ece1638 linear systems and control - university of...
TRANSCRIPT
ECE1638 Linear Systems and Control
Coursenotes by Bruce Francis
Spring Term, 2006
Contents
Chapter 1 Function Spaces in the Time Domain
Chapter 2 Basic System Concepts
Chapter 3 Function Spaces in the Frequency Domain
Chapter 4 Time-Invariant Systems
Chapter 5 The Continuous-Time Setup
Chapter 6 Control Systems: Introduction
Chapter 7 Stability of Feedback Systems
Chapter 8 H2-Optimal Control
These notes are intended to be more-or-less self-contained for the mythical “mathemat-ically mature engineering student.” You need to know linear algebra and linear systems atthe undergraduate level. Below are some reference texts. Most are quite advanced, but youmay be able to mine them for some useful nuggets.
Chapter 1
[1] P. R. Halmos, Finite-Dimensional Vector Spaces — the very best, beautifully writtentext on linear algebra
[2] P. R. Halmos, Introduction to Hilbert Space — for beginners
[3] J. Conway, A Course in Functional Analysis — a much deeper treatment, certainlymuch more than you need for this course, but useful as a reference
Chapter 2
[4] A. Feintuch and R. Saeks, System Theory: A Hilbert Space Approach — a real messbut it has some useful stuff
Chapter 3
[5] P. R. Halmos, A Hilbert Space Problem Book — beautiful, but quite advanced
[6] P. Duren, Theory of Hp Spaces — a standard reference on Hardy spaces
[7] M. Rosenblum and J. Rovnyak, Hardy Classes and Operator Theory
Chapter 4
[8] S. Power, Hankel Operators on Hilbert Space
Chapters 5, 6, and 7 are taken from
[9] B. Francis, A Course in H∞ Control Theory
Chapter 8
[10] M. Vidyasagar, Control System Synthesis: A Factorization Approach — also agood reference for Chapter 7
The state-space material in Chapter 8 is taken from the paper
[11] J. Doyle, K. Glover, P. Khargonekar, B. Francis, “State-space solutions to standardH2 and H∞ control problems,” IEEE Trans. Auto. Control, 1989, pp. 831-847.
History
To see where this course fits into the general scheme of things, here’s a brief history oflinear control system synthesis for dynamical systems.
1930’s: Bode, Black, Nyquist at Bell Labs. Design of feedback amplifiers using frequency-domain methods. See Bode’s book.
1940’s: Military applications at MIT. Wiener and optimal filtering (see Wiener’s book, TimeSeries). Phillips applied Wiener’s method to design H2 optimal controllers with realparameters. See Nichols, James, Phillips.
1950’s: Book by Newton, Gould, Kaiser; Q-parametrization of stabilizing controllers; H2 op-timal control for stable siso plants. The Soviets developed optimal control theory fordifferential equation models. Pontryagin’s maximum principle. Minimum-time prob-lems; minimum-fuel problems. Introduction of state-space models. Kalman filter.
1960’s: Heyday of state-space theory. Controllability and observability: Kalman, Gilbert, etal. Observers: Luenberger. Pole assignment theorem: Wonham. Decoupling: Falb-Wolovich, Wonham-Morse. Geometric linear systems theory. LQG: Athans et al.
1970’s: More state-space. Regulator theory: Pearson, Wonham, Davison. Algebraic meth-ods in the s-domain: Rosenbrock, Wolovich. Generalized Nyquist plots: MacFarlane.Design by diagonal dominance: Rosenbrock. Design by parametrization and math pro-gramming: Polak. Introduction of operator-theoretic methods into systems; broadbandmatching: Helton. Optimal gain margin via Nevanlinna-Pick interpolation: Tannen-baum. Sensitivity minimization: Zames. Lack of robustness of LQG: Doyle. Adaptivecontrol. Controller parametrization: Youla et al.
1980’s: Special issue on multivariable control. Robustness issue. Classical-modern blend:Doyle, Stein; singular values, robust stability, LTR. H∞ optimization: Zames andlots more. H2/H∞ unification via state-space methods: ONR/Honeywell workshop. µ:Doyle. Simplification of computations: Doyle, Glover.
1990’s: Numerical methods to improve the design procedures, LMI.
What is this course about?
Linear systems can be modeled, in the continuous-time domain, via either differentialequations, especially state equations, or input-output equations. Both viewpoints have some-thing to offer. I would advocate that certain system concepts, such as time-invariance, causal-ity, and certainly input-output stability, are more transparent in the input-output model.On the other hand, computations can be readily done using state model representations andMATLAB.
This course develops linear system theory in the input-output domain. The mathematicalframework is operator theory in both the time domain and frequency domain. We begin indiscrete time because it’s easier.
Here’s an example development in the course to give you the flavour. Consider a discrete-time system with input u(k) and output y(k). Assume the system is linear. To talk aboutbounded input, bounded output (BIBO) stability we need a norm for inputs and outputs. InRn, all norms are equivalent, but not in signal space, which is infinite dimensional. The inputnorm ‖u‖ can be based on energy, peak value, or something else. Once we settle on a normfor u and y (actually, they don’t have to have the same norm), then BIBO stability means ‖y‖is finite whenever ‖u‖ is finite and also ‖y‖ ≤ B‖u‖ for some constant B. This is equivalentto saying that the mapping from u to y is a bounded operator. If this operator is time-invariant, then there’s a transfer function, and BIBO stability translates to some property ofthe transfer function. We haven’t yet assumed that the system is finite dimensional, so thetransfer function could be irrational. If it is rational, so there’s a state model, then BIBOstability is independent of the norm adopted and it can be checked in the familiar way bypole locations or eigenvalue locations.
Why might you take this course? Because you like mathematical system theory. To learninput-output system theory as opposed to state-space theory. To learn some deep resultssuch as the parametrization of all stabilizing feedback controllers.
On the other hand you should realize that, although the course was research-orientedwhen it was first developed in 1988, now it represents a mature subject.
Chapter 1
Function Spaces in the Time Domain
The goal of this brief, elementary chapter is to introduce Banach and Hilbert spaces and theparticular example h2 of square-summable discrete-time signals.
Some motivation
We motivate the subject by some examples about system gain.Let a ∈ R and define the mapping
F : R −→ R, Fx = ax
Think of F as a “system”: input x, output y = ax. Graph F . Define
‖F‖ = max{|Fx| : |x| ≤ 1}
It’s clear that ‖F‖ = |a|.Again let a ∈ R but define the “system”
F : Rn −→ Rn, Fx = ax
Take the Euclidean norm on Rn. Then define
‖F‖ = max{‖Fx‖ : ‖x‖ ≤ 1}
What is ‖F‖ in terms of a?What about this harder problem: Let A ∈ Rn×n and define the “system”
F : Rn −→ Rn, Fx = Ax
Take the Euclidean norm on Rn. Then define the norm of F :
‖F‖ = max{‖Fx‖ : ‖x‖ ≤ 1}
What is ‖F‖ in terms of A?
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What about this even harder problem: Let
G(s) =10s− 1
s2 + s + 10
and define the system F that maps an input u(t) to an output y(t) where y(s) = G(s)u(s).Take the norm on ‖u‖ = supt |u(t)|. Then define the norm of F :
‖F‖ = max{‖Fu‖ : ‖u‖ ≤ 1}What is ‖F‖ in terms of G(s)?
These kinds of questions motivate us to study function spaces, norms, operators, etc.
1.1 Vector space
You are assumed to be familiar with the notion of a vector space over a field, for us eitherthe real field R or the complex field C. Familiar examples of vector spaces are
Rn, Cn, Rn×m, Cn×m
Perhaps a less familiar example is s(Cn), the space of sequences written like this:
x =
x0
x1...
, xi ∈ Cn
It’s convenient to write x as an infinite column vector, as shown, so that later we can multiplyit by a matrix. Sometimes we’ll write x = (xk). You can think of x as a multivariate discrete-time signal, with k as the time variable. If Cn is understood or irrelevant, we may abbreviates(Cn) by s.
Let X be a vector space. A subspace V of X is a subset that is closed under addition andscalar multiplication. Thus V is a vector space too. For example,
{x ∈ s : x0 = 0}is a subspace of s—the space of signals that start at 0.
For two subspaces V and W define
V +W := {v + w : v ∈ V , w ∈ W}
V ∩W := {x : x ∈ V , x ∈ W}
It’s easy to check that this sum and intersection are subspaces.Suppose X and Y are vector spaces. There’s a natural way to combine them to get a
third vector space, called their external direct sum, denoted X ⊕ Y . As a set, X ⊕ Y is theCartesian product that consists of all ordered pairs, written[
xy
]Addition and scalar-multiplication are performed componentwise.
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1.2 Normed and inner-product spaces
Let X be a complex vector space. A norm on X is a function x 7→ ‖x‖ from X to R havingthe four properties
1. ‖x‖ ≥ 0
2. ‖x‖ = 0 iff x = 0
3. ‖cx‖ = |c|‖x‖, c ∈ C
4. ‖x + y‖ ≤ ‖x‖+ ‖y‖
Then the pair (X , ‖ ‖), or just X , is called a (complex) normed space. Similarly for realnormed space.
An inner-product 〈, 〉 on X is a function from X × X to C having the four properties
1. 〈x, x〉 is real and ≥ 0
2. 〈x, x〉 = 0 iff x = 0
3. for each x the function y 7→ 〈x, y〉 from X to C is linear
4. 〈y, x〉 = 〈x, y〉
Then the pair (X , 〈, 〉), or just X , is called a (complex) inner-product space. Similarly forreal inner-product space. Such an inner-product induces a norm, namely ‖x‖ := 〈x, x〉1/2.Examples of inner-product spaces are
the space Cn with 〈x, y〉 := x∗y :=∑
xiyi
the space Cn×m with 〈A, B〉 := traceA∗B
If X and Y are two inner-product spaces, their external direct sum has a natural inner-product, namely⟨[
x1
y1
],
[x2
y2
]⟩:= 〈x1, x2〉+ 〈y1, y2〉
Two vectors x and y in an inner-product space are orthogonal, symbolized x⊥y, if
〈x, y〉 = 0
The following two statements hold in an inner-product space:
(Pythagorean theorem) if x⊥y, then ‖x + y‖2 = ‖x‖2 + ‖y‖2
(Cauchy-Schwarz inequality) |〈x, y〉| ≤ ‖x‖‖y‖
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An important example for us is the space h2(Cn), or just h2 when Cn is understood. It’sthe subspace of s(Cn) of sequences x = (xk) that are square-summable, i.e.,
∞∑k=0
‖xk‖2 < ∞
Here ‖xk‖ is the Cn-norm of xk. The inner-product on h2 is defined to be
〈x, y〉 :=∞∑0
x∗kyk
We should verify that the series∑∞
0 x∗kyk does indeed converge when both∑‖xk‖2 and∑
‖yk‖2 converge. It suffices to prove absolute convergence of∑
x∗kyk. (Absolute convergence
in C implies convergence.) Set ck :=∑k
0 |x∗i yi|. Then ck is an increasing real sequence, so itsuffices to show it’s bounded. But
ck = |x∗0y0|+ · · ·+ |x∗kyk|
≤ ‖x0‖‖y0‖+ · · ·+ ‖xk‖‖yk‖ by the Cauchy-Schwarz inequality in Cn
≤ (k∑0
‖xi‖2)1/2(k∑0
‖yi‖2)1/2 by the Cauchy-Schwarz inequality in Rk+1
≤ (∞∑0
‖xi‖2)1/2(∞∑0
‖yi‖2)1/2
The last quantity is a bound for ck.The norm on h2 is denoted ‖x‖2, i.e.,
‖x‖2 := (∞∑0
‖xk‖2)1/2
Now let X be a general normed space. A sequence xk in X converges to x in X if thesequence ‖x − xk‖ of real numbers converges to zero. In this case xk is convergent and x isits limit. A subspace V of X is closed if every sequence in V that converges in X actuallyhas its limit in V , i.e., V contains the limits of all its convergent sequences.
For example, every subspace of Cn is closed. To see this, let V be a subspace andlet {v1, . . . , vm} be a basis for it. We may as well suppose these vectors are orthonormal;otherwise, orthonormalize them via Gram-Schmidt. Now extend to get a basis
{v1, . . . , vm, vm+1, . . . , vn}
for all of Cn. Again, we can suppose this basis is orthonormal. Now let {xk} be a sequencein V converging to some x in Cn. We must show that x ∈ V . Being in V , each xk has theform
xk = a1kv1 + · · ·+ amkvm
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Similarly,
x = a1v1 + · · ·+ amvm + · · ·+ anvn
By orthonormality
‖xk − x‖2 = |a1k − a1|2 + · · ·+ |amk − am|2 + |am+1|2 + · · ·+ |an|2
This converges to zero as k →∞. Thus
am+1 = · · · = an = 0
i.e., x ∈ V .Two subspaces V , W of an inner-product space X are orthogonal, written V⊥W , if v⊥w
for every v in V and w in W . Then we write their sum as V⊕W . Note that their intersectionconsists solely of the zero vector. The orthogonal complement of V is
V⊥ := {x ∈ X : x⊥v ∀ v ∈ V}
If V is a closed subspace, then X = V ⊕ V⊥ (proof omitted). For a simple example, take
V := {x ∈ h2 : x0 = 0}
Then
V⊥ = {x ∈ h2 : xk = 0 ∀ k > 0}
1.3 Banach and Hilbert spaces
Let X be a normed space. A sequence xk in X is a Cauchy sequence if
(∀ε > 0)(∃N)k, l > N ⇒ ‖xk − xl‖ < ε
We say X is complete if every Cauchy sequence in X converges in X . A complete normedspace is called a Banach space and a complete inner-product space is called a Hilbert space.Examples of Hilbert spaces are Rn, Cn, h2, and the external direct sum of two Hilbert spaces.
1.4 Exercises
1. Let V := {x ∈ h2(Cn) : x0 = 0}. Prove that V is closed. Give an example of a subspaceof h2(Cn) that isn’t closed.
2. Let X be an inner-product space, V a subspace, not necessarily closed. Prove that V⊥ isa closed subspace.
3. Prove that every closed subspace of a Banach space is complete.
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Chapter 2
Basic System Concepts
In this chapter we look at four basic concepts: linearity, causality, time-invariance, andstability.
2.1 Linearity
Let U and Y be vector spaces. A function F from U to Y is a linear transformation if
F (a1u1 + a2u2) = a1Fu1 + a2Fu2, ai ∈ C, ui ∈ U
Recall that every linear transformation F : Cn → Cm has a matrix representation [F ] withrespect to the standard basis: Let {ei} denote the standard basis for Cn; define fj := Fej, avector in Cm. Then the jth column of [F ] is by definition fj. It is then routine to prove thatfor every x in Cn
Fx = [F ]x
The left-hand side is the linear transformation F applied to the vector x, while the right-handside is the matrix [F ] multiplying the n-tuple x.
The formal definition of a linear system is this: a complex linear system is a triple (F,U ,Y),where U , the input space, is s(Cm), Y , the output space, is s(Cp), and F : U → Y is a lineartransformation. Real linear system is defined with the obvious change. The picture goingalong with this definition is
F- -
uk yk
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Thus uk and yk are the input and output at time k. They are vector-valued, i.e., the systemis multi-input and multi-output. For the input sequence u the output sequence is defined tobe y := Fu. We shall usually refer to F alone as a linear system.
Let’s formally define matrix representation for a linear system F . We shall do it forthe single-input, single-output case, i.e., m = p = 1, the generalization being routine. Thestandard basis vectors for s are defined to be e0, e1, . . .: the ith component of ei is 1, all theothers are 0. Then [F ] is defined by saying that its jth column is Fej.
Example 1 Moving average model.Let the input-output relation be the equation
yk =∞∑0
Fklul, k ≥ 0 (2.1)
where Fkl ∈ Cp×m. To obviate the need for a discussion of convergence, it is assumed that foreach k only finitely many of Fk0, Fk1, . . . are nonzero; that is, the sum in (2.1) has only finitelymany nonzero terms. Thus in (2.1) the output at each time is a finite linear combination ofthe inputs at all times. Evidently, the matrix [F ] in this case is F00 F01 · · ·
F10 F11 · · ·...
...
An example is the (backward) shift S: the defining equations are
y0 = 0
yk = uk−1, k ≥ 1
Thus S acts as a delay of one time unit. The matrix [S] has identity blocks all along the firstsub-diagonal:
0 0 · · ·I 0 · · ·0 I · · ·...
...
Another example is the truncation projection Pk, defined for k ≥ 0 by
yi =
{ui, i ≤ k0, i > k
The corresponding matrix has k + 1 identity blocks:
block diag (I, . . . , I, 0, . . .)
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Example 2 ARMA model.Integration by the trapezoidal rule leads to a linear system modelled by the difference
equation
y0 = 0
yk = yk−1 +1
2(uk + uk−1), k ≥ 1
Here the output is a linear combination of past outputs and the present and past inputs.The matrix is
0 0 0 · · ·1/2 1/2 0 · · ·1/2 1 1/2 · · ·1/2 1 1 · · ·...
