lectures on multivariable feedback...

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Lectures on Multivariable Feedback Control Ali Karimpour Department of Electrical Engineering, Faculty of Engineering, Ferdowsi University of Mashhad (September 2009) Chapter 4: Stability of Multivariable Feedback Control Systems 4-1 Well-Posedness of Feedback Loop 4-2 Internal Stability 4-3 The Nyquist Stability Criterion 4-3-1 The Generalized Nyquist Stability Criterion 4-3-2 Nyquist arrays and Gershgorin bands 4-4 Coprime Factorizations over Stable Transfer Functions 4-5 Stabilizing Controllers 4-6 Strong and Simultaneous Stabilization

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Page 1: Lectures on Multivariable Feedback Controlprofsite.um.ac.ir/~karimpor/multi/Multivariable_chapter4.pdf · Lectures on Multivariable Feedback Control ... Chapter 4 Lecture Notes of

Lectures on Multivariable Feedback Control Ali Karimpour

Department of Electrical Engineering, Faculty of Engineering, Ferdowsi University of

Mashhad (September 2009)

Chapter 4: Stability of Multivariable Feedback Control Systems

4-1 Well-Posedness of Feedback Loop

4-2 Internal Stability

4-3 The Nyquist Stability Criterion 4-3-1 The Generalized Nyquist Stability Criterion

4-3-2 Nyquist arrays and Gershgorin bands

4-4 Coprime Factorizations over Stable Transfer Functions

4-5 Stabilizing Controllers

4-6 Strong and Simultaneous Stabilization

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Chapter 4 Lecture Notes of Multivariable Control

2

This chapter introduces the feedback structure and discusses its stability properties. The

arrangement of this chapter is as follows: Section 4-1 introduces feedback structure and describes

the general feedback configuration and the well-posedness of the feedback loop is defined. Next,

the notion of internal stability is introduced and the relationship is established between the state

space characterization of internal stability and the transfer matrix characterization of internal

stability in section 4-2. The stable coprime factorizations of rational matrices are also introduced

in section 4-3. Section 4-4 discusses how to achieve a stabilizing controller. Finally section 4-5

introduces strong and simultaneous stabilization.

4-1 Well-Posedness of Feedback Loop

We will consider the standard feedback configuration shown in Figure 4-1. It consists of the

interconnected plant P and controller K forced by command r, sensor noise n, plant input

disturbance di, and plant output disturbance d. In general, all signals are assumed to be

multivariable, and all transfer matrices are assumed to have appropriate dimensions.

Assume that the plant P and the controller K in Figure 4-1 are fixed real rational proper transfer

matrices. Then the first question one would ask is whether the feedback interconnection makes

sense or is physically realizable. To be more specific, consider a simple example where

1,21

=+−

−= KssP

are both proper transfer functions. However,

idsdnrsu3

1)(3

2 −+−−

+=

i.e., the transfer functions from the external signals r, n, d and di to u are not proper. Hence, the

feedback system is not physically realizable!

Definition 4-1

A feedback system is said to be well-posed if all closed-loop transfer matrices are well-defined

and proper.

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Chapter 4 Lecture Notes of Multivariable Control

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Figure 4-1 Standard feedback configuration

Now suppose that all the external signals r, n, d and di are specified and that the closed-loop

transfer matrices from them to u are respectively well-defined and proper. Then, y and all other

signals are also well-defined and the related transfer matrices are proper. Furthermore, since the

transfer matrices from d and n to u are the same and differ from the transfer matrix from r to u by

only a sign, the system is well-posed if and only if the transfer matrix from di and d to u exists

and is proper.

Theorem 4-1

The feedback system in Figure 4-1 is well-posed if and only if

)()( ∞∞+ PKI is invertible. 4-1

Proof. As we explain, the system is well-posed if and only if the transfer matrix from di and d to u

exists and is proper. The transfer matrix from di and d to u is:

( ) [ ] ⎥⎦

⎤⎢⎣

⎡+−= −

idd

KPKKPIu 1

Thus well-posedness is equivalent to the condition that ( ) 1−+ KPI exists and is proper. But this is

equivalent to the condition that the constant term of the transfer matrix )()( ∞∞+ PKI is invertible.

