Title: Output feedback stabilization of linear systems
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Permanent Link: http://ufdc.ufl.edu/UF00099384/00001
 Material Information
Title: Output feedback stabilization of linear systems
Alternate Title: Linear systems, Output feedback stabilization of
Physical Description: vii, 80 leaves : ill. ; 28 cm.
Language: English
Creator: Chen, Raymond, 1949-
Copyright Date: 1979
Subject: Stability   ( lcsh )
Mathematics thesis Ph. D
Dissertations, Academic -- Mathematics -- UF
Genre: bibliography   ( marcgt )
non-fiction   ( marcgt )
Statement of Responsibility: by Raymond Chen.
Thesis: Thesis--University of Florida.
Bibliography: Bibliography: leaves 77-79.
General Note: Typescript.
General Note: Vita.
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Bibliographic ID: UF00099384
Volume ID: VID00001
Source Institution: University of Florida
Holding Location: University of Florida
Rights Management: All rights reserved by the source institution and holding location.
Resource Identifier: alephbibnum - 000092343
oclc - 05951063
notis - AAK7749


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To My CMo thic,


I wish to express my sincere appreciation to all those

who contributed in various aspects toward the completion of

this work.

I am particularly grateful to the chairman of my super-

visory committee, Professor V. M. Popov. During the past

few years, he has been my constant source of encouragement

and help, not to mention his invaluable guidance. I would

like to extend my gratitude to my committee members Dr. Brooks,

Dr. Bullock, Dr. Block and Dr. Varma who have influenced

this work from various aspects.

The understanding and tolerance shown by my wife and my

family have been indispensable for this dissertation. This

work would not exist, were it not for their love and encour-

agement. To my typist Sharon Bullivant, I wish to thank her

for a job well done.



ACKNOWILEDGEMENT .......................................... iii

ABSTRACT ................................................ vi


I. INTRODUCTION ................ ......... ..... 1

n a
2.1 Convex lulls of Sn and RS .. .............. 9
n a a
2.2 Criteria of S -Stabilizability of
Hyperplanes.. ............................ 21
2.3 Bounds of Stabilizability................... 25

III. C STABILIZABLE HYPEPPLANES ................... 32

3.1 The Convex Hulls, co(Cn ) and co(RCn )... 33
3.2 Cn -Stabilizable Hlyperplanes .............. 45


4.1 Convex Hull of D ............... .......... 49

4.2 0 -Stabilizable Hyperplancs................ 57
4.3 An Example .................... ............ 63

V. CONCLUDING REMARKS ............................... 66

APPENDIX I............................. .................. 72

APPENDIX II.................................... ........... 74

IIBLIOGRAPHY .............................................. 77

BIOGRAPHICAL SKETCH ............. ....................... 80



Figure 2.1......... ................ .... ............... 11

Figure 3.1................. ............................. 44

Figure 4.1........... ......... .... ...................... 50

Figure 4.2...... ........................................ 53

Figure 4.3.......... ................ .................... 62

Abstract of Dissertation Presented to the Graduate Council
of the University of Florida in Partial Fulfillment of the
Requirements for the Degree of Doctor of Philosophy



Raymond Chen

June 1979

Chairman: Vasile M. Popov
Major Department: Mathematics

In this dissertation, the output feedback stabilization

problem is studied from amoregeneral framework. For any

S let Sn and Cn denote {(a, a2,....,a) Rn xn +

n =laxn-" = 0 implies Re(x) < a} and {(al,a2,..,an) e Rn

xn n=la xn-" 0 implies Ix (a+B)/2[ < (B-a)/2) respec-

tively. In chapter one, we illustrate that under the general

setting, the output feedback stabilization problem of a

system with single control can be converted to a problem of

examining whether a given affine set (depending on the
system) intersects SO. Main results of this dissertation are

also mentioned. In chapter two, we obtain easy necessary

and sufficient conditions for a hyperplane H to intersect

n. We also show that H n S n I if and only if II n RS S

where RS = {(al,a 2...',a) R"n all the roots of xn +

Z =laxn-3 = 0 are real, distinct and strictly less than a}.

The results are based on the following property: "The convex
n n
hulls of RSO and SO are both equal to the positive orthant


of R ." ri chI mpter three, w; clihar-icLcr ze i yp,rrpjI an(cs whiich

intersect Ca for arbitrary fixed ( < B. We also show that

co(C, 6) is an open simplex in Rn generated by some special

vertices. In chapter four, we investigate the output feedback

stabilization problem of a special delayed system. Chapter

five contains a sufficient condition for a straight line to
intersect Sn and a computer aided algorithm for determining

the Sa-stabilizability of a given straight line. Some dis-

cussion of the directions for future research is included.


