ, K̂)∥∞ < 1.)
It remains to show the necessity....
2
This theorem shows that one of the full information H∞ controllers is a static
state feedback plus a static disturbance feedforward, similar to the optimal H2
full information control.
10.3
H∞ Filtering Problem
10.4
General H∞ Optimization
z
w
G
y
u
- K
Figure 10.3: The standard setup
Consider Figure 10.3. Assume that G is given. The general H∞ optimization problem is to design K so that the overall system is internally stable and
the H∞ norm of the transfer function from w to z
∥F l (Ĝ, K̂)∥∞
is minimized.
182
CHAPTER 10. H∞ OPTIMIZATION
Bibliography
[1] P. Brunovsky, A Classification of Linear Controllable Systems, Kybernetika, 6(3): 173–188, 1970.
[2] E. Bonilla, J. J. Loiseau, and S. Baquero, A Geometric Proof of Rosenbrock’s Theorem on Pole Assignment, Kybernetika, 33(4): 357–370, 1997.
[3] D. S. Flamm, A New Proof of Rosenbrock’s theorem on Pole Assignment,
IEEE Transactions on Automatic Control, 25(6): 1128–1133, 1980.
[4] H. H. Rosenbrock, State Space and Multivariable Theory, Wiley, New York,
1970.
183
184
BIBLIOGRAPHY
Appendix A
Mathematical Preliminaries
A.1
Algebraic Structures
Basic set theory is assumed.
A.1.1
Groups, Rings, and Fields
A binary operation “◦” in a set S is a map S × S → S.
Definition A.1 A nonempty set G with a binary operation “◦”, denoted by
(G, ◦), is said to be a group if
1. (a ◦ b) ◦ c = a ◦ (b ◦ c) for all a, b, c ∈ G,
(associative law)
2. there exists e ∈ G such that e ◦ a = a ◦ e = a for all a ∈ G, (existence of
identity)
3. for each a ∈ G, there exists x ∈ G such that a ◦ x = x ◦ a = e. (existence
of inverse)
Definition A.2 A group (G, ◦) is said to be abelian if a ◦ b = b ◦ a for all
a, b ∈ G.
Example A.1
1. Let Rn×m denote the set of n × m real matrices. Then (Rn×m , +) is an
abelian group. The identity element is the zero matrix and the inverse is
just the negation. Similarly (Cn×m , +), the set of n × m complex matrices
together with the addition operation, is an abelian group with the identity
being the zero matrix and the inverse being the negation.
2. Let GL(n, R) and GL(n, C) denote the sets of n×n nonsingular real matrices and complex matrices respectively. Then (GL(n, R), ×) (GL(n, C), ×)
are groups. In both groups, the identity element is the identity matrix
and the inverse is the usual matrix inverse. These two groups are called
the general linear groups.
185
186
APPENDIX A. MATHEMATICAL PRELIMINARIES
3. Let O(n) denote the set of n×n real orthogonal matrices. Then (O(n), ×)
is a group. The identity element is the identity matrix and the inverse is
the usual matrix inverse (which is equal to matrix transpose). Similarly,
let U(n) denote the set of n×n complex unitary matrices. Then (U(n), ×)
is a group. The identity element is the identity matrix and the inverse is
the usual matrix inverse (which is equal to matrix conjugate transpose).
The groups O(n) and U(n) are called the orthogonal group and the unitary
group respectively.
4. Let SO(n) denote the set of n × n real orthogonal matrices with determinant equal to 1. Then (SO(n), ×) is a group. This group is called the
special orthogonal group. Clearly, SO(n) is a subgroup of O(n) which is in
turn a subgroup of GL(n, R). Similarly, let SU(n) denote the set of n × n
complex unitary matrices with determinant equal to 1. Then (SU(n), ×)
is a group. This group is called the special unitary group. Clearly, SU(n)
is a subgroup of U(n) which is in turn a subgroup of GL(n, C).
5. The set of all subsets of set S is denoted by 2S . Then (2S , ∪) and (2S , ∩)
satisfy the associative law and the existence of identity. The identity
elements are empty set ∅ and the whole set S respectively. However, they
do not satisfy the existence of inverse. Hence neither of them is a group.
Definition A.3 A nonempty set R with binary operation “+” and binary operation multiplication “·” is said to be a ring if
1. (R, +) is an abelian group,
2. (ab)c = a(bc) for all a, b, c ∈ R,
(associative law of multiplication)
3. a(b + c) = ab + ac, (b + c)a = ba + ca for all a, b, c ∈ R. (distributive law)
A ring R is said to be commutative if the multiplication is commutative,
i.e., ab = ba for all a, b ∈ R.
A ring is said to be a ring with unity if there exists u ∈ R such that
ua = au = a for all a ∈ R.
Example A.2
1. The set of all n × n real (or complex) matrices is a ring with unity.
2. The set of integers Z is a ring with unity.
3. The set of all univariate polynomials with the usual addition and multiplication is a commutative ring with unity.
Definition A.4 A commutative ring F with unity is said to be a field if every
nonzero element of F has a multiplicative inverse in F.
Example A.3
1. The set of real numbers R, the set of complex numbers C, the set of
rational numbers Q are all fields.
2. The set of all univariate rational functions is a field.
A.1. ALGEBRAIC STRUCTURES
A.1.2
187
Equivalence Relations
A relation in a set S is a map S × S → {0, 1}.
Definition A.5 A relation “≡” in a set S is said to be an equivalence relation
if
1. a ≡ a for all a ∈ S,
(reflexive)
2. a ≡ b implies b ≡ a.
(symmetric)
3. if a ≡ b and b ≡ c imply a ≡ c.
(transitive)
Example A.4
1. Let S be the set of all individuals in a country, a city, or a university.
Then having the same surname defines an equivalence relation, i.e., we
can say a ≡ b if a and b have the same surname. Also having the same
gender defines another equivalence relation, i.e., we can say a ≡ b if a and
b are both male or both female.
2. Let F be either R or C. In the set Fn×n of n × n matrices, matrices A
and B are said to be similar, denoted by A ∼ B, if there is a nonsingular
matrix T ∈ GL(n, F) such that A = T −1 BT . Then ∼ is an equivalence
relation.
3. Denote the set of all n × n real symmetric matrices by Sn . Two matrices
A, B ∈ Sn are said to be congruent, denoted by A ∼
= B, if there is a
nonsingular matrix T ∈ GL(n, R) such that A = T ′ BT . Here the prime
means matrix transpose. Then ∼
= is an equivalence relation. Similarly,
denote the set of all n × n complex Hermitian matrices by Hn . Two
matrices A, B ∈ Hn are said to be congruent, denoted by A ∼
= B, if there
is a nonsingular matrix T ∈ GL(n, C) such that A = T ∗ BT . Here the star
means matrix conjugate transpose. Then ∼
= is an equivalence relation.
4. Two matrices A, B ∈ Rn×n are said to be orthogonally similar, denoted by
A ≡ B, if there is an orthogonal matrix U ∈ O(n) such that A = U ∗ BU .
Then ∼ is an equivalence relation. Two matrices A, B ∈ Cn×n are said
to be unitarily similar, denoted by A ≡ B, if there is a unitary matrix
U ∈ U(n) such that A = U ∗ BU . Then ∼ is an equivalence relation.
Apparently A ≡ B implies A ∼ B.
A subset of S containing all equivalent elements is called an equivalent class.
Each pair of equivalent classes are either identical or disjoint, and S is simply the
union of all equivalent classes. In other words, the set S can be partitioned into
disjoint subsets called equivalent classes by grouping the equivalent elements
together. The set of all equivalent classes is called the quotient of S with respect
to “≡”, and is sometimes denoted by S/ ≡.
Example A.5
1. Let S be the set of all individuals in a country and let the equivalence
relation be having the same gender, i.e., we say a ≡ b if a and b are
188
APPENDIX A. MATHEMATICAL PRELIMINARIES
both male or both female. Then S is divided into two equivalence classes
containing all male individuals and all female individuals respectively.
