Transfer Functions

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Transfer Functions

Olof Staffans

March 10, 2009

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Chapter 1 Introduction

1.1 State Space Model of a BB

Let’s consider a BB = “Black Box”.

- -

input u output y

Black box

A standard engineering assumption is that the dynamics of the black box can be described by

˙x(t) = Ax(t) + Bu(t), x(0) = x0

y(t) = Cx(t) + Du(t), (1.1)

where

x is an n-dimensional vector, the “state” (“tillst˚and” eller “fas”), u is a p-dimensional vector, the “input” (“insignal”),

y is a q-dimensional vector, the “output” (“utsignal”),

and A ∈ R^n×n, B ∈ R^n×p,

C ∈ R^q×n, D∈ R^q×p

are matrices. Often p = 1 (“single input”) and q = 1 (“single output”). We denote

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SISO = “single input, single output”, MIMO = “multiple input, multiple output”

etc. If n = ∞, then we have a “Distributed Parameter System” (treated in the previous course).

The solution of (1.1) is obtained (from standard theory of ODE:s) x(t) = e^Atx0+

Z t 0

e^A(t−s)Bu(s) ds, y(t) = Ce^Atx0+

Z t 0

Ce^A(t−s)Bu(s) ds + Du(t).

The system is exponentially stable if all the eigenvalues of A have strictly negative real parts. In this case there exist constants M <∞ and ε > 0 such that

ke^Atk ≤ Me^−εt. (The constant ε is called the “stability margin”.)

The “initial state” corresponds to “excitations” of the system due to past inputs or “initialization terms” (which pop up when you switch on the current to the box). If the system is “at rest” at the time t = 0, then x0 = 0. In that case

y(t) = Z t

0

Ce^A(t−s)Bu(s) ds + Du(t) = Du(t) + Z t

0

K(t− s)u(s) ds, where K(t) = Ce^AtB. We call the “distribution”

Dδ + K

(where δ is the “Delta-function”) the impulse response of the box. Here D is the feedthrough or feedforward and K is the strictly proper part of the impulse response.

1.2 An Extended BB

We group the input and output vectors into two parts.

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- - -

u₁ u2

y₁ y2

Plant = P

Controller = Q

Both P and Q are black boxes,

u1 ∈ R^p¹ = external input u2 ∈ R^p² = control input

y1 ∈ R^q¹ = external output y2 ∈ R^q² = measured output and

p1+ p2 = p = total input dimension q₁ + q₂ = q = total output dimension.

Task (uppgift): Design a controller Q in such a way that y1 behaves “in the desired way” as u1 varies. The “desired way” may include

• tracking, the output follows the input (e.g. a power amplifier)

• disturbance rejection, y¹ ≈ 0 independently of how u¹ behaves

• L²-optimal control, minimize the effect of initial transients

• L¹-optimal control, minimize the maximal value of the output

• H^∞-optimal control, minimize the amount of energy which passes from u1 to y1.

Classes of permitted controllers:

• the same type as P (black box)

• digital implementation (very common).

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1.3 A Distributed Parameter Example

Consider a bar with insulation on the sides. The end point is heated. Heat evaporates through the end point by radiation. A sensor measures the radiation, and it gives us the temperature at the end point.

Let

T0+ T (t, x)

be the temperature of the bar at the point x ∈ [0, ∞) at time t ∈ [0, ∞). The constant T0 is the surrounding temperature. Standard PDE-theory gives









T_t(t, x) = T_xx(t, x), x≥ 0, t ≥ 0, Tx(t, 0) = αT (t, 0)

| {z }

radiation

− u(t)|{z}

heat source

, t≥ 0,

T (0, x) = 0 (initial rest temperature).

The solution is given by “Green’s formula”, T (t, x) = 1

√π Z t

0

√ 1

t− se⁻^4(t−s)^x2 (u(s)− αT (s, 0)) ds.

Thus, if we know T (s, 0), then we can compute the rest of the temperature distribution from this formula. Substitute x = 0 above, and denote

y(t) = T (t, 0), k(t) = 1

√πt. Then

y(t) = Z t

0

k(t− s)(u(s) − y(s)) ds.

- - -

6

d r

+ −

u u− y y

k ∗

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This can be interpreted as a feedback connection of the BB which maps u into

(k∗ u)(t) = Z t

0

k(t− s)u(s) ds.

This function k cannot be written in the form k(t) = Ce^AtB for any matrices C, A, B (of finite size).

General questions:

• What is a reasonable requirement on the impulse response of a BB?

• How do we deal with unstable systems (like an atomic reactor or a modern combat air plane like JAS Gripen)?

• What do we do when we have many inputs and outputs? (MIMO) Answers later.

1.4 A General Black Box Model

Let us make a number of “natural” assumptions, and use these to derive a model of a general BB:

- -

u y = P u

P

(dimension p) (dimension q)

Hyp. 1 Causality: Future inputs have no influence on past outputs, if u1(s) = u2(s) for s≤ T , then

(P u1)(s) = (P u2)(s) for s≤ T.

Hyp. 2 Time independence: The response of P is the same today as tomor- row. If the input signal arrives h seconds later, then the response is the same, except that it also is delayed by h seconds. If u2(s) = u1(s + h) for some constant h, then

(P u2)(s) = (P u1)(s + h), s∈ R

(this hypothesis can be removed, the model then becomes more complicated).

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Hyp. 3 Continuity or well-posedness: Small changes in the input causes small changes in the output. For all ε > 0 there is a δ > 0 such that if (i) u1(s) = u2(s) for s≤ S

(ii) ku¹(s)− u²(s)k ≤ δ for S ≤ s ≤ T , then

(iii) k(P u1)(s)− (P u2)(s)k ≤ ε for S ≤ s ≤ T . (This hypothesis can be slightly weakened.)

Hyp. 4 Linearity: Two signals do not interfere with each other, and if we increase the amplitude of the input, then the amplitude of the output increases by the same scale. If u(t) = λ1u1(t) + λ2u2(t), then

(P u)(t) = λ1(P u1)(t) + λ2(P u2)(t).

(Standard requirement for all amplifiers: We want to avoid distortion of the original signals, and different frequency components should not interact.) (True in a limited amplitude range, typically.)

Hyp. 5 Now pure dealys: If we feed in a step signal u(t) =

1, t≥ 0 0, t < 0,

then the output (P u)(t) is an (absolutely) continuous function (we allow a jump discontinuity at zero). (This hypothesis can be removed, at the expense of a more complex model.)