Consider an unstable continuous linear time invariant

control system:

(C) x = Fx + Gu, y = Hx

where F e Rnn, G Rn1m, H R n. One of the problems which

have been receiving a great deal of attention in control

theory is the Output Feedback Stabilization problem. It is

to determine the existence of an output feedback u = Ky or

even to find such K c Rmxr so that the closed loop system

becomes stable. In other words, it is to find a matrix

K e Rmxr such that all the roots of the following equation

have negative real parts.

det(xl I GKH) = 0

Wonham Fl] had proved the following theorem.

Theorem 1.1.1. For a control system defined as in (C), let

H = I. Then the pair (F,G) is controllable, namely, rank

(G, 1G,...,I''' n- ) = n (see Kalinn et al. I l ) if and only if

for every choice of (al,a2',..." a ) R", there is a matrix
n- sc n nI
K c Rmr such that det(xl GKI) = x + a xn-j

Note that in this theorem, physically, the assumption H = I

implies that all the states are measurable. This is unlikely

to most practical situations.

A considfi r-rab 1 amount of research in the Ii L t-rtre hs.

been devoted to finding a controller designed fr-om the a-vail-

able state measurements to stabilize the system. One approach

is to obtain estimates of the inaccessible states from a

Luenberger Observer (Luenberger [1], [2], 13]) and use these

estimates along with the available state measurements to

construct a feedback control. Along the same line, Brasch

and Pearson [1] showed that arbitrary pole placement can be

obtained by adding a compensator to the system. Both are

powerful methods for stabilizing a system; yet, one pays the

price of increasing the dimension of the system. Hence, it is

still desirable to know whether a constant gain feedback

can be designed directly from the available state measurements.

Most of the efforts of the researchers such as Miller, Cochran

and Howze [1], Luus [13, McBrinn and Roy [11, and Sirisena and

Choi [1] were devoted to the study of numerical schemes for

finding the feedback gain matrix k. In a paper by Anderson,

Bose and Jury [1], the problem of output feedback stabilization

was related to the decidability problems (see Jacobson [11)

and these authors showed how the decision methods of Tarski

[1] and Seidenberg [1] can be applied to solve the stabilization


In this dissertation, we shall look at the output feedback

stabilization problem from a more ('ncral point of view

described in the following.

Definition 1: ;iven a set IK, ,in n- imn .nsioinal iin.'r outpjut

feedback system with parameters in K is defined to be an

ordered pair (n,K)n, where n is a mapping from K into Rnxn x

R m x Rmxn for some fixed m, which associates with each para-

meter K c K a triple (A(K), B(K), C(K)) with A(K) Rnxm,

B(K) c Rnm and C(K) c Rmn. For brevity, in the sequel we

simply say "a system (n,K)n with parameters in K."

For a fixed m and n, let X be a mapping from Rn x Rn x

R into R which assigns each (A,B,C) a vector (al,a2 ....

a ) c Rn where a 's satisfy xn 1x Z n-a = det(xT-A-BC).


Sn = {(ala2' .."an) e Rn x ~n Ela xjn- = 0 implies

Re(x) < a}

RSo = {(al,a2'...,a) Rn all the roots of xn + Z= ajx.n-j = 0

are real, distinct and strictly less than a}

Definition 2: For a fixed a E R, an n-dimensional linear

output feedback system (T,K)n with parameters in K, is said

to be S -stabilizable if X7(K) n S A system ( with
0a A system (n,K)n aith
parameters in K is said to be RS -stabilizable if XT(K) n RS n

If one lets K = Rm and defines n(K) = (F,G,KH) for each

K E K where F,G,H are the matrices given in (C), then X7r(K)

is equal to the coefficients of the characteristic polynomial

of (C) with output feedback u = Ky. Note that (C) is stabiliz-

able if and only if Xn(K) n SO 1,, or equivalently, by

definition, (n,K)n is S0-stabilizable. Suppose H = I and

(F,G) forms a completely controllable pair. Under the same n

as above, we observe that the well known Irsul t (ti -or(m 1. 1 .1)

of Wonham [1] means then Xn(K) = Rn.

For any system (C) with single control i.e., m = 1, it

was shown that (see Byrnes [1], Kamen [ 11)

det(xI-F-GKll) = det(xI-F) + kli adj(xI-F)G (1.1)

With K = Rr and the same n as above, the relation (1.1)

implies that )n(K) = XT (0) + KM where M is a r x n matrix

depending on F,G,H. Hence, XTi(K) forms an affine set in Rn.

Now consider the class of all systems (T,K)n for which

XT(K) forms an affine set. This is a large class which includes

all systems (C) with single control as the discussion in the

above shows. Observe that for a fixed a c R if one can

characterize affine sets which intersect S, then Sn-stabili-

zability of a system (n,K)n for which T(K) forms an affine set

is characterized. In particular, the problem of output feed-

back stabilization for systems with single control is resolved.