......
Example 3 State-space model.
The input-to-output mapping is defined via
xk+1 = Akxk + Bkuk, x0 = 0
yk = Ckxk + Dkuk
Here Ak, Bk, Ck, and Dk are complex matrices.
From the definition of [F ] it follows that
Fx = [F ]x (2.2)
for every standard basis vector x of s. Again, the left-hand side is the linear transformationF applied to x, while the right-hand side is the matrix [F ] multiplying the vector x. It’s notalways true that (2.2) holds for every x in s, but for the remainder of this chapter we assumethat it does.
Finally, we want to be able to connect compatible linear systems in parallel and series.The sum of (F,U ,Y) and (G,U ,Y) is (F + G,U ,Y):
F
G-
-
j?6
-
8
Note that
[F + G] = [F ] + [G]
The composition of (F,U ,Y) and (G,Y ,W) is (GF,U ,W):
F G- - -
Note that
[GF ] = [G][F ]
2.2 Causality
The idea of causality is that the output at time k depends only on inputs up to time k; inother words, if two inputs are equal up to time k, then the two corresponding outputs shouldbe equal up to time k. The latter description leads to the formal definition, the linear systemF is causal if
(∀k ≥ 0)(∀u, u ∈ U)Pku = Pku ⇒ PkFu = PkFu
Introduce the notation
AV := {Av : v ∈ V}
for a function A and set V in A’s domain.
Theorem 1 For a linear system F , the following conditions are equivalent:
(i) F is causal
(ii) PkF (I − Pk) = 0
(iii) [F ] is lower (block) triangular
The linear transformation I −Pk annihilates the first k + 1 components of a signal, so wecan think of the subspace (I−Pk)U as the future input space starting at time k+1; similarlyfor (I − Pk)Y . Thus (ii) says that “F leaves the future invariant.” Notice that if F is causaland (Fij) is its matrix, then the input-output equation y = Fu is equivalent to
yk =k∑
l=0
Fklul
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Proof of Theorem 1 Assume (i). To prove that (ii) holds start with the definition ofcausality and use linearity of F and Pk to get
(∀k ≥ 0)(∀u ∈ U)Pku = 0 ⇒ PkFu = 0 (2.3)
But Pku = 0 iff u ∈ (I − Pk)U and PkFu = 0 iff Fu ∈ (I − Pk)Y . Thus (2.3) implies (ii).That (ii) implies (i) follows by reversing this argument.
Now assume (ii). To prove that (iii) holds write the matrix of F as
[F ] =
F00 F01 · · ·F10 F11 · · ·...
...
Condition (ii) for k = 0 is
F (I − P0)U ⊂ (I − P0)Y (2.4)
The matrix of F (I − P0) is F00 F01 · · ·F10 F11 · · ·...
...
0 0 · · ·
0 I · · ·...
...
=
0 F01 · · ·0 F11 · · ·...
...
(2.5)
Now (2.4) implies that the first row in (2.5) equals zero, i.e., the upper right corner[F01 F02 · · ·
]in the matrix [F ] equals zero. Similarly, the other conditions in (ii) imply that the otherupper right corners in [F ] equal zero. Again, the implication (iii) ⇒ (ii) follows by thereverse argument. �
Examples of causal systems are S, Pk, and the state-space model.
2.3 Time-invariance
The previous section introduced the backward shift S. A related linear system is the forwardshift, denoted S∗ for reasons soon to become apparent. The defining equation is
yk = uk+1, k ≥ 0
It’s easy to verify that [S∗] equals the transpose of [S] (so S∗ is not causal), S∗S = I (S∗ isa left inverse of S), and SS∗ = I − P0.
Let F be a linear system. The idea of time-invariance is this: if an input {u0, u1, · · ·}produces the output {y0, y1, · · ·}, then the input {0, u0, u1, · · ·} produces an output of theform {w, y0, y1, · · ·} for some w. Note that w will equal zero if the system is causal. So
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roughly speaking, time-invariance means that shifting the input shifts the output. Theformal definition is that F is time-invariant if S∗FS = F .
A matrix A = (Aij), where Aij may be complex blocks, is Toeplitz if it’s constant alongdiagonals, i.e.,
Ai+k,j+k = Aij
Theorem 2 A linear system F is time-invariant iff [F ] is Toeplitz.
Proof The result follows immediately upon noting that
[S∗][F ][S] = [F ] ⇔ [F ] is Toeplitz �
When F is time-invariant we can define Fi = Fi0, and then y = Fu is equivalent to theconvolution equation
yi =∞∑
j=0
Fi−juj
Moreover
[F ] =
F0 F−1 F−2 · · ·F1 F0 F−1 · · ·F2 F1 F0 · · ·...
......
Examples of time-invariant linear systems are S, S∗, and the state-space model with Ak,
Bk, Ck, Dk all constant matrices.
2.4 Stability
Let X and Y be normed spaces and let F : X → Y be a linear transformation. Then F isbounded if
(∃c ∈ R)(∀x ∈ X )‖Fx‖ ≤ c‖x‖
The least such constant c is called the norm of F and is denoted ‖F‖. A bounded lineartransformation is called an operator. Alternative expressions for the norm are as follows:
‖F‖ = inf{c : (∀x)‖Fx‖ ≤ c‖x‖}
= inf{c : (∀x 6= 0)‖Fx‖‖x‖
≤ c}
= supx6=0
‖Fx‖‖x‖
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= sup‖x‖=1
‖Fx‖
= sup‖x‖≤1
‖Fx‖
If the dimension of X is finite, then every linear transformation X → Y is bounded. LetF : Cn → Cm be linear and let [F ] denote its matrix with respect to the standard basis. Thesingular values of [F ] are the square roots of the eigenvalues of the Hermitian matrix [F ]∗[F ].It is a fact that
‖F‖ = maximum singular value of [F ]
In infinite dimensions not all linear transformations are bounded.A linear system F is h2-stable if F is an operator h2 → h2. The motivation for this defi-
nition is as follows. For a discrete-time signal u = (uk), we interpret ‖uk‖2 as instantaneouspower and ‖u‖2
2 =∑‖uk‖2 as (total) energy. Thus h2-stability means there exists a constant
c such that
(output energy) ≤ c× (input energy)
It’s convenient to introduce adjoint operators at this point.
Theorem 3 Let X and Y be Hilbert spaces and F : X → Y an operator. There exists aunique operator F ∗ : Y → X satisfying the equation
< Fx, y >=< x, F ∗y >, x ∈ X , y ∈ Y
The proof of this standard result in functional analysis is omitted. The operator F ∗ iscalled the adjoint of F . The adjoint of the shift S is the operator we’ve been denoting S∗.
2.5 Exercises
1. Write down [F ] for the state-space model in Example 3 in Section 1.
2. A linear system is strictly causal if the output is initially zero and the output at time kdepends only on inputs up to time k − 1. Give a formal definition of strict causality, andstate and prove the analogous theorem.
3. Consider the state-space model where the four matrices Ak, Bk, Ck, Dk are all periodic,of period N . Then it’s not true that S∗FS = F , but what is true?
4. Suppose F is a linear system. Show that F is causal and time-invariant iff it commuteswith S.
5. A linear system F is memoryless if its matrix is (block) diagonal; thus the output at timek depends only on the input at time k. Show that F is memoryless and time-invariant iff itcommutes with both S and S∗.
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6. For this exercise we shall redefine some earlier notation. Consider sequences defined forall time, −∞ < k < ∞. Let s denote the vector space of such sequences taking values in Cn.Define the backward and forward shifts S and S∗ on s in the obvious way. Finally, let F bea linear transformation on s. Show that the following three conditions are equivalent:
S∗FS = F
F commutes with S
F commutes with both S and S∗
7. Let F : Cn → Cm be linear. Prove that
‖F‖ = maximum singular value of [F ]
8. Suppose F is an operator from h2 to h2. Show that the matrix of F ∗ is the complex-conjugate transpose of [F ].
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Chapter 3
Function Spaces in the FrequencyDomain
In this chapter we study spaces of complex-valued functions. The treatment is descriptive andmostly without proofs. Sections 1 and 2 treat scalar-valued functions; the matrix extensionsare presented in Section 3. Section 4 summarizes the basic transform method relating thetime and frequency domains.
3.1 Some Hilbert spaces
The Lebesgue space L2 consists of all complex-valued functions defined and square-integrableon the unit circle; that is, a complex-valued function f(ejθ), defined for almost all θ in [0, 2π),is in L2 iff∫ 2π
0
|f(ejθ)|2dθ < ∞
It’s a fact that L2 is a Hilbert space under the inner-product
< f, g >:=1
2π
∫ 2π
0
f(ejθ)g(ejθ)dθ
An important example is that of real-rational functions in L2. Let R[λ] and R(λ) denote,respectively, the ring of polynomials and the field of rational functions in λ with real coef-ficients. The intersection of L2 and R(λ) is denoted RL2. Here and in the sequel, a prefixR signifies “real-rational.” Concretely, RL2 consists precisely of the real-rational functionshaving no poles on the unit circle.
The generic complex variable is taken to be1 λ. The Hardy space H2 consists of allcomplex-valued functions f(λ) defined and analytic on the open unit disc and having theuniform square-integrability property
sup0≤r<1
∫ 2π
0
|f(rejθ)|2dθ < ∞
1Normally it would be z, but we can’t use z for a reason to be stated later.
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Thus, there is a uniform bound for the integral of |f |2 around every circle inside the openunit disc.
An important fact about H2 is the existence of radial limits. For f in H2 the limit
f(ejθ) := limr→1
f(rejθ)
exists for almost all θ in [0, 2π) and the boundary function f(ejθ) belongs to L2. By identifyingf and its boundary function we can regard H2 as a subspace of L2; as such, it is closed.
The space RH2 consists precisely of the real-rational functions that are analytic in theclosed unit disc. To see this, let f(λ) ∈ R(λ); suppose for simplicity f has only simple poles.Then f is a linear combination of terms of the form
f1(λ) = λm, m ≥ 0
f2(λ) =1
λ
f3(λ) =1
1− aλ, |a| < 1
f4(λ) =1
1− aλ, |a| ≥ 1
It is routine to check using the definition of H2 that f1, f3 ∈ RH2, while f2, f4 6∈ RH2.Cauchy’s formula holds for H2: if f ∈ H2 and |a| < 1, then
f(a) =1
2πj
∮f(λ)
λ− adλ
We saw above that an H2-function has a boundary value in L2, and this function has amodulus. Sometimes it is useful to be able to go the other way—to start with a positivefunction on the unit circle and make it the modulus of the boundary value of some H2-function. This is possible if the positive function satisfies a certain condition. Let φ be apositive L2-function such that log φ is absolutely integrable on the unit circle. For λ in theunit disc define
f(λ) := exp
{1
2π
∫ 2π
0
ejθ + λ
ejθ − λlog φ(ejθ)dθ
}Then f ∈ H2 and |f(ejθ)| = φ(ejθ) for almost all θ. (Actually, this construction gives f astronger property: It is outer, meaning that fH2 is dense in H2.)
Being a closed subspace of L2, H2 has an orthogonal complement, H⊥2 . The properties of
functions in H⊥2 can be derived from those in H2 using the fact that
f(λ) ∈ H⊥2 ⇔
1
λf
(1
λ
)∈ H2
Suppose X is a Hilbert space and V a closed subspace. Then X = V⊥⊕V , so for every xin X there exist unique u in V⊥ and v in V such that x = u + v. Thus the operator ΠV fromX to X taking x to v is well-defined. It’s called the orthogonal projection of X onto V andhas these properties:
15
Im ΠV = V
Ker ΠV = V⊥
Π2V = ΠV
Π∗V = ΠV
‖ΠV‖ = 1 (unless V = 0)
For convenience, from now on we shall make the abbreviations
Π1 := ΠH⊥2
Π2 := ΠH2
As an example, consider the rational function
f(λ) =λ3 + λ
−2λ2 − 5λ + 3
Having no poles on the unit circle, f belongs to RL2. The projections f1 := Π1f andf2 := Π2f are uniquely determined by the properties
f = f1 + f2
f1 is strictly proper and analytic in |λ| ≥ 1
f2 is analytic in |λ| < 1
To get these functions, first write f as the sum of a polynomial and a strictly proper rationalfunction. Then do a partial-fraction expansion of the latter. The polynomial goes into f2.There results
f1(λ) =−0.0893
λ− .5
f2(λ) = −0.5λ + 1.25− 4.2857
λ + 3
3.2 Some Banach spaces
Elements of L∞ are functions mapping the unit circle into C whose moduli are essentiallybounded. Thus, for a function f in L∞ the modulus |f(ejθ)| is uniformly bounded for almostall θ from 0 to 2π. The least such bound is the L∞-norm of f :
‖f‖∞ := supθ|f(ejθ)|
For example, RL∞ consists of the real-rational functions having no poles on the unit circle;thus RL∞ = RL2.
16
It is a fact that L∞ is a Banach space. Another useful fact is
R[λ] ⊂ L∞ ⊂ L2
Both of this inclusions are strict. The second inclusion is a consequence of the inequality
‖f‖2 ≤ ‖f‖∞, f ∈ L∞
The Hardy space H∞ consists of complex-valued functions that are analytic and ofbounded modulus on the open unit disc. Just as H2-functions have radial limits in L2,so functions in H∞ have radial limits and the limiting function belongs to L∞. In this way,H∞ is a subspace of L∞, in fact a closed subspace and hence a Banach space on its own. Itis quite easy to check that RH∞ = RH2. The inclusions analogous to those above are
R[λ] ⊂ H∞ ⊂ H2
and again both are strict.
3.3 Matrix spaces
All the frequency-domain spaces in the previous two sections consisted of complex-valued,i.e., scalar-valued, functions. When we want to express this fact we shall write L2(C) forL2, and similarly for the others. In this section we extend these spaces to matrix-valuedfunctions.
First the space L2(Cn×m). Its elements are n×m complex matrix-valued functions eachentry of which is a scalar-valued L2-function. The inner-product on L2(Cn×m) is
< F, G >:=1
2π
∫ 2π
0
trace F (ejθ)∗G(ejθ)dθ
Here superscript ∗ means complex-conjugate transpose. It follows that the L2-matrix normis
‖F‖2 = (∑i,j
‖fij‖22)
1/2
The vector-valued space L2(Cn) is the special case m = 1. The subspaces H2(Cn×m) andH2(Cn×m)⊥ are defined in the obvious way.
Finally, the space L∞(Cn×m) consists of matrices whose entries are scalar-valued L∞-functions, and the matrix norm is
‖F‖∞ := supθ‖F (ejθ)‖
Recall that the latter norm is maximum singular value. The subspace of matrices analyticin the open unit disc is H∞(Cn×m).
When the dimensions are irrelevant, L2(Cn×m) etc. are written simply L2 etc.
17
3.4 The λ-transform
Recall from Chapter 1 that h2 is the space of square-summable sequences defined for non-negative times. We shall need two additional spaces. Define h⊥2 to be the space of square-summable sequences defined for negative times, k = . . . ,−2,−1. A signal x in h⊥2 is written ...
x−2
x−1
and has the property
−1∑−∞
|xk|2 < ∞
This space is a Hilbert space under the inner-product
< x, y >=−1∑−∞
xkyk
Also, define l2 to be the external direct sum of h⊥2 and h2:
l2 := h⊥2 ⊕ h2
Elements of l2 will be written[xy
], x ∈ h⊥2 , y ∈ h2
With the inherited inner-product, l2 is a Hilbert space, h2 is a closed subspace, and h⊥2 is itsorthogonal complement.
Recall that the (two-sided) z-transform of x in l2 is
∞∑k=−∞
xkz−k
For us it’s more convenient to use λ := z−1. So define the λ-transform of x to be
x(λ) :=∞∑
k=−∞
xkλk
The theorem coming up, a combination of the Riesz-Fischer theorem and Parseval’s equal-ity, connects the time-domain Hilbert spaces to the frequency-domain Hilbert spaces. It iscompactly stated via the notion of isomorphism. Let X and Y be Hilbert spaces. An iso-morphism from X to Y is a linear transformation having the two properties
18
it is surjective
it preserves inner-products
Such a function automatically has the further properties
it preserves norms
it is injective
it is bounded
it has a bounded inverse
If such an isomorphism exists, then X and Y are isomorphic.
Theorem 1 The λ-transform is an isomorphism from l2 onto L2; it maps h2 onto H2 andh⊥2 onto H⊥
2 .
3.5 Summary
Hilbert spaces
L2: square-integrable functions on the unit circle
H2: subspace of L2 of functions analytic in the open unit disc
H⊥2 : subspace of L2 of functions analytic in the exterior of the closed unit disc and zero
at infinity
Banach spaces
L∞: essentially bounded functions on the unit circle
H∞: subspace of functions analytic in the open unit disc
real-rational spaces
RL2 = RL∞: real-rational functions with no poles on the unit circle
RH2 = RH∞: real-rational functions with no poles in the closed unit disc (include thepolynomials)
RH⊥2 : strictly-proper real-rational functions with no poles on the unit circle or outside
the unit disc
19
3.6 Exercises
1. Fix a in the open unit disc. Find a function f in H2 such that for every g in H2
< f, g >= g(a)
2. Show that RH⊥2 consists precisely of the strictly proper real-rational functions that are
analytic in |λ| ≥ 1.