Theorem 1.4.1 Assume Hyp. 1-5. Then there is a matrix-valued function k ∈ L¹loc(R⁺; R^q×p) which represents P in the following sense. For every continuous input signal u which satisfies u(t) = 0 for all t ≤ T (for some T ∈ R) we have

(P u)(t) = Du(t) + Z t

T

k(t− s)u(s) ds, t ∈ R

= Du(t) + Z t

−∞

k(t− s)u(s) ds

.

Proof. Bypassed because of lack of time. Not terribly difficult, but one needs some knowledge of integration theory. Instead we prove the discrete time version of this result.

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1.5 A Discrete Time BB Model

Consider a discrete time variable t∈ {. . . , −2, −1, 0, 1, 2, . . . } = Z.

- -

u(k) y(k) = (P u)(k)

BB

We still have p inputs and q outputs, so

u(k)∈ R^p, y(k)∈ R^q, k ∈ Z, and u and y are vector-valued sequences,

u ={u(k)}^∞_k=−∞, y ={y(k)}^∞_k=−∞. We still make the same hypothesis as in the last section.

Hyp. 1 Causality: If u1(k) = u2(k) for k≤ K, then (P u₁)(k) = (P u₂)(k) for k ≤ K.

Hyp. 2 Time independence: If u2(k) = u1(k + h) for some h ∈ Z and all k ∈ Z, then

(P u2)(k) = (P u1)(k + h), k∈ Z.

Hyp. 3 Continuity or well-posedness: To every given input u which vanishes for all k ≤ K (for some K) there is a unique output y.

Hyp. 4 Linearity: If u = λ1u1+ λ2u2, then

P u = λ1P u1+ λ2P u2. Hyp. 5 Properness: Is not needed.

Theorem 1.5.1 If Hyp. 1-4 hold, then there exists a matrix-valued sequence {K(k)}^∞k=0 (each K(k)∈ R^q×p) which represents P in the following sense. If u(k) = 0 for all k < M (for some M ∈ Z), then

(P u)(k) = Xk l=M

K(k− l)u(l) = Xk l=−∞

K(k− l)u(l)

!

. (1.2)

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Note 1.5.2 If (1.2) holds, then it is easy to show that Hyp. 1-4 are satisfied.

Thus, Hyp. 1-4 are necessary and sufficient for the representation (1.2).

Definition 1.5.3 We call{K(k)}^∞k=0the (discrete time) impulse response of P .

Proof of Theorem 1.5.1. Step 1. Find K(0): Suppose that u(k) = 0 for k < 0, and look at (P u)(0). By Hyp. 1, 3 and 4, this is a linear function of u(0) only. But every linear map R^p 7→ R^q has a matrix representation, there is a matrix K(0)∈ R^q×p such that

(P u)(0) = K(0)u(0), if u(k) = 0 for k < 0.

Step 2. Find K(1): Suppose that u(k) = 0 for all k ≤ 0, k 6= −1. Consider the value (P u)(0). As above we conclude that there is a matrix in R^q×p, which we denote by K(1), such that

(P u)(0) = K(1)u(−1), if u(k) = 0 for all k ≤ 0, except k = −1.

Step 3. Find K(m) m ≥ 2: Take u(k) = 0 for all k ≤ 0 except for k = −m.

As above we find a matrix K(m)∈ R^q×p such that (P u)(0) = K(m)u(−m), if u(k) = 0 for k ≤ 0, except k = −m.

Step 4. Find (P u)(0) for general u: Suppose that u(k) = 0 for all k ≤ M, where M ∈ Z, M ≤ 0. We can then write {u(k)}⁰k=M in the form

{u(M), u(M + 1), . . . , u(−1), u(0)} = {u(M), 0, . . . , 0, 0}

+ {0, u(M + 1), 0, . . . , 0}

...

+ {0, . . . , 0, u(−1), 0}

+ {0, . . . , 0, 0, u(0)}.

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Each piece is of the type considered in Step 3. By Hyp. 4 (linearity), (P u)(0) = (P (first seq.))(0) + (P (second seq.))(0) + . . .

= K(−M)u(M) + K(1 − M)u(M + 1) + . . .

· · · + K(1)u(−1) + K(0)u(0)

= X0 l=M

K(−l)u(l)

= X0 l=−∞

K(−l)u(l) (where all terms are zero if l < M).

Note. So far we have not used the time invariance, Hyp. 2.

Step 5. Find (P u)(k) for all K: Define v(l) = u(k + l), l ∈ Z (a shifted sequence). The time invariance gives

(P u)(k) = (P v)(0)

= X0 l=−∞

K(−l)v(l)

= X0 l=−∞

K(−l)u(k + l) (put k + l = m)

= Xk m=−∞

K(k− m)u(m)

which proves (1.2).

Definition 1.5.4 We call the sum above the convolution of K and u, and denote it by K∗ u. Thus,

(P u)(k) = (K∗ u)(k) = Xk l=−∞

K(k− l)u(l)

(where all but finitely many terms are zero).

Comment 1.5.5 Without any further knowledge of K, we may compute (1.2) by using the FFT, so it is easy to implement a Discrete BB of this type on a computer. Note that K(1), K(2), . . . represent delayed terms (we have time to compute them), but K(0) is needed immediately (implement by a direct coupling).

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1.6 Transfer Functions

Definition 1.6.1 The transfer function of the discrete time system described in Theorem 1.5.1 is the Z-transform of the impulse response, i.e.,

K(z) =ˆ X∞

k=0

K(k)z^−k

(defined for those z ∈ C for which the sum converges, and analytically extended to a domain which is as big as possible).

Definition 1.6.2 The transfer function of the continuous time system in Theorem 1.4.1 is given by

G(s) = D + ˆk(s), where ˆk is the Laplace transform of k,

k(s) =ˆ Z ∞

0

e^−stk(t) dt

(defined for those s ∈ C for which the integral converges, and extended by analyticity).

Note 1.6.3 Typically ˆK(z) is defined and analytic for|z| large enough (close to complex infinity). Typically G(s) is defined for Re(s) large enough (in some right half-plane). In the rational case they can be extended to all but finitely many points in C.

Note 1.6.4 Some people replace z^−k by z^k in Definition 1.6.1.