As a first step in this direction we consider mainly, the

case of hyperplanes in this dissertation.

Chapter 2 is devoted to the study of characterizing hyper-

n nn
planes which intersect S and RS We prove that a hyperplane

H = {(ala2,... an) c Rnl A + n 0) intersects S if
n n n
and only if II n IP1 where R+ is the positive orthant of I ), i.e.,

(al,a2,.. an) Rn a. > 0). lurLthierore, it is shown that

II n S is equivalent to 11 n RS 7 We obtain also the

criterion which states H n SO if and only if A A. < 0
D~ i i

for some i / j, i,j = 0,1,2,..., n. The simplicity of these

properties results from the following lemma:

n n
"The convex hulls of SO and RSn are equal to the positive
n n
orthant R of R i.e., co(S ) = co(RS ) = R .

The author would like to acknowledge that for the case

of n = 4, the lemma was first shown by Professor Popov. Based

on the idea from that proof, this author extended the lemma

for arbitrary n. Also, the proof for general n, was greatly

simplified by a suggestion from Professor Popov.

The results mentioned above are generalized to characterize

all hyperplanes which intersect Sn or RSn for an arbitrary
a (
fixed a c R. Observe that for any system (i,K)n with param-

eters in K, there exists always a E R such that X (I() n S

From the point of view of stability, it is desirable to have

that a "as negative as possible" to reduce the transient time.
However, if X7(K) forms only a hyperplane in R then one can

not expect the relation X( (K) n Sn x to hold for any a c R.

We show that given a hyperplane H the value inf{lI H n Sn 4}

is equal to the negative of the maximal real root of a specially

constructed polynomial. Some computation aspects such as how

to find a (ala2,...,a) c H n S if H n Sn r are dis-

cussed. An example chosen from Anderson ct al. [11 is


Given a discrete linear time invariant control system


(S) xk+1 = Fxk + Gu, yk

(S) is stabi] liable with output frdbaick if and only if Li'cre

exist K c R' r such that the modulus of all the roots of

det(xI F GKH) = 0 are less than one. Given a < 6, let

C = ((al,a2 ,a ) e Rn xn + axn = 0 implies

x (B+a)/21 < (8-a)/2)

RC = {(al,a2 ...a) c Rn all the roots of xn + Zn a jn-j 0

are real,distinct and strictly between a and G).

By taking K = Rmxr and T(K) = (F, G, KIH) for each K e K, one

can regard (S) with output feedback as a system (rn,K)n with

parameters in K. (S) is stabilizable by an output feedback

if and only if Xn(K) n C 5 n. The main objective of

chapter 3 is to find a necessary and sufficient condition for

a hyperplane to have nonempty intersection with C For

the reason of mathematical interest, we obtain necessary and

sufficient conditions for a hyperplane to intersect C for

general a < 8. It is interesting to point out that given
n n
a < 8, the CS and RC are both equal to the open n simplex

in Rn generated by the n+l vertices (qil,qi2 ... ,i) c Rn
_n n n-j (xcn-i (x_)1i
i = 0,1,... ,n where x + Lj= lqij = (x-c) (x-)

In chapter 4, we consider the stabilization of a special

kind of system with delay:

(P) x(t) = x(t) + bx(t-l) + u, y = cx(t) + dx(t-l), u = ky

With a,b,c,d fixed, one wishes to find k c R such that all the

roots of + (a+kc) + (b+kd)e = 0 have negative real parts.

Given a e R, let

0( a = ( ( a, ) W ( I U 4 1 he -- 0 l)O J i I j C) s ( '5 X < (t

This stabilization problem turns out to be a problem of

determining whether a given straight line intersects V,. We

obtain a necessary and sufficient condition for a straight

line to intersect DV for general a c R. It is interesting to

note that for any given hyperplane [I, the number inf{al H n Sn

} is always bounded from below (theorem 2.3.1). However,

in this delay case, there are lines H c R2 (hyperplane in R2)

for which the number inf{aj H n Da ) = (theorem 4.2.4).

One of the problems concerning the stabilization of

systems with delay is that "Can one stabilize a system with

delay by a feedback without delay?" In the end of the

chapter 4 we show with an example, that this can not be done

in general, despite the fact that this can be done for (0).

Despite all the easy criteria for determining different

kinds of stabilizabilities obtained for hyperplanes in chapters

2, 3 and 4, we are unable to find an easy necessary and

sufficient condition for affine sets with dimension strictly

lower than n-1 to intersect Sn. As an attempt,in chapter 5,

we obtain a sufficient condition for a straight line to

intersect S". Also, a computer aided algorithm is proposed

for determining the S -stabilizability of a straight line.

Some directions for future research are discussed.