The quotient set S/ ≡ contains two members: the male class and the
female class.
2. For Cn×n with equivalence relation ∼, it follows from the Jordan canonical
form theorem that each equivalent class consists matrices with the same
Jordan canonical form. The quotient set Cn×n / ∼ consists all possible
n × n Jordan matrices.
3. For Sn with equivalence relation
inertia that each equivalent class
same inertia. The quotient Sn / ∼
=
∼
=, it follows from Sylvester’s law of
consists symmetric matrices with the
is the set of all possible inertias.
4. For a set S, the power set S n consists of all ordered n elements of S. Now
let us define a relation ≃ in S n in the following way: x ≃ y if they contain
the same elements of S possibly ordered in different way. Apparently ≃
is an equivalence relation. The quotient set S n / ≃ is the set of bags of n
elements, possibly with repetitions, of S.
A.1.3
Partial Order
Definition A.6 A relation “≼” in a set S is said to be a partial order if
1. a ≼ a for all a ∈ S,
(reflexive)
2. a ≼ b and b ≼ a imply a = b,
(anti-symmetric)
3. a ≼ b and b ≼ c imply a ≼ c.
(transitive)
A set with a partial order is called a partially ordered set or poset, denoted
by (S, ≼). We read a ≼ b as a precedes b or b supercedes a.
Example A.6
1. For A, B ∈ S n , define A ≤ B if x∗ Ax ≤ x∗ Bx for all x ∈ Rn . Then
(Sn , ≤) is a poset. For A, B ∈ Hn , define A ≤ B if x∗ Ax ≤ x∗ Bx for all
x ∈ Cn . Then (Hn , ≤) is a poset.
2. For A = [aij ], B = [bij ] ∈ Rn×m , we say A ≼ B if aij ≤ bij for all
1 ≤ i ≤ n, 1 ≤ j ≤ m. Then (Rn×m , ≼) is a poset.
3. For x, y ∈ Rn , we say that x is majorized by y, denoted by x ≺ y, if
max xi ≤ max yi
1≤i≤n
max
1≤ii <i2 ≤n
1≤i≤n
x i1 + x i2 ≤
max
1≤ii <i2 ≤n
yi1 + yi2
..
.
max
1≤ii <i2 <···<in−1 ≤n
xi1 + xi2 + · · · + xin−1 ≤
max
1≤ii <i2 <···<in−1 ≤n
yi1 + yi2 + · · · + yin−1
x1 + x2 + · · · + xn = y1 + y2 + · · · + yn .
A.1. ALGEBRAIC STRUCTURES
189
We say that x is weakly majorized by y, denoted by x ≺w y, if the last
′
equality above is changed to inequality “≤”. For example, 2 2 2 ≺
′
1 2 3 . We can see that ≺ orders the level of fluctuation when
′
′
the average is the same. Also notice that 1 2 3 ≺w 3 3 3 .
This shows that ≺w orders the level of fluctuation and the average in an
combined way. Neither the majorization ≺ nor the weak majorization
≺w is a partial order in Rn since neither is anti-symmetric. Nevertheless,
both ≺ and ≺w are partial orders in Rn / ≃.
4. In a set S with two equivalence relations ≡1 and ≡2 , we say that ≡1 is
finer than ≡2 , denoted by ≡1 ≼≡2 , if a ≡1 b implies a ≡2 b. When ≡1 is
finer than ≡2 , then we also say that ≡2 is coarser than ≡1 . For example,
in Fn×n , equivalence relation ≡ is finer than ∼. In Hn , ≡ is finer than
∼
=. For a set S with a set E of equivalence relations, ≼ defines a partial
order, i.e., (E, ≼) is a poset.
Let T be a subset of a poset (S, ≼). An element a ∈ S is said to be a lower
bound of T if a ≼ b for all b ∈ T . Similarly, an element a ∈ S is said to be an
upper bound of T if b ≼ a for all b ∈ T . A lower bound a of T is said to be the
greatest lower bound or infimum of T if c ≼ a for any other lower bound c of
T . The greatest lower bound of T is denoted by inf T . An upper bound a of
T is said to be the least upper bound or supremum of T if a ≼ c for any other
up bound c of T . The least upper bound of T is denoted by sup T . It is easy
to see that inf T and sup T , if exist, are unique.
Let a, b ∈ S. Then inf{a, b} is also denoted by a ∧ b, called the meet of a, b;
and sup{a, b} is also denoted by a ∨ b, called the join of a, b.
Definition A.7 A poset (S, ≼) is called a lattice if a ∧ b and a ∨ b exist for all
a, b ∈ S.
Example A.7
1. Let N be the set of natural numbers {1, 2, . . .}. We say a|b if a divides b.
Then “|” is a partial order and (N, |) is a lattice. Here a ∧ b is the greatest
common divisor and a ∨ b is the least common multiple.
2. Another partial order in N is given by “≤”, the usual order in terms of
magnitude. This partial order is special since for a, b ∈ N, either a ≤ b
or b ≤ a. Hence this partial order is actually a total order. In this case,
a ∧ b = min{a, b} and a ∨ b = max{a, b}.
3. Let P be the set of all nonzero monic polynomials. We say a|b if a divides
b. Then “|” is a partial order and (S, |) is a lattice. Here a ∧ b is the
greatest common divisor and a ∨ b is the least common multiple.
4. For a given set S, (2S , ⊂) is a lattice. Here the meet and join are the
intersection and union respectively.
5. For the set of real-valued continuous functions on [0, 1], denoted by C[0, 1],
we say f ≤ g if f (x) ≤ g(x) for all x ∈ [0, 1]. (C[0, 1], ≤) is a lattice with
(f ∧ g)(x) = min{f (x), g(x)} and (f ∨ g)(x) = max{f (x), g(x)}.
190
APPENDIX A. MATHEMATICAL PRELIMINARIES
6. (S n , ≤) and (Hn , ≤) are not lattices. We leave it as an exercise to check
this.
7. For x, y ∈ Rn / ≃, if their entries have different total sums, they it is
impossible to find their meet or join under partial order ≺. Hence (Rn / ≃
, ≺) is not an lattice. On the other hand, (Rn / ≃, ≺w ) is an lattice. We
leave it as an exercise to find the meet and join under partial order ≺w
for each pair in Rn / ≃.
A.1.4
Linear Spaces and Linear Transformation
Definition A.8 A set X is said to be a linear space (vector space) over a field
F, denoted by (X , F), if there is a binary operation “+”, called addition, in X
and there is a map F × X → X , called multiplication by scalars, such that
1. (X , +) is an abelian group,
2. (αβ)x = α(βx) for all α, β ∈ F and x ∈ X ,
3. α(x + y) = αx + αy for all α ∈ F and x, y ∈ X
4. (α + β)x = αx + βx for all α, β ∈ F and x ∈ X ,
5. 1x = x for all x ∈ X , where 1 is the unity in F.
Example A.8
1. The set of n-vectors over F, denoted by Fn , is a linear space over F.
2. The set of n × m matrices with entries in F, denoted by Fn×m , is a linear
space over F.
3. The set of all polynomials with coefficients in F is a linear space.
4. The set of bilateral sequences of the form
{. . . , x(−1), |x(0), x(1), . . .}
is a linear space.
A subset U of a linear space is said to be a subspace if itself is a linear space.
Let U and V be subspaces of X . Then the intersection and the sum of U and
V:
U ∩ V = {x : x ∈ U and x ∈ V}
U + V = {u + v : u ∈ U and v ∈ V}
are also subspaces of X . These operations make the set of subspaces of X ,
partially ordered by the set inclusion “⊂”, a lattice.