1.7 Realizations of Rational Transfer Func- tions

Recall the engineering model of Section 1.1:





˙x(t) = Ax(t) + Bu(t), y(t) = Cx(t) + Du(t), x(0) = x0,

(1.3)

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whose solution is (if x0 = 0 and t≥ 0) x(t) =

Z t 0

e^A(t−s)Bu(s) ds, t≥ 0 y(t) = Du(t) +

Z t 0

Ce^A(t−s)Bu(s) ds.

The impulse response is

k(t) = D δ(t)

|{z}

δ-function

+Ce^AtB.

Problem 1.7.1 Find the transfer function.

Solution: Take Laplace transforms in (1.3) to get

sˆx(s) = x(0) + Aˆx(s) + B ˆu(s), ˆ

y(s) = C ˆx(s) + Dˆu(s), (Re(s) large enough). This implies

(s− A)ˆx(s) = x(0) + Bˆu(s) and so

ˆ

x(s) = (s− A)⁻¹(x(0) + B ˆu(s)), ˆ

y(s) = C(s− A)⁻¹(x(0) + B ˆu(s))Dˆu(s).

If x(0) = 0, then ˆ

y(s) = C(s− A)⁻¹B ˆu(s) + Dˆu(s).

On the other hand, by standard Laplace transform theory, if y(t) = Du(t) +

Z t 0

k(t− s)u(s) ds, then

ˆ

y(s) = Dˆu(s) + ˆk(s)ˆu(s).

Comparing these two to each other we get the following Theorem.

Theorem 1.7.2 The transfer function of the continuous time system (1.3) is

G(s) = D + C(s− A)⁻¹B. (1.4)

In particular, each component of this function is rational.

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Proof. We derived (1.4) above. This function is rational, because we may write (by Cramer’s rule)

(s− A)⁻¹ = adj(s− A) det(s− A),

where det(s− A) is a polynomial of degree n (dimension of state space), and each element of adj(s− A) is a polynomial of degree n − 1 (it is ± the sub determinant which you get by dropping one row and one column of (s− A)).

Corollary 1.7.3 If the dimension of the state space is n, then the transfer function has at most n poles (counted according to their multiplicity).

Proof. det(s− A) is a polynomial of degree n in s. A similar result is true in discrete time. A standard engineering assumption is that we can describe the input/output relation by using a system of difference

equations, 





x(k + 1) = Ax(k) + Bu(k), k∈ Z⁺ y(k) = Cx(k) + Du(k),

x(0) = x0,

(1.5)

where Z⁺ ={0, 1, 2, . . . }. We multiply this by z^−k and add, X∞

k=0

z^−kx(k + 1) = A X∞ k=0

z^−kx(k) + B X∞ k=0

z^−kBu(k),

where the left hand side is z

" _∞ X

k=0

z^−k−1x(k + 1) + x(0)− x(0)

#

= z

" _∞ X

k=0

z^−kx(k)− x⁰

#

= z[ˆx(z)− x⁰].

Thus,

z(ˆx(z)− x0) = Aˆx(z) + B ˆu(z), so

(z− A)ˆx(z) = zx0+ B ˆu(z) which implies

ˆ

x(z) = (z− A)⁻¹(zx0+ B ˆu(z)), ˆ

y(z) = C(z− A)⁻¹(zx0 + B ˆu(z)) + Dˆu(z).

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If again x0 = 0, then we get ˆ

y(z) = [D + C(z− A)⁻¹B]ˆu(z).

On the other hand, by taking z-transforms (multiply by z^−k and add) in formula (1.2) we get

ˆ

y(z) = ˆK(z)u(z), so we arrive at the following theorem.

Theorem 1.7.4 The transfer function of the discrete time system (1.5) is K(z) = D + C(zˆ − A)⁻¹B.

In particular, it is rational.

There is also a converse to Theorems 1.7.2 and 1.7.4.

Theorem 1.7.5 Every proper rational matrix-valued function with rational elements can be interpreted as the transfer function of a continuous time system of the type described in Theorem 1.7.2, and also as a transfer function of a discrete time system of the type described in Theorem 1.7.4. In other words it has a representation of the form

K(s) = D + C(sˆ − A)⁻¹B

for suitably chosen matrices A, B, C and D (and a state space Rⁿ of suitable dimension).

Note 1.7.6 By Corollary 1.7.3, the dimension of the state space is at least as large as the number of poles of the given rational transfer function (counted according to their multiplicity).

Definition 1.7.7 The minimal dimension of the state space in Theorem 1.7.5 is called the Mac-Millan degree of the given rational matrix-valued function.

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Chapter 2 Different Stability Notions

2.1 Exponential Stability

Definition 2.1.1 The continuous time system





˙x(t) = Ax(t) + Bu(t), t≥ 0 y(t) = Cx(t) + Du(t), t≥ 0 x(0) = x0

(2.1)

is exponentially stable if ke^Atk → 0 exponentially, i.e., there exist constants ε > 0 and M <∞ such that ke^Atk ≤ Me^−εt for all t≥ 0.

Theorem 2.1.2 The system (2.1) is exponentially stable if and only if all the eigenvalues of A lie in the open left half-plane. In particular, (s− A)⁻¹ then has no poles in the closed right half-plane.

Proof (outline). Based on Jordan’s canonical form. After a similarity transform in the state space we get a number of independent Jordan blocks, and it suffices to prove this for one single block.

Corollary 2.1.3 If (2.1) is exponentially stable, then its transfer function D + C(s− A)⁻¹B has no poles in the closed right half-plane.

Proof. The function (s− A)⁻¹ has no poles there. Comment 2.1.4 The converse is not true. Even if the transfer function has no poles in the right half-plane, the system need not be exponentially stable (unless it is minimal, i.e., both controllable and observable).

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2.2 Power Stability

Definition 2.2.1 The discrete time system





x(k + 1) = Ax(k) + Bu(k), k∈ Z⁺ y(k) = Cx(k) + Du(k),

x(0) = x0,

(2.2)

is power stable if the solution with u(k)≡ 0 tends to zero as k → ∞ with a geometrical rate,

kx(k)k ≤ Mη^k for some M < ∞ and 0 ≤ η < 1.

Theorem 2.2.2 The system (2.1) is power stable if and only if the eigen- values of A lie strictly inside the unit disc, |λ| < 1 for every eigenvalue λ of A.