For a fixed n, let (n,K)n be a n dimensional system with

parameters in K. Recall that K is a set of parameters and n

is a mapping from K into Rnxn x Rnxm x Rmxn for some fixed m,

which associates each parameter K e K, a triple denoted by

(A(K), B(K), C(K)). Let X be the mapping from Rnn x Rnxm x

Rmxn into Rn which assigns each (A,B,C) c Rn x Rnx x Rmxn

a vector (al,a2,...,a) e Rn where xn + : lajxn-j = det(xI -

A BC).

The main purpose of this chapter is to characterize the

S -stabilizability and RSn-stabilizability of systems (w,K)n

for which XT(K) forms a hyperplane in Rn. In other words, one

wishes to characterize hyperplanes which intersect Sn and RS"
a a
Recall that for a given a c R,

Sn = ((ala2, ...a ) Rn xn + Z a.xn- = 0 implies Re(x) j=1
S=l 3x-O
RSo = {(al,a2,...,an) Rn all the roots of xn + axn-=0

are real, distinct and strictly less than a).

Definition: Given a r R, a hyperplane H rRn is said to bIe S-

stabilizable if H n Sn x 4. A hyperplane H is said to be

RSn-stabilizable if H n RS n .
0( c

The tec(tniqiue which we use in hii.re is biiecd on a simple

property that a hyperplane in Rn intersects an open connected

set in Rn if and only if the hyperplane intersects the convex

hull of that open connected set. In lemma 2.1.5, we prove

that co(RSn) = co(Sn) = R By an invertible affine transfor-

mation Ta and lemma 2.1.5, we obtain the descriptions of

co(Sn) and co(RS ). Section 2 contains some criteria for
a a
determining the S -stabilizability of a hyperplane. We show
n n
that the S -stabilizability and RS -stabilizability are equiv-
U a
alent for hyperplanes. In particular, we find that a hyperplane
is S0-stabilizable if and only if the hyperplane intersects

the positive orthant. Finally, in section 3, the value

inf{aJH n Sn 5 4} for a given hyperplane H c R is computed

and some consideration of actually finding a point (al,a2,...,

a ) c H n Sn 4 is included.
n a

n n
2.1 Convex Hulls of Sn and RS".

Let us begin with some basic properties of Sn and RSn

which will be needed later. Recall that Sn = (E a.ei E Rn

n n n-3
x n+ nl a xn-3 implies Re(x) < ad and RSn = an P n

all the roots of x of xn + ._ a.x n- = 0 are real, distinct
j=1 j
and x < a).

Lemma 2.1.1. (i) RSn and Sn are open connected sets in Rn
[- a
for any a c R;

(ii) Sn c Sn if and only if a < B;

(iii) RSn c RSn if and only if a < B;
(iv) RS c S c for ver < 0.
(iv) RS c S c Rn for every a < 0.
a a +

The proof of (i) will be pr:;entuc ed in .iippcndix> I

the others follow straightforwardly from the definitions of

Sn and RS" together with the fact that SO c R+
a a 0 +
Since Hermite [l], enormous research had been devoted to

n n
the study of the set SO. Algebraically, SO can be characterized

by the well known criteria of Routh and Ilurwitz or Lidnard

and Chipart (Gantmacher [1]), in terms of system of nonlinear

inequalities. From these criteria, it follows that Sn is
subset of R Since the roots of a polynomial depend contin-

uously on the coefficients, the boundary of Sn consists of

coefficients of polynomials which have either a zero root or

a pair of purely imaginary roots. Indeed, the D-decomposition

method of Yu Naimark says that the hypersurfaces HI = ((al,a2'

E. la xn-3 = 0 has a pair of purely imaginary roots) divide

the coefficient space R" into nHl disjoint regions; and the

polynomials with coefficients chosen from the same region will

have the same number of roots with positive real parts. The

example below illustrates tie case n = 2.

Example: For n = 1, SO = I(al,a2) al > 0, a2 > 0) and
2 2
RSO = {(al,a2) a > 0, a2 > 0, a1 > 4a2}. On the curve
a1 = 4a2, the polynomials will have identical roots. In

figure 2.1, p represents the number of roots with positive

real parts.
n n
The sets Sn for a c R, as exemplified by SO, are deter-

mined by collections of nonlinear inequalities for a 's. In


Figure 2.1

general, for a given hyperplanei II Rn, it is not ;easy to

determine if H is S -stabilizable by considering directly the

intersection of H with Sn. However, because Sn is open

connected in Rn, the following lemma shows that one can

consider instead the intersection of IH with co(Sn), a set

which is larger and more regular than Sn itself.

Lemma 2.1.2. Let be a hyperplane in Rn and let S c R" be an

open connected set. Then H n S z if and only if H n co(S) v .

Proof: (1). Trivial, since S c co(S).