It is straightforward to show that (4-1) is equivalent to either one of the following two conditions:

⎥⎦

⎤⎢⎣

⎡∞−

∞IP

KI)(

)( is invertible

)()( ∞∞+ KPI is invertible 4-2

e

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Chapter 4 Lecture Notes of Multivariable Control

4

The well-posedness condition is simple to state in terms of state-space realizations. Introduce

realizations of P and K:

⎥⎦

⎤⎢⎣

⎡≅

DCBA

P

⎥⎥⎦

⎢⎢⎣

⎡≅

DC

BAK ))

ˆˆ

4-3

Then DP =∞)( and DK ˆ)( =∞ . For example, well-posedness in (4-2) is equivalent to the

condition that

⎥⎦

⎤⎢⎣

− IDDI)

is invertible 4-4

Fortunately, in most practical cases we will have D = 0, and hence well-posedness for most

practical control systems is guaranteed.

4-2 Internal Stability Consider a system described by the standard block diagram in Figure 4-1 and assume the system

is well-posed. Furthermore, assume that the realizations for P(s) and K(s) given in equation (4-3)

are stabilizable and detectable.

Let x and x̂ denote the state vectors for P and K, respectively, and write the state equations in

Figure 4-1 with r, d, di and n set to zero:

yDxCu

yBxAx

DuCxyBuAxx

ˆˆˆ

ˆˆˆˆ

−=

−=

+=+=

&

&

4-5

Definition 4-2

The system of Figure 4-1 is said to be internally stable if the origin (x, x̂ ) = (0, 0) is

asymptotically stable, i.e., the states (x, x̂ ) go to zero from all initial states when r = 0, d=0, di=0

and n=0.

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5

Note that internal stability is a state space notation. To get a concrete characterization of internal

stability, solve equations (4-5) for y and u:

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

−=⎥

⎤⎢⎣

⎡−

xx

CC

IDDI

yu

ˆ0

ˆ0ˆ 1

4-6

Note that the existence of the inverse is guaranteed by the well-posedness condition. Now

substitute this into (4-5) to get

⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

⎡⎟⎟

⎜⎜

⎛⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

−⎥⎦

⎤⎢⎣

−+⎥

⎤⎢⎣

⎡=⎥

⎤⎢⎣

⎡−

xx

Axx

CC

IDDI

B

B

A

A

x

~ˆ0

ˆ0ˆˆ0

0ˆ0

0

ˆ

1

&

&4-7

Where

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

−⎥⎦

⎤⎢⎣

−+⎥

⎤⎢⎣

⎡=

0

ˆ0ˆˆ0

0ˆ0

0~1

CC

IDDI

B

B

A

AA 4-8

Thus internal stability is equivalent to the condition that A~ has all its eigenvalues in the open left-

half plane. In fact, this can be taken as a definition of internal stability.

Theorem 4-2

The system of Figure 4-1 with given stabilizable and detectable realizations for P and K is

internally stable if and only if A~ is a Hurwitz matrix (All eigenvalues are in open left half plane).

It is routine to verify that the above definition of internal stability depends only on P and K, not

on specific realizations of them as long as the realizations of P and K are both stabilizable and

detectable, i.e., no extra unstable modes are introduced by the realizations.

The above notion of internal stability is defined in terms of state-space realizations of P and K. It

is also important and useful to characterize internal stability from the transfer matrix point of

view. Note that the feedback system in Figure 4-1 is described, in term of transfer matrices, by

⎥⎦

⎤⎢⎣

⎡−

=⎥⎦

⎤⎢⎣

−⎥⎦

⎤⎢⎣

⎡− r

de

uIPKI ip 4-9

Note that we ignore d and n as inputs since they produce similar transfer matrices as equation 4-9.

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Now it is intuitively clear that if the system in Figure 4-1 is internally stable, then for all bounded

inputs (di, -r), the outputs (up, - e) are also bounded. The following theorem shows that this idea

leads to a transfer matrix characterization of internal stability.

Theorem 4-3

The system in Figure 4-1 is internally stable if and only if the transfer matrix

⎥⎥⎦

⎢⎢⎣

++

+−+−=⎥

⎤⎢⎣

⎡− −−

−−−

11

111

)()()()(

PKIPPKIPKIKPPKIKI

IPKI

4-10

from (di, -r) to (up, - e) be a proper and stable transfer matrix.