Proof. Based on the Jordan decomposition of A, and the fact that x(k) = A^kx0, k ∈ Z⁺,

if u(k) = 0 for all k.

Corollary 2.2.3 If (2.1) is power stable, then the transfer function of (2.1) has no poles in the closed exterior of the unit disc, i.e. in the region

{z ∈ C | |z| ≥ 1}.

Proof. The function (z− A)⁻¹ has no poles there. Comment 2.2.4 Comment 2.1.4 applies also to the discrete case.

2.3 Input/Output Stability

If we have no state space representation of an impulse response, then the preceding stability definitions are not valid. Instead we must look directly at the input/output map. Several different stability notions exist.

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Definition 2.3.1 Consider the discrete time input/output map (assuming that u(k) = 0 for k < 0)

y(n) = Xn k=0

K(n− k)u(k), (n ≥ 0).

This map is

(i) ℓ^∞- or BIBO-stable (Bounded Input Bounded Output) if there is a constant M < ∞ such that for all u,

sup

n ky(n)k ≤ M sup

n ku(n)k.

(ii) energy stable, or ℓ² stable, if there is a constant M so that X

n

ky(n)k²

!1/2

| {z }

ℓ²-norm of y

≤ M X

n

ku(n)k²

!1/2

| {z }

ℓ²-norm of u

.

(iii) ℓ¹ stable if there is a constant M so that X

n

ky(n)k ≤ MX

n

ku(n)k.

In this course we only study energy stability, which we simply call input/output stability. The same notions are also used in continuous time, with the sums replaced by integrals.

2.4 The Cayley Transform

We will now review a simple method for solving a given continuous time system. We transform the system into a discrete system in such a way that the discrete trajectory can be considered a numerical approximation of the continuous trajectory.

Let h > 0 to make {tⁿ}^∞n=0, tn = nh, an equidistant discretisation of the interval [0,∞). By integrating the standard input equation

˙x(t) = Ax(t) + Bu(t)

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h

(h,x(h))

(nh,x(nh))

t (nh, (n))x

(0,x(0)= (0))x

(h, (1))x ε_n

nh

(t,x(t))

Figure 2.1: The internal Cayley transform interpreted as a discrete approximation of the original system. The points nh, x_n

do not lie on the curve t, x(t)

, but the error ε[t/h] is proportional to h².

from tn to tn+1 we get

x(t_n+1)− x(tn) = Z tn+1

tn

Ax(t) + Bu(t) dt.

If the continuous integrand is approximated by a straight line, this becomes x(tn+1)− x(tⁿ)≈ h

2 Ax(tn+1) + Ax(tn) + Bu(tn+1) + Bu(tn)

. (2.3) Introduce α := 2/h > 0 and un= u(tn+1) + u(tn)

/√

2α. If α∈ ρ(A), then (2.3) can be solved for x(tn+1):

x(tn+1)≈ (α − A)⁻¹(α + A)x(tn) +√

2α(α− A)⁻¹Bun, Thus define x_n by

x0 = x(0), xn+1 = (α− A)⁻¹(α + A)xn+√

2α(α− A)⁻¹Bun (2.4) in order to make x_nan approximation of x(t_n). If d²u/dt² exists for all t≥ 0, then kx^[t/h]− x(t)k ≤ C(t)h². See Figure 2.1 for an illustration. The scaling of the input has been chosen so that the ℓ²-norm of the input sequence uk is approximately equal to the L²-norm of the original input function u.

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The above transform (together with a similar discretization of the output) is called the Cayley transform. The numerical properties of this transform has also been studied recently in [MH04]. The above discussion is taken from [KS05].

The Cayley transform is also used is used a lot in “theoretical proofs”, to convert a theorem concerning continuous time to discrete time and back.

However, in that connection one uses a different correspondence between the continuous time functions x, u, and y and their discrete counterparts. Instead of sampling the continuous time functions one defines the coefficients of the discrete time functions to be the Fourier exponents of the continuous time functions with respect to a Laguerre basis. In this version the correspondence between a continuous time signal and a discrete time signal is not causal, and it cannot be used in on-line computations. The place to read more about this is Chapter 11 of [Sta05].

Definition 2.4.1 We consider the continuous time system





˙x(t) = Ax(t) + Bu(t), y(t) = Cx(t) + Du(t), x(0) = x0.

(2.5)

Let Re(α) > 0 and suppose that (α−A) is invertible (α is not an eigenvalue).

Then the discrete time system





xn+1 = A(α)xn+ B(α)un, yn= C(α)xn+ D(α)un, x0 = x0

(2.6)

is called the Cayley transform of (2.5) with parameter α. Here A(α) = (α + A)(α− A)⁻¹

B(α) = p

2Re(α)(α− A)⁻¹B C(α) = p

2Re(α)C(α− A)⁻¹ D(α) = G(α),

and G(s) is the transfer function of (2.5).

Note. The most popular values of α are α = 1 and α = 1/2 (in the latter case p

2Re(α) = 1).

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Lemma 2.4.2 There are many interesting connections between the original system and it’s Cayley transform (2.6). The most interesting ones in this course are

(i) (2.5) is exponentially stable if and only if (2.6) is power stable (here it is important that the dimension of the state space is finite).

(ii) The transfer functions of the two systems are related as follows: If G is the transfer function of (2.5), and Q is the transfer function of (2.6), then

G(s) = Q(z) if we choose s and z to satisfy

z = α + s

α− s ⇐⇒ s = αz− α z + 1 . (Thus, G(s) = Q(^α+s_α−s) and Q(z) = G(^αz−α_z+1 ).)

Proof. A lengthy computation (manipulation of matrices and their in- verses).

Comment 2.4.3 The same transform is also used in the case of the general models described in Theorem 1.4.1 and Theorem 1.5.1. We say that the discrete time impulse response {K(j)}^∞j=0 is the Cayley transform with parameter α of the continuous time impulse response Dδ + k if their respective transfer functions satisfy

G(s) = ˆK(z),

s and z as above. (G is the Laplace transform of Dδ + k and ˆK is the Z-transform of {K(n)}^∞n=0.)

2.5 Time Discretization

The Cayley transform behaves mathematically in a nice way, but in its

“mathematical” Laguerre version it cannot be used for online control pur- poses (because of the lack of causality). In digital control one often uses direct “sampling” of the input/output map instead. This method has the advantage that it does not need a state space representation of the system.