(:). Let E1 and E2 be the open half spaces corresponding

to the hyperplane H. Assume the contrary that H n S = -.

Then, by the hypothesis that S is open and connected, we have

either S c E, or S c E2. Without loss of generality, let us

assume S c E Since E is convex and S c El, it follows that

H n co(S) c H n co(EC) = H E1 = This contradicts the

hypothesis H n co(S) z 6. Therefore, one concludes that

H n S S

Using this lemma, the Sn-stabilizable hyperplanes can be

characterized as those which intersect co(S ). However, if

one wishes to take the full advantage of this characterization

of Sn-stabilizable hyperplanes, one must know what co(Sn)
a u
looks like. This will be the aim of the rest of this section.

First, we shall prove the main lemma mentioned in the intro-

duction (i.e., co(S ) = co(RS ) = R) Then, we shall use

this lemma to obtain descriptions of co(S ) and co(RSn) for
ar y
arbitrary a c R.

In ordir to prove the maiin ](nina, s-i I'ce c(' (S ), (co(PS )
and R, are open convex shots, it suffices to show that their

closures are equal i.e., co(S) = co(RS) = R (see e.g.,

Rockafellar [11, p. 45). This is what we establish in the

following lemmas.

Lemma 2.1.3. The origin 6 = (0,0,...,0) = E j ce is contained

in co(RS ) .

Proof: Let 6 be a positive number. Consider the following


n n
T (x+j6) = xn + Z P.(6)xn-J
j=l j= 1

Observe that for each j, P.(6) = c.63 where c. is a positive

real number. Therefore, for any c > 0, one can find 6 > 0

such that P (r) < n-1/2c for every j = 1,2,...n. Note that

HI. (x+j6) = 0 has only stable roots x = -j6, j = 1,2,...,n.
Thus, by the definition of RSg,

E P.(5)e. E RS (2.1)

Also, one has

n n n 2 1/2 n 2 1/2
II E P (6)ej Z Oe .l = ( Z P (6)) < ( ) =
j=0 j=1 j=1 j=l n

n n
By (2.1), it follows that E P p ()e,. N (0) n RS

Hence,, RS, co(RS ). I
0 0

Let ({c.)n be the standard basis of R .
] ]=1

Lmuma 2.1.4. For any r > 0 and ,ny post iv_ iicier k, the

point rek is contained in co(RS ).

Proof: In this proof j will be used to represent positive

integers. Take r and k as specified in the lemma.

For arbitrary q > 0 and 6. > 0, i = 1,2,...,n where
6i x 6j if i t j, let P (q,61,62 ...'6 ) be the polynomial of

q and 6 's (also written P (q,6i)) obtained from the following

polynomial expansion

k n n
i (x+q+6 ) (r+6i) = x + EP (q,6 ,6 6 )xn
i=l i=k+l j=1

Note that for every j < k, the highest degree of variable q

in P (q,6i) is j. In particular,

Pk(q,6i) = qk + k(q,6i) (2.2)

The degree of q of P (q,6 ) is at most k-1. Fix 6. 6 > 0,

i = 1,2,...,n. For every c > 0, choose q0 > 0 large enough

such that for all j, 1 j < k-i

kT Pj(q06162' '6n < Vn (2.3)


r Pk(0'61 .2' 6 < n (2.4)
k k'
q 0

as well as

max{2,r) < q0 (2.5)

This is posr ;ible l, 'cause the ,lcjrres of q i n P.(, e,6.) -re l:ss

than or equal to k-1 for all j k-1 and the sie property is

true for Pk(q,i ). Since for any j > k, P j(qC,O,O,....,0) = 0

and since P (qo,61,62 .....6 ) are polynomials of 6 's with

positive coefficients, one can find 0 < i < 6., i = 1,2,...,n,

where 6. 6. if i j, such that for each j > k
1 ]

SP(q 0'61,62 6 '6) < (2.6)

Also, since for each i, 6i < 6i, by the property that Pj(q,6i),

j = 1,2,...,n have positive coefficients, one obtains for

each j < k-i

r r
r P (q0,61,6 6 < r P4(q0, 62, 6n) < n n (2.7)
j O' i 2'2... n < 0,1, ...6n
q0 q0

and similarly,

r k 6 r (q ,, (
k Pk(q ,61,62,.''''' n k P( 61' 62q.... 6n) < ,/ (2.8)
90 0

Now, consider the following polynomial equation

k n n
1 (x+q0+6 ) 1 (x+6 ) n = + P (q0', 6 ,6 )xn-3
j=l j=k+l j=1

It has only stable roots x = -(q+6i) < 0, i = 1,2,...,k and

x = -6 < 0, i = k+l,...,n. This, by the definition of RSO'

implies that the point

z = E Pj q0,61 62'.... n)e (2.9)

belongs to I'. I
belongs to S tl ce, using Ihe relrition i S c o(lr; ), we


+ n
z e co(RS

From Lemma 2.1.4, e= c co(RS ). Since co(RS ) is

convex, the set co(RSO) also contains the line segment with
the origin and ze as end points. In particular, (r/q0)zE.