3. Define f(λ) = 2λ− 1 and V = fH2. Show that V is a closed subspace of H2. What is thedimension of V⊥, the orthogonal complement in H2? Find a basis for V⊥.
4. Find the projections in H2 and H⊥2 of[
λ3(λ+3)2λ2−5λ+2
1λ2−2λ
]
5. Define
F (λ) :=
[λ 00 2λ− 1
]Let V = FH2 and let V⊥ be its orthogonal complement in H2. Find a basis for V⊥.
6. Compute the L∞-norm of[λ2+1λ+2
1λ
λ2−λ−6λ
]
7. Prove that if F ∈ H∞(Cn×m), then
‖F‖2 ≤√
m‖F‖∞
so that F ∈ H2(Cn×m).
20
Chapter 4
Time-Invariant Systems
In this chapter we study time-invariant linear systems and their transfer functions. Transferfunctions of causal and non-causal stable systems are characterized, and the problem istreated of finding the distance from a non-causal stable system to the nearest causal stableone. All the systems in this chapter are linear and time-invariant.
4.1 Transfer functions
Suppose F is a time-invariant linear system with matrix
[F ] =
F0 F−1 · · ·F1 F0 · · ·...
...
The corresponding convolution equation is
yk =∞∑l=0
Fk−lul, k ≥ 0 (4.1)
where
yk ∈ Cp, uk ∈ Cm, Fk ∈ Cp×m
The transfer function (tf) of F is defined to be
F (λ) :=∞∑−∞
Fkλk
Associated with the tf is a region of convergence (roc): Let
r2 := radius of convergence of∞∑0
Fkλk = (lim sup ‖Fk‖1/k)−1
21
1
r1
:= radius of convergence of∞∑1
F−kzk
i.e.,
r1 = lim sup ‖F−k‖1/k
Then the roc for F (λ) is defined to be the annulus
r1 < |λ| < r2
Thus F exists only if r1 < r2, and then F (λ) is analytic in the roc. We assume from now onthat F exists. Note that if F is causal, then the roc is a disc, |λ| < r2. Also, if F is strictlycausal, then F (0) = 0.
These definitions are illustrated by the following examples. For the first take
Fk =
{1, k ≥ −10, k < −1
Then r1 = 0, r2 = 1, and
F (λ) =1
λ(1− λ)
This system is not causal, and this is reflected in the fact that the roc is not a disc.The second example is
Fk =
{0, k ≥ 02k, k < 0
Here r1 = 1/2, r2 = ∞, and
F (λ) =1
2λ− 1Again, the system is not causal.
The third example is the state-space model
xk+1 = Axk + Buk
yk = Cxk + Duk
The tf is
F (λ) = D + λCB + λ2CAB + · · ·= D + λC(I + λA + · · ·)B= D + λC(I − λA)−1B
The latter rational matrix is denoted[A BC D
]To get the roc of the series I + λA + · · · we invoke Gelfand’s formula:
limk→∞
‖Ak‖1/k = ρ(A) := spectral radius of A
Thus the roc is the disc |λ| < 1/ρ(A).
22
4.2 Response to white noise
We continue with the system of the previous section, but with the following assumptions:
(i) F is causal and F ∈ H2
(ii) the input u is standard white noise
The second assumption means that uk is a random vector with zero mean and unity covariancematrix and that uk and ul are uncorrelated for k 6= l. Under these assumptions the outputsignal y has the following properties.
Theorem 1 y is (wide-sense) stationary, it has zero mean, and its root-mean-square valueequals ‖F‖2.
Proof The first two properties are routine to prove. For the third, we first compute thecovariance matrix. Let E denote the expectation operator. Then
Eyky∗k = E
(∑i
Fk−iui
)(∑j
Fk−juj
)∗
= E∑
i
∑j
Fk−iuiu∗jF
∗k−j
=∑
i
Fk−iF∗k−i
=∑
i
FiF∗i
Thus the mean-square value of y is
Ey∗kyk = trace Eyky∗k
= trace∑
i
FiF∗i
=∑
i
traceF ∗i Fi
The last expression equals the square of the h2-norm of the sequence F0, F1, . . . . But thisequals ‖F‖2
2 by Parseval’s equality. �
There is an analogous deterministic result in the single-input, single-output case. Supposeu is the unit pulse, i.e., u(λ) = 1, and, as above, F ∈ H2. Then y ∈ h2 and ‖y‖2 = ‖F‖2.
23
In words: the H2-norm of the tf equals the square-root of the energy of the output when theinput is the unit pulse.
We complete this section with two ways to compute the H2-norm of a rational tf. Firstis the familiar method of residues. Suppose F is a scalar-valued function in RH2. Then
‖F‖22 =
1
2π
∫ 2π
0
|F (ejθ)|2dθ =1
2πj
∮1
λF (λ)F
(1
λ
)dλ
The latter expression equals the sum of the residues of
1
λF (λ)F
(1
λ
)at its poles in the open unit disc.
Second is the state-space method. Suppose
F (λ) =
[A BC D
]where ρ(A) < 1. Define the controllability grammian
Lc :=∞∑i=0
AiBB′A′i
Then
‖F‖2 = [trace (DD′ + CLcC′)]1/2
The proof of this is left as an exercise.
4.3 tfs for causal systems
Consider a system F with transfer function F . If F is causal, then its matrix is lowertriangular and, consequently, the roc of F is some disc. It turns out that if F is h2-stable,then the closure of the roc contains the open unit disc and F ∈ H∞.
Theorem 2 The function F 7→ F maps the space of time-invariant causal h2-stable systemsinto H∞. This function is linear, bijective, and norm-preserving.
Proof Some parts of this theorem are fairly hard to prove. To avoid getting bogged downin technicalities, we’ll take some shortcuts.
To see that the co-domain of the mapping F 7→ F is H∞, we’ll simplify by assuming thatthe tf is real-rational and scalar-valued. Apply the unit pulse as input. Then the outputequals the first column of [F ] and belongs to h2. The λ-transform of this output equals F (λ),which therefore belongs to RH2. So F ∈ RH2 = RH∞.
Linearity and injectivity are easy.
24
To prove surjectivity, start with F in H∞. Thus F has a power series at λ = 0:
F (λ) = F0 + λF1 + · · ·
It is claimed that FH2 ⊂ H2. To prove this, let u ∈ H2 and define y := F u. Since u andF are analytic in the open unit disc, so is y. Moreover,
‖y‖22 =
1
2π
∫ 2π
0
‖F (ejθ)u(ejθ)‖2dθ
≤ ‖F‖2∞‖u‖2
2 (4.2)
Thus y ∈ L2. Being analytic and square-integrable, y ∈ H2. This proves the claim.Define the linear transformation F on s u0
u1...
7→ F0 0 · · ·
F1 F0 · · ·...
...
u0
u1...
In view of the claim, F maps h2 to h2. Moreover, from (4.2) ‖F‖ ≤ ‖F‖∞, so F is bounded,i.e., h2-stable. This proves surjectivity.
Finally, to prove that the function is norm-preserving, it remains to show that ‖F‖ ≥‖F‖∞. Here again we’ll take a shortcut by assuming that F is scalar-valued and continuouson the unit circle.
Let ε and δ be positive numbers. Choose an interval (θ1, θ2) such that
θ ∈ (θ1, θ2) ⇒ |F (ejθ)| ≥ ‖F‖∞ − ε (4.3)
Then let u be a function in H2 having the boundary magnitude
|u(ejθ)| ={
c, θ ∈ (θ1, θ2)δ, else
(4.4)
(Such a function exists as discussed in Section 3.1.) Here c is chosen so that ‖u‖2 = 1, i.e.,
1
2π
{c2(θ2 − θ1) + δ2[2π − (θ2 − θ1)]
}= 1 (4.5)
Then
‖F u‖22 ≥
1
2π
∫ θ2
θ1
|F (ejθ)u(ejθ)|2dθ
≥ 1
2πc2(θ2 − θ1)(‖F‖∞ − ε)2 from (4.3) and (4.4)
=
{1− δ2 2π − (θ2 − θ1)
2π
}(‖F‖∞ − ε)2 from (4.5)
25
Since the last expression is independent of u, we get
‖F‖2 ≥{
1− δ2 2π − (θ2 − θ1)
2π
}(‖F‖∞ − ε)2
But δ was arbitrary, so
‖F‖ ≥ ‖F‖∞ − ε
Since ε was arbitrary too, ‖F‖ ≥ ‖F‖∞. �
Some points raised in the above proof are worth emphasizing. Let F be a causal h2-stablesystem and F its tf. The input and output are related by either of the equivalent equations
time-domain y = Fu
frequency-domain y = F u
In other words, F u = F u. This is equivalent to commutativity of the diagram
-
-
? ?
h2 h2
H2H2
MF
F
The two vertical arrows stand for λ-transformation, and MF is the operator of multiplication
by F :
MF g := F g
Since λ-transformation preserves norms, ‖F‖ = ‖MF‖. Thus from the theorem we have
‖F‖∞ = ‖F‖ = ‖MF‖
We conclude this section with a remark about causal but unstable systems. In general theroc for F is a disc, |λ| < r2. If r2 < 1, then F 6∈ H∞, so by the theorem F is not h2-stable.For example, if F is rational, then F is h2-stable iff F has no poles in the closed unit disc.It is emphasized that this assumes F is causal.
26
4.4 tfs for non-causal systems
If F is non-causal, then its matrix is not lower triangular and the roc of F is only an annulus,not a disc. However if F is h2-stable, then the closure of this annulus contains the unit circle.
Theorem 3 The function F → F maps the space of time-invariant h2-stable systems intoL∞. This function is linear, bijective, and norm-preserving.
The proof is similar to the one in the previous section.Let F be h2-stable. Its matrix looks like
[F ] =
F0 F−1 · · ·F1 F0 · · ·...
...
and the corresponding convolution equation is
yk =∞∑l=0
Fk−lul, k ≥ 0
Extend this equation backwards in time and allow u to belong to l2 = h⊥2 ⊕ h2 instead ofjust h2. We get
yk =∞∑
l=−∞
Fk−lul, −∞ < k < ∞
This equation corresponds to an extended operator Fe with matrix
[Fe] =
...
...· · · F−1 F−2 · · ·· · · F0 F−1 · · ·· · · F1 F0 · · ·
......
This operator, an extension of F , maps l2 to itself. It is easily checked that the two equations
time-domain y = Feu
frequency-domain y = F u
are equivalent, i.e., Feu = F u. The corresponding commutative diagram is
27
-
-
? ?
l2 l2
L2L2
MF
Fe
Again, the two vertical arrows stand for λ-transformation. We have the following equalities:
‖F‖∞ = ‖F‖ = ‖Fe‖ = ‖MF‖
The first equality is from the theorem, the third from the commutative diagram, and thesecond is left as an exercise.
This section concludes with a caution. If the tf is given and it belongs to L∞ but notH∞, does it represent an unstable causal system or a stable non-causal system? The answeris that we don’t know. We must also be given the roc to make this decision. Consider forexample the tf
F (λ) =1
2λ− 1
having a pole in the unit disc at λ = 1/2. If the roc is the disc |λ| < 1/2, then F is causalbut not h2-stable; if the roc is the (semi-infinite) annulus |λ| > 1/2, then F is h2-stable butnot causal.
4.5 Nehari’s theorem
Consider the two questions: What is the distance from a given stable non-causal system tothe nearest stable causal one? What is the distance from a given unstable causal systemto the nearest stable causal one? Suitably posed, these two questions are equivalent. Thissection presents time- and frequency-domain versions of a solution.
4.5.1 Time-domain
Suppose F is h2-stable. As in the previous section, bring in its matrix
[F ] =
F0 F−1 · · ·F1 F0 · · ·...
...
(4.6)
28
and its extension Fe. The Hankel operator corresponding to F is
ΓF : h2 → h⊥2
ΓF u := Πh⊥2Feu
The matrix of this operator is
[ΓF ] =
......
F−2 F−3 · · ·F−1 F−2 · · ·
(4.7)
Comparison of (4.6) and (4.7) shows that [ΓF ] is the limit of the upper right-hand corners of[F ]. It’s instructive to look at the convolution equations corresponding to the three operatorsF , Fe, and ΓF :
F : yk =∑∞
0 Fk−lul, k ≥ 0Fe : yk =
∑∞−∞ Fk−lul, −∞ < k < ∞
ΓF : yk =∑∞
0 Fk−lul, k < 0
The last equation gives the output before time 0 in terms of the input at and after time 0.In this way we can think of the Hankel operator as mapping present and future inputs topast outputs. Consequently, if F is causal, then ΓF = 0.
If F and G are h2-stable systems, the distance between them is ‖F −G‖.
Theorem 4 The distance from an h2-stable system F to the nearest causal h2-stable systemequals ‖ΓF‖. Moreover, the distance is attained.
For an outline of a proof, we need some notation. Suppose A and B are operators onHilbert spaces, having the same co-domain, say
A : X → Z, B : Y → Z
Then there’s a natural operator mapping the external direct sum X ⊕ Y to Z, namely[xy
]7→ Ax + By
This operator is denoted[
A B]. There’s a similar construction for two operators having
the same domain.
Lemma 1 Suppose A, B, and C are operators on Hilbert space. Then
minX
∥∥∥∥[ A BX C
]∥∥∥∥ = max
{∥∥[ A B]∥∥ ,
∥∥∥∥[ BC
]∥∥∥∥}
29
Here the minimum is over all Hilbert space operators X.A general proof of this lemma would require too many extra tools. To give some compre-
hension of the lemma, below is a simple proof in the very special case that the Hilbert spaceequals R, the reals. So let a, b, c be real numbers. We want to show that
minx
∥∥∥∥[ a bx c
]∥∥∥∥ = max
{∥∥∥∥[ ab
]∥∥∥∥ ,
∥∥∥∥[ bc
]∥∥∥∥}where the minimum is over all real numbers x. Thus three elements of a 2 × 2 real matrixare fixed, the fourth is to be chosen to minimize the norm, and the minimum equals themaximum norm of the fixed row and the fixed column. Assume without loss of generalitythat ∥∥∥∥[ b
c
]∥∥∥∥ ≥ ∥∥∥∥[ ab
]∥∥∥∥i.e., |c| ≥ |a|. Since∥∥∥∥[ a b
x c
]∥∥∥∥ ≥ ∥∥∥∥[ bc
]∥∥∥∥for every x, it suffices to construct an x to achieve equality. To do this, simply choose x sothat [
ax
]⊥[
bc
](4.8)
Looking at these two vectors in the plane, i.e.,
AAAAAA
����
�����
–
–
| |
x
c
a b
we see that (because |c| ≥ |a|)∥∥∥∥[ ax
]∥∥∥∥ ≤ ∥∥∥∥[ bc
]∥∥∥∥ (4.9)
30
Now we have∥∥∥∥[ a bx c
]∥∥∥∥2
=
∥∥∥∥[ a xb c
] [a bx c
]∥∥∥∥=
∥∥∥∥[ a2 + x2 00 b2 + c2
]∥∥∥∥ by (4.8)
= b2 + c2 by (4.9)
=
∥∥∥∥[ bc
]∥∥∥∥2
Proof of theorem From Exercise 5, F and Fe have equal norm. Hence F and Fe|h2, therestriction of Fe to h2, have equal norm. The matrix of the latter operator is
[Fe|h2] =
...
...F−1 F−2 · · ·F0 F−1 · · ·F1 F0 · · ·...
...
So it suffices to prove that the minimum of ‖(Fe −Ge)|h2‖ over all causal h2-stable systemsG equals ‖ΓF‖. For such G we have
[(Fe −Ge)|h2] =
...
...F−1 F−2 · · ·
F0 −G0 F−1 · · ·F1 −G1 F0 −G0 · · ·
......
The upper part of this matrix is [ΓF ]. Hence
‖(Fe −Ge)|h2‖ ≥ ‖ΓF‖
To achieve equality, construct G by specifying first G0, then G1, etc. The idea is to movedown the above matrix, [(Fe−Ge)|h2], row by row without increasing the norm. First, chooseG0 such that the norm of the operator with matrix ...
...F−1 F−2 · · ·
F0 −G0 F−1 · · ·
31
equals ‖ΓF‖; this is possible by virtue of the lemma. Next choose G1 such that the norm ofthe operator with matrix
......