Proof. Let stabilizable and detectable realizations of P and K defined as 4-3. Then we have the

state space equation for the system in Figure 4-1 is

⎥⎦

⎤⎢⎣

⎡−

+⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+⎥

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

⎥⎦

⎤⎢⎣

−⎥⎦

⎤⎢⎣

−+⎥

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

rd

uy

II

e

u

e

u

D

Dxx

C

Cuy

e

u

B

Bxx

A

A

x

x

ip

p

p

00

ˆ0

0ˆˆ0

0

ˆ0

0ˆˆ0

0&̂

&

The deleting ⎥⎦

⎤⎢⎣

⎡uy

from last two equations we can rewritten a

⎥⎦

⎤⎢⎣

⎡−⎥

⎤⎢⎣

−+⎥

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

−=⎥

⎤⎢⎣

−⇒⎥

⎤⎢⎣

⎡−

+⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎤⎢⎣

−⎥⎦

⎤⎢⎣

−−

rd

IDDI

xx

CC

IDDI

eu

rd

xx

CC

eu

IDDI ipip

11 ˆˆ0

ˆ0ˆˆ0

ˆ0ˆ

By substituting this in the states space equation we have

⎥⎦

⎤⎢⎣

⎡−⎥

⎤⎢⎣

−+⎥

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

−=⎥

⎤⎢⎣

⎥⎦

⎤⎢⎣

⎡−⎥

⎤⎢⎣

−⎥⎦

⎤⎢⎣

−+⎥

⎤⎢⎣

⎡⎟⎟

⎜⎜

⎛⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

−⎥⎦

⎤⎢⎣

−+⎥

⎤⎢⎣

⎡=⎥

⎤⎢⎣

−−

−−

rd

IDDI

xx

CC

IDDI

eu

rd

IDDI

B

Bxx

CC

IDDI

B

B

A

A

x

x

ip

i

11

11

ˆˆ0

ˆ0ˆ

ˆˆ0

0ˆ0

ˆ0ˆˆ0

0ˆ0

0&̂

&

Now suppose that this system is internally stable. So the eigenvalues of

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

−⎥⎦

⎤⎢⎣

−+⎥

⎤⎢⎣

⎡=

0

ˆ0ˆˆ0

0ˆ0

0~1

CC

IDDI

B

B

A

AA

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Chapter 4 Lecture Notes of Multivariable Control

7

are in the open left-half plane, it follows that the transfer matrix from (di, -r) to (up,- e) given in

(4-10) is stable.

Conversely, suppose that )( PKI + is invertible and the transfer matrix in (4-10) is stable. Then, in

particular, 1)( −+ PKI is proper which implies that )ˆ())()(( DDIKPI +=∞∞+ is invertible.

Therefore,

⎥⎦

⎤⎢⎣

− IDDI ˆ

is nonsingular. Now routine calculations give the transfer matrix from (di, -r) to (up, -e) in terms

of the state space realizations: 1

ˆ0

0)~(

0

ˆ0ˆ −

−⎥⎦

⎤⎢⎣

−⎥⎦

⎤⎢⎣

−−⎥

⎤⎢⎣

−−

+⎥⎦

⎤⎢⎣

− IDDI

B

BAsI

CC

IDDI

Since the above transfer matrix is stable, it follows that

⎥⎦

⎤⎢⎣

⎡−⎥

⎤⎢⎣

−− −

B

BAsI

CC

ˆ0

0)~(

0

ˆ0 1

as a transfer matrix is stable. Finally, since (A, B, C) and )ˆ,ˆ,ˆ( CBA are stabilizable and detectable,

⎟⎟⎠

⎞⎜⎜⎝

⎛⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

0

ˆ0,ˆ0

0,~

CC

B

BA

is stabilizable and detectable. It then follows that the eigenvalues of A~ are in the open left-half

plane.

Note that to check internal stability, it is necessary (and sufficient) to test whether each of the four

transfer matrices in (4-10) is stable. Stability cannot be concluded even if three of the four

transfer matrices in (4-10) are stable. For example, let an interconnected system transfer function

be given by

11,

11

−=

+−

=s

KssP

Then it is easy to compute

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Chapter 4 Lecture Notes of Multivariable Control

8

⎥⎦

⎤⎢⎣

⎡−

⎥⎥⎥⎥

⎢⎢⎢⎢

++

+−

+−+

−++

=⎥⎦

⎤⎢⎣

− rd

ss

ss

sss

ss

e

u ip

21

21

)2)(1(1

21

which shows that the system is not internally stable although three of the four transfer functions

are stable. This can also be seen by calculating the closed-loop A-matrix with any stabilizable and

detectable realizations of P and K.