co(RSQ), since 0 < r < q0. By (2.9) and (2.2), we have

r8 r r 0 ))2 r i(q )2
I k -I Pj(q0F)) + (0 i( )K
q0 j=1 q0 q0

+ E (- P g (0 i))2 }1 2 (2.10)
j=k+l q0

We now prove that llrek (r/ql )z < c. Let RllS denote the

right hand side of (2.10). Then, by (2.6), (2.7), and (2.8)

one has

2 2 2
mHS < {(k-1)- + + (n-k) -}2 =
n n n

Therefore, we have shown that for every E > 0, there exists
r )r o n.c
a point -7 z E Neje1k) n co(RS ). Hence, rek e co(RS) 0.n

Now we state the main lemma.

Lemma 2.1.5. For every positive intcqer n, co(RS) = co (
0 0

Proof: First, observe that R" ={,n r(re) r > 0,
+ l a 21, rk k co
kr 1, rk > 0). Since, by lemmuna 2.1.4, re E co(RSn)
k 1 k k


for every r 0 and every k = 1,2,...,n. l'-nce, it follows
,n co n (Sn) lyb
that R_ co(co(S0S ) = co(RS) Conversely, by Icra 2.1.1

we have co(RSn) co(S) c co(R ) = R Therefore, R -
we have co(+S0 +_ +

co(RS ) = co(S ). Since for any n dimensional convex set

S c R, interior (S) = interior (S) (see Rockafellar 'i],

p. 46) also, since co(RS ), co(S ) and R are open, one con-
0 0 n
cludes that co(RS ) = co(S ) Rn U

Before we set out to find RS" and S, let us mention the

following remark.

Remark: If T is an invertible affine transformation of Rn

then for any S c R,

T(co(S)) = co(T(S))

This follows easily from the following facts:

(1) Every point z c co(S) can be represented as a con-

vex combination of points from S, i.e., z = Zi ks. with s e S

and k > 0, Z.k = 1;
1 1 1
(2) T(i.k.s.) = i.k.T(s.) since T is invertible affine.

Using this remark, one observes that if there exists an

invertible affine transformation T on Rn such that T (Sn) = SO,


T (co(S" ) = co(T (S")) = co(S") =
a a c x 0 +

Hence, the convex hull of Sn, namely, co(Sa) will simply be

the set of all points (al,a"2...,an) e Rn whose image under

T belongs to R+.

Such a Lrainsformation does indeed exist, as we will show

in the next lemma.

For any fixed a c R, let T be the transformation on Rn

such that for every (al,a2, ... a) Rn,

T (ala2 .. .,a ) = (ala2 .... an

where a 's are obtained from the following polynomial expansion:

n n
(x+a)n + a.x+n- = xn + a.xn-j (2.11)
j=l 1 j=l ]

We remark that this transformation was considered in a

paper by Hughes and Nguyen [1], and the explicit expression

of a. was also given there. For completion we include the


Lemma 2.1.6. T is an invertible affine transformation on Rn

such that

T (RSn) = RS" and T (Sn) = S"
a a 0 a a 0

Proof: (i) T is invertible and affine.

We first observe that, by Taylor's theorem, every polyno-

mial p(y) = yn + =lajyn- can be expressed as

n (j)
p(y) = p (y-n) (2.12)
j=0 j'

Thus, by substituting y = x + a, (2.12) lecomlis

n n (n-j)
Z a.(x+-) = p(x+) = E ( x
j=0 3 j=0 (n-j)

with a0 = 1. In view of (2.11), it follows tI it, for e.ach j,

1 5 j < n

(n-j) ji (2.
a. = P- =_ ) 1 ( ) (2.13)
3 Tn-l) i=0

To show that T is affine, let (ai,' ,...,a') and (a"

a, ...,an) be arbitrary two points of Rn. Then, for each i,

1 < i < n and for 0 < A < 1,

~ n-i j -n-ii a I +
a. = E ( )a [Aa' + (l- )a'.'] = A[ 2 (n 1) 'a ] +
S i=0 j-i i=O -i 1


Hence, T is affine. Represented in matrix form,

T n(a 2 n)(n)
T (ala2, ...an = (al'a2 ...- an + (( ) (2) ( )
(. n-i j-
where Q, = (q .) is an n x n matrix with entries qij. ( 3-

i = 1,2,...,n, j = ,2,...,n. Observe that Qa is an upper

triangular matrix with all diagonal entries equal to one.