If U ∩ V = {0}, then U and V are said to be linearly independent. In this
case, we write U + V as U ⊕ V. More generally, a number of linear subspaces
U 1 , · · · , U m of X are said to be linearly independent if
x1 + x2 + · · · + xm = 0
A.1. ALGEBRAIC STRUCTURES
191
for some xi ∈ U i , i = 1, 2, . . . , m, implies xi = 0 for all i. If U i , i = 1, 2, . . . , m,
are linearly independent, we write
U1 + U2 + · · · + Um
or
m
X
Ui
i=1
as
U1 ⊕ U2 ⊕ · · · ⊕ Um
or
m
M
U i.
i=1
Several vectors x1 , x2 , . . . , xm ∈ X are set to be linearly independent if
α1 x1 + α2 x2 + · · · + αm xm = 0
implies α1 = α2 = · · · = αm = 0. The largest number of linearly independent
vectors in X is called the dimension of X . The dimension can be infinity. For
a finite dimensional space, let the dimension be n. Then a set of n linearly
independent vectors {x1 , x2 , . . . , xm } is called a basis of X .
Let X and Y be linear spaces over F. A map A from X to Y is a linear
transformation if
A(α1 x1 + α2 x2 ) = α1 Ax1 + α2 Ax2 .
The null space or kernel of A is
N (A) = {x ∈ X : Ax = 0}.
The range or image of A is
R(A) = {y ∈ Y : y = Ax for some x ∈ X }.
Clearly, A is injective or one-to-one iff N (A) = {0}, and A is surjective or
onto iff R(A) = Y. Putting together, we see that A is bijective or a one-one
correspondence iff N (A) = {0} and R(A) = Y.
Let U be a subspace of X . Denote
AU = {Au : u ∈ U}
and call it the image of U under A. Let V be a subspace of Y. Denote
A−1 V = {x ∈ X : Ax ∈ V}
and call it the inverse image of V. Here by no means we imply that A is
invertible. Using this notation, we see that R(A) = AX and N (A) = A−1 {0}.
It is not hard to verify the following relations: (Cf: Exercise A.6.1)
A
A(U 1 + U 2 ) = AU 1 + AU 2
(A.1)
A(U 1 ∩ U 2 ) ⊂ AU 1 ∩ AU 2
(A.2)
−1
(V 1 + V 2 ) ⊃ A
−1
A
(V 1 ∩ V 2 ) = A
−1
−1
V1 + A
−1
−1
V1 ∩ A
V2
(A.3)
V 2.
(A.4)
192
APPENDIX A. MATHEMATICAL PRELIMINARIES
One can also show that (A.2) becomes an equality if and only if
N (A) ∩ (U 1 + U 2 ) = N (A) ∩ U 1 + N (A) ∩ U 2
(A.5)
and (A.3) becomes an equality if and only if
R(A) ∩ (V 1 + V 2 ) = R(A) ∩ V 1 + R(A) ∩ V 2
(A.6)
(Cf: Exercise A.6.2).
Now consider a linear transformation A : X → X . A subspace U ⊂ X is
said to be A-invariant if AU ⊂ U. In particular, {0}, X , N (A) and R(A) are
A-invariant subspaces. The set of A-invariant subspaces is a lattice, under the
partial order ⊂. Hence it is a sublattice of the set of all subspaces of X .
Let U be an A-invariant subspace. Then A can also be considered as a map
from U to U. This map is called the restriction of A on U and is denoted by
A|U .
A.1.5
Normed Linear Space
Consider a linear space X over a field F. For us F is either the real field R or
the complex field C. A norm ∥ · ∥ in X is a function X → R satisfying
1. ∥x∥ > 0 for all x ̸= 0,
2. ∥αx∥ = |α|∥x∥ for all x ∈ X and α ∈ F.
3. ∥x + y∥ ≤ ∥x∥ + ∥y∥ for all x, y ∈ X .
Common norms in Fn are the Hölder p-norms, 1 ≤ p ≤ ∞, defined by
x1
∥ ... ∥p =
xn
n
X
!1/p
|xi |
p
i=1
for 1 ≤ p < ∞ and
x1
∥ ... ∥∞ = max |xi |.
1≤i≤n
xn
It is not a trivial matter to show ∥ · ∥p indeed satisfies the third requirement
above. We leave it to the reader as an exercise.
In the space of matrices, several classes of norms are used in different situations. First, Hölder p-norms can be extended to the space of matrices Fn×m :
1/p
a11 · · · a1m
n X
m
X
.. ||| =
||| ...
|aij |p .
. p
i=1 j=1
an1 · · · anm
A.1. ALGEBRAIC STRUCTURES
193
Here the reason why ||| · ||| instead of ∥ · ∥ is used for the norms is purely a
notational matter and is to avoid confusion with other classes of matrix norms.
Another way to define ||| · |||p is by introducing the so called “vec” operator:
vec A1 A2 · · · Am =
A1
A2
..
.
.
Am
Here Ai are the columns of A. Then
|||A|||p = ∥vecA∥p .
The case when p = 2 is of particular interest. The norm ||| · |||2 is called the
Frobenius norm, and is also denoted by ∥ · ∥F .
Secondly, let the singular values of a matrix A ∈ Fn×m be σ1 (A), σ2 (A), . . . ,
σmin{n,m} (A), ordered nonincreasingly. Define
σ1 (A)
..
.
|A|p = ∥
∥p .
σmin{n,m} (A)
It is nontrivial to show that | · |p is indeed a norm, which is left as an exercise.
The third class of matrix norms are also used. For A ∈ Fn×m , define
∥A∥p =
∥Ax∥p
= sup ∥Ax∥p .
x∈Fm ,x̸=0 ∥x∥p
∥x∥p =1
sup
This norm is called the induced p-norm. It is not hard to show
∥A∥1 =
n
X
|aij |
(A.7)
∥A∥2 = σ1 (A)
m
X
∥A∥∞ = max
|aij |.
(A.8)
max
1≤j≤m
1≤i≤n
i=1
j=1
Let ℓp (Z), 1 ≤ p ≤ ∞, be the set of bilateral sequences
x = {· · · , x(−1), |x(0), x(1), · · · } ⊂ F
satisfying
∞
X
|x(k)|p < ∞
k=−∞
for p < ∞ and
sup
−∞<k<∞
|x(k)| < ∞
(A.9)
194
APPENDIX A. MATHEMATICAL PRELIMINARIES
for p = ∞. Define the norm in ℓp (Z) by
∞
X
∥x∥p =
!1/p
|x(k)|
p
k=−∞
for p < ∞ and
∥x∥∞ =
sup
|x(k)|.
−∞<k<∞
A.1.6
Inner Product Space
An inner product on a linear space X is a function ⟨·, ·⟩ : X × X → F such that
1. ⟨x, x⟩ > 0 for all x ̸= 0.
2. ⟨x, y⟩ = ⟨y, x⟩ for all x, y ∈ X .
3. ⟨x + y, z⟩ = ⟨x, z⟩ + ⟨y, z⟩ for all x, y, z ∈ X .
4. ⟨x, αy⟩ = α⟨x, y⟩ for all x, y ∈ X and α ∈ F.
A linear space with an inner product is called an inner product space.
Proposition A.1 (Cauchy-Schwarz inequality)
p
|⟨x, y⟩| ≤ ⟨x, x⟩⟨y, y⟩.
Proof The case when ⟨x, y⟩ = 0 is trivial. Assume now that ⟨x, y⟩ =
̸ 0. For
all α ∈ F,
0 ≤ ⟨x + αy, x + αy⟩
= ⟨x, x⟩ + ⟨x, αy⟩ + ⟨αy, x⟩ + ⟨αy, αy⟩
= ⟨x, x⟩ + α⟨x, y⟩ + α⟨x, y⟩ + ∥α∥2 ⟨y, y⟩
= ⟨x, x⟩ + 2Re(α⟨x, y⟩) + ∥α∥2 ⟨y, y⟩.
Let α =
⟨x,y⟩
|⟨x,y⟩| t.
Then for all t ∈ R,
0 ≤ ⟨x, x⟩ + 2|⟨x, y⟩|t + ⟨y, y⟩t2 .
This can happen only if
(2|⟨x, y⟩|)2 − 4⟨x, x⟩⟨y, y⟩ ≤ 0,
which yields
|⟨x, y⟩|2 ≤ ⟨x, x⟩⟨y, y⟩.