F−1 F−2 · · ·F0 −G0 F−1 · · ·F1 −G1 F0 −G0 · · ·
equals ‖ΓF‖; this is possible for the same reason. Continue ad infinitum. �
4.5.2 Frequency domain
Let F ∈ L∞. Then FH2 ⊂ L2 (cf. the proof of the theorem in Section 4.3). The Hankeloperator corresponding to F is
ΓF : H2 → H⊥2
ΓF g := Π1F g
(Recall that Π1 is the orthogonal projection from L2 onto H⊥2 .) Sometimes F is called the
symbol of its Hankel operator. Note that if F ∈ H∞, then ΓF = 0. The relationship betweenthe two Hankel operators is exhibited in this commutative diagram:
-
-
? ?
h2 h⊥2
H2H⊥2
ΓF
ΓF
The distance between two functions in L∞ is the L∞-norm of their difference.
Theorem 5 The distance from a function F in L∞ to the nearest function in H∞ equals‖ΓF‖ and the distance is attained.
4.6 State-space computation of Hankel-norm
The subject of this section is how to compute numerically the norm of a Hankel operatorwith real-rational symbol. So let F ∈ RL∞. The H⊥
2 -component of F has the form
Π1F (λ) = · · ·+ 1
λ2F−2 +
1
λF−1
32
and then
[ΓF ] =
......
F−2 F−3 · · ·F−1 F−2 · · ·
Being real-rational, Π1F has a minimal realization:
Π1F (λ) = C(λ− A)−1B
Let n denote the dimension of A and observe that ρ(A) < 1 because Π1F is analytic in theexterior of the unit disc. Then we have
[ΓF ] =
......
CAB CA2B · · ·CB CAB · · ·
=
...CAC
[ B AB · · ·]
This equation leads us to define two auxiliary operators: the controllability operator
Ψc : h2 → Cn, [Ψc] =[
B AB · · ·]
and the observability operator
Ψo : Cn → h⊥2 , [Ψo] =
...CAC
Clearly
ΓF = ΨoΨc (4.10)
Now bring in the controllability and observability gramians:
Lc :=∞∑0
AiBB′A′i ∈ Cn×n
Lo :=∞∑0
A′iC ′CAi ∈ Cn×n
These matrices are the unique solutions of the Lyapunov equations
Lc = ALcA′ + BB′
Lo = A′LoA + C ′C
It is immediate from the definitions that
[ΨcΨ∗c ] = Lc, [Ψ∗
oΨo] = Lo (4.11)
33
Lemma 2 The operator ΓF∗ΓF and the matrix LcLo share the same nonzero eigenvalues.
Proof Let µ be a nonzero eigenvalue of ΓF∗ΓF and u in h2 a corresponding eigenvector.
Thus
ΓF∗ΓF u = µu
This and (4.10) imply
Ψ∗cΨ
∗oΨoΨcu = µu (4.12)
Pre-multiply by Ψc:
ΨcΨ∗cΨ
∗oΨo(Ψcu) = µ(Ψcu)
This and (4.11) yield
LcLo(Ψcu) = µ(Ψcu)
Now Ψcu 6= 0, from (4.12). Hence µ is an eigenvalue of LcLo.The converse is similar. �
To proceed we need two extra facts. First, for every operator Φ on a Hilbert space,‖Φ‖ = ‖Φ∗Φ‖1/2. Secondly, the norm of ΓF
∗ΓF equals its maximum eigenvalue. (This is nottrue in general, but is a consequence of the fact that this operator has finite rank.)
In summary, here is a procedure to compute the norm of the Hankel operator with symbolF in RL∞.
Step 1 Compute a minimal realization C(λ− A)−1B of Π1F .
Step 2 Solve the Lyapunov equations
Lc = ALcA′ + BB′
Lo = A′LoA + C ′C
Step 3 Then ‖ΓF‖ equals the square root of the maximum eigenvalue of LcLo.
34
4.7 The big commutative diagram
The main constructions of this chapter can be summarized in a grand commutative diagram,as shown below. The upper half pertains to the time-domain, the lower half to the frequency-domain. Start with F , time-invariant and h2-stable, but not necessarily causal. Let Fe be itstime-invariant extension to l2, F its tf, MF the multiplication operator, and ΓF and ΓF theHankel operators, time- and frequency-domain respectively. The vertical arrows representthe following maps: h2 → l2 and H2 → L2 are subspace insertions; l2 → h⊥2 and L2 → H⊥
2
are orthogonal projections; and l2 → L2 is λ-transformation. The outer two curved arrowsare restrictions of λ-transformation. Finally, when F is real-rational, Ψc and Ψo are thecontrollability and observability operators from a minimal realization of the H⊥
2 -componentof F .
h2 h!
2
l2 l2
L2 L2
H2 H!
2
!c !o
!F
Fe
MF
!F
4.8 Exercises
1. Suppose
F (λ) =
[A BC D
]ρ(A) < 1
Lc :=∞∑i=0
AiBB′A′i
35
Prove that
‖F‖2 = [trace (DD′ + CLcC′)]1/2
2. Continuing with the previous exercise, prove that Lc equals the unique solution of theLyapunov equation
Lc = ALcA′ + BB′
3. Compute the H2-norm of[λ2+1λ+2
1λ
λ2−λ−6λ
]
4. Consider the linear time-invariant causal system with matrix0 0 0 · · ·1 0 0 · · ·
1/2 1 0 · · ·1/3 1/2 1 · · ·...
......
Is it h2-stable?
5. For the set-up in Section 4, prove that ‖F‖ = ‖Fe‖.
6. Consider an h2-stable time-invariant system with tf
F (λ) =λ3
λ2 + 3.5λ + 1.5
The roc is the annulus .5 < |λ| < 3. Find [F ]. Compute the distance from F to the nearestcausal h2-stable time-invariant system.
36
Chapter 5
The Continuous-Time Setup
The purpose of this chapter is to collect the continuous analogs of some of the material inChapters 2-4.
5.1 Time-domain spaces
Consider a signal x(t) defined for all time, −∞ < t < ∞, and taking values in Cn. Thus xis a function
(−∞,∞) 7→ Cn
Restrict x to be square-(Lebesgue) integrable:∫ ∞
−∞‖x(t)‖2dt < ∞ (5.1)
The norm in (5.1) is our previously defined norm on Cn. The set of all such signals is theLebesgue space L2(−∞,∞). (To simplify notation we suppress the dependence of this spaceon the integer n.) This space is a Hilbert space under the inner product
< x, y >:=
∫ ∞
−∞x(t)∗y(t)dt
Then the norm of x, denoted ‖x‖2, equals the square-root of the left-hand side of (5.1).The set of all signals in L2(−∞,∞) that equal zero for almost all t < 0 is a closed
subspace, denoted L2[0,∞). Its orthogonal complement (zero for almost all t > 0) is denotedL2(−∞, 0].
5.2 Frequency-domain spaces
Consider a function x(jω) that is defined for all frequencies, −∞ < ω < ∞, takes values inCn, and is square-(Lebesgue) integrable with respect to ω. The space of all such functions is
38
denoted L2 and is a Hilbert space under the inner product
< x, y >:=1
2π
∫ ∞
−∞x(jω)∗y(jω)dω
The norm on L2 will be denoted ‖x‖2. The space RL2, the real-rational functions in L2,consists of n-vectors each component of which is real-rational, strictly proper, and withoutpoles on the imaginary axis.
Next, H2 is the space of all functions x(s) that are analytic in Re s > 0, take values inCn, and satisfy the uniform square-integrability condition
‖x‖2 :=
[supξ>0
1
2π
∫ ∞
−∞‖x(ξ + jω)‖2dω
]1/2
< ∞
(We have used the same norm symbol for L2(−∞,∞), L2, and H2. Context determineswhich is intended.) This makes H2 a Banach space. Functions in H2 are not defined a priorion the imaginary axis, but we can get there in the limit.
Theorem 1 If x ∈ H2, then for almost all ω the limit
x(jω) := limξ→0
x(ξ + jω)
exists and x belongs to L2. Moreover, the mapping x 7→ x from H2 to L2 is linear, injective,and norm-preserving.
It is customary to identify x in H2 and its boundary function x in L2. So henceforth wedrop the tilde and regard H2 as a closed subspace of the Hilbert space L2. The space RH2
consists of real-rational n-vectors that are stable and strictly proper.The orthogonal complement H⊥
2 of H2 in L2 is the space of functions x(s) with thefollowing properties: x(s) is analytic in Re s < 0; x(s) takes values in Cn; the supremum
supξ<0
∫ ∞
−∞‖x(ξ + jω)‖2dω
is finite. Again, we identify functions in H⊥2 and their boundary functions in L2.
Now we turn to two Banach spaces. First, an n×m complex-valued matrix F (jω) belongsto the Lebesgue space L∞ iff ‖F (jω)‖ is essentially bounded (bounded except possibly ona set of measure zero). The norm just used for F (jω) is the norm on Cn×m introduced inSection 2.4 (largest singular value). Then the L∞-norm of F is defined to be
‖F‖∞ := supω‖F (jω)‖
This makes L∞ a Banach space. It is easily checked that F ∈ RL∞ iff F is real-rational,proper, and without poles on the imaginary axis.
The final space is H∞. It consists of functions F (s) that are analytic in Re s > 0, takevalues in Cn×m, and are bounded in Re s > 0 in the sense that
sup{‖F (s)‖ : Re s > 0} < ∞
39
The left-hand side defines the H∞-norm of F . There is an analog of Theorem 1 in whichH2 and L2 are replaced by H∞ and L∞ respectively: each function in H∞ has a uniqueboundary function in L∞, and the mapping from H∞-function to boundary L∞-function islinear, injective, and norm-preserving. So henceforth we regard H∞ as a closed subspace ofthe Banach space L∞. Finally, RH∞ consists of those real-rational matrices that are stableand proper.
Let’s recap in the real-rational case:
RL2: vector-valued, strictly proper, no poles on imaginary axis
RH2: vector-valued, strictly proper, stable
RH⊥2 : vector-valued, strictly proper, no poles in Re s < 0
RL∞: matrix-valued, proper, no poles on imaginary axis
RH∞: matrix-valued, proper, stable.
5.3 Connections
This section contains statements of two basic theorems relating the spaces just introduced.The first, a combined Plancherel and Paley-Wiener theorem, connects the time-domainHilbert spaces and the frequency-domain Hilbert spaces. A mapping from one Hilbert spaceto another is a Hilbert space isomorphism if it is a linear surjection that preserves innerproducts. (Such a mapping is continuous, preserves norms, is injective, and has a continuousinverse.)
Theorem 2 The Fourier transform is a Hilbert space isomorphism from L2(−∞,∞) ontoL2. It maps L2[0,∞) onto H2 and L2(−∞, 0] onto H⊥
2 .
This important theorem says in particular that H2 is just the set of Laplace transformsof signals in L2[0,∞), i.e., of signals on t ≥ 0 of finite energy.
The second theorem connects the Hilbert space H2 with the Banach spaces L∞ and H∞.
Theorem 3 (i) If F ∈ L∞, then FL2 ⊂ L2 and
‖F‖∞ = sup{‖Fx‖2 : x ∈ L2, ‖x‖2 = 1}
= sup{‖Fx‖2 : x ∈ H2, ‖x‖2 = 1}
(ii) If F ∈ H∞, then FH2 ⊂ H2 and
‖F‖∞ = sup{‖Fx‖2 : x ∈ H2, ‖x‖2 = 1}
40
5.4 Hankel operators
Let’s look at some operators on the spaces we’ve just introduced.
Example 1The Fourier transform is an operator from L2(−∞,∞) to L2. Theorem 2 says that its
norm equals 1.
Example 2Introduce the direct sum
L2(−∞,∞) = L2(−∞, 0]⊕ L2[0,∞)
Each function f in L2(−∞,∞) has a unique decomposition f = f1 + f2 with f1 ∈ L2(−∞, 0]and f2 ∈ L2[0,∞):
f1(t) = f(t), f2(t) = 0, t ≤ 0
f1(t) = 0, f2(t) = f(t), t > 0
The function f 7→ f1 from L2(−∞,∞) to L2(−∞, 0] is an operator, the orthogonal projectionof L2(−∞,∞) onto L2(−∞, 0]. It’s easy to prove that its norm equals 1.
In the same way we have
L2 = H⊥2 ⊕H2.
The orthogonal projection from L2 onto H⊥2 will be denoted Π1 and from L2 onto H2 by Π2.
Example 3Let F ∈ L∞ and define the function ΛF from L2 to L2 via
ΛF g := Fg
Thus the action of ΛF is multiplication by F . Obviously ΛF is linear. Theorem 3 says that‖ΛF‖ = ‖F‖∞, so ΛF is bounded. This operator is called a Laurent operator and F is calledits symbol; so ΛF is the Laurent operator with symbol F .
A related operator is ΛF |H2, the restriction of ΛF to H2, which maps H2 to L2. Theorem3 says that its norm also equals ‖F‖∞.
Observe that if F ∈ H∞ then, also by Theorem 3,
ΛFH2 ⊂ H2
The converse is true too: if ΛFH2 ⊂ H2, then F ∈ H∞.
Example 4This is the time-domain analog of the previous example. Recall that convolution in
the time-domain corresponds to multiplication in the frequency-domain. Suppose F (s) is amatrix-valued function that is analytic in a vertical strip containing the imaginary axis and
41
that belongs to L∞. Taking the region of convergence to be this strip, let f(t) denote theinverse bilateral Laplace transform of F (s). Now define the convolution operator Ξf fromL2(−∞,∞) to L2(−∞,∞) via
y = Ξfu
y(t) =
∫ ∞
−∞f(t− τ)u(τ)dτ
This system is linear, but not necessarily causal because f(t) may not equal zero for negativetime. Note that the system is causal iff Ξf maps L2[0,∞) into L2[0,∞), i.e., “Ξf leaves thefuture invariant”. The operators Ξf and ΛF are intimately related via the Fourier transform.(Draw the commutative diagram.)
Example 5Again let F ∈ L∞. The Toeplitz operator with symbol F , denoted ΘF , maps H2 to H2
and is defined as follows: for each g in H2, ΘF g equals the orthogonal projection of Fg ontoH2. Thus
ΘF = Π2ΛF |H2
As a concrete example consider the scalar-valued function F (s) = 1/(s− 1) in RL∞. For gin H2 we have
Fg = g1 + g2
g1 ∈ H⊥2 , g2 ∈ H2
g1(s) = g(1)/(s− 1)
g2(s) = [g(s)− g(1)]/(s− 1)
Thus ΘF maps g to g2.
Example 6For F in L∞ the Hankel operator with symbol F, denoted ΓF , maps H2 to H⊥
2 and isdefined as
ΓF := Π1ΛF |H2
For the example F (s) = 1/(s − 1), ΓF maps g(s) in H2 to g(1)/(s − 1) in H⊥2 . Note that
ΓF = 0 if F ∈ H∞.The relationship between the three operators ΛF , ΘF , and ΓF can be described as follows.
We have
ΛF : H⊥2 ⊕H2 → H⊥
2 ⊕H2
42
and correspondingly we can regard ΛF as a 2× 2 matrix with operator entries:
ΛF =
[Λ11 Λ12
Λ21 Λ22
]For example
Λ11 = Π1ΛF |H⊥2
It follows from the definitions that Λ12 = ΓF and Λ22 = ΘF . Thus
ΛF |H2 =
[ΓF
ΘF
]
Example 7In this example we study the Hankel operator with the special symbol
F (s) =
[A BC 0
]:= C(s− A)−1B
where A is antistable (all eigenvalues in Re s > 0). Suppose A is n× n. Such F belongs toRL∞. The inverse bilateral Laplace transform of F (s) is
f(t) = −CeAtB, t < 0
f(t) = 0, t ≥ 0
The time-domain analog of the Hankel operator, denoted Γf , maps a function u in L2[0,∞)to the function y in L2(−∞, 0] defined by
y(t) =
∫ ∞
0
f(t− τ)u(τ)dτ, t < 0
= −CeAt
∫ ∞
0
e−AτBu(τ)dτ, t < 0 (5.2)
Define two auxiliary operators: the controllability operator
Ψc : L2[0,∞) → Cn
Ψcu := −∫ ∞
0
e−AτBu(τ)dτ
and the observability operator
Ψo : Cn → L2(−∞, 0]
(Ψox)(t) := CeAtx, t < 0
43
From (5.2) we have that
Γf = ΨoΨc
There is a systemic interpretation of Γf in terms of the usual state-space equations
x = Ax + Bu (5.3)
y = Cx (5.4)
To see the action of Γf , solve these equations in the following way. First, apply an input uin L2[0,∞) to equation (5.3) with initial condition x(0) = x0 and such that x(t) is boundedon [0,∞). Then
x(t) = eAtx0 + eAt
∫0
tote−AτBu(τ)dτ, t ≥ 0
so that
x0 = −∫ ∞
0
e−AτBu(τ)dτ = Ψcu
Now solve (5.3) and (5.4) backwards in time starting at t = 0 and noting that u(t) = 0 fort < 0. The solution is
y(t) = CeAtx0 = (Ψox0)(t), t < 0
In this way Γf maps future input to initial state to past output.The adjoint of a Laurent operator ΛF can be obtained explicitly as follows. Introduce the
notation
F∼(jω) := F (jω)∗ (5.5)
If F ∈ RL∞, then we shall interpret F∼ as
F∼(s) := F (−s)′
which is consistent with (5.5). If g and h belong to L2, then
< ΛF g, h >=< g, Λ∗F h >
and
< ΛF g, h >=1
2π
∫ ∞
−∞g(jω)∗F (jω)∗h(jω)dω =< g, F∼h >
We conclude that Λ∗F equals the Laurent operator with symbol F∼.Similarly, the adjoint of a Hankel operator ΓF can be shown to be
Γ∗F = Π2Λ∗F |H⊥
2
For example if F (s) = 1/(s−1), then Γ∗F maps h(s) in H⊥2 to −h(−1)/(s+1) in H2. (Verify.)