Therefore, Qa is invertible. This implies that T is


(ii) We now prove that T (Sn) Sn and Ta(RSn) = RS.

From the way in which T was defined, it follows that for
n n n-j
every number vx such that xg + n-xj=lajx 0, the number
n n 0 and vice
Y- = x0 + a satisfies p(yo) y0 + ja 0 = 0 and vice

versa. Notice that Re(x0) < 0 if and only if Rc(y0) < a.

Hence, (a1,a2,...,a) c Sn is equivalent to T (al,a2...,an)
n n n
(a ,a2"... a) E SO. In other words, T (S ) = S. The same

proof holds for T (RSn) = RSn
C% C%0'

Remark: For veryy e- ., 1t q (x) xn a xn -j An
j 1 J
elementary calculation shows that the inverse of Tc is

defined by
-1 n X 2
Ta (la2,...,an al'aQ(-a),
.a. .. = ) (a1,a2 an) n + ( 1( ) (-a) ( ) (_a)2 .....

() (-,) n)

n-i j-i
where Q ( ij) is an n x n matrix with qij (j) (-a)
1-1 ni
Furthermore, T satisfies the relations, T (S ) = Sn and
a C1 0 n
T (RSn) = RS.

Theorem 2.1.7. Given any ac c R, the convex hulls of Sn and

RS" are both equal to the polygonal convex set formed by the

intersection of the following half spaces:

i n-ij-i
i [( ). -la > 0
i= 0 1i

i = 1,2,...,n

Proof: Let T be the transformation defined in lenma 2.1.6.
The theorem follows readily from the discussion preceding
lemma 2.1.6 and (2.13). The same argument holds for RSP. U
In terms of the usual standard basis of R one may write
n n 1(SO) Sn and co(S0 =n
R = { =lrk k I k > 0}. Since Tl = S and co(S

it follows that

co(S ) co(T-1(Sn) -1 co(S

-1 n -o = n -
( T n) =T k) rk )(-)

=n l =lkjj]rklr > 01

{ -( ) (-a)3 + rk r > 0} n (-Z)j +
S ( '-k > 0k
Ek=l j-k (-a) Jrklejrkr 0] (2.14)

where for Cr :h i, qk .e is the ith jow of matrix 0 .
j=13 k3
Hence, we obltii in another way otf oxp1i, rssinl co(S[)

Theorem 2.1.8. co(Sn) and co(RS") are both equal to the open
a a

convex cone with vertex E. ( ) (-a)3Je, generated by the row
n- j=l-
vectors of matrix Q = (qij) ; explicitly, qij (n-i) (-)

Note that the entries of the row vector ((n) (-u) (n) (-) ...,

(n)(-a)n) represent the coefficients of the polynomial (x-c)n

which has n identical roots x = a.

2 2
Example: For n = 2 and a fixed real number a, co(Sc ) = co(RS )

{(al,a2)I 2a + al > 0 and a + ala + a2 > 0 It is the

shaded area in figure 2.1.

As an easy generalization of theorem 2.1.8, one has

Theorem 2.1.9. Given a c R, suppose F is a set of complex

numbers which satisfies

{xI x < a)} c F {zl Re(z) < a] (2.15)

Let n(F) denote {naij I[a xn + Zn lax"n- = 0 implies x c F}.

Then, co(H(F)) = co(S ) = co(RS ).

Proof: (2.15) implies that co(RSn) c co(n(r)) c co(Sn). By

the relation co(RSn) = co(S), one concludes that co((C) =
co(RSn) = co(Sn).
(X 1

2.2 Criteria of S -SLabi lizabiliLy of ilyporplancs

First, let us prove the following fact.

Lernma 2.2.1. Given a hyper-plane Ii = { (a,i ,. ,a ) r' I

AO + j=1jal 01 the following proplirt i s are yjuivalcnt.

(i) IH n R
(ii) At least two of the numbers A., j = 0,1,2,...,n

have strictly opposite signs.

Proof: (i) (ii) This follows easily from the observation

that the sum + i=1 A aj will never equal zero for any

(al,a2 ,a ) e Rn if all Aj, j = 0,1,2,...,n are of the

same sign.

(ii) (i). Suppose that there exist i x j such that

AiAj < 0. Without loss of generality, let us assume A0 > 0.

Let n = {j A] < 0} and = {j| A. > 0}. Then, H n R ,

since H n R contains at least the following point

Z. A.
jt- A. j 3d e
z Z jc jj j A jjej

Using this lemma, a criterion for determining Sn-stabili-

zability for hyperplanes can be easily formulated.

Theorem 2.2.2. Given a hyperplane H = {(a,,a2,..., an) c Rn"

A + Z0IAja = 01, let h(x) be the polynomial
0 jljj .
=A n )x n- Let = n =0A ) '- i = n,1,2,...,n.
j-ln n-j j 0 j -

Then, the following statements are equivalenLt.