2
p Proposition A.1 implies that an inner product space induces a norm ∥x∥ =
⟨x, x⟩. Another consequence of Proposition A.1 is that the inner product is
a continuous function in both variables.
The norm induced by an inner product satisfies the so-called parallelogram
law.
A.1. ALGEBRAIC STRUCTURES
195
Theorem A.1 (Parallelogram law)
∥x + y∥2 + ∥x − y∥2 = 2∥x∥2 + 2∥y∥2 .
It is quite trivial to verify this, but its opposite question is more interesting:
given a norm in a linear space, how can one know if the norm is induced by
an inner product? The following theorem answers this question, whose proof is
left as an exercise (Exercise A.14).
Theorem A.2 ∥ · ∥ is induced by an inner product if
∥x + y∥2 + ∥x − y∥2 = 2∥x∥2 + 2∥y∥2 .
In Fn , the standard inner product is
⟨x, y⟩ = x∗ y.
The norm induced by this inner product is the Hölder 2-norm ∥ · ∥2 .
In Fn×m , let us define
⟨A, B⟩ = tr(A∗ B).
The norm induced by this inner product is the Frobenius norm ∥ · ∥F .
In ℓ2 (Z), let us define
⟨x, y⟩ =
∞
X
x(k)y(k).
k=−∞
It is an easy exercise to show that ⟨x, y⟩ is well defined for all x, y ∈ ℓ2 (Z)
and ⟨·, ·⟩ satisfies the requirements for an inner product. The induced norm is
exactly ∥ · ∥2 .
Two vectors x, y are said to be orthogonal, denoted by x ⊥ y, if ⟨x, y⟩ = 0.
Several vectors xi , i = 1, 2, . . . , m, are said to be orthogonal if they are pairwise
orthogonal. They are said to be orthonormal if they are orthogonal and ∥xi ∥ = 1
for all i.
Let x1 , x2 , . . . , xm be a set of linearly independent vectors. Construct a new
set of vectors in the following way
i−1
i−1
X
X
ui = xi −
⟨uj , xi ⟩uj / xi −
⟨uj , xi ⟩uj , i = 1, 2, . . . , m.
j=1
j=1
One can check in a straightforward way that u1 , u2 , . . . , um are orthonormal.
This process of getting a set of orthonormal vectors from a set of linearly independent vectors is called the Gram-Schmidt orthonormalization.
The orthogonal complement of a vector x ∈ X , denoted by x⊥ , is the set of
vectors in X orthogonal to x, i.e.,
x⊥ = {y ∈ X : x ⊥ y}.
The orthogonal complement of a set S of vectors in X , denoted by S ⊥ , is the
set of vectors in X orthogonal to every member of S, i.e.,
S ⊥ = {y ∈ X : x ⊥ y for all x ∈ S}.
If U is a subspace, it is easy to see that
U ⊕ U⊥ = X .
196
APPENDIX A. MATHEMATICAL PRELIMINARIES
A.2
Matrix Analysis
A.2.1
Matrix Operations
Basic matrix operations such as addition, multiplication, and inverse are assumed. Also assumed are matrix functions such as determinant and trace.
Two special matrix operations will be introduced in this subsection.
The first is the Schur complement. Let
A11 A12
A=
∈ F(n1 +n2 )×(m1 +m2 )
A21 A22
be a partitioned matrix. If A11 is invertible, then the Schur complement of A11
in A is defined as
A/11 = A22 − A21 A−1
11 A12 .
If A12 is invertible, then the Schur complement of A12 in A is defined as
A/12 = A21 − A22 A−1
12 A11 .
If A21 is invertible, then the Schur complement of A21 in A is defined as
A/21 = A12 − A11 A−1
21 A22 .
If A22 is invertible, then the Schur complement of A22 in A is defined as
A/22 = A11 − A12 A−1
22 A21 .
The following theorem can be easily verified.
Theorem A.3 Assume the existence of the inverses required in each identities.
1. det(A) = det(A11 ) det(A/11 ).
2. det(A) = det(A22 ) det(A/22 ).
−1
A11 + A−1
A12 (A/11 )−1 A21 A−1
−A−1
A12 (A/11 )−1
−1
11
11
11
3. A =
.
−(A/11 )−1 A21 A−1
(A/11 )−1
11
(A/22 )−1
−(A/22 )−1 A12 A−1
−1
22
4. A =
−1 A−1 + A−1 A (A/ )−1 A A−1 .
−A−1
22
12 22
22 A21 (A/22 )
22
22 21
(A/22 )−1 (A/12 )−1
5. A−1 =
.
(A/21 )−1 (A/11 )−1
The second matrix operation is the linear fractional transformation. Let
A11 A12
A=
∈ F(n1 +n2 )×(m1 +m2 ) , X ∈ Fm2 ×n2 .
A21 A22
The lower linear fractional transformation is defined as
Fl (A, X) = A11 + A12 X(I − A22 X)−1 A21 .
A.2. MATRIX ANALYSIS
197
The upper linear fractional transformation is defined as
Fu (A, X) = A22 + A21 X(I − A11 X)−1 A12 .
Now let
B=
B11 B12
B21 B22
∈ F(n2 +n3 )×(m2 +m3 ) ,
Y ∈ Fm3 ×n3 .
Then what is the composition
Fl (A, Fl (B, Y ))?
Let us define the star product of A and B be
F l (A, B11 )
A12 (I − B11 A22 )−1 B12
A⋆B =
.
B21 (I − A22 B11 )−1 A21
F u (B, A22 )
Then it is straightforward to see
Fl (A, Fl (B, Y )) = Fl (A ⋆ B, Y ).
A.2.2
Matrix Eigenvalue Problem
In this subsection, we will be concerned with square matrices A ∈ Cn×n .
Definition A.9 The characteristic polynomial of A is defined to be
cA (z) = det(zI − A).
Proposition A.2 Let A ∈ Cn×n and λ ∈ C. The following statements are
equivalent:
1. cA (λ) = 0.
2. There exists nonzero x ∈ Cn such that Ax = λx.
3. There exists nonzero y ∈ Cn such that y ∗ A = λy ∗ .
A complex number λ satisfying item 1 of Proposition A.2 is called an eigenvalue of A. The vectors x and y satisfying the items 2 and 3 of Proposition A.2
are called respectively the right and left eigenvectors of A corresponding to the
eigenvalue λ. The set of eigenvalues of A, called the spectrum of A, is denoted
by σ(A). The maximum modulus of the eigenvalues of A, called the spectral
radius of A, is denoted by ρ(A).
We can see that A ∈ Cn×n has n eigenvalues including multiplicity. Hence
σ(A) lies in Cn / ≃.
The matrix function (zI − A)−1 is called the resolvent of A. Using linear
algebra knowledge, we know
(zI − A)−1 =
Adj(zI − A)
cA (z)
where Adj(zI − A) means the adjugate of zI − A.
198
APPENDIX A. MATHEMATICAL PRELIMINARIES
Theorem A.4 (Leverrier-Souriau-Faddeeva-Frame) Let
cA (z) = z n + a1 z n−1 + a2 z n−2 + · · · + an
Adj(zI − A) = B1 z n−1 + B2 z n−2 + · · · + Bn .
Then a1 , . . . , an and B1 , . . . , Bn can be computed using the following recursive
formulas:
B1 = I,
1
ai = − tr(Bi A), i = 1, 2, . . . , n,
i
Bi+1 = Bi A + ai I, i = 1, 2, . . . , n − 1.
Since
Proof
cA (z)I = Adj(zI − A)(zI − A),
we have
(z n + a1 z n−1 + a2 z n−2 + · · · + an )I
= B1 z n + (B2 − B1 A)z n−1 + (B3 − B2 A)z n−2 + · · · − Bn A.
This gives B1 = I, Bi+1 = Bi A + ai I for i = 1, 2, . . . , n − 1, and an I + Bn A = 0.