Remark: It should be noted that internal stability is a basic requirement for a practical feedback

system. This is because all interconnected systems may be unavoidably subject to some nonzero

initial conditions and some (possibly small) errors, and it cannot be tolerated in practice that such

errors at some locations will lead to unbounded signals at some other locations in the closed-loop

system. Internal stability guarantees that all signals in a system are bounded provided that the

injected signals (at any locations) are bounded. However, there are some special cases under

which determining system stability is simple.

Theorem 4-4

Suppose K is stable. Then the system in Figure 4-1 is internally stable iff PPKI 1)( −+ is stable.

Proof. The necessity is obvious. To prove the sufficiency, it is sufficient to show that if

PPKIQ 1)( −+= is stable, the other three transfer matrices are also stable since:

QKIPKPKIIIPKIIPKIQKIKPKIK

KQIPPKIKI

−=+−=−++=+

−−=+−

−=+−

−−−

111

1

1

)()()()()(

)(

This theorem is in fact the basis for the classical control theory where the stability is checked only

for one closed-loop transfer function with the implicit assumption that the controller itself is

stable.

Theorem 4-5

Suppose P is stable. Then the system in Figure 4-1 is internally stable iff 1)( −+ PKIK is stable.

Proof. The necessity is obvious. To prove the sufficiency, it is sufficient to show that if 1)( −+= PKIKQ is stable, the other three transfer matrices are also stable since:

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9

PQIPKIPKIIPKIIPKIPPQIPPKIQPIPPKIKI

−=+−=−++=+

−=+

−=+−

−−−

111

1

1

)()()()()(

)(

Theorem 4-6

Suppose P and K are both stable. Then the system in Figure 4-1 is internally stable iff 1)( −+ PKI

is stable.

Proof. The necessity is obvious. To prove the sufficiency, it is sufficient to show that if 1)( −+= PKIQ is stable, the other three transfer matrices are also stable since:

KQPKIKQPPPKI

KQPIPPKIKI

=+

=+

−=+−

1

1

1

)()(

)(

4-3 The Nyquist Stability Criterion Let G(s) be a square, rational transfer matrix. This G(s) will often represent the series connection

of a plant with a compensator, and we must assume that there are no hidden unstable modes in the

system represented by this transfer function. In this section we wish to examine the stability of the

negative-feedback loop created by inserting the compensator kI in to the loop (i.e. an equal gain k

into each loop), for various real values of k.

4-3-1 The Generalized Nyquist Stability Criterion Suppose G(s) is a square, rational transfer matrix. We are going to check the stability of Figure 4-

2 for various real values of k.

Let )](det[ skGI + have 0P poles and cP zeros in the closed loop right half plane. Then, just as

with SISO systems, we have, by the principle of the argument, that the Nyquist plot of

))(det()( skGIs +=φ (the image of )(sφ as s goes once round the Nyquist contour in the clockwise

direction) encircles the origin, oc PP − times in the counter-clockwise direction. Nyquist contour is

shown in Figure 4-3.

Remark: For stable system )](det[ skGI + has no zero in the closed loop right half plane, so 0=cP .

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10

Figure 4-2 System with constant controller

Figure 4-3 Nyquist contour

Note that the argument so far has exactly paralleled that for SISO feedback. However, if we

stopped here we would have to draw the Nyquist locus of ))(det()( skGIs +=φ for each value of k

in which we were interested, whereas the great virtue of the classical Nyquist criterion is that we

draw a locus only once, and then infer stability properties for all values of k.

It is easy to show that, if )(siλ is an eigenvalue of )(sG , then )(sk iλ is an eigenvalue of )(skG ,

and )(1 sk iλ+ is an eigenvalue of )(1 skG+ . Consequently (since the determinant is the product of

the eigenvalues) we have

∏ +=+i

i skskGI )](1[)](det[ λ 4-11

ProcessG(s)

Controller kI

-

+

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Chapter 4 Lecture Notes of Multivariable Control

11

and hence

{ } ( )⎭⎬⎫

⎩⎨⎧

+=+ ∑i

i skskGI )(1)](det[ λϑϑ 4-12

where ( ))(sfϑ is the total number of encirclements of the origin made by the graphs of )(sf . So,

we see that we can infer closed-loop stability by counting the total number of encirclements of the

origin made by the graphs of )(1 sk iλ+ , or equivalently, by counting the total number of

encirclements of -1 made by the graphs of )(sk iλ . In this form, the criterion has the same

powerful character as the classical criterion, since it is enough to draw the graphs once, usually

for 1=k .

The graphs of )(siλ (as s goes once around the Nyquist contour) are called characteristic loci.