The rank of an operator Φ : X → Y is the dimension of the closure of its image spaceΦX .
Our interest is in Hankel operators with real-rational symbols, Example 7 being a specialcase.
44
Theorem 4 If F ∈ RL∞, then ΓF has finite rank.
Proof There is a unique factorization (by partial-fraction expansion, for example) F =F1 + F2, where F1 is strictly proper and analytic in Re s ≤ 0 and F2 is proper and analyticin Re s ≥ 0, i.e., F2 ∈ RH∞. Since ΓF = ΓF1 , we might as well assume at the start that Fis strictly proper and analytic in Re s ≤ 0. Introduce a minimal realization:
F (s) =
[A BC 0
]The operator ΓF and its time-domain analog have equal ranks. As in Example 7 the latteroperator equals ΨoΨc. By controllability and observability Ψc is surjective and Ψo is injective.Hence ΨoΨc has rank n, so ΓF does too. �
For the remainder of this section let F ∈ RL∞. The self-adjoint operator Γ∗F ΓF maps H2
to itself and its rank is finite by Theorem 4. This property guarantees that it does in facthave eigenvalues. We state without proof the following fact.
Theorem 5 The eigenvalues of Γ∗F ΓF are real and nonnegative and the largest of them equals‖Γ∗F ΓF‖.
This theorem impies that ‖ΓF‖ equals the square root of the largest eigenvalue of Γ∗F ΓF .So we could compute ‖ΓF‖ if we could compute the eigenvalues of Γ∗F ΓF . How to do thislatter computation is the last topic of this section.
We continue with the notation introduced in Example 7 and the proof of Theorem 4. Theself-adjoint operators ΨcΨ
∗c and Ψ∗
oΨo map Cn to itself. Thus they have matrix representa-tions with respect to the standard basis on Cn. Define the controllability and observabilitygramians
Lc :=
∫ ∞
0
e−AtBB′e−A′tdt (5.6)
Lo :=
∫ ∞
0
e−A′tC ′Ce−Atdt (5.7)
It is routine to show that Lc and Lo are the unique solutions of the Lyapunov equations
ALc + LcA′ = BB′ (5.8)
A′Lo + LoA = C ′C (5.9)
Theorem 6 The operator Γ∗F ΓF and the matrix LcLo share the same nonzero eigenvalues.
In summary, the norm of ΓF for F in RL∞ can be computed as follows. First, find aminimal realization (A, B, C) of the antistable part of F (s), i.e.,
F (s) =
[A BC 0
]+ (a matrix in RH∞)
Next, solve the Lyapunov equations (5.8) and (5.9) for Lc and Lo. Then ‖ΓF‖ equals thesquare-root of the largest eigenvalue of LcLo.
45
5.5 Nehari’s theorem
In this section we look at the problem of finding the distance from an L∞-matrix R to H∞:
dist (R,H∞) := inf{‖R−X‖∞ : X ∈ H∞}
In systemic terms we want to approximate, in L∞-norm, a given unstable transfer matrix bya stable one. Nehari’s theorem is an elegant solution to this problem.
A lower bound for the distance is easily obtained. Fix X in H∞. Then
‖R−X‖∞ = ‖ΛR − ΛX‖
≥ ‖Π1(ΛR − ΛX)|H2‖
= ‖ΓR − ΓX‖
= ‖ΓR‖
The last equality is due to the fact that ΓX = 0. Thus ‖ΓR‖ is a lower bound for the distancefrom R to H∞. In fact it equals the distance.
Theorem 7 There exists a closest H∞-matrix X to a given L∞-matrix R, and ‖R−X‖ =‖ΓR‖.
In general there are many X’s nearest R. Interpreted in the time-domain Theorem 7 statesthat the distance from a given noncausal system to the nearest causal one (the systems beinglinear and time-invariant) equals the norm of the Hankel operator; in other words the normof the Hankel operator is a measure of noncausality.
5.6 Exercises
1. In the scalar-valued case prove that RL2 equals the set of all real-rational functions thatare strictly proper and have no poles on the imaginary axis.
2. Let
F (s) =s− 1
s + 1
Prove that FH∞ is closed in H∞.
3. Let
F (s) =s
s + 1
Prove that FH2 is not closed in H2.
46
4. Show that Ψc is surjective if (A, B) is controllable and that Ψo is injective if (C, A) isobservable.
5. Show that the adjoints of Ψc and Ψo are as follows:
Ψ∗c : Cn → L2[0,∞)
(Ψ∗cx)(t) = −B′e−A′tx, t ≥ 0
Ψ∗o : L2(−∞, 0] → Cn
Ψ∗oy =
∫ 0
−∞eA′tC ′y(t)dt
6. Prove that the matrix representations of ΨcΨ∗c and Ψ∗
oΨo are Lc and Lo respectively.
47
Chapter 6
Control Systems: Introduction
In this chapter we look at a general control system and some specific examples. All thecomponents in these systems are assumed to be linear, causal, and time-invariant, and aremodeled by their transfer functions that are assumed to be proper. Similarly, all signals arein the frequency domain.
The general setup is shown in Figure 1. In this figure w, u, z, and y are vector-valuedsignals: w is the exogenous input, typically consisting of command signals, disturbances,and sensor noises; u is the control signal; z is the output to be controlled, its componentstypically being tracking errors, filtered actuator signals, etc.; and y is the measured output.The transfer function G represents a generalized plant, the fixed part of the system, and Krepresents a controller. Partition G as
G =
[G11 G12
G21 G22
]Then Figure 1 stands for the algebraic equations
z = G11w + G12u
y = G21w + G22u
u = Ky
K
G
� �
�
-
z w
y u
Figure 1: Basic control system
48
6.1 Examples
Some common special cases of the general setup are described below.
Example 1: Filtering
Km m- - - -?
6
xn
y x z−
Figure 2: Filtering example
Figure 2 shows noise n added to a signal x to produce an observed signal y, that is passedthrough a filter K to yield an estimate x of x. The filtering error is z. The figure can beredrawn as Figure 1 by defining
w =
(nx
)u = x
G =
[ [0 I
]−I[
I I]
0
]
Example 2: A feedback system
F
K
W1
P W2m m
m
- - - - -
��
?
?��
6
y
uf
u
d
v
n
−
Figure 3: Single-loop feedback system
49
Figure 3 shows a standard single-loop feedback system with plant P , controller K, andfeedback sensor F . There are two exogenous inputs: a disturbance d and a noise n corruptingthe plant output. Also shown are two filters, W1 and W2, generating a filtered control signal,uf , and a filtered plant output, v. It is supposed that uf and v are both to be controlled. Toconvert to Figure 1 define
w =
(dn
)
z =
(vuf
)
G =
[
W2P 0
0 0
] [W2PW1
][−FP −F
]−FP
Example 3: A tracking control system
K2
PK1m- - - -
�
6
w u v
Figure 4: Tracking control system
Shown in Figure 4 is a plant P and a two degree-of-freedom controller, with componentsK1 and K2. Supposing the signal to be controlled is merely the tracking error between thereference input w and the plant output v, we get Figure 1 by defining
z = w − v
y =
(wv
)
G =
I −P[I0
] [0P
] K =
[K1 K2
]
50
Example 4: A model-matching system
T3 Q T2
T1
m-
?
- - -6
-w z
−
Figure 5: Model-matching system
Finally, Figure 5 shows a model, T1, that is to be approximated by the cascade of threeother systems, T2, Q, T3. The three components Ti, i=1,2,3, are assumed fixed, whereas Q isdesignable. The signal to be controlled is the error z between the two outputs for a commoninput w. We arrive at Figure 1 by setting
G =
[T1 −T2
T3 0
]
K = Q
Example 5: Bilateral hybrid telerobot
The setup is shown in Figure 6.1. Two robots, a master, Gm, and a slave, Gs, are controlled
Gm Gs
K
j j- - - � � �
6 6
?fh
fm fs
vm vs
fe− −
Figure 6.1: Bilateral hybrid telerobot.
by one controller, K. A human provides a force command, fh, to the master, while theenvironment applies a force, fe, to the slave. The controller measures the two velocities, vm
and vs, together with fe via a force sensor. In turn it provides two force commands, fm andfs, to the master and slave. Ideally, we want motion following (vs = vm), a desired mastercompliance (vm a desired function of fh), and force reflection (fm = fe).
We shall design K for two test inputs, namely, fe(t) is the finite-width pulse
fe(t) =
{10, 0 ≤ t ≤ 0.20, t > 0.2,
(6.1)
51
indicating an abrupt encounter between the slave and a stiff environment, and fh(t) is thetriangular pulse
fh(t) =
2t, 0 ≤ t ≤ 1
−2t + 4, 1 ≤ t ≤ 20, t > 2,
(6.2)
to mimic a ramp-up, ramp-down command.The generalized error vector z is taken to have four components: the velocity error vm−vs;
the compliance error fh−vm (for simplicity, the desired compliance is assumed to be vm = fh);the force-reflection error fm−fe; and the slave actuator force. The last component is includedas part of regularization, that is, to penalize excessive force applied to the slave. Introducingfour scalar weights to be decided later, we arrive at the generalized error vector
z =
αv(vm − vs)αc(fh − vm)αf (fm − fe)
αsfs
.
The Laplace transforms of fe and fh are not rational:
Fe(s) =10
s
(1− e−0.2s
), Fh(s) =
2
s2
(1− e−s
)2.
To get a tractable problem, we shall use second- and third-order Pade approximations,
e−Ts ≈[1− Ts
2+
(Ts)2
12
]/[1 +
Ts
2+
(Ts)2
12
]and
e−Ts ≈[1− Ts
2+
(Ts)2
10− (Ts)3
120
]/[1 +
Ts
2+
(Ts)2
10+
(Ts)3
120
].
Using the third-order approximation for Fe(s) and the second-order one for Fh(s), we get
Fe(s) ≈ 20
[0.2
2+
0.23s2
120
]/[1 +
0.2s
2+
(0.2s)2
10+
(0.2s)3
120
]=: Ge(s)
fh(s) ≈ 2
/(1 +
s
2+
s2
12
)2
=: Gh(s).
Incorporating these two prefilters into the preceding block diagram leads to Figure 6.2. Thetwo exogenous inputs wh and we are unit impulses. The vector of exogenous inputs is therefore
w =
[wh
we
].
52
Gh Gm Gs Ge
K
j j- - - - � � � �
6 6
?wh
fm fs
vm vs we
− −
Figure 6.2: Telerobot with prefilters.
P
K
z
y
w
u
� �
�
-
Figure 6.3: Telerobot configured in the standard form.
The control system is shown in Figure 6.3, where z and w are as above and
y =
vm
vs
fe
, u =
[fm
fs
].
Now for the problem. Begin with generic state models for Gh, Gm, Gs, Ge, namely,[Ah Bh
Ch 0
],
[Am Bm
Cm 0
],
[As Bs
Cs 0
],
[Ae Be
Ce 0
],
with corresponding states xh, xm, xs, xe. Use the interconnections in Figure 6.2 and definethe overall state
x =
xm
xs
xe
xh
.
In this way get a state model for P .
53
6.2 Well-posedness
We now wish to define what it means for the system in Figure 1 to be well-posed, and tocharacterize this condition. For this purpose, introduce two additional inputs, v1 and v2, toget Figure 6.
K
G
� �
m
m
� �
-?
6
-
z w
y
u
v2
v1
Figure 6: Well-posedness problem
Well-posedness means that the nine transfer functions from the three inputs (w, v1, v2)to the three signals (z, u, y) all exist and are proper. Since (w, u, v2) uniquely determine (z,y), it suffices to look at the three transfer functions from (w, v1, v2) to u.
Theorem 1 The system in Figure 1 is well-posed iff the matrix[I −K
−G22 I
](6.3)
has a proper inverse.
Proof Substitute
y = v2 + G21w + G22u
into
u = v1 + Ky
to get
u = v1 + K(v2 + G21w + G22u)
Hence
(I −KG22)u = v1 + K(v2 + G21w)
Thus the transfer function from v1 to u exists and is proper iff I−KG22 has a proper inverse.The latter condition then implies existence and properness of the other two transfer functionsfrom v2 and w to u. We conclude that the system is well-posed iff I − KG22 has a properinverse. But this is equivalent to proper invertibility of (6.3). �
A useful sufficient condition is provided by the following corollary.
Corollary 1 The system in Figure 1 is well-posed if G22 is strictly proper.
Strict properness of G22 means that in Figure 1 there’s attenuation at infinite frequencygoing around the internal loop from u to y and back to u again.
54
6.3 Exercises
1. Prove the corollary.
2. Apply the corollary to get sufficient conditions for well-posedness for the four examples inSection 6.1.
55
Chapter 7
Stability of Feedback Systems
This chapter continues with the general setup introduced in Chapter 6. The main objectiveis to parametrize all K’s that stabilize a given G.
7.1 Preliminaries
7.1.1 Controller parametrization
Consider the familiar unity-feedback setup
r e
d
u
!
K(s) P (s)
with the plant and controller being SISO, and with P (s) strictly proper and K(s) proper.There are two equivalent definitions of stability of the feedback system.
1. The first is a state definition. Take two minimal realizations of P and K; let the statevectors be xP , xK . Then xP (t) and xK(t) converge to zero as t tends to ∞ for everyinitial condition and with r = d = 0. One can test stability by finding the closed-loopA-matrix and checking if all its eigenvalues are in Re s < 0.
2. The second is a BIBO definition. For every bounded r, d, we have e, u bounded. Onecan test stability this way: Find coprime polynomials N, M and coprime polynomialsU, V such that
P =N
M, K =
U
V;
then check that the roots of the characteristic polynomial
NU + MV (7.1)
53
are in Re s < 0.
Let’s consider the simpler case where P is already stable and we want to characterize all Kthat maintain feedback stability.
Lemma 1 Assume P is stable. Then the feedback system is stable iff
(∃Q ∈ RH∞)K =Q
1− PQ.
Proof (=⇒) Let Q be the transfer function from r to u.(⇐=) We have[
eu
]=
[1− PQ −P (1− PQ)
Q 1− PQ
] [rd
].
�
Thus the formula
K =Q
1− PQ, Q ∈ RH∞
parametrizes all controllers that stabilize the feedback loop.When P is not stable, one could try to stabilize like this: Start with coprime numera-
tor/denominator P = N/M ; then find U, V to satisfy (7.1); then set K = U/V . This won’tquite work, because the resulting K is not guaranteed to be proper. The trick is to take thefactors in RH∞.
Before we see that, let’s look at Euclid’s algorithm for fun.
7.1.2 Euclid’s algorithm
Euclid’s algorithm is a way to find the GCD of two integers, f, g. The algorithm producesintegers x, y and d—the GCD—satisfying the equation
fx + gy = d.
We’re interested only in the case where f, g are coprime, in which case d = 1.Example The integers f = 66, g = 35 are coprime. It follows that there exist integers x, ysatisfying the equation
fx + gy = 1.
This is called variously a Bezout or Diophantine equation. The construction of x, y (they’renot unique) is as follows: First, divide g into f :
66
35= 1 +
31
35, 66 = 35 + 31.
54
Next, divide 31 into 35:
35
31= 1 +
4
31, 35 = 31 + 4.
Keep going until the remainder is 1. Divide 4 into 31:
31
4= 7 +
3
4, 31 = 7× 4 + 3.
Divide 3 into 4:4
3= 1 +
1
3, 4 = 3 + 1.
Now, backsubstitute; the underlined numbers (remainders) are replaced:
4− 3 = 1
4− (31− 7× 4) = 1
4× 8− 31 = 1
(35− 31)× 8− 31 = 1
35× 8− 31× 9 = 1
35× 8− (66− 35)× 9 = 1
−9× 66 + 17× 35 = 1.
Thus x = −9, y = 17.
Example The algorithm extends to polynomials, for example
f = s2 + 2s + 1, g = s3 + 4s2 + s− 6.
Divide f into g:
g
f= s + 2 +
−4s− 8
f.
Divide −4s− 8 into f :
f
−4s− 8= −1
4s +
1
−4s− 8.
Go backwards:
f − (4s + 8)1
4s = 1
f + [g − (s + 2)f ]1
4s = 1
f
(−1
4s2 − 1
2s + 1
)+ g
1
4s.
So
x = −1
4s2 − 1
2s + 1, y =
1
4s.