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If we in the continuous time formula y(t) =

Z t 0

k(t− s)y(s) ds

take t to be a multiple of h, t = nh (h > 0 is the sampling interval), and approximate the integral by

y(nh)≈ h Xn

j=0

k(nh− jh)u(jh),

then we get a discrete time impulse response.





u(jh)→ uj

y(jh)→ yj

hk(jh)→ K^j

yn= Xn

j=0

K_n−juj

The discrete time transfer function ˆK is K(z) =ˆ

X∞ j=0

hk(nj)z^−j.

This can also be interpreted as an approximation of the original continuous time kernel. We replace K(t) by a sum of δ-functions located at the points 0, h, 2h, . . . .

k(t)≈ h[K(0)δ0+ k(h)δh+ k(2h)δ2h+ . . .

Obviously, this transformation cannot be inverted exactly (since we loose information). The Laplace transform of the sum of δ-functions is

k(s) = h(k(0) + k(h)eˆ ^−hs+ k(2h)e^−2hs+ . . . )

= h(K0+ K1e^−hs+ K2(e^−hs)²+ K3(e^−hs)³+ . . . )

= h X∞

j=0

Kj(e^hs)^−j

= h ˆK(e^hs)

= h ˆK(z),

where z = e^hs. We have for all integers n that e^2πn= 1, so e^hs = e^h(s+2πn/h)

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for all n. That is, the new continuous transfer function is periodic in the direction of the imaginary axis, with period 2π/h. If we are only interested in the frequency band

−π

h < Im(s)≤ π s,

then we can map this part of the plane one-to-one onto the discrete time frequency variable z = e^hs. Note that

Re(s)≥ 0 ⇐⇒ |z| > 1,

so the appropriate part of the “unstable” part of the continuous time frequency domain is mapped into the “unstable” part of the discrete time frequency domain.

Note. If the original transfer function is rational, then the transformed function is not. We loose the state space representation (the function s→ e^hs is not rational). More on time discretization in the course on wavelets.

2.6 The Z-transform Versus the Fourier Se- ries

The Z-transform is very closely related to the Fourier series. The Z-transform of the sequence {u(k)}^∞k=0 is

ˆ u(z) =

X∞ k=0

u(k)z^−k.

Put z = re^2πiφ, where r = |z| and 2πφ, 0 ≤ φ < 1 is the “argument” of z.

Then

ˆ u(z) =

X∞ k=0

u(k)r^−ke^2πikφ

is the Fourier series (a function defined for 0 ≤ φ < 1) of the sequence u(k)r^−k. In particular, taking r = 1 we find that we get the Fourier series of a sequence {u(k)}^∞k=0 by evaluating it’s Z-transform on the unit circle

|z| = 1.

Therefore, you can use any book on Fourier series (the inverse of the finite Fourier transform) to learn more about the Z-transform.

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Chapter 3 The Frequency Domain

Terminology: We work in the frequency domain when we study ei- ther how the Laplace transforms (continuous time) or Z-transforms (discrete time) of the inputs and outputs behave.

3.1 Notations

R= (−∞, ∞) R⁺ = [0,∞)

Z={0, ±1, ±2, . . . } Z⁺ = N₀ ={0, 1, 2, . . . } N={1, 2, 3, . . . }

C={all complex numbers} C⁺ ={z ∈ C | Re(z) ≥ 0}

◦

C⁺={z ∈ C | Re(z) > 0}

D={z ∈ C | |z| ≤ 1} D^◦={z ∈ C | |z| < 1}

E={z ∈ C | |z| ≥ 1} E^◦={z ∈ C | |z| > 1}

R^n×m= n× m-dimensional matrix with real elements C^n×m= n× m-dimensional matrix with complex elements Rⁿ= R^n×1

Cⁿ= C^n×1

Note. Most mathematicians like to work in the interior D^◦ of the unit disc in discrete time. Most control engineers like to work in the exterior E of the unit disc. The “mathematical” Z-transform is

X∞ k=0

u(k)z^k,

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and the “control” Z-transform is X∞ k=0

u(k)z^−k.

Here we use the control version throughout the course. (Replace z by 1/z to get the “mathematical” version.)

3.2 L

^p

-Spaces

3.2.1 The Discrete Time Case

Definition 3.2.1 • ℓ^p(Z; C^m×n) = The set of sequences K(k), k ∈ Z, that are (m× n)-matrix valued, and satisfy

X∞ k=−∞

kK(k)k^p <∞.

Here 1≤ p < ∞, and kK(k)k is some norm in C^m×n. The norm in ℓ^p is

kKkp =

X∞ k=−∞

kK(k)k^p

!1/p

.

• ℓ^∞(Z; C^m×n) = The set of bounded C^m×n-valued sequences with norm kKk∞ = sup

k∈ZkK(k)k.

• ℓ^p(Z⁺; C^m×n) = As above, but k≥ 0.

• ℓ^p(N; C^m×n) = As above, but k ≥ 1.

Note. In the last two cases we typically define K(k) = 0 for k ≤ −1 respectively k ≤ 0.

Note. We typically require that the norm in C^m×n or R^m×n is “adapted” to the norm in C^m and Cⁿ etc. If we use the Euclidean norms in Rⁿ and Cⁿ, then

kK(k)k = sup

x∈Rⁿ x6=0

kK(k)xk^R^m kxk^Rⁿ . (This is the “operator” norm of K(k).)

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3.2.2 The Continuous Time Case

Definition 3.2.2 • L^p(R; C^m×n) = The set of (m× n)-matrix valued functions on R which are “measurable” and satisfy

kKkp =

Z ∞

−∞

kK(t)k^p dt

1/p

<∞.

Here 1≤ p < ∞.

• L^∞(R; C^m×n) = The set of (m× n)-matrix valued “measurable” functions which are “essentially bounded”, with norm

kKk∞= (ess)sup_t∈RkK(t)k.

• L^p(R⁺; C^m×n) and L^∞(R⁺; C^m×n) same as above, but t∈ R⁺ instead.

The variables k (discrete time) and t (continuous time) live in the time domain (which is R or Z). The appropriate transforms (Laplace or Z) live in the frequency domain.

Interpretation:

L² and ℓ²: energy

L^∞ and ℓ^∞: maximal amplitude L¹ and ℓ¹: total mass.