Remark: The eigenvalues of a rational matrix are usually irrational functions, so, we may have

problem counting encirclements. It turns out that in practice there is no problem, since together

the characteristic loci forms a closed curve.

Theorem 4-7 (Generalized Nyquist theorem)

If G(s) with no hidden unstable modes, has 0P unstable (Smith-McMillan) poles, then the closed-

loop system with return ratio )(skG− is stable if and only if the characteristic loci of )(skG , taken

together, encircle the point -1, 0P times anticlockwise.

Proof: Desoer and Wang proved the theorem in 1980. “On the generalized Nyquist stability

criterion” IEEE Transaction on Automatic control, AC-25

Example 4-1

In the Figure 4-2

⎥⎦

⎤⎢⎣

⎡−−

−++

=26

1)2)(1(25.1

1)(s

ssss

sG

Suppose that G(s) has no hidden modes, check the stability of system for different values of k.

Solution: We need to find the eigenvalues of G(s). So we let

0)( =− sGIλ

then after some manipulation we find

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Chapter 4 Lecture Notes of Multivariable Control

12

)2)(1(5.224132

1 ++−−−

=ss

ssλ and )2)(1(5.2

241322 ++

−+−=

ssssλ

The characteristic loci of this transfer function are shown in Figure 4-4. Since G(s) has no

unstable poles, we will have closed loop stability if these loci give zero net encirclements of -1/k

when a negative feedback kI is applied. From Figure 4-4 it can be seen that for 8.0/1 −<−<∞− k ,

for 0/14.0 <−<− k and for ∞<−< k/153.0 there will be no encirclements, and hence closed loop

stability will be obtained. For 4.0/18.0 −<−<− k there is one clockwise encirclements, and hence

closed loop instability, while for 53.0/10 <−< k there are two clockwise encirclements, and

therefore closed loop instability again.

4-3-2 Nyquist arrays and Gershgorin bands The Nyquist array of G(s) is an array of graphs (not necessarily closed curve), the thji ),( graph

being the Nyquist locus of )(sgij . The key to Nyquist array methods is:

Theorem 4-8 (Gershgorin’s theorem)

Let Z be a complex matrix of dimensions mm× . Then the eigenvalues of Z lie in the union of the

m circles, each with center iiz and radius

mizm

ijj

ij ,....,1,1,1

=∑≠=

4-13

They also lie in the union of the circles, each with center iiz and radius

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13

Figure 4-4 The two characteristic loci for the system defined in example 4-1

mizm

ijj

ji ,....,1,1,1

=∑≠=

4-14

Proof: See “Advanced engineering mathematics” Kreyszig(1972)

Consider the Nyquist array of some square G(s). On the loci of )( ωjgii superimpose, at each

point, a circle of radius

∑≠=

m

ijj

ij jg1

)( ω or ∑≠=

m

ijj

ji jg1

)( ω 4-15

(making the same choice for all the diagonal elements at each frequency). The ‘bands’ obtained in

this way are called Gershgorin bands, each is composed of Gershgorin circles. Gershgorin

bands for a sample system are illustrated in Figure 4-5.

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14

By Gershgorin’s theorem we know that the union of the Gershgorin bands ‘traps’ the union of the

characteristic loci. Furthermore, it can be shown that if the Gershgorin bands occupy distinct

regions, then as many characteristic loci are trapped in the region as the number of Gershgorin

Figure 4-5 A Nyquist array, with Gershgorin bands for a sample system

bands occupying it. So, if all the Gershgorin bands exclude the point -1, then we can assess

closed-loop stability by counting the encirclements of -1 by the Gershgorin bands, since this tells

us the number of encirclements made by the characteristic loci.

If the Gershgorin bands of G(s) exclude the origin, then we say that G(s) is diagonally dominant

(row dominant or column dominant, if applicable). Note that to assess stability we need )]([ sGI +

to be diagonally dominant.

The greater the degree of dominant (of G(s) or I+G(s)) – that is, the narrower the Gershgorin

bands- the more closely does G(s) resembles m non-interacting SISO transfer function.

4-4 Coprime Factorizations over Stable Transfer Functions

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Recall that two polynomials )(sm and )(sn , with, for example, real coefficients, are said to be

coprime if their greatest common divisor is 1 (equivalent, they have no common zeros). It follows

from Euclid's algorithm that two polynomials )(sm and )(sn are coprime if there exist

polynomials )(sx and )(sy such that 1)()()()( =+ snsysmsx ; such an equation is called a Bezout

identity. Similarly, two transfer functions )(sm and )(sn in the set of stable transfer functions are

said to be coprime over stable transfer functions if there exists )(sx and )(sy in the set of stable

transfer functions such that

1)()()()( =+ snsysmsx 4-16

More generally, we have the following definition for MIMO systems.