Observe
d
cA (z) = nz n−1 + (n − 1)a1 z n−2 + (n − 2)a2 z n−3 + · · · + an−1
dz
z − a11 −a12 · · · −a1n
−a21 z − a22 · · · −a2n
d
=
det
..
..
..
..
dz
.
.
.
.
−an1
−an2
· · · z − ann
0
−a2n
..
.
1
0
···
−a21 z − a22 · · ·
= det .
..
..
..
.
.
−an1 −an2 · · · z − ann
z − a11 −a12 · · · −a1n
0
1
···
0
+ det
..
..
..
.
.
.
.
.
.
−an1 −an2 · · · z − ann
z − a11 −a12 · · · −a1n
−a21 z − a22 · · · −a2n
+ · · · + det
..
..
..
..
.
.
.
.
0
0
···
1
= tr[Adj(zI − A)]
= tr(B1 )z n−1 + tr(B2 )z n−2 + · · · tr(Bn ).
This shows (n−i)ai = tr(Bi+1 ) for i = 1, 2, . . . , n−1. Plugging Bi+1 = Bi A+ai I
in, we get (n − i)ai = tr(Bi A) + nai for i = 1, 2, . . . , n − 1. Rearrangement gives
ai = − 1i tr(Bi A) for i = 1, 2, . . . , n − 1. Finally an = − n1 tr(Bn A) follows from
0 = Bn A + an I.
2
A.2. MATRIX ANALYSIS
199
Theorem A.5 (Cayley-Hamilton) cA (A) = 0.
Proof
From the proof of Theorem A.4,
0 = Bn A + an I
= Bn−1 A2 + an−1 A + an I
= ···
= An + a1 An−1 + · · · + an−1 A + an I.
2
Assume that A has l distinct eigenvalues λ1 , λ2 , . . . , λl . Then
cA (z) = (z − λ1 )n1 (z − λ2 )n2 · · · (z − λl )nl
P
where ni ≥ 1 and li=1 ni = n. Here ni is called the (algebraic) multiplicity of
eigenvalue λi .
Consider
Ei = N (A − λi I)
and
Ẽi = N [(A − λi I)ni ]
Here the space E i is called the eigenspace associated with eigenvalue λi and is
the set of eigenvectors associated with eigenvalue λi , together with the origin.
The space Ẽi is called the generalized eigenspace associated with eigenvalue λi .
The following observations are quite obvious:
O1 Ei and Ẽi are A-invariant subspaces.
O2 Ei ⊂ Ẽi .
O3 1 ≤ dim Ei ≤ dim Ẽi .
Here comes a less obvious observation.
Proposition A.3 dim Ẽi ≤ ni .
Proof Suppose that dim Ẽi = d > ni for a fixed i. Find subspace F i , a
complement of Ẽi , such that
Ẽi ⊕ F i = Cn .
We have to have dim F i = n − d. Choose a matrix E ∈ Cn×d whose columns
form a basis of E i and a matrix F ∈ Cn×(n−d) whose columns form a basis of
F i . Then P = E F is a nonsingular matrix. Since Ẽi is A-invariant, there
is a matrix Ã11 ∈ Cd×d such that AE = E Ã11 . Hence
Ã11
Ã12
P −1 AP =
.
0(n−d)×d Ã22
Let µ be an eigenvalue of Ã11 . Then there exists nonzero x1 ∈ Cd such that
Ã11 x1 = µx1 .
200
APPENDIX A. MATHEMATICAL PRELIMINARIES
This means AEx1 = µEx1 . Since Ex1 belongs Ẽi , it follows that (A−λi I)ni Ex1 =
(µ − λi )ni Ex1 = 0, which forces µ = λi . This shows that all eigenvalues of Ã11
are equal to λi . Consequently,
zId − Ã11
−Ã12
det(zI − A) = det
0(n−d)×d zIn−d − Ã22
= det(zId − Ã11 ) det(zIn−d − Ã22 )
= (z − λi )d det(zIn−d − Ã22 )
which contradicts the fact that the multiplicity of λi is ni .
Because of Proposition A.3, observation O3 above can be amended as
2
O3’ 1 ≤ dim Ei ≤ dim Ẽi ≤ ni .
Theorem A.6 For each A ∈ Cn×n , it holds
Ll
i=1 Ẽi
= Cn .
Proof Using partial fractional expansion, we see that there exist polynomials
qi , i = 1, 2, . . . , l, such that
l
X qi (z)
1
.
=
cA (z)
(z − λi )ni
i=1
Multiplying both sides by cA (z), we obtain
1=
l
X
qi (z)pi (z)
i=1
where pi (z) =
cA (z)
(z−λi )ni .
Thus,
I=
l
X
qi (A)pi (A).
i=1
For x ∈ Cn , let
xi = qi (A)pi (A)x.
Then
x = x1 + x2 + · · · + xl .
Since
(A − λi I)ni xi = (A − λi I)ni qi (A)pi (A)x = qi (A)cA (A)x = 0,
P
ni
it follows that xi ∈ Ẽi . This shows that Cn = m
i=1 N [(A − λi I) ].
To show that this sum is a direct sum, we need to show that if
0 = x1 + x2 + · · · xl
with xi ∈ Ẽi , then xi = 0 for each i. Assume on the contrary that x1 ̸= 0. Then
x1 = −x2 − · · · − xl .
A.2. MATRIX ANALYSIS
201
Since x1 ∈ Ẽ1 , we have (A − λ1 I)n1 x1 = 0. Since xi ∈ Ẽi for i ≥ 2, it follows
that p1 (A)x1 = 0. We know that (z − λ1 )n1 and p1 (z) are coprime, there exist
polynomials h2 (z) and h2 (z) such that
h1 (z)(z − λ1 )n1 + h2 (z)p1 (z) = 1.
Therefore,
x1 = [h1 (A)(A − λ1 I)n1 + h2 (A)p1 (A)]x1 = 0.
This is a contradiction.
2
Because of Theorem A.6, observation O3’ above can be further amended as
O3” 1 ≤ dim Ei ≤ dim Ẽi = ni .
Since Ei ⊂ Ẽi for each i = 1, 2, . . . , l, Theorem A.6 also implies that the
eigenspaces E1 , E2 , . . . , El are linearly independent. However, dim Ei can be anywhere from 1 through dim Ẽi .
Example A.9
For matrix
2 1 0
A = 0 2 1 ,
0 0 2
the only distinct eigenvalue is λ1 = 2. We have dim E1 = 1 and dim Ẽ1 = 3.
For matrix
2 1 0
A = 0 2 0 ,
0 0 2
the only distinct eigenvalue is again λ1 = 2.
For matrix
2 0
A= 0 2
0 0
We have dim E1 = 2 and dim Ẽ1 = 3.
0
0 ,
2
the only distinct eigenvalue is still λ1 = 2. We have dim E1 = 3 and dim Ẽ1 = 3.
We see that Ei ⊂ Ẽi in general and the eigenspaces do not span the whole
space Cn in general. The gap between the eigenspace Ei and the generalized
eigenspace Ẽi is the cause of all the troubles in matrix eigenstructure analysis.
For matrix A, if there is a gap between the eigenspace and the generalized
eigenspace corresponding to one of its eigenvalues, then A is said to be defective,
otherwise A is said to be nondefective.
If A has n distinct eigenvalues, i.e., ni = 1 for all i, then dim E i = dim Ẽi
for all i, i.e., A is always nondefective. Hence a defective matrix must have
repeated eigenvalues.
For a matrix P ∈ GL(n, C), the transformation A → P −1 AP is called a
similarity transformation. Two matrices A and B are said to be similar if there
exists a P ∈ GL(n, C) such that B = P −1 AP . Two similar matrices have the
same eigenvalues. A matrix A ∈ Cn×n is said to be diagonalizable if there exists
a P ∈ GL(n, C) such that P −1 AP is a diagonal matrix.
202
APPENDIX A. MATHEMATICAL PRELIMINARIES
Theorem A.7 A ∈ Cn×n is diagonalizable if and only if A is nondefective.