55
7.2 Coprime factorization
Recall that two polynomials f(s) and g(s), with, say, real coefficients, are said to be coprimeif their greatest common divisor is 1 (equivalently, they have no common zeros). It followsfrom Euclid’s algorithm that f and g are coprime iff there exist polynomials x(s) and y(s)such that
fx + gy = 1 (7.2)
We are going to take the practical route and define two functions f and g in RH∞ to becoprime (over RH∞) if there exist x, y in RH∞ such that (7.2) holds. The more primitive,but equivalent, definition is that f and g are coprime if every common divisor of f and g isinvertible in RH∞, i.e.,
h, fh−1, gh−1 ∈ RH∞ ⇒ h−1 ∈ RH∞
More generally, two matrices F and G in RH∞ are right-coprime (over RH∞) if theyhave equal number of columns and there exist matrices X and Y in RH∞ such that
[X Y ]
[FG
]= XF + Y G = I
This is equivalent to saying that the matrix[FG
]is left-invertible in RH∞.
Similarly, two matrices F and G in RH∞ are left-coprime (over RH∞) if they have equalnumber of rows and there exist X and Y in RH∞ such that
[F G]
[XY
]= FX + GY = I
Equivalently, [F G] is right-invertible in RH∞.Now let G be a proper real-rational matrix. A right-coprime factorization of G is a
factorization G = NM−1 where N and M are right-coprime RH∞-matrices. Similarly, a left-coprime factorization has the form G = M−1N where N and M are left-coprime. Of courseimplicit in these definitions is the requirement that M and M be square and nonsingular.We shall require special coprime factorizations, as described in the next lemma.
Lemma 2 For each proper real-rational matrix G there exist eight RH∞-matrices satisfyingthe equations
G = NM−1 = M−1N (7.3)[X −Y
−N M
] [M YN X
]= I (7.4)
56
Equations (7.3) and (7.4) together constitute a doubly-coprime factorization of G. Itshould be apparent that N and M are right-coprime and N and M are left-coprime; forexample, (7.4) implies
[X − Y ]
[MN
]= I
proving right-coprimeness.It’s useful to prove Lemma 1 constructively by deriving explicit formulas for the eight ma-
trices. The formulas use state-space realizations, and hence are readily amenable to computerimplementation.
We start with a state-space realization of G,
G(s) =
[A BC D
](7.5)
where A, B, C, D are real matrices with (A, B) stabilizable and (C, A) detectable. Nowintroduce state, input, and output vectors x, u, and y respectively so that y = Gu and
x = Ax + Bu (7.6)
y = Cx + Du (7.7)
Next, choose a real matrix F such that AF := A + BF is stable, i.e., all eigenvalues in theopen left half-plane, and define the vector v := u−Fx and the matrix CF := C +DF . Thenfrom (7.6) we get
x = AF x + Bv
u = Fx + v
y = CF x + Dv
Evidently from these equations the tf from v to u is
M(s) :=
[AF BF I
](7.8)
and that from v to y is
N(s) :=
[AF BCF D
](7.9)
Therefore
u = Mv, y = Nv
so that y = NM−1u, i.e., G = NM−1.
57
Similarly, by choosing a real matrix H so that AH := A + HC is stable and defining
BH := B + HD
M(s) :=
[AH HC I
](7.10)
N(s) :=
[AH BH
C D
](7.11)
we get G = M−1N . (This can be derived as above by starting with G′ instead of G.) Thuswe’ve obtained four matrices in RH∞ satisfying (7.3).
Formulas for the other four matrices to satisfy (7.4) are as follows:
X(s) :=
[AF −HCF I
](7.12)
Y (s) :=
[AF −HF 0
](7.13)
X(s) :=
[AH −BH
F I
](7.14)
Y (s) :=
[AH −HF 0
](7.15)
The explanation of where these latter four formulas come from is deferred to Section 7.5.
7.3 Stability
We begin this section with a definition of stability for the general setup shown in Figure 1.
K
G
� �
�
-
z w
y u
Figure 1: Basic setup
58
As before, G and K are proper real-rational tf’s. To guarantee well-posedness, we shallassume throughout this chapter that G22 is strictly proper. Again, bring in additional inputsas in Figure 2.
K
G
� �
m
m
� �
-?
6
-
z w
y
u
v2
v1
Figure 2: With additional inputs
The system is said to be internally stable, or K stabilizes G, provided the nine tf’s from(w, v1, v2) to (z, u, y) all belong to RH∞. An equivalent definition in terms of state modelsis this: take minimal state-space realizations of G and K and in Figure 1 set w = 0; then Kstabilizes G iff the state vectors of G and K converge to zero from every initial condition.
We would like a test for when K stabilizes G. Introduce left- and right-coprime factor-izations
G = NM−1 = M−1N (7.16)
K = UV −1 = V −1U (7.17)
Also, introduce two constant matrices,
Eu :=
[0I
]of size (dim w + dim u) × dim u, and
Ey :=
[0I
]of size (dim z + dim y) × dim y.
Theorem 1 The following are equivalent statements about K:
(i) K stabilizes G
(ii)
[M EuU
E ′yN V
]is invertible in RH∞
(iii)
[M NEu
UE ′y V
]is invertible in RH∞
59
The idea underlying the equivalence of the first two conditions is simply that the deter-minant of the matrix in the second is the least common denominator (in RH∞) of all thetf’s from w, v1, v2 to z, u, y; hence the determinant must be invertible for all these tf’s tobelong to RH∞, and conversely.
The proof of Theorem 1 requires a preliminary result. Insert the factorizations (7.9) intoFigure 2, split apart the factors, and introduce two new signals ξ and η to get Figure 3.
U V −1
M−1 Nm
m
- -
--
-
?���
6v1 u
w
η
ξ
y
z
v2
Figure 3: For Lemma 2
Lemma 3 The nine tf ’s in Figure 2 from w, v1, v2 to z, u, y belong to RH∞ iff the six tf ’sin Figure 3 from w, v1, v2 to ξ, η belong to RH∞.
Proof (If) This direction follows immediately from the equations[zy
]= Nξ +
[0v2
]u = Uη + v1
which in turn follow from Figure 3.(Only if) By right-coprimeness there exist RH∞-matrices X and Y such that
XM + Y N = I
Hence
ξ = XMξ + Y Nξ (7.18)
But from Figure 3.
Mξ =
[wu
]
Nξ =
[z
y − v2
]Substitution into (7.18) gives
ξ = X
[0u
]+ Y
[zy
]+ X
[w0
]− Y
[0v2
]60
Hence the three transfer matrices from w, v1, v2 to ξ belong to RH∞.A similar argument works for the remaining three transfer matrices to η. �
Proof of Theorem 1 We shall prove the equivalence of the first and second conditions.First, let’s see that the matrix displayed in the second is indeed nonsingular, i.e., its inverseexists as a rational matrix. We have[
M EuUE ′
yN V
]=
[I EuK
E ′yG I
] [M 00 V
]=
I 0 00 I K
G21 G22 I
[M 00 V
](7.19)
Now [M 00 V
]is nonsingular because both M and V are. Also, since G22 is strictly proper, we have that I 0 0
0 I KG21 G22 I
is nonsingular when evaluated at s =∞: its determinant equals 1 there. Thus both matriceson the right-hand side of (7.19) are nonsingular.
The equations corresponding to Figure 3 are[M −EuU
−E ′yN V
] [ξη
]=
wv1
v2
Thus by Lemma 3 K stabilizes G iff[
M −EuU−E ′
yN V
]is invertible in RH∞. But this is equivalent to the second condition. �
7.4 Stabilizability
Let’s say that G is stabilizable if there exists a (proper real-rational) K that stabilizes it.Not every G is stabilizable; an obvious non-stabilizable G is G12 = 0, G21 = 0, G22 = 0,G11 unstable. In this example, the unstable part of G is disconnected from u and y. Interms of a state-space model G is stabilizable iff its unstable modes are controllable from u(stabilizability) and observable from y (detectability). The next result is a stabilizability testin terms of left- and right-coprime factorizations
G = NM−1 = M−1N
61
Theorem 2 The following conditions are equivalent:
(i) G is stabilizable
(ii) M , E ′yN are right-coprime and M , Eu are left-coprime
(iii) M , NEu are left-coprime and M , E ′y are right-coprime
The proof requires some preliminaries. The reader will recall the following fact. For eachreal matrix F there exist real matrices G and H such that
F = G
[I 00 0
]H
The matrices G and H may be obtained by elementary row and column operations, andthe size of the identity matrix equals the rank of F . The following analogous result forRH∞-matrices is stated without proof.
Lemma 4 For each matrix F in RH∞ there exist matrices G, H, and F1 in RH∞ satisfyingthe equation
F = G
[F1 00 0
]H
and having the properties that G and H are invertible in RH∞ and F1 is diagonal andnonsingular.
This result is now used to prove the following useful fact that if M and N are right-
coprime, then the matrix
[MN
]can be filled out to yield a square matrix that is invertible
in RH∞.
Lemma 5 Let M and N be RH∞-matrices with equal number of columns. Then M and Nare right-coprime iff there exist matrices U and V in RH∞ such that[
M UN V
]is invertible in RH∞.
Proof (If) Define[X Y? ?
]:=
[M UN V
]−1
where a question mark denotes an irrelevant block. Then[X Y
] [MN
]= I
62
so M and N are right-coprime.(Only if) Define
F :=
[MN
]and bring in matrices G, H, and F1 as per Lemma 4. Since F is left-invertible in RH∞ (byright-coprimeness), it follows that[
F1 00 0
]is left-invertible in RH∞ too. But then it must have the form[
F1
0
]with F−1
1 in RH∞. Defining
K := G
[F1H 0
0 I
]we get [
MN
]= K
[I0
]Thus the definition[
UV
]:= K
[0I
]gives the desired result, that[
M UN V
]= K
is invertible in RH∞. �
The obvious dual of Lemma 5 is that M and N are left-coprime iff there exist U and Vsuch that[
M NU V
]is invertible in RH∞.
Proof of Theorem 2 We shall prove equivalence of (i) and (ii).Suppose G is stabilizable. Then by Theorem 1 there exist U and V in RH∞ such that[
M EuUE ′
yN V
]63
is invertible in RH∞. This implies by Lemma 5 and its dual that M , E ′yN are right-coprime
and M , EuU are left-coprime. But the latter condition implies left-coprimeness of M , Eu.Conversely, assume the second condition in the theorem. Choose, by right-coprimeness
and Lemma 5, matrices X and Y in RH∞ such that[M X
E ′yN Y
]is invertible in RH∞. Also, choose, by left-coprimeness, matrices R and T in RH∞ suchthat
[M Eu]
[RT
]= I (7.20)
Now define
U := TX (7.21)
V := Y − E ′yNRX (7.22)
Then we have from (7.20), (7.21), and (7.22) that[M X
E ′yN Y
] [I −RX0 I
]=
[M EuU
E ′yN V
](7.23)
The two matrices on the left in (7.23) have inverses in RH∞, hence so does the matrix onthe right.
The next step is to show that V is nonsingular. We have[M EuU
E ′yN V
]=
[I EuU
E ′yG V
] [M 00 I
]
=
I 0 00 I U
G21 G22 V
[M 00 I
](7.24)
Evaluate all the matrices in (7.24) at s = ∞; then take determinants of both sides notingthat G22 is strictly proper and the matrix on the left-hand side of (7.24) is invertible in RH∞.This gives
0 6= detV (∞)detM(∞)
Thus detV (∞) 6= 0, i.e., V −1 exists. Hence we can define K := UV −1.Next, note that U and V are right-coprime (this follows from invertibility in RH∞ of the
matrix on the right-hand side of (7.23)). We conclude from Theorem 1 that K stabilizes G.�
Hereafter, G will be assumed to be stabilizable. Intuitively, this implies that G and G22
share the same unstable poles (counting multiplicities), so to stabilize G it is enough tostabilize G22. Let’s define the latter concept explicitly: K stabilizes G22 if in Figure 4 thefour tf’s from v1 and v2 to u and y belong to RH∞.
64
K
G22m
m
- -
?��
6v1 u
y v2
Figure 4: For Theorem 3
Theorem 3 K stabilizes G iff K stabilizes G22.
The necessity part of the theorem follows from the definitions. To prove sufficiency weneed a result analogous to Lemma 3.
Lemma 6 The four transfer matrices in Figure 4 from v1, v2 to u, y belong to RH∞ iff thefour transfer matrices in Figure 3 from v1, v2 to ξ, η belong to RH∞.
The proof is omitted, it being entirely analogous to that of Lemma 3.
Proof of Theorem 3 Suppose K stabilizes G22. To prove that K stabilizes G it suffices toshow, by Lemma 3, that the six transfer matrices in Figure 3 from w, v1, v2 to ξ, η belongto RH∞. But by Lemma 6 we know that those from v1, v2 to ξ, η do. So it remains to showthat the two from w to ξ, η belong to RH∞.
Set v1 = 0 and v2 = 0 in Figure 3 and write the corresponding equations:
Mξ =
[wUη
](7.25)
V η = E ′yNξ (7.26)
By left-coprimeness there exist matrices R and T in RH∞ such that
[M Eu]
[RT
]= I (7.27)
Post-multiply (7.27) by
[w0
]to get
MR
[w0
]+ EuT
[w0
]=
[w0
](7.28)
Now subtract (7.28) from (7.25), rearrange, and define
ξ1 := ξ −R
[w0
](7.29)
v1 := T
[w0
](7.30)
65
to get
Mξ1 = Eu(v1 + Uη) (7.31)
Also, rearrange (7.26) and define
v2 := E ′yNR
[w0
](7.32)
to get
V η = E ′yNξ1 + v2 (7.33)
The block diagram corresponding to (7.31) and (7.33) is Figure 5:
U V −1
M−1 Nm
m
- -
--
-
?���
6v1
0
η1
ξ
v2
Figure 5: For proof of Theorem 3
By Lemma 5 and the fact that K stabilizes G22 we know that the tf’s in Figure 5 fromv1, v2 to ξ1, η belong to RH∞. But by (7.30) and (7.32) those from w to v1, v2 belong toRH∞. Hence those from w to ξ1, η belong to RH∞. Finally, we conclude from (7.29) thatthe tf from w to ξ belongs to RH∞. �
7.5 Parametrization
This section contains a parametrization of all K’s that stabilize G22. To simplify notationslightly, in this section the subscript 22 on G22 is dropped. The relevant block diagram isFigure 6.
K
Gm
m
- -
?��
6v1 u
y v2
Figure 6: Basic loop
66
Bring in a doubly-coprime factorization of G,
G = NM−1 = M−1N[X −Y
−N M
] [M YN X
]= I (7.34)
and coprime factorizations (not necessarily doubly-coprime) of K,
K = UV −1 = V −1U
The first result is analogous to Theorem 1; the proof is omitted.
Lemma 7 The following are equivalent statements about K:
(i) K stabilizes G
(ii)
[M UN V
]is invertible in RH∞
(iii)
[V −U
−N M
]is invertible in RH∞
The main result of this chapter is the following.
Theorem 4 The set of all (proper real-rational) K’s stabilizing G is parametrized by theformulas
K = (Y −MQ)(X −NQ)−1 (7.35)
= (X −QN)−1(Y −QM) (7.36)
Q ∈ RH∞
Proof Let’s first prove (7.36). Let Q ∈ RH∞. From (7.34) we have[I Q0 I
] [X −Y
−N M
] [M YN X
] [I −Q0 I
]= I
so that[X −QN −(Y −QM)
−N M
] [M Y −MQN X −NQ
]= I (7.37)
Equating the (1,2)-blocks on each side in (7.37) gives
(X −QN)(Y −MQ) = (Y −QM)(X −NQ)
which is equivalent to (7.36).
67
Next, we show that if K is given by (7.35), it stabilizes G. Define
U := Y −MQ, V := X −NQ
U := Y −QM, V := X −QN
to get from (7.37) that[V −U
−N M
] [M UN V
]= I (7.38)
It follows from (7.38) that U , V are right-coprime and U , V are left-coprime (Lemma 5).Also from (7.38)[
M UN V
]is invertible in RH∞. So from Lemma 7 K stabilizes G.