Definition 4-3 Two matrices M and N in the set of stable transfer matrices are right coprime over

the set of stable transfer matrices if they have the same number of columns and if there exist

matrices rX and rY in the set of stable transfer matrices such that

[ ] INYMXNM

YX rrrr =+=⎥⎦

⎤⎢⎣

⎡ 4-17

Similarly, two matrices M~ and N~ in the set of stable transfer matrices are left coprime over the

set of stable transfer matrices if they have the same number of rows and if there exist two

matrices lX and lY in the set of stable transfer matrices such that

[ ] IYNXMYX

NM lll

l =+=⎥⎦

⎤⎢⎣

⎡ ~~~~ 4-18

Note that these definitions are equivalent to saying that the matrix ⎥⎦

⎤⎢⎣

⎡NM

is left invertible in the set

of stable transfer matrices and the matrix [ ]NM ~~ is right-invertible in the set of stable transfer

matrices. Equations 4-17 and 4-18 are often called Bezout identities.

Now let P be a proper real-rational matrix. A right-coprime factorization (rcf) of P is a

factorization 1−= NMP where N and M are right-coprime in the set of stable transfer matrices.

Similarly, a left-coprime factorization (lcf) has the form NMP ~~ 1−= where N~ and M~ are left-

coprime over the set of stable transfer matrices. A matrix P(s) in the set of rational proper transfer

matrices is said to have double coprime factorization if there exist a right coprime factorization

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1−= NMP , a left coprime NMP ~~ 1−= and lrr XYX ,, and lY in the set of stable transfer matrices

such that

IXNYM

MN

YX

l

lrr =⎥⎦

⎤⎢⎣

⎡ −⎥⎦

⎤⎢⎣

− ~~ 4-19

Of course implicit in these definitions is the requirement that both M and M~ be square and

nonsingular.

Theorem 4-9 Suppose P(s) is a proper real-rational matrix and

⎥⎦

⎤⎢⎣

⎡≅

DCBA

sP )(

is a stabilizable and detectable realization. Let F and L be such that A+BF and A+LC are both

stable, and define

⎥⎥⎥

⎢⎢⎢

+

−+≅⎥

⎤⎢⎣

⎡ −

IDDFCIF

LBBFA

XNYM

l

l 0 4-20

⎥⎥⎥

⎢⎢⎢

+−+≅⎥

⎤⎢⎣

− IDCIF

LLDBLCA

MN

YX rr 0)(

~~

4-21

Figure 4-6 Feedback representation of right coprime factorization

Then 1−= NMP and NMP ~~ 1−= are rcf and lcf, respectively.

v

+

u + x C

D

+

y

+B

A

∫+

+

F

x&

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17

The coprime factorization of a transfer matrix can be given a feedback control interpretation. For

example, right coprime factorization comes out naturally from changing the control variable by a

state feedback. Consider the state space equations for a plant P in the Figure 4-6:

DuCxyBuAxx

+=++&

4-22

Next, introduce a state feedback and change the variable

Fxvu += 4-23

where F is such that A + BF is stable. Then we get

DvxDFCyvFxu

BvxBFAx

++=+=

++=

)(

){&

4-24

Evidently from these equations, the transfer matrix from v to u is

⎥⎦

⎤⎢⎣

⎡ +≅

IFBBFA

sM )( 4-25

and that from v to y is

⎥⎦

⎤⎢⎣

⎡++

≅DDFCBBFA

sN )( 4-26

Therefore we have )()()( svsMsu = and )()()( svsNsy = so

)()()()()()()()()()()( 11 sMsNsPsusPsusMsNsvsNsy −− =⇒=== .

4-5 Stabilizing Controllers

In this section we introduce a parameterization known as the Q-parameterization or Youla-

parameterization, of all stabilizing controllers for a plant. By all stabilizing controllers we mean

all controllers that yield internal stability of the closed loop system. We first consider stable

plants for which the parameterization is easily derived and then unstable plants where we make

use of the coprime factorization.

The following theorem forms the basis for parameterizing all stabilizing controllers for stable

plants.