3.3 Plancherel’s Theorem

To move back and forth between the time domain and the frequency domain we need two versions of Plancherel’s theorem (or “Parseval”), one for continuous time and one for discrete time.

Lemma 3.3.1 (Continuous time Plancherel.) Let u∈ L¹(R; Cⁿ)∩L²(R; Cⁿ).

Then the Fourier transform of u,

˜ u(ω) =

Z _∞

−∞

e^−iωtu(t) dt

belongs to L²(R; Cⁿ), and if we use the Euclidean norm in Cⁿ, that is ku(t)k^Cⁿ =X

|ui(t)|²1/2

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where u(t) =





 u1(t) u2(t)

...

un(t)





, then

kuk²2= Z _∞

−∞

ku(t)k² dt = 1 2π

Z _∞

−∞

k˜u(ω)k² dω = 1 2πk˜uk²2.

Note 3.3.2 Throughout the rest of this course we use the Euclidean norm in Rⁿ and Cⁿ.

Lemma 3.3.3 (Discrete time Plancherel.) Let u∈ ℓ¹(Z; Cⁿ) (then also u∈ ℓ²(Z; Cⁿ)). Then the (discrete) Fourier transform

˜ u(φ) =

X∞ k=−∞

e^−ikφu(k), 0≤ φ ≤ 2π,

satisfies ˜u∈ C[0, 2π], and kuk²2 =

X∞ k=−∞

ku(k)k² = 1 2π

Z 2π 0

|˜u(φ)|² dφ = 1 2πk˜uk²2.

Thus, the Fourier transform maps L²(R)7→ L²(R) in continuous time, ℓ²(Z)7→ L²([0, 2π]) in discrete time.

3.4 H

^p

-Spaces

From the preceding two lemmas we can derive certain results for Laplace and Z-transforms. We again begin with continuous time.

ˆ u(s) =

Z _∞

0

e^−stu(t) dt

is the Laplace transform of u. Define u(t) = 0 for t < 0, and write s = α+iω.

Then

ˆ

u(α + iω) = Z ∞

0

e^−iωte^−αtu(t) dt

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which is the Fourier transform of the function

e^−αtu(t), t≥ 0 0, t < 0.

By Lemma 3.3.1, we have for all α > 0, 1

2π Z _∞

−∞

kˆu(α + iω)k² dω = Z _∞

0 ke^−αtu(t)k² dt = Z _∞

0

e^−2αtku(t)k² dt

≤ Z ∞

0 ku(t)k² dt =kuk²2.

Thus, there is a constant (= 2πkuk²2) such that Z ∞

−∞kˆu(α + iω)k²dω ≤ K

for all α > 0 (with equality as α→ 0+). In addition ˆu(s) is analytic inC^◦⁺={s ∈ C | Re(s) > 0}.

This is true whenever u∈ L²(R⁺; Cⁿ).

Definition 3.4.1 Let 1 ≤ p < ∞. The space H^p(C⁺; C^m×n) consists of all C^m×n-valued functions f which are analytic in C^◦⁺ and satisfy the condition

kfk^H^p^(C⁺⁾ =

sup

α>0

Z ∞

−∞kf(α + iω)k^p dω

1/p

.

The space H^∞(C⁺; C^m×n) consists of all C^m×n-valued bounded and analytic functions in C^◦⁺, with norm

kfkH^∞(C⁺) = sup

Re(s)>0kf(s)k.

The computation on the preceding page proves

Theorem 3.4.2 If u ∈ L²(R⁺; Cⁿ), then the Laplace transform ˆu of u sat- isfies

ˆ

u∈ H²(C⁺; Cⁿ) (with equality of norms).

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Let us also mention without proof the following theorem.

Theorem 3.4.3 Every function in H²(R⁺; Cⁿ) is the Laplace transform of some function u∈ L²(R⁺; Cⁿ).

Thus, we get a one-to-one unitary (i.e. norm preserving) map between L²(R⁺; Cⁿ) and H²(C⁺; Cⁿ):

-

L²(R⁺; Cⁿ) H²(C⁺; Cⁿ)

√2π× Inverse Laplace

√1

2π× Laplace

We continue with discrete time.

Let {u(k)}^∞k=0 ∈ ℓ²(Z⁺; Cⁿ). Then

ˆ u(z) =

X∞ k=0

z^−ku(k).

By writing z = re^iφ we get ˆ

u(re^iφ) = X∞ k=0

e^−ikφr^−ku(k)

which is the discrete Fourier transform of the sequence {r^−kz(k)}^∞k=0 (we define u(k) = 0 for k < 0).

By Lemma 3.3.2, if r ≥ 1, then 1

2π Z 2π

0

kˆu(re^iφ)k² dφ = X∞

k=0

r^−2kku(k)k² ≤ X∞ k=0

ku(k)k² =kuk²2.

Thus, there is a constant (=kuk²2) such that for all r > 1, the “total energy” of ˆu on a circle of radius r is bounded, i.e.,

1 2π

Z 2π

0 |ˆu(re^iφ)|² dφ≤ K

(with equality as r → 1+). In addition ˆu is analytic in E^◦= {z ∈ C| |z| > 1}, and also at infinity.

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Analytic at infinity means that the function ˆu(1/z) is analytic at zero. Thus, ˆ

u(1/z) is analytic in D^◦ . Definition 3.4.4 Let 1≤ p < ∞.

• The space H^p(D; C^m×n) consists of all C^m×n-valued analytic functions in D^◦={|z| < 1} which satisfy

kfkH^p(D) = sup

0≤r<1

1 2π

Z 2π

0 kf(re^iφ)k^p dφ

^1/p

<∞.

• The space H^∞(D; C^m×n) consists of all bounded analytic C^m×n-valued functions, with norm

kfkH^∞(D) = sup

z∈

◦

D

kf(z)k.

• H^∞(E; C^m×n) as above but D is replaced by E^◦= {z ∈ C | |z| > 1}.

(Here f must also be analytic at infinity.)

• H^p(E; C^m×n) as above, but r > 1 instead of r < 1 (and f is analytic also at infinity).

Thus, the argument on the preceding page proves one half of the following theorem.

Theorem 3.4.5 If u ∈ ℓ²(Z⁺; Cⁿ), then ˆu ∈ H²(E ; Cⁿ), and conversely, every f ∈ H²(E ; Cⁿ) is the Z-transform of some u ∈ ℓ²(Z ; Cⁿ). The Z- transform is a unitary map of ℓ²(Z⁺; Cⁿ) onto H²(E ; Cⁿ) (unitary means that it is one-to-one onto, and preserves norms and inner products).