Finally, suppose K stabilizes G. We must show K satisfies (7.35) for some Q in RH∞.Let K = UV −1 be a right-coprime factorization. From (7.34) and defining D := MV − NUwe have[
X −Y
−N M
] [M UN V
]=
[I XU − Y V0 D
](7.39)
The two matrices on the left in (7.39) have inverses in RH∞, the second by Lemma 7. HenceD−1 ∈ RH∞. Define
Q := −(XU − Y V )D−1
so that (7.39) becomes[X −Y
−N M
] [M UN V
]=
[I −QD0 D
](7.40)
Pre-multiply (7.40) by[M YN X
]and use (7.34) to get[
M UN V
]=
[M YN X
] [I −QD0 D
]Therefore[
UV
]=
[(Y −MQ)D(X −NQ)D
]68
Substitute this into K = UV −1 to get (7.35). �
As a special case suppose G is already stable, i.e., G ∈ RH∞. Then in (7.34) we maytake
N = N = G
X = M = I, X = M = I
Y = 0, Y = 0
in which case the formulas in the theorem become simply
K = −Q(I −GQ)−1
= −(I −QG)−1Q
There is an interpretation of Q in this case: −Q equals the tf from v2 to u in Figure 6.We can now explain the idea behind the choice of X, Y , X, Y in Section 7.2. Recall that
the state-space equations for G were
x = Ax + Bu
y = Cx + Du
that
AF := A + BF, AH := A + HC
were stable, and that we defined
BH := B + HD, CF := C + DF
Let’s find a stabilizing K by observer theory. The familiar state-space equations for Kare
˙x = Ax + Bu + H[Cx + Du− y]
u = Fx
or equivalently
˙x = Ax + By
u = Cx
where
A := A + BF + HC + HDF = AF + HCF
69
B := −H
C := F
Thus
K(s) =
[A B
C 0
]By observer theory K stabilizes G.
Now find coprime factorizations of K in the same way as we found coprime factorizationsof G in Section 7.2. To get a right-coprime factorization K = Y X−1 we first choose F so thatAF := A + BF is stable. It is convenient to take F := CF , so that AF = AF . By analogywith (7.7) we get K = Y X−1, where
X(s) :=
[AF B
F I
]=
[AF −HCF I
]Y (s) :=
[AF B
C 0
]=
[AF −HF 0
]A similar derivation leads to a left-coprime factorization K = X−1Y , where
X(s) :=
[AH −BH
F I
]Y (s) :=
[AH −HF 0
]These formulas coincide with (7.8).
By Lemma 7 we know that[M YN X
]and [
X −Y
−N M
]are invertible in RH∞. Hence the product[
X −Y
−N M
] [M YN X
]is too. The only surprise is that the product equals the identity matrix, as is verified byalgebraic manipulation.
70
7.6 Closed-loop tf’s
Now we return to the standard setup of Figure 1. Theorem 4 gives every stabilizing K as atransformation of a free parameter Q in RH∞. The objective in this section is to find the tffrom w to z in terms of Q.
In the previous section we dropped the subscript on G22; now we must restore it. Bringin a doubly-coprime factorization of G22:
G22 = N22M−122 = M−1
22 N22[X22 −Y22
−N22 M22
] [M22 Y22
N22 X22
]= I (7.41)
Then the formula for K is
K = (Y22 −M22Q)(X22 −N22Q)−1 (7.42)
= (X22 −QN22)−1(Y22 −QM22) (7.43)
Now define
T1 := G11 + G12M22Y22G21 (7.44)
T2 := G12M22 (7.45)
T3 := M22G21 (7.46)
Theorem 5 The matrices Ti (i = 1–3) belong to RH∞. With K given by (7.42) the transfermatrix from w to z equals T1 − T2QT3.
Proof The first statement follows from the realizations to be given below. For the secondstatement we have
z = [G11 + G12(I −KG22)−1KG21]w (7.47)
Substitute G22 = N22M−122 and (7.43) into (I −KG22)
−1 and use (7.41) to get
(I −KG22)−1 = M22(X22 −QN22)
Thus from (7.43) again
(I −KG22)−1K = M22(Y22 −QM22)
Substitute this into (7.47) and use the definitions of Ti to get z = (T1 − T2QT3)w. �
Compare Theorem 5 with the model-matching example of Chapter 4: the tf from w to zis the same in both cases, namely T1− T2QT3. The conclusion is that, when the controller isparametrized as in (7.42), the general setup reduces to the model-matching setup.
71
For computations it is useful to have explicit realizations of the transfer matrices Ti. Startwith a minimal realization of G:
G(s) =
[A BC D
]Since the input and output of G are partitioned as[
wu
] [zy
]the matrices B, C, and D have corresponding partitions:[
B1 B2
][
C1
C2
][
D11 D12
D21 D22
]Then
G11(s) =
[A B1
C1 D11
]G12(s) =
[A B2
C1 D12
]
G21(s) =
[A B1
C2 D21
]G22(s) =
[A B2
C2 D22
]Note that D22 = 0 because G22 is strictly proper. It can be proved that stabilizability of G(an assumption from Section 7.4) implies that (A, B2) is stabilizable and (C2, A) is detectable.
Next, find a doubly-coprime factorization of G22 as developed in Section 7.2. For thischoose F and H so that
AF := A + B2F, AH := A + HC2
are stable. Then the formulas are as follows:
M22(s) =
[AF B2
F I
]
N22(s) =
[AF B2
C2 0
]
M22(s) =
[AH HC2 I
]
N22(s) =
[AH B2
C2 0
]72
X22(s) =
[AF −HC2 I
]
Y22(s) =
[AF −HF 0
]
X22(s) =
[AH −B2
F I
]
Y22(s) =
[AH −HF 0
]Finally, substitution into (7.44), (7.45), and (7.46) yields the following realizations:
T1(s) =
[A BC D11
]
A =
[AF −B2F0 AH
]
B =
[B1
B1 + HD21
]C =
[C1 + D12F −D12F
]T2(s) =
[AF B2
C1 + D12F D12
]
T3(s) =
[AH B1 + HD21
C2 D21
]It can be observed that Ti ∈ RH∞ as claimed in Theorem 5.
The results of this chapter can be summarized as follows. The matrix G is assumed to beproper, with G22 strictly proper. Also, G is assumed to be stabilizable. The formula (7.42)parametrizes all K’s that stabilize G. In terms of the parameter Q the tf from w to z equalsT1 − T2QT3. Such a function of Q is called affine.
7.7 Exercises
1. Prove equivalence of (i) and (iii) in Theorem 1.
2. Prove equivalence of (i) and (iii) in Theorem 2.
3. Suppose G11 = G12 = G21 = G22. Is G stabilizable?
73
4. In Figure 6 take G(s) = 1/[s(s− 1)]. Consider a controller of the form
K =−Q
1−GQ
where Q is real-rational. Find necessary and sufficient conditions on Q in order that Kstabilize G.
5. Consider Figure 6 again with both G and K strictly proper. Start with a stabilizable,detectable realization of G. Get realizations for the eight matrices in a doubly-coprimefactorization of G as in Section 7.2. Parametrize K as in Theorem 4. Now reconfigure K as
Q
J
� �
�
-
u y
r q
Derive a realization of J .
6. Continue with the previous exercise as follows. Suppose Q is strictly proper and stable.Take a minimal realization,
Q(s) =
[Aa Ba
Ca 0
]Define the augmentation
Ga(s) =
[Aa 00 0
]and then define the extended plant Ge = G + Ga (take the state of Ge to be the stack of thesummand states). Show that K is an observer-based compensator for Ge.
Conclusion: every internal stabilization amounts to adding stable dynamics and thenstabilizing the extended plant by an observer-based compensator.
7. Consider a single-loop feedback system with plant P , controller C, negative-unity feedback.Set
P (s) =s + 1
(s− 1)(s− 2)
Parametrize all proper C’s for which the feedback system is stable. Choose a parameter sothat the dc gain from reference input to tracking error equals 0. Compute the resulting C.
74
Chapter 8
H2-Optimal Control
This chapter gives two different approaches to an H2-optimal control problem: one basedon the controller parametrization of the previous chapter, and the traditional state-spaceapproach.
8.1 Preliminary example
An LQR example via H2 optimal control. The problem is
x = −x + u, x(0−) = 1
J =
∫ ∞
0
(x2 + u2)dt.
You can check via LQR that the optimal control is u = Fx, F = −1/(1 +√
2).Move the initial condition to an input:
x = −x + u + w, w = δ.
Define a measured output y = x and a controller output
z =
[z1
z2
]=
[xu
].
Then J = ‖z‖22.The block diagram is
P (s)
K(s)
w x
u
z1
z2
!
74
where P (s) = 1/(s + 1).Then J equals the square of the H2-norm from w to z. Parametrize K(s):
K =Q
1− PQ, Q ∈ RH∞.
We have
Tzw =
P
1 + PKPK
1 + PK
=
[P (1− PQ)
PQ
].
Thus
J = ‖P (1− PQ)‖22 + ‖PQ‖22and the problem reduces to
minQ∈RH∞
‖P − P 2Q‖22 + ‖PQ‖22.
Bring in the notation F (s) = F (−s). So for example
〈FR, Q〉 = 〈R, FQ〉.
Then we get
J = 〈P − P 2Q,P − P 2Q〉+ 〈PQ,PQ〉
J = 〈P, P 〉 − 〈P, P 2Q〉 − 〈P 2Q,P 〉+ 〈P 2Q,P 2Q〉+ 〈PQ,PQ〉
J = ‖P‖2 − 〈P 2P, Q〉 − 〈Q, P 2P 〉+ 〈P 2P 2Q,Q〉+ 〈PPQ,Q〉
J = ‖P‖2 − 〈P 2P, Q〉 − 〈Q, P 2P 〉+ 〈(P 2P 2 + PP )Q,Q〉.
Let’s try to factor P 2P 2 + PP as T2T2, with T2 stable and minimum phase. This is calledspectral factorization. In our example we have
P 2P 2 + PP =1
1− s2
[1
1− s2+ 1
]=
1
1− s2
2− s2
1− s2,
so
T2(s) =s +√
2
(s + 1)2.
So we have
J = ‖P‖2 − 〈P 2P, Q〉 − 〈Q, P 2P 〉+ 〈T2Q, T2Q〉.
75
Now let’s try to write J as
J = c + ‖T1 − T2Q‖22.This is completing the square. We require
‖P‖2 − 〈P 2P, Q〉 − 〈Q, P 2P 〉+ 〈T2Q, T2Q〉 = c + ‖T1‖2 − 〈T2T1, Q〉 − 〈Q, T2T1〉+ 〈T2Q, T2Q〉,so it suffices to take
P 2P = T2T1,
which gives
T1(s) =1
(1 + s)(√
2− s).
Thus T1 ∈ RL2. The projection of T1 onto RH2 is
1
1 +√
2
1
s + 1=: T1a(s).
So now we have
J = const. + ‖T1a − T2Q‖22.In this case T1a/T2 ∈ RH∞, so that’s the optimal Q:
Q(s) =1
1 +√
2
1 + s√2 + s
.
Finally, the optimal K is
K =Q
1− PQ=
1
1 +√
2.
Conclusion: We got the optimal state feedback without a Riccati equation. What replacedit is spectral factorization.
8.2 Problem statement
We begin with the standard setup:
K
G
� �
�
-
z w
y u
It is assumed that G is proper and stabilizable and G22 is strictly proper. The input w isstandard white noise. The problem is to design a proper K that stabilizes G and minimizesthe rms value of z, or equivalently, from the continuous analog of Theorem 4.1, minimizesthe H2-norm of the transfer function from w to z.
76
8.3 Solution via parametrization
In this section only a sketch of the solution is given. The method of solution involves thefollowing steps:
Step 1 Get a doubly-coprime factorization of G.
Step 2 Parametrize as in Section 7.5 all proper K’s that stabilize G. Then the tf from w to zhas the form T1 − T2QT3.
Step 3 Choose Q in RH∞ to minimize
‖T1 − T2QT3‖2
Step 4 Get K from Q by back-substitution.
The only new step is the third. In view of Example 4 in Chapter 6, we call this theH2-model matching problem: given Ti ∈ RH∞, i = 1, 2, 3, find Q in RH∞ to minimize
‖T1 − T2QT3‖2
We’ll solve this first when the Ti’s are 1× 1.
8.3.1 Scalar-valued case
We may as well suppose T3 = 1. The problem is
minQ∈RH∞
‖T1 − T2Q‖2
where T1, T2 are given functions in RH∞.For the solution we need the concepts of inner and outer functions. For a real-rational
function F define F∼(s) := F (−s). Now let F ∈ RH∞. Then F is inner if F∼F = 1 andouter if it has no zeros in Re s > 0. It is routine to derive this basic fact: Every functionin RH∞ has an inner-outer factorization, i.e., for every F in RH∞ there exist an inner Fi
and an outer Fo such that F = FiFo. Moreover, if F−1 is proper and has no poles on theimaginary axis, then F−1
o ∈ RH∞.For a solution to the model-matching problem we shall assume that T−1
2 is proper andanalytic on the imaginary axis, i.e., T−1
2 ∈ RL∞. The idea of the solution is easy to explainwhen T1 ∈ RH2. We want to find the distance, in H2-norm, from T1 to the subspace T2RH2,and also a closest function in this subspace. The assumption just made assures that thesubspace T2RH2 is closed, and this assures in turn that there exists a closest function. Thisclosest function is unique and is obtained by taking the orthogonal projection from T1 ontoT2RH2.
Fix Q in RH∞ such that T1 − T2Q belongs to RH2. Such Q exists, for example,
Q(s) ≡ T1(∞)/T2(∞)
77
Now do an inner-outer factorization
T2 = UiUo
By the assumption on T2, Uo is invertible in RH∞. Then
‖T1 − T2Q‖2 = ‖Ui(T1U∼i − UoQ)‖2
Using the following trivial fact
F ∈ RL2, U inner ⇒ ‖UF‖2 = ‖F‖2we get
‖T1 − T2Q‖2 = ‖T1U∼i − UoQ‖2
Define
R := T1U∼i , X := UoQ
and note that the mapping
Q 7→ X : RH∞ → RH∞
is bijective. Also, R ∈ RL∞ and R−X ∈ RL2, so R(∞) = X(∞). Write uniquely
X = R(∞) + X1, X1 ∈ RH2
Recall that Π1 and Π2 are the orthogonal projections from L2 onto, respectively, H⊥2 and
H2. Then
‖T1 − T2Q‖22 = ‖R−X‖22= ‖R−R(∞)−X1‖22= ‖Π1[R−R(∞)−X1]‖22 + ‖Π2[R−R(∞)−X1]‖22= ‖Π1[R−R(∞)]‖22 + ‖Π2[R−R(∞)]−X1‖22
Conclusion: the optimal X1 is
Π2[R−R(∞)]
so the optimal X equals
R(∞) + Π2[R−R(∞)]
i.e., X equals the stable proper part of R.
Theorem 1 Under the assumption that T−12 is proper and analytic on the imaginary axis,
there exists a unique optimal Q, namely,
Q = U−1o {R(∞) + Π2[R−R(∞)]}
and for this Q the model-matching error is
‖T1 − T2Q‖2 = ‖Π1[R−R(∞)]‖2
78
8.3.2 Matrix-valued case
The method is the same but some of the technicalities are harder. The treatment will besomewhat terse.
For a real-rational matrix F define F∼(s) := F (−s)′. A matrix F in RH∞ is inner ifF∼F = I and outer if the rank of F (s) equals its number of rows at all points in Re s > 0.Every matrix F in RH∞ has an inner-outer factorization. If F (jω) has constant rank for all0 ≤ ω ≤ ∞, then its outer factor has a right-inverse in RH∞.
Assume T2(jω) and T3(jω) have constant rank for all 0 ≤ ω ≤ ∞. Do inner-outerfactorizations:
T2 = UiUo, T ′3 = ViVo
Then Uo and Vo are both right-invertible in RH∞.Necessary and sufficient conditions for the existence of Q in RH∞ such that T1−T2QT3 ∈
RH2 are
Im T2(∞) ⊂ Im T1(∞), Ker T3(∞) ⊂ Ker T1(∞)
Assume these and fix such Q. Then
T1 − T2QT3 = T1 − UiUoQV ′oV
′i
Define X := UoQV ′o . The mapping
Q 7→ X : RH∞ → RH∞
is surjective. So the model-matching problem reduces to
minX∈RH∞
‖T1 − UiXV ′i ‖2
You can check that the matrix
W1 :=
[U∼
i
I − UiU∼i
]has the property W∼
1 W1 = I. This impies that
‖W1G‖2 = ‖G‖2, G ∈ RL2
Hence
‖T1 − UiXV ′i ‖2 = ‖W1(T1 − UiXV ′
i )‖2
Similarly, with
W2 :=
[V ∼
i
I − ViV∼i
]
79
we get
‖T1 − UiXV ′i ‖2 = ‖W1(T1 − UiXV ′
i )W′2‖2
But
W1Ui =
[I0
]V ′
i W′2 =
[I 0
]and
W1T1W′2 =
[U∼
i T1V∼′i ?
? ?
](“?” denotes irrelevant). Defining
R := U∼i T1V
∼′i
we get
‖T1 − UiXV ′i ‖22 =
∥∥∥∥[R−X ?
? ?
]∥∥∥∥2
2
= ‖R−X‖22 + ‖?‖22 + ‖?‖22 + ‖?‖22
We conclude that the unique optimal X is
X = R(∞) + Π2[R−R(∞)]
You can check that X again equals the stable proper part of R.
Theorem 2 Under the assumptions that T2(jω) and T3(jω) have constant rank for all 0 ≤ω ≤ ∞ and that
Im T2(∞) ⊂ Im T1(∞), Ker T3(∞) ⊂ Ker T1(∞)
there exists an optimal Q.