Theorem 4-10

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18

Suppose P is stable. Then the set of all stabilizing controllers in Figure 4-1 can be described as 1)( −−= PQIQK 4-27

for any Q in the set of stable transfer matrices and )()( ∞∞− QPI nonsingular.

Proof. First we know that if 1)( −−= PQIQK exist then 11 )()()( −− +=⇒=−⇒−= PKIKQQPQIKPQIQK

Since Q is stable so 1)( −+ PKIK is stable so by theorem 4-5 the system in Figure 4-1 is stable.

We must show that every stabilizing controller can be shown as 4-27. Consider that the system in

the Figure 4-1 is stable, since P is also stable, so by theorem 4-5 1)( −+ PKIK is stable. Define

KKPIPKIKQ 11 )()( −− +=+= so we have KKPQQ =+ . Clearly since )()( ∞∞− QPI is

nonsingular K is defined by 1)( −−= PQIQK

Example 4-2

For the plant

)2)(1(1)(

++=

sssP

Suppose that it is desired to find an internally stabilizing controller so that y asymptotically tracks

a ramp input.

Solution: Since the plant is stable the set of all stabilizing controller is derived from 1)( −−= PQIQK , for any stable Q such that, )()( ∞∞− QPI is nonsingular.

Let

3++

=s

basQ

We must define the variables a and b such that that y asymptotically tracks a ramp input, so the

sensitivity transfer function S must have two zeros at origin (L must be a type two system).

)3)(2)(1()()3)(2)(1(

)3)(2)(1(11)(11 1

++++−+++

=+++

+−=−=+−=−= −

sssbassss

sssbasPQPKIPKTS

So we should take 6,11 == ba . This gives

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19

3611

++

=ssQ

)6()22/12)(2)(1(11

2 ++++

=ss

sssK .

However if P(s) is not stable, the parameterization is much more complicated. The result can be

more conveniently stated using state-space representations. The following theorem shows that a

proper real-rational plant may be stabilized irrespective of the location of its RHP-poles and

RHP-zeros, provided the plant does not contain unstable hidden modes.

Theorem 4-11

Let P be a proper real-rational matrix and NMNMP ~~ 11 −− == be corresponding rcf and lcf over

the set of stable transfer matrices. Then there exists a stabilizing controller rrll YXXYK 11 −− ==

with lrr XYX ,, and lY in the set of stable transfer matrices ( IXMYNINYMX llrr =+=+ ~~, ).

Proof. See “Multivariable Feedback Design” By Maciejowski J.M. (1989).

Furthermore, suppose

⎥⎦

⎤⎢⎣

⎡≅

DCBA

sP )(

is a stabilizable and detectable realization of P and let F and L be such that A+BF and A+LC are

stable. Then a particular set of state space realizations for these matrices can be given by 4-20 and

4-21.

The following theorem forms the basis for parameterizing all stabilizing controllers for a proper

real-rational matrix.

Theorem 4-12

Let P be a proper real-rational matrix and NMNMP ~~ 11 −− == be corresponding rcf and lcf over

the set of stable transfer matrices. Then the set of all stabilizing controllers in Figure 4-1 can be

described as

)~()~( 1 MQYNQXK rrrr +−= − 4-28

or 1))(( −−+= llll NQXMQYK 4-29

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20

where rQ is any stable transfer matrices and )(~)()( ∞∞−∞ NQX rr is nonsingular or lQ is any

stable transfer matrices and )()()( ∞∞−∞ ll QNX is nonsingular too.

Proof. See “Multivariable Feedback Design” By Maciejowski J.M. (1989).

Example 4-3:

For the plant

)2)(1(1)(

−−=

sssP

Find a stabilizing controller.

Solution: First of all it is clear that

⎥⎥⎥

⎢⎢⎢

⎡−≅

001132010

)(sP

is a stabilizable and detectable realization. Now let ]51[ −=F and [ ]TL 237 −−= clearly A+BF

and A+LC are stable. Since the plant is unstable we use theorem 4-11 to derive coprime

factorization parameters.