Ll
n
Proof If A is nondefective, then dim E i = ni and
i=1 E i = C . Choose
basis in E i to form a matrix Pi ∈ Cn×ni . Then APi = λi Pi . Define
P = P1 P2 · · · Pl .
Then
AP = P
λ1 In1
λ2 In2
..
,
.
λl Inl
i.e., P −1 AP is diagonal.
If A is diagonalizable, then there exists a P ∈ GL(n, C) such that
λ1 In1
λ2 In2
−1
P AP =
.
..
.
λl Inl
where λ1 , . . . , λl are distinct distinct eigenvalues of A, each with multiplicity
ni . Even if initially the diagonal elements are not ordered so that equal ones
are grouped together, one can reorder them by modifying P . Partition P as
P = P1 P2 · · · Pl
such that Pi ∈ Cn×ni . Then it is easy to see that E i = N (A − λi I) = R(Pi )
and dim E i = ni . This implies that A is nondefective.
2
In case when A is not diagonalizable, what is the best we can do? Choose
arbitrary basis in Ẽi . Put these basis vectors together to form an n × n matrix
P . Note that these basis vectors, i.e., the columns of P , form a basis of Cn .
Under this basis, A has matrix representation
A1
A2
−1
P AP =
.
..
.
Al
Clearly, the eigenvalues of each Ai are all equal to λi . Hence A can be transformed into a block diagonal matrix. The eigenvalues of each diagonal block
are the same. The size of each diagonal block is equal to the multiplicity of
that particular eigenvalue.
If we really wish to simplify matrix A further by similarity transformation,
then the best we can achieve is the so-called Jordan canonical form.
Theorem A.8 For each A ∈ Cn×n , there exists P ∈ GL(n, C) such that
J1
J2
−1
−1
P AP = Pi Ai Qi =
..
.
Jl
A.2. MATRIX ANALYSIS
where
203
Ji1
Ji2
Ji =
..
∈ Cni ×ni
.
Jimi
where
Jij =
λi
1
λi
1
..
.
..
.
.
λi 1
λi
A matrix in the form of Jij is called a Jordan block. The number of Jordan blocks mi in matrix Ji , all having the same diagonal elements λi , can be
anywhere from 1 through nk . If it has nk Jordan blocks, then it is actually
diagonal.
Finding the Jordan canonical form is generally not an easy task, especially
in terms of numerical computation. It is then often not to ask for the simplest
form, but for a form which is simple enough to explicitly demonstrate some
properties, e.g., eigenvalues. This will be the topic of the following subsection.
A.2.3
Matrix Factorizations
The standard inner product of x, y ∈ Fn is defined as
⟨x, y⟩ = x∗ y.
The norm in Fn induced by this inner product is the Hölder 2-norm:
p
√
∥x∥2 = ⟨x, x⟩ = x∗ x.
Let now x1 , x2 , . . . , xm be a set of linearly independent vectors in Fn . The
Gram-Schmidt orthonormalization process gives a set of orthonormal vectors
u1 , u2 , . . . , um by
i−1
i−1
X
X
ui = xi −
⟨uj , xi ⟩uj / xi −
⟨uj , xi ⟩uj , i = 1, 2, . . . , m.
j=1
j=1
Notice that ui only depends on x1 , . . . , xi . In matrix form, this can be written
as
u1 u2 · · · um = x1 x2 · · · xm T
where T = [tij ] ∈ Fm×m satisfies tij = 0 for i > j.
A matrix T = [tij ] ∈ Fn×n is said to be upper triangular if tij = 0 for
i > j, i.e., all elements below the main diagonal are zero. The set of triangular
matrices is closed under addition, multiplication, and multiplication by scalars.
So it is called an algebra. A matrix U ∈ Fn×m is said to be an isometry if
U ∗ U = I, i.e., all columns of U are orthonormal. For an isometry U , we have
204
APPENDIX A. MATHEMATICAL PRELIMINARIES
⟨U x, U y⟩ = ⟨x, y⟩, i.e., isometry preserves the inner product. In particular,
∥U x∥ = ⟨U x, U x⟩1/2 = ⟨x, x⟩1/2 = ∥x∥, i.e., isometry also preserves the norm.
A matrix U ∈ Fn×m is said to be an co-isometry if U U ∗ = I, i.e., all rows of U
are orthonormal. A matrix which is both an isometry and a co-isometry is said
to be unitary. For a unitary matrix U , we have U −1 = U ∗ . The set of unitary
matrices is closed under multiplication only, so it is a group.
Proposition A.4 (QR factorization) Let A ∈ Fn×m with n ≥ m. Then there
exist an isometry Q and an upper triangular matrix R such that A = QR.
Proof We will only prove for the case when A has full column rank. The
general case is left as an exercise. Let
A = x1 x2 · · · xm .
Then x1 , x2 , . . . , xm are linearly independent. By the Gram-Schmidt orthonormalization, we can find orthonormal vectors u1 , u2 , . . . , um and upper triangular
matrices T such that
u1 u2 · · · um = x1 x2 · · · xm T.
Since
i−1
X
tii = xi −
⟨uj , xi ⟩uj
−1
,
j=1
the
matrix T is nonsingular
and T −1 is also upper triangular. Let Q =
u1 u2 · · · um , which is an isometry, and R = T −1 . Then A = QR
is the desired factorization.
2
Theorem A.9 (Schur) For each A ∈ Fn×n with eigenvalues λ1 , λ2 , . . . , λn in
any prescribed order, there exists unitary matrix U ∈ Cn×n such that
U ∗ AU = T
is an upper triangular matrix with diagonal entries tii = λi .
Proof Let x1 be an eigenvector of A corresponding to eigenvalue λ1 . Aug
ment x1 to form a basis in Cn : {x1 , x2 , . . . , xm }. Let X1 = x1 x2 · · · xn .
Carry out QR factorization X1 = Q1 R1 where Q1 is unitary and R1 is upper triangular. Then the first column of Q1 is still an eigenvector of A corresponding
to λ1 . Thus
λ1 ∗
∗
Q1 AQ1 =
0 A1
where A1 ∈ C(n−1)×(n−1) and it has eigenvalues λ2 , · · · , λn . Using the same
procedure, we can find unitary matrix Q2 ∈ C(n−1)×(n−1) such that
λ2 ∗
∗
Q2 A1 Q2 =
.
0 A2
A.2. MATRIX ANALYSIS
205
Let
U2 =
Then
1 0
0 Q2
.
λ1 ∗
∗
U2∗ Q∗1 AQ1 U2 = 0 λ2 ∗ .
0 0 A2
Continue this process to produce unitary matrices Qi ∈ C(n−i+1)×(n−i+1) such
that
λi ∗
∗
Qi Ai−1 Qi =
0 Ai
and ui ∈ Cn×n with
Ui =
I 0
0 Qi
.
Then the matrix
U = Q1 U2 · · · Un−1
is unitary and U ∗ AU gives the desired form.
2
Theorem A.9 is the basis for numerical eigenvalue computation. Based
on Theorem A.9, one can also derive the extremely important singular value
decomposition (SVD). The detailed derivation is left as an exercise.
Theorem A.10 (SVD) For A ∈ Fn×m , there exist unitary matrices U ∈ Fn×n ,
V ∈ Fm×m , and a nonnegative diagonal matrix S ∈ Rn×m such that
A = U SV ∗ .
The diagonal entries of S are called singular values of matrix A and are
usually denoted by σ1 (A), σ2 (A), . . . , σmin{m,n} (A), ordered nonincreasingly.
A.2.4
Real Symmetric and Hermitian Matrices
Let us also denote S n by Hn (R) and Hn by Hn (C). The study of matrices
in Hn (F) is often associated with the study of quadratic forms or quadratic
functions Fn → R of the form
q(x) = x∗ Ax
where A ∈ Hn (F). Put it in another way: there is a one-one correspondence
between a quadratic form and a Hermitian matrix.