3.5 Frequency Domain Stability

We discussed different versions of input/output stability in Section 2.3. Only one of them has a simple frequency domain interpretation, namely the “energy” version (L² or ℓ²).

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Theorem 3.5.1 A continuous time input/output map u7→ y(t) = Du(t) +

Z t 0

k(t− s)u(s) ds

is (energy) input/output stable if and only if its transfer function G satisfies G∈ H^∞(C⁺; C^q×p).

Theorem 3.5.2 A discrete time input/output map u7→ y(n) =

Xn k=0

K(n− k)u(k)

is (energy) input/output stable if and only if its transfer function Q satisfies Q∈ H^∞(E ; C^q×p).

Continuous time: “No poles in the right half plane”.

Discrete time: “No poles outside the unit disc”.

Proof (partial) of Theorem 3.5.2. Suppose that Q is analytic and bounded in E. Let u ∈ ℓ²(Z⁺; C^p). Then ˆu ∈ H²(E ; C^p). The Z-transform of the output y is

ˆ

y(z) = Q(z)ˆu(z) (|z| > 1)

(the Z-transform maps convolution into pointwise multiplication like the Laplace transform). Therefore we find that ˆy is analytic in E^◦, and that for all r ≥ 1,

1 2π

Z 2π

0 kˆy(re^iφ)k² dφ = 1 2π

Z 2π

0 kQ(reîφ)û(reîφ)k² dφ

≤ 1

2π Z 2π

0 kQ(re| {z^iφ)k}²

≤M²

kˆu(re^iφ)k² dφ

≤ M² 2π

Z 2π

0 kˆu(re^iφ)k² dφ

≤ M²

2π kuk²2 (by the computation on p. 28).

Therefore ˆy∈ H²(E ; Cⁿ), and by Theorem 3.4.5, the inverse Z-transform of ˆ

y, i.e., y itself, must therefore belong to ℓ²(Z⁺; Cⁿ).

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Thus, input/output (energy) stable ⇐⇒

transfer function is in H^∞.

Proof of converse: Take u(t) = e^stu0, t≤ 0.

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Chapter 4 The Standard Problem

The results presented in this section are also found in [Fra87], Chapter 3.

4.1 The General Setting

- -

-

w

G

z

u

K

y

All signals are vectors,

w = external input (disturbance?) ∈ R^p¹

u = control input ∈ R^p²

z = external output ∈ R^q¹ y = measured output ∈ R^q² A more detailed picture

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G₁₁ G₁₂

G21 G22

@+@

K w -

-

u

y

z -





z = G11w + G12u y = G₂₁w + G₂₂u u = Ky

Think about these as frequency domain signals. G11, G12, G21 and G22 are rational transfer functions. To study the stability of this system we add two more disturbance signals

v1 = actuator noise (reglerarens brus) v₂ = sensor noise (m¨atfel)

- -

-

-d

r -

-

r d

w

G

z u

K

y v1

v2

+ +

This is identical to the preceding picture if v1 = v2 = 0, but in the new setting we have

(external) inputs: w, v1, v2 (internal input: u) (external) outputs: u, z, y

(out of which the output u is fed back into the system). This gives us the system of equations:





z = G11w + G12u y = G₂₁w + G₂₂u + v₂ u = Ky + v1.

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This system defines (implicitly) u, z, y as functions of w, v1, v2. We can solve it, e.g., as follows.

u = Ky + v1 = K(G21w + G22u + v2) + v1 ⇐⇒

(1− KG²²)u = KG21w + Kv2+ v1 ⇐⇒

u = (1− KG²²)⁻¹(KG21w + Kv2+ v1). (4.1) Substitute into the equations above to get y and z.

If this is to be used as an industrial process, then it must be “stable” in some sense. To simplify the mathematics we use “energy stability”: For all inputs w, v₁, v₂ ∈ L²(R⁺) (with appropriate dimensions), the signals produced by the system, i.e., u, y, z should also belong to L²(R⁺) (and the “energy amplification” should be finite).

We may solve u, z, y explicitly by using the method described above, and get a formula of the type



 z u y



 =



 F11 F12 F13

F₂₁ F₂₂ F₂₃ F31 F32 F33







 w v₁ v2



 .

By the discussion in section 3.5, this means that we need

Every Fij must belong to H^∞(C^◦⁺) (if continuous time) or H^∞(E^◦) (if discrete time).

One especially important part of F is F₂₂ = (1− KG22)⁻¹ (see (4.1)).

If we denote the feedthrough operators by

feedthrough of K = K(∞), feedthrough of G22= G22(∞),

then F22(∞) = (1 − K(∞)G22(∞)), so a necessary condition is that 1− K(∞)G22(∞) is invertible.

Let us suppose this invertibility condition in the sequel. True, e.g., if G22(∞) = 0 (which is the “standard assumption” made in many text books), or if K(∞) = 0.

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4.2 H

^∞

-minimization Problem

Find the “compensator” K which makes the closed loop system stable, and minimizes the H^∞-norm of F₁₁

(this is the transfer function from w to z).

By checking the computations leading to Theorem 3.5.1 and Theorem 3.5.2 we find that we will minimize the “energy amplification” of the input/output map of the time domain u to the time domain z (which is the norm of this operator as a mapping L²(R⁺) 7→ L²(R⁺) or ℓ²(Z⁺) 7→ ℓ²(Z⁺)). Let’s compute this transfer function.

z = G11w + G12u

= [G11+ G12(1− KG²²)⁻¹KG21]w + terms which depend on v1 and v2

⇒ F₁₁ = G₁₁+ G₁₂(1− KG22)⁻¹KG₂₁

Theorem. In the case of a rational matrix-valued transfer function this problem has a solution, which can be computed (although the solution is rather complicated).

There are MATLAB toolboxes that solve this:

• Robust Control Toolbox

• µ-systems toolbox

• LMI-Toolbox

4.3 The Model Matching Problem

Find the “compensator” Q which minimizes the H^∞-norm of the following system.

T3 Q T2

T1

T3

- - -

6 d r

+

−

w z

u y

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



y = T3w z = T1w− T2u u = Qy

⇐⇒





z y

=

T1 −T² T3 0

w u

u = Qy Here T1, T2, T3, Q∈ H^∞ (“stable”).