2

2

222 )1(389,

)1(72108,

)1()1)(2(,

)1(1

+++

=+−

=+

−−=

+=

sssX

ssY

sssM

sN ll

2

2

222 )2(389,

)2(72108,

)2()1)(2(~,

)2(1~

+++

=+−

=+

−−=

+=

sssX

ssY

sssM

sN rr

Now let 0== lr QQ then by theorem 4-12 one of the stabilizing controllers is:

38972108

211

++−

=== −−

sssYXXYK rrll

Example 4-4

For the plant in Figure 4-1

)2)(1(1)(

−−=

sssP

The problem is to find a controller that

1. The feedback system is internally stable.

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21

2. The final value of y equals 1 when r is a unit step and d=0.

3. The final value of y equals zero when di is a sinusoid of 10 rad/s and r=0.

Solution: The set of all stabilizing controller is:

)~()~( 1 MQYNQXK rrrr +−= −

where from the example 4-3 we have

2

2

22222 )2(389,

)2(72108,

)1()1)(2(,

)1(1,

)2()1)(2(~,

)2(1~

+++

=+−

=+

−−=

+=

+−−

=+

=s

ssXs

sYs

ssMs

Ns

ssMs

N rr

Clearly for any stable Q the condition 1 satisfied.

To met condition 2 the transfer function from r to y ( )~( MQYN rr + ) must satisfy

38)0(1))0(~)0()0()(0( =⇒=+ rrr QMQYN

To met condition 3 the transfer function from d to y ( )~( NQXN rr − ) must satisfy

jjQjNjQjXjN rrr 9062)10(0))10(~)10()10()(10( +−=⇒=−

Now define

2321 )1(1

11)(

++

++=

sx

sxxsQr

And then set x1, x2 and x3 to meet the requests.

4-6 Strong and Simultaneous Stabilization

Practical control engineers are reluctant to use unstable controllers, especially when the plant

itself is stable. Since if a sensor or actuator fails, and the feedback loop opens, overall stability is

maintained if both plant and controller individually are stable. If the plant itself is unstable, the

argument against using an unstable controller is less compelling. However, knowledge of when a

plant is or is not stabilizable with a stable controller is useful for another problem namely,

simultaneous stabilization, meaning stabilization of several plants by the same controller.

The issue of simultaneous stabilization arises when a plant is subject to a discrete change, such as

when a component burns out. Simultaneous stabilization of two plants can also be viewed as an

example of a problem involving highly structured uncertainty.

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22

Say that a plant is strongly stabilizable if internal stabilization can be achieved with a controller

itself is a stable transfer matrix. Following theorem shows that poles and zeros of P must share a

certain property in order for P to be strongly stabilizable.

Theorem 4-13

P is strongly stabilizable if and only if it has an even number of real poles between every pairs of

real RHP zeros( including zeros at infinity).

Proof. See “Linear feedback control By Doyle”.

Example 4-5

Which of the following plant is strongly stabilizable?

32

22

21 )1()2()1()1()(

)2(1)(

+−+−−

=−−

=ss

ssssPssssP

Solution: P1 is not strongly stabilizable since it has one pole between z=1 and ∞=z but P2 is

strongly stabilizable since it has two pole between z=1 and ∞=z .

Exercises

4-1 Find two different lcf's for the following transfer matrix.

⎥⎦⎤

⎢⎣⎡

+−

+−

=22

11)(

ss

sssG

4-2 Find a lcf's and a rcf's for the following transfer matrix.

⎥⎦

⎤⎢⎣

⎡−

−+

=)1(25.4

412

1)(s

ss

sG

4-3 Find a lcf's and a rcf's for the following transfer matrix.

⎥⎦

⎤⎢⎣

⎡−

−−+

=)1(25.4

41)3)(2(

1)(s

sss

sG

4-4 Derive a feedback control interpretation for the left coprime factorization.

4-5 Show that the equation of all stabilizing controller for stable plants (eq. 4-27) is a special case

of the equation of all stabilizing controller for proper real-rational plants (eq. 4-28 or 4-29).

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4-6 In example 4-4 find x1, x2 and x3.

4-7 In example 4-4 find the controller and then find the step response of the system. Then

suppose the input is zero but a sinusoid with frequency of 10 rad/s applied as disturbance, find the

response of system.

4-8 Derive a representation for left coprime factorization. (the same as right coprime

interpretation in figure 4-6)

4-9 By use of MIMO rule in chapter 2 show that the figure 4-7 introduces the controller in

equation 4-28.

4-10 By use of MIMO rule in chapter 2 show that the figure 4-8 introduces the controller in

equation 4-29.

Figure 4-7 Representation of the stabilizing controller 4-28

Figure 4-8 Representation of the stabilizing controller 4-29

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References

Skogestad Sigurd and Postlethwaite Ian. (2005) Multivariable Feedback Control: England, John

Wiley & Sons, Ltd.

Maciejowski J.M. (1989). Multivariable Feedback Design: Adison-Wesley.