Here’s the procedure for computing an optimal Q:
Step 1 Do inner-outer factorizations:
T2 = UiUo, T ′3 = ViVo
Step 2 Compute
R := U∼i T1V
∼′i
X = R(∞) + Π2[R−R(∞)]
80
Step 3 Solve the equation
UoQV ′o = X
for Q in RH∞.
This procedure is at a high level. Considerable work is involved in reducing it to imple-mentable steps.
For the state-space solution we need some preliminaries about Lyapunov and Riccatiequations.
8.4 Lyapunov equation
The equation
A′X + XA + Q = 0
is called a Lyapunov equation. Here A, Q, X are all square matrices, say n × n, with Qsymmetric.
One situation is where A and Q are given and the equation is to be solved for X. Existenceand uniqueness are easy to establish in principle. Define the linear map
Φ : Rn×n → Rn×n, Φ(X) = A′X + XA
Then the Lyapunov equation has a solution X iff Q ∈ Im Φ; if this condition holds, thesolution is unique iff Φ is injective, hence bijective. It can be shown that
σ(Φ) = {λ1 + λ2 : λ1, λ2 ∈ σ(A)}So the Lyapunov equation has a unique solution iff A has the property that no two of itseigenvalues add to zero. For example, if A is stable, the unique solution is
X =
∫ ∞
0
eA′tQeAtdt
We’ll be more interested in another situation: where we want to infer stability of A.
Theorem 3 Suppose A, Q, X satisfy the Lyapunov equation, (Q,A) is detectable, and Qand X are positive semi-definite. Then A is stable.
Proof For a proof by contradiction, suppose A has some eigenvalue λ with Re λ ≥ 0. Let xbe a corresponding eigenvector. Pre-multiply the Lyapunov equation by x∗ and post-multiplyby x to get
(2Re λ)x∗Xx + x∗Qx = 0
Both terms on the left are ≥ 0. Hence x∗Qx = 0, which implies that Qx = 0 since Q ≥ 0.Thus [
A− λQ
]x = 0
By detectability we must have x = 0, a contradiction. �
81
8.5 Riccati equation
Let A, Q, R be real n× n matrices with Q and R symmetric. Define the 2n× 2n matrix
H :=
[A RQ −A′
]A matrix of this form is called a Hamiltonian matrix.
It is claimed that σ(H) is symmetric about the imaginary axis. To prove this, introducethe 2n× 2n matrix
J :=
[0 −II 0
]having the property J2 = −I. Then
J−1HJ = −JHJ = −H ′
so H and −H ′ are similar. Thus λ is an eigenvalue iff −λ is.Now assume H has no eigenvalues on the imaginary axis. Then it must have n in Re
s < 0 and n in Re s > 0. Thus the two spectral subspaces X−(H) and X+(H) both havedimension n. Let’s focus on X−(H). Finding a basis for it, stacking the basis vectors up toform a matrix, and partitioning the matrix, we get
X−(H) = Im
[X1
X2
]where X1, X2 ∈ Rn×n. If X1 is nonsingular, i.e. if the two subspaces
X−(H), Im
[0I
]are complementary, we can set X := X2X
−11 to get
X−(H) = Im
[IX
]Notice that X is then uniquely determined by H, i.e. H 7→ X is a function. We shall denotethis function by Ric and write X = Ric(H).
To recap, Ric is a function R2n×2n → Rn×n that maps H to X where
X−(H) = Im
[IX
]The domain of Ric, denoted dom Ric, consists of Hamiltonian matrices H with two properties,namely, H has no eigenvalues on the imaginary axis and the two subspaces
X−(H), Im
[0I
]are complementary.
Some properties of X are given below.
82
Lemma 1 Suppose H ∈ dom Ric and X = Ric(H). Then
(i) X is symmetric
(ii) X satisfies the algebraic Riccati equation
A′X + XA + XRX −Q = 0
(iii) A + RX is stable
Proof (i) Let X1, X2 be as above. It’s claimed that
X ′1X2 is symmetric (8.1)
To prove this, note that there exists a stable matrix H− in Rn×n such that
H
[X1
X2
]=
[X1
X2
]H−
(H− is a matrix representation of H|X−(H).) Pre-multiply this equation by[X1
X2
]′J
to get [X1
X2
]′JH
[X1
X2
]=
[X1
X2
]′J
[X1
X2
]H− (8.2)
Now JH is symmetric; hence so is the left-hand side of (8.2); hence so is the right:
(−X ′1X2 + X ′
2X1)H− = H ′−(−X ′
1X2 + X ′2X1)
′
= −H ′−(−X ′
1X2 + X ′2X1)
This is a Lyapunov equation. Since H− is stable, the unique solution is
−X ′1X2 + X ′
2X1 = 0
This proves (8.1).We have XX1 = X2. Pre-multiply by X ′
1 and then use (8.1) to get that X ′1XX1 is
symmetric. Since X1 is nonsingular, this implies that X is symmetric too.(ii) Start with the equation
H
[X1
X2
]=
[X1
X2
]H−
and extract X1 to get
H
[IX
]X1 =
[IX
]X1H−
83
Post-multiply by X−11 :
H
[IX
]=
[IX
]X1H−X−1
1 (8.3)
Now pre-multiply by [X − I]:
[X − I]H
[IX
]This is precisely the Riccati equation.
(iii) Pre-multiply (8.3) by [I 0] to get
A + RX = X1HX−11
Thus A + RX is stable because H− is. �
The following result gives verifiable conditions under which H belongs to dom Ric.
Theorem 4 Suppose H has the form
H =
[A −BB′
−C ′C −A′
]with (A, B) stabilizable and (C, A) detectable. Then H ∈ dom Ric and Ric(H) ≥ 0. If (C, A)is observable, then Ric(H) > 0.
Proof We’ll first show that H has no imaginary eigenvalues. Suppose, on the contrary, that
jω is an eigenvalue and
(xz
)a corresponding eigenvector. Then
Ax−BB′z = jωx
−C ′Cx− A′z = jωz
Re-arrange:
(A− jω)x = BB′z (8.4)
−(A− jω)∗z = C ′Cx (8.5)
Thus
〈z, (A− jω)x〉 = 〈z, BB′z〉 = ‖B′z‖2
−〈x, (A− jω)∗z〉 = 〈x, C ′Cx〉 = ‖Cx‖2
and hence
〈z, (A− jω)x〉 = ‖B′z‖2
84
〈(A− jω)x, z〉 = −‖Cx‖2
Thus 〈(A− jω)x, z〉 is real and
−‖Cx‖2 = 〈(A− jω)x, z〉 = ‖B′z‖2
Therefore B′z = 0 and Cx = 0. So from (8.4) and (8.5)
(A− jω)x = 0
(A− jω)∗z = 0
Combine the last four equations to get
z∗[A− jω B] = 0[A− jω
C
]x = 0
By stabilizability and detectability it follows that x = z = 0, a contradiction.Next, we’ll show that
X−(H), Im
[0I
]are complementary. This requires a preliminary step. As in the proof of Lemma 1 bring inX1, X2, H− so that
X−(H) = Im
[X1
X2
]
H
[X1
X2
]=
[X1
X2
]H− (8.6)
We want to show that X1 is nonsingular, i.e. Ker X1 = 0. First, it is claimed that Ker X1 isH−-invariant. To prove this, let x ∈ Ker X1. Pre-multiply (8.6) by [I 0] to get
AX1 −BB′X2 = X1H− (8.7)
Pre-multiply by x′X ′2, post-multiply by x, and use the fact that X ′
2X1 is symmetric (see(8.1)) to get
−x′X ′2BB′X2x = 0
Thus B′X2x = 0. Now post-multiply (8.7) by x to get X1H−x = 0, i.e. H−x ∈ Ker X1. Thisproves the claim.
Now to prove that X1 is nonsingular, suppose on the contrary that Ker X1 6= 0. ThenH|Ker X1 has an eigenvalue, λ, and a corresponding eigenvector, x:
H−x = λx (8.8)
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Re λ < 0, 0 6= x ∈ Ker X1
Pre-multiply (8.6) by [0 I]:
−C ′CX1 − A′X2 = X2H− (8.9)
Post-multiply by x and use (8.8):
(A′ + λ)X2x = 0
Since B′X2x = 0 too from above, we have
x∗X ′2[A + λ B] = 0
Then stabilizability implies X2x = 0. But if X1x = 0 and X2x = 0, then x = 0, a contradic-tion. This concludes the proof of complementarity.
Now set X := Ric(H). We’ll show that X ≥ 0. The Riccati equation is
A′X + XA−XBB′X + C ′C = 0
or equivalently
(A−BB′X)′X + X(A−BB′X) + XBB′X + C ′C = 0
Noting that A−BB′X is stable (Lemma 1), we have
X =
∫ ∞
0
e(A−BB′X)′t(XBB′X + C ′C)e(A−BB′X)tdt (8.10)
Since XBB′X + C ′C is positive semi-definite, so is X.Finally, suppose (C, A) is observable. We’ll show that if x′Xx = 0, then x = 0; thus
X > 0. Pre-multiply (8.10) by x′ and post-multiply by x:
x′Xx =
∫ ∞
0
‖B′Xe(A−BB′X)tx‖2dt +
∫ ∞
0
‖Ce(A−BB′X)tx‖2dt
Thus if x′Xx = 0, then Xx = 0 and
Ce(A−BB′X)tx = 0, ∀t ≥ 0
But this implies that x belongs to the unobservable subspace of (C, A) and so x = 0. �
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8.6 State-space solution
Our solution is for a special case of the problem stated in Section 1. The realization of thetransfer matrix G is taken to be of the form
G(s) =
A B1 B2
C1 0 D12
C2 D21 0
The following assumptions are made:
(i) (A, B1) is stabilizable and (C1, A) is detectable
(ii) (A, B2) is stabilizable and (C2, A) is detectable
(iii) D′12
[C1 D12
]=
[0 I
](iv)
[B1
D21
]D′
21 =
[0I
]Assumption (ii) is necessary and sufficient for G to be internally stabilizable. Assumption(i) is for a technical reason: together with (ii) it guarantees that two Hamiltonian matrices(H2 and J2 below) belong to dom(Ric).
Assumption (iii) means that C1x and D12u are orthogonal and that the latter controlpenalty is nonsingular and normalized. In the conventional LQG setting this means thatthere is no cross weighting between the state and control input, and that the control weightmatrix is the identity . Other nonsingular control weights can easily be converted to thisproblem with a change of coordinates in u. Relaxing the orthogonality condition introducesa few extra terms in the controller formulas.
Finally, assumption (iv) is dual to (iii) and concerns how the exogenous signal w entersG: the plant disturbance and the sensor noise are orthogonal, and the sensor noise weightingis normalized and nonsingular. Two additional assumptions that are implicit in the assumedrealization for G is that D11 = 0 and D22 = 0. Relaxing these assumptions complicates theformulas substantially.
By Theorem 4 the Hamiltonian matrices
H2 :=
[A −B2B
′2
−C ′1C1 −A′
], J2 :=
[A′ −C ′
2C2
−B1B′1 −A
]belong to dom(Ric) and, moreover, X2 := Ric(H2) and Y2 := Ric(J2) are positive semi-definite. Define
F2 := −B′2X2, L2 := −Y2C
′2
AF2 := A + B2F2, C1F2 := C1 + D12F2
AL2 := A + L2C2, B1L2 := B1 + L2D21
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A2 := A + B2F2 + L2C2
Gc(s) :=
[AF2 IC1F2 0
], Gf (s) :=
[AL2 B1L2
I 0
]and let Tzw denote the tf from w to z.
Theorem 5 The unique optimal controller is
Kopt(s) :=
[A2 −L2
F2 0
]Moreover,
min ‖Tzw‖22 = ‖GcB1‖22 + ‖F2Gf‖22
The first term in the minimum cost, ‖GcB1‖22, is associated with optimal control withstate feedback and the second, ‖F2Gf‖22, with optimal filtering. These two norms can easilybe computed as follows:
‖GcB1‖22 = trace (B′1X2B1)
A′F2
X2 + X2AF2 + C ′1F2
C1F2 = 0
‖F2Gf‖22 = trace (F2Y2F′2)
AL2Y2 + Y2A′L2
+ B1L2B′1L2
= 0
The controller Kopt has the well-known separation structure: the controller equations canbe written as
˙x = Ax + B2u + L2(C2x− y)
u = F2x
F2 is the optimal feedback gain were x directly measured; L2 is the optimal filter gain; x isthe optimal estimate of x.
Proof of Theorem 5 Let K be a proper, stabilizing controller. Start with the systemequations
x = Ax + B1w + B2u
z = C1x + D12u
and define a new control variable, v := u− F2x. The equations become
x = AF2x + B1w + B2v
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z = C1F2x + D12v
or in the frequency-domain
z = GcB1w + Uv
This implies that
Tzw = GcB1 + UTvw
You will prove in an exercise the following fact: U is inner and U∼Gc belongs to RH⊥2 .
This implies that GcB1 and UTvw are orthogonal matrices in RH2 (Tvw belongs to RH2 byinternal stability). So from the previous equation
‖Tzw‖22 = ‖GcB1‖22 + ‖Tvw‖22
Look at how v is generated:
K
A B1 B2
−F2 0 IC2 D21 0
� �
�
-
v w
y u
Note that K stabilizes G iff K stabilizes the above system (the two closed-loop systems haveidentical A-matrices). So
minK‖Tzw‖22 = ‖GcB1‖22 + min
K‖Tvw‖22
and therefore the theorem will be proved once we show the following: for the setup in theprevious block diagram, the unique optimal controller is[
A + B2F2 + L2C2 −L2
F2 0
]and the minimum value of ‖Tvw‖2 equals ‖F2Gf‖2. Notice in this setup that A + B2F2 isstable.
By the assignment C1 ← −F2, the previous statement becomes this: for
G(s) =
A B1 B2
C1 0 IC2 D21 0
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with A−B2C1 stable, the unique optimal controller is[A−B2C1 + L2C2 L2
C1 0
]and the minimum cost is ‖C1Gf‖2.
The dual of the last statement is this: for
G(s) =
A B1 B2
C1 0 D12
C2 I 0
with A−B1C2 stable, the unique optimal controller is[
A + B2F2 −B1C2 B1
F2 0
]and the minimum cost is ‖GcB1‖2.
To prove this, apply the controller and let x denote its state. Then the system equationsare
x = Ax + B1w + B2u
z = C1x + D12u
y = C2x + w
˙x = (A + B2F2 −B1C2)x + B1y
u = F2x
so
˙x = Ax + B2u + B1(y − C2x)
Defining e := x− x, we get
e = (A−B1C2)e
It’s now easy to infer internal stability from stability of A + B2F2 and A − B1C2. For zeroinitial conditions on x, x, we have e(t) ≡ 0. Hence
u = F2x = F2x (8.11)
For every proper, stabilizing controller the equation
‖Tzw‖22 = ‖GcB1‖22 + ‖Tvw‖22is still valid, showing that
‖Tzw‖2 ≥ ‖GcB1‖2But for the present controller, (8.11) implies that v ≡ 0, i.e., Tvw = 0. Thus the presentcontroller is optimal and the minimum cost is ‖GcB1‖2. Finally, for uniqueness it can beshown (an exercise) that the unique solution of Tvw = 0 is the controller above. �
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8.7 Exercises
1. Define
T1(s) =s + 1
s + 2, T2(s) =
s2 − s + 1
(s + 1)2
Find a function Q in RH∞ to minimize ‖T1 − T2Q‖2. Compute the minimum norm.
2. Give an interesting (i.e., nontrivial) example of a 2× 1 inner matrix.
3. Consider
x = Ax + Bu, x(0) = x0
with A stable. Prove true or false: for every u in L2[0,∞), x(t) tends to 0 as t tends to ∞.
4. Suppose u and y are scalar-valued signals and the tf from u to y is 1/s2. For the standardcanonical realization (A, B, C) consider the optimization problem
minu=Fx
∫ ∞
0
ρy(t)2 + u(t)2 dt
where ρ is positive. Find the optimal F . Study the eigenvalues of A + BF as ρ→ 0 and asρ→∞.
5. Prove that U is inner and U∼Gc ∈ RH⊥2 .
6. Prove uniqueness in Theorem 5.
7. You know that right half-plane zeros place definite performance limitations on the controlof a system. This exercise illustrates this fact in the present context.
Consider the system
x = Ax + Bu, x(0) = x0
z = Cx
Then
z = C(s− A)−1x0 + C(s− A)−1Bu
If A is stable, we might like to see how small we can make ‖z‖2 by suitable choice of u inH∞. In particular, we might like to know if ‖z‖2 can be made arbitrarily small.
Let
C(s− A)−1B =s− 1
(s + 2)(s + 3)
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Compute
infu∈H∞
‖C(s− A)−1x0 + C(s− A)−1Bu‖2
as a function of x0. For what values of x0 is the infimum equal to zero.Repeat for
C(s− A)−1B =s + 1
(s + 2)(s + 3)
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