A matrix A ∈ Hn (F) is said to be
positive definite if x∗ Ax > 0 for all nonzero x ∈ Fn ;
positive semi-definite if x∗ Ax ≥ 0 for all x ∈ Fn ;
negative definite if x∗ Ax < 0 for all nonzero x ∈ Fn ;
negative semi-definite if x∗ Ax ≤ 0 for all x ∈ Fn ;
206
APPENDIX A. MATHEMATICAL PRELIMINARIES
indefinite if it is none of above.
We use notation A > 0, A ≥ 0, A < 0, A ≤ 0 to mean A being positive
definite, positive semi-definite, negative definite, and negative semi-definite,
respectively. The set of all n × n positive semi-definite matrices is denoted by
Pn .
We often use A ≤ B to mean B − A ≥ 0. Then the relation ≤ is a partial
order in Hn (F). The partially ordered set (Hn (F), ≤), however, is not a lattice
as we have argued.
What is the easiest way to test whether a given matrix A ∈ Hn (F) is positive
definite, positive semi-definite, negative definite, or negative semi-definite? The
following theorem gives an answer.
Theorem A.11 The following statements are equivalent:
1. A > 0.
2. σ(A) > 0.
a11 · · · a1i
.. > 0 for all i = 1, 2, . . . , n.
3. det ...
.
ai1 · · · aii
If our purpose is to test whether A is positive semi-definite, can we simply
replace all “>” in the above theorem by “≤”? The answer is a big no. This is
an example where the intuition based on ”continuity” is often misleading.
More generally, a matrix A ∈ Hn (F) can have eigenvalues distributed in
the real line. We define the triple of the numbers of negative eigenvalues,
zero eigenvalues, and positive eigenvalues, all with multiplicity counted, as the
inertia of A, denoted by
π(A) = {π− (A), π0 (A), π+ (A)}
where π− (A), π0 (A), π+ (A) are the numbers of negative eigenvalues, zero eigenvalues, and positive eigenvalues of A respectively. For example, an n×n positive
definite matrix A has π(A) = {0, 0, n}. The signature of matrix A is defined to
be π+ (A) − π− (A). Also note that the rank of A is π+ (A) + π− (A).
In the quadratic form x∗ Ax, if we make a variable change x = P z where
P ∈ GL(n, F), then the quadratic form becomes z ∗ P ∗ AP z. The corresponding
matrix changes from A to P ∗ AP . The transformation from A to P ∗ AP is called
a congruence transformation. If P happens to be unitary, i.e., if P ∗ = P −1 , then
the congruence transformation is also a similarity transformation. This means
that a unitary congruence transformation is also a unitary similarity transformation. An immediate consequence of the Schur decomposition theorem is that
for each A ∈ Hn (F), there exists a P ∈ GL(n, F) such that
Iπ+ (A)
0
0
.
0
0π0 (A)
0
P ∗ AP =
(A.10)
0
0
−Iπ− (A)
A.2. MATRIX ANALYSIS
207
To see this more clearly, one can first apply a unitary congruence transformation
to transform A to a diagonal matrix with the eigenvalues on the diagonal and
then follow by another diagonal congruence transformation to normalize the
nonzero diagonal elements. The matrix on the right hand side of (A.10) is
called a signature matrix, whose trace gives the signature of A. One can then
immediately show the following Sylvester inertia theorem.
Theorem A.12 π(P ∗ AP ) = π(A) for each P ∈ GL(n, F).
This theorem says that the inertia is invariant under congruence transformation. It leads to another important theorem, whose proof is left as an exercise.
A C
Theorem A.13 Let M =
∈ H(n1 +n2 ) and A is invertible. Then
B C∗
π(M ) = π(A) + π(M/11 ). In particular, M > 0 if and only if A > 0 and M/11 .
Let f : Rl → Hn (F) be an affine map. In particular, f maps x = [x1 · · · xl ]′ ∈
Rl to
f (x) = H0 + x1 H1 + x2 H2 + · · · xl Hl
for some given H0 , H1 , . . . , Hl ∈ Hn (F). An inequality of the form
f (x) ≥ 0
is called a Linear Matrix Inequality (LMI). An LMI problem is as follows:
Given H0 , H1 , · · · , Hl ∈ Hn (F), find x so that the LMI is satisfied, i.e., find x
in f −1 (Pn ).
Theorem A.14 f −1 (Pn ) is a closed convex set.
Because of this theorem and the special structure of LMIs, there are efficient algorithms to solve an LMI problem. Many engineering problems, especially linear system analysis and synthesis problems, can be converted into LMI
problems.
Exercises
A.1 Give an example for each of the following algebraic structures: group,
ring, field, equivalence relation, poset, linear space, normed linear space,
inner product space.
A.2 Show that the poset of real symmetric matrices (Sn , ≤) and Hermitian
matrices (Hn , ≤) are not lattices when n > 1.
A.3 Let R, S, T be subspaces of Fn .
1. Show that in general
R ∩ (S + T ) ̸= R ∩ S + R ∩ T ,
i.e., the lattice of all subspaces of Fn is not modular.
208
APPENDIX A. MATHEMATICAL PRELIMINARIES
2. Show that if S ⊂ R, then
R ∩ (S + T ) = R ∩ S + R ∩ T .
3. Show that if any one of the following is true, then all are true.
R ∩ (S + T ) = R ∩ S + R ∩ T
S ∩ (R + T ) = S ∩ R + S ∩ T
T ∩ (R + S) = T ∩ R + T ∩ S.
A.4 Show that (Rn / ≃, ≺w ) is a lattice. How can the meet and join be computed.
A.5 The space of complex n × n Hermitian matrices Hn is a linear space over
R. What is its dimension? Find a basis for this linear space.
A.6 Verify (A.1-A.4) and also verify that (A.2) and (A.3) become equalities
when (A.5) and (A.6) hold respectively.
A.7 Show that the set of invariant subspaces of linear transformation A : X →
X is a lattice, i.e., show that if U and V are A-invariant, so are U ∩ V and
U + V.
A.8 Let A : X → X be a linear transformation where dim X = n.
1. Show that in general
R(A) + N (A) ̸= X .
2. Show that the following three statements are equivalent
(a) R(A) + N (A) = X .
(b) A−1 U = U + N (A) for all A-invariant subspace U.
(c) AV = R(A) ∩ V for all A-invariant subspace V.
3. Show that
R(An ) + N (An ) = X
always holds. Hence A−n U = U + N (An ) and An V = R(An ) ∩ V for
all A-invariant subspaces U and V.
A.9 Let A : X → X be a linear transformation and U ⊂ X be a subspace.
Prove that A(A−1 U) = U ∩ R(A) and A−1 (AU) = U + N (A).
A.10 Assume A, B be matrices with the same number of rows. Prove that
R(A) ⊂ R(B) if and only if N (B ∗ ) ⊂ N (A∗ ). Also show the following
equalities:
1. R(A) = N (A∗ )⊥ .
2. R(A) = R(AA∗ ) and N (A) = N (A∗ A).
A.11 Verify (A.7)-(A.9).
A.2. MATRIX ANALYSIS
209
A.12 Prove or disprove that a continuous function f : X → Y, where X
and Y are real normed linear spaces, satisfying f (x1 + x2 ) = f (x1 ) +
f (x2 ), ∀x1 , x2 ∈ X is a linear transformation.
A.13 Prove the Pythagorean Theorem: If x, y ∈ Rn and x ⊥ y, then ∥x∥22 +
∥y∥22 = ∥x + y∥22 .
A.14 Prove that a norm in a linear space satisfying the parallelogram law
∥x + y∥2 + ∥x − y∥2 = 2∥x∥2 + 2∥y∥2
is induced by an inner product. (Hint: the proof for a linear space over
C is slightly different from that for a linear space over R. Let us prove it
for both cases.)
A.15 Assume compatibility and existence of all involved inverses.
1. Show all identities in Theorem A.3.
2. Show that
(A − BD−1 C)−1 = A−1 + A−1 B(D − CA−1 B)−1 CA−1 .