Homework? Reformulate this so that it becomes a special case of the standard setup presented on p. 35. (I.e., find G11, G12, G21, G22.)

4.4 Tracking Problem

Find the compensators C1 and C2 which solve the following minimization problem.

C1

C₂

- - - P

6

d r -

+ +

w u v

Here we use a “two degree of freedom” compensator pair C1 and C2. The cost of the control is

Z ∞ 0

kw(t) − v(t)k²

| {z }

quality of tracking, want v(t) ≈ w(t)

+ ku(t)k| {z }²

cost of control

dt

To turn this into a minimization problem we replace the original output v by a new output z, and interpret also u as an output.

C₁

C₂

- - - P

6

6 -

- -

d

r r

r +

+

−

w u

v z

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Write the compensator in the form

K = [C1, C2]

(these have the same output dimension, can be combined into a larger matrix, also z, w and v have the same dimension).

Summary:

Controller K = [C1 C2] External input: w Control input: u External output:

z u

= ˜z Control output:

w v

= y The equations become

z˜ y

=





 z u w v





 =







1 −P

0 1

1 0

0 P







| {z }

= 2 4

G11 G12

G21 G22 3 5

w u

=







w− P u u w P u







u = Ky =

C1 C2

| {z }

K

w v

We realize that this is a special case of the H^∞ minimization problem with G =

G₁₁ G₁₂ G21 G22

given above.

These are minimax problems. For example, model matching is minQ max

Re s>0kT1(s)− T2(s)Q(s)T3(s)k.

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Chapter 5 Coprime Factorizations

Idea: Factor an unstable transfer function into a numerator and a denomi- nator with no common factors (a “minimal” factorization in some sense).

5.1 The SISO Case

5.1.1 Discrete Time

In the SISO case the transfer function is scalar (matrix of dimension 1× 1).

Let us suppose that k(z) is a rational transfer function. It is analytic at infinity, so it can be written as a quotient

k(z) = p(z)˜

˜

q(z) = a₀+ a₁z +· · · + amz^m b0+ b1z +· · · + bnzⁿ ,

where bn 6= 0 and n ≥ m (since k is analytic at infinity). From a stability point of view, this is not a good representation. We can without loss of generality take m = n (define ak = 0 for k > m). Neither ˜p(z) = a0+ a1z +

· · ·+anzⁿnor ˜q(z) = b0+b1z +· · ·+bnzⁿis a proper transfer function (proper

= bounded at infinity). We get a better representation by writing (divide both ˜p(z) and ˜q(z) by zⁿ)

k(z) = p(1/z)

q(1/z) = a0+ a1z⁻¹+· · · + aⁿz⁻ⁿ b0+ b1z⁻¹+· · · + bnz⁻ⁿ,

where b0 6= 0. (I have changed bk to b_n−k and akto a_n−k and denoted z⁻ⁿp(z)˜ by p(1/z) and z⁻ⁿq(z) by q(1/z). Note that p and q are polynomials.)˜

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Basic stability requirement: We cannot allow p(1/z) and q(1/z) to have any common zeros in E (outside the unit disk). Equivalently, p(z) and q(z) must not have common zeros in D (inside unit disk).

Motivation: When the system gets older the parameters change slightly, and the zeros move, and we suddenly get an unstable pole (which is very close to an unstable zero) from nowhere, and the system goes unstable.

Principle: We do not allow unstable zero-pole cancellations.

This is not a problem with stable zero-pole cancellations. Even if these change slightly, the input/output behavior does not change much.

Near zero-pole cancellations: An unstable zero of p(1/z) which almost cancels an unstable zero of q(1/z) causes the system to be badly behaved (difficult to control).

Note. The function k is (energy) stable if and only if q(1/z) has no zeros in E (assuming no unstable zero-pole cancellations).

Theorem 5.1.1 If the polynomials p and q have no common zeros, then there exist unique polynomials x and y such that

p(z)x(z) + q(z)y(z) ≡ 1 (z ∈ C). (5.1) Proof. Standard “division algorithm” for polynomials taught in school.

Note that (5.1) also can be written in “transfer function form” as

p(1/z)x(1/z) + q(1/z)y(1/z)≡ 1 (z 6= 0). (5.2)

Definition 5.1.2 (i) Two polynomials p and q are coprime (relativa prim- polynom, relativt odelbara) if they have no common zeros (anywhere in C).

(ii) Equation (5.1) is called the Bezout identity.

Comment 5.1.3 The terminology comes from prime number theory. Two positive integers are coprime (relativa primtal) if they have no common factors. Two polynomials have a common polynomial as a factor if and only if they have a common zero.

Definition 5.1.4 The polynomials p(z) and q(z) are coprime with respect to D (the unit disk) if they have no common zeros in D. (Then p(1/z) and q(1/z) have no common zeros in E.)

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5.1.2 Continuous Time

We copy the same idea to the continuous time setting, and write the rational (transfer) function k(s) in the form

k(s) = p(s) q(s),

where p and q are no longer polynomials, but they are rational transfer functions in RH^∞(C^◦⁺). Here RH^∞ = “rational H^∞-function”.

Definition 5.1.5 Two functions in RH^∞(C^◦⁺) are coprime (relativt odelbara) (with respect to RH^∞(C^◦⁺)) if they have no common zeros in the closed right half-plane C⁺.

Theorem 5.1.6 Two functions p and q in RH^∞(C^◦⁺) are coprime if and only if there exist two functions x and y in RH^∞(C^◦⁺) such that

p(s)x(s) + q(s)y(s)≡ 1, Re(s) ≥ 0. (5.3)

Proof. (Outline.) If (5.3) holds, then p and q cannot have any common zeros in Re(s)≥ 0 (including the imaginary axis). The converse direction is more complicated, straightforward but long. It is based on Theorem 5.1.1

(polynomial division).

5.2 The MIMO Case

What does a “zero” of a matrix-valued function mean? The answer does exist, but it is too complicated to be presented here (the Smith-MacMillan canonical form). Instead we use the Bezout identity as a definition of co- primeness.

Definition 5.2.1 (i) Assume that F and G are two matrix-valued functions in RH^∞(C^◦⁺), with the same input dimension (number of columns).

Then F and G are right coprime (with respect to RH^∞(C^◦⁺)) if

Transfer Functions