A.16 Consider the star product in F2n×2n :
A11 A12
B11 B12
A⋆B =
⋆
A21 A22
B21 B22
Fl (A, B11 )
A12 (I − B11 A22 )−1 B12
=
.
B21 (I − A22 B11 )−1 A21
Fu (B, A22 )
1. The star identity is a matrix J such that
J ⋆A=A⋆J =A
for all A. Find J.
2. The star inverse of A is a matrix A⊖ , if exists, such that
A ⋆ A⊖ = A⊖ ⋆ A = J.
Find A⊖ .
A11 A12
A.17 Let A =
and X be unitary constant matrices with I −A22 X
A21 A22
invertible. Show that Fl (A, X) is also a unitary matrix.
A.18 A matrix A is said to be normal if A∗ A = AA∗ .
1. Show that Hermitian matrices, skew-Hermitian matrices, and unitary matrices are normal.
2. Show that a normal matrix is diagonalizable by a unitary similarity
transformation (using the Schur theorem).
210
APPENDIX A. MATHEMATICAL PRELIMINARIES
3. If an n × n normal matrix A has eigenvalues λ1 , . . . , λn , what are its
singular values?
A.19 Prove the following statements:
1. The eigenvalues of a nilpotent matrix are all zero.
2. The eigenvalues of a unitary matrix have unit absolute values.
3. The eigenvalues of a Hermitian matrix are real.
4. The eigenvalues of a skew-Hermitian matrix are imaginary.
A.20 A matrix A ∈ Fn×n is said to be idempotent if A2 = A.
1. Show that an idempotent matrix is diagonalizable.
2. Show that if A ∈ Fn×n is idempotent, then N (A) ⊕ R(A) = Fn .
3. What are the possible eigenvalues of an idempotent matrix?
A.21 Two matrices A, B ∈ C n×n are said to be simultaneously diagonalizable
if there exists P ∈ GL(n, C) such that P −1 AP and P −1 BP are diagonal.
They are said to commute if AB = BA.
1. Show that if A, B are simultaneously diagonalizable, then they commute.
2. Show that if A, B are diagonalizable and commute, then they are
simultaneously diagonalizable. (Hint: One may first prove for the
case when all eigenvalues of A are distinct and then extend to the
general case.)
A.22 Using the Schur decomposition theorem to show the following:
1. For A ∈ Fn×m , there exist unitary matrices U, V , and a positive
diagonal matrix S such that
A = U SV ∗ .
(This decomposition is called the singular value decomposition (SVD).)
2. For each A ∈ Fn×n , there exist a positive semidefinite matrix P and
a unitary matrix U such that
A = P U.
(This decomposition is called the polar decomposition.)
3. A positive definite matrix A can be factorized as A = T ∗ T where T
is upper triangular. (This factorization is called Cholesky factorization.)
4. For a 2n × 2n unitary matrix W , there exist n × n unitary matrices
U1 , U2 , V1 , V2 such that
∗
U1 0
C −S
V1
0
W =
0 U2
S C
0 V2∗
A.2. MATRIX ANALYSIS
211
where
cos θ1
C=
..
.
,
S=
sin θ1
cos θn
..
.
sin θn
for some θi ∈ [0, π2 ], i = 1, . . . , n. (This decomposition is called CS
decomposition.)
A.23 Let A be a 2n×2n reversed diagonal matrix with reversed diagonal entries
d1 , . . . , d2n ∈ R ordered from top-right to bottom-left:
d1
.
.
..
A=
d2n
What are the eigenvalues and singular values of A?
A.24 For A ∈ Cm×n with singular values σ1 (A), σ2 (A), . . . , σmin{m,n} (A), what
are the eigenvalues of
0 A
?
A∗ 0
A.25 Let A ∈ Cn×n .
1. Show that there exist diagonalizable matrix D and nilpotent matrix
N such that A = D + N .
2. Show that the eigenspace of A corresponding to eigenvalue λi is equal
to N (D − λi I).
3. Show that N (A − λi I) = N (D − λi I) ∩ N (N ).
A.26 Write a MATLAB program to implement the LSFF algorithm. The input
of the program should be the matrix A and the output should be a vector
containing the coefficients of cA and a three dimensional array containing
the coefficients of Adj(zI − A). Use your program to compute cA and
Adj(zI − A) for
−4 −6 −4 −1
1
0
0
0
.
A=
0
1
0
0
0
0
1
0
A.27 Familiarize yourself with MATLAB commands qr, schur, svd, polar, and
chol.
A.28 Show the following:
1. Let A ∈ Rn×m and B ∈ Rn×p . Then there exists C ∈ Rp×m such
that BC = A if and only if R(A) ⊂ R(B).
2. Let A ∈ Rn×m and B ∈ Rp×m . Then there exists C ∈ Rn×p such
that CB = A if and only if N (B) ⊂ N (A).
212
APPENDIX A. MATHEMATICAL PRELIMINARIES
A.29 Show that the set of complex Hermitian matrices Hn is a not a linear space
over C, but rather a linear space over R. What is its dimension? Find
an orthonormal basis under the standard matrix inner product ⟨A, B⟩ =
trA∗ B.
A.30 Prove Theorem A.13.
A.31 Use Theorem A.13 to prove Theorem A.11.
A.32 Show the following
1. det(I + AB) = det(I + BA);
2. trAB = trBA.
A.33 Consider linear equation
Ax = b
where A ∈ Fm×n , b ∈ Fm , and x ∈ Fn . If A is square and nonsingular, we
can get the unique solution of the equation easily as x = A−1 b. However
we are often interested in the case when A is rectangular or singular.
1. Assume rankA = m. Then there may be infinity many solutions.
Show that the minimum norm solution, i.e., the solution with the
smallest 2-norm, is A∗ (AA∗ )−1 b.
2. Assume rankA = n. Then there may not be solutions. Show that the
so-called least square solution, i.e., the x which minimizes ∥Ax − b∥2 ,
is (A∗ A)−1 A∗ .
3. In general, we are interested in the minimum norm least square solution, which is the one with the smallest 2-norm among all x that
minimizes ∥Ax − b∥2 . Find the minimum norm least square solution
of the linear equation.
4. For linear equation
1 2
1
x=
,
3 6
2
find its minimum norm least square solution.
A.34 Let the singular values of A ∈ Cm×n be σ1 (A), σ2 (A), . . . , σmin{m,n} (A).
Prove that
v
umin{m,n}
u X
σi2 (A).
∥A∥F = t
i=1
A.35 Show that for each A ∈ Pn , there exists a unique B ∈ Pn such that
1/2
A = B 2 . This
√ B
is called the square root of A, denoted by A . For
2
2
A= √
, compute A1/2 .
2 3
A.36 Show that for A, B ∈ Hn (F), with A > 0, there exists P ∈ GL(n, F) such
that P ∗ AP and P ∗ BP are simultaneously diagonal.
A.2. MATRIX ANALYSIS
213
A.37 Prove the following statements about matrices A, B ∈ Hn (F):
1. If either A > 0 or B > 0, then the eigenvalues of AB are real.
2. If both A > 0 and B > 0, then the eigenvalues of AB are positive.
A.38 The linear space Rn×n has an inner product ⟨A, B⟩ = trA′ B and has two
subspaces:
Sn = {A ∈ Rn×n : A′ = A}
Tn = {A ∈ Rn×n : A′ = −A}
which are called symmetric matrix space and skew-symmetric matrix
space.
1. Show that Sn ⊥ Tn and Sn ⊕ Tn = Rn×n .
2. If L is a linear transformation on Rn×n mapping X to A′ XA − X.
Show that Sn and Tn are invariant subspaces of L.
A.39 The spectral radius of A ∈ Fn×n , denoted by ρ(A), is the maximum
modulus of all its eigenvalues. Show that ∥A∥p ≥ ρ(A), where ∥ · ∥p is
the induced matrix 2-norm. Also show that for each ϵ > 0, there exists a
P ∈ GL(n, F) such that ∥P −1 AP ∥2 < ρ(A) + ϵ.