(i) II is RSn-stabi]izable.

(ii) H is Sn-stabilizable.
(iii) At least two of the numbers A., i = 0,1,2,...,n, are

of strictly opposite signs.

(iv) 'T er' ox L non q iI i V i n I / k ,' th..it

[h (- ) [h(k) (-(0) < 0 where h ) (-() and
h ()(-) are respectively, the j h and kth deriva-

tive of h(x) at x = -a.

Proof: (i) (ii) Recall that Sn and RS are open connected

sets in Rn. Hence, by lemma 2.1.2, H n Sn 4( if and only if
H n co(Sn) 4. Also, H n RSn 4 if and only if H n co(RSn)
1 a a

Since co(RSn) = co(Sn), one concludes that the property

H n S is equivalent to H n RS .

(ii) (iii). By lemma 2.1.2, we know that the property

H n Sn 4 4 is equivalent to

H n co(Sn) 4 } (2.16)

Recall from (2.14) that,

co(Sn) = { l1[( (-c)j + zn n-i-) r r > 0
a 3t i(l (ji ( Ii3$ I

Thus, (2.16) is true if and only if there exist r. > 0,

i = 1,2,...,n such that

n ~n n
Ao + E L()(-o) + (_. ) (- V)3- r ]A = LA + ()(-a)3]
j=l i= 1-i 3 j=l

nn n n
+ S [ (n 1) (-a) A.r. = A + Z A r. = 0 (2.17)
i=l j= 1- i= 1

By lemma 2.2.1, the existence of such r 's which satisfy

(2.17) is equivalent to the property that at least two of the

numbers, A., j = 0,1,2,...,n are of strictly opposite signs.

(iii) (iv) This equivalence follows readily from

the observation

(i) n
h (-() = A. for each i = 0,1,2,...n []

A special case worth mentioning is when a = 0.

Corollary 2.2.3. A hyperplane H = {(al'a2, .' '. an) Rn|

Ag + iiaa = 01 is S0-stabilizable if and only if A A < 0

for some integers j z k where 0 < j, k < n.

Proof: The proof follows readily from the fact that A = A.
for all i, when a = 0. 1

It is interesting to note that the equivalence between

(i) and (ii) of theorem 2.2.2 means that on any Sn-stabilizable

hyperplane not only can one find points (al,a2,...,an) e H

such that x" + =lajxn-3 = 0 has only roots with real parts

smaller than a but one also can find points (al,a2 ... an) e H

for which all the roots of the polynomial equation

x+ = a xn- = 0 are real, distinct and less than -. In

fact, theorem 2.2.2 can be generalized to the following:

Theorem 2.2.4. Let F be a set of complex numbers satisfying

the condition

{x x < a r c {z[ Re(z) < a}

Let 1(') = {(a ,a2.... a) E Rn" xn + :l a.xn-J = 0 implies

x e r}. Then, the property H n 1n() T I is equivalent to

H n RSn d.

I'roof Ob; r-rv t ha L IS c II() 1 F 'Ni ]-(, t, H p( I 1

i plics H n II 1( ) '1'. Conversely, if 11 n 1 (r) L then

HI n co(n(i)) F ,. Since, by theorem 2.1.10, co(2(f)) = co RSn),

it follows that Ii n co(RSn) x ,. This implies that

if n RSn x because RSn is open connected in Rn
a 0

2.3 Bounds of Stabilizability

Let H = {(al,a2 ...an) e Rnl A0 + 1ja = 0 be a
n n n
hyperplane in Rn. Since S c S if and only if a < E, it is
obvious that there exists always an a0 such that H n Sn

for all a > aO. Hence, the set {a[ H n Sn I f} is always

unbounded from above. The interesting question is "what is

the inf{aI H n Sn q}?" The next theorem gives the answer.

Theorem 2.3.1. Given a hyperplane H = {(ala2 ...,an) Rnl

A + Z" A.a. = 0) in Rn, let h(x) = n )A "n- and
0 i=1 ] i =0 n-j n-j
k = max{ji A. 0). Also, let n = -max{aJ 11 h0 i (a) = 0 .
] 1=0
Then, inf{ci| H1 n Sn r ) = r.

Proof: Observe that h(x) and its derivatives h) (x), j= 1,

2,...,n,have the same leading coefficient Ak. Without loss of

generality, we assume that Ak > 0. Then, for every a < ri,

h i) (-a) > 0 for all i = 0,1,2,...,n. In view of theorem 2.2.2,

this implies that II n S = ct for every a < ?I. Hl nce,

inf{t II i Sn I/ n} n (2.18)

Conversely, by the definition of r, there exists some positive

integer i < k-l such that h i)(-) = 0. Also, since h (x)

i s a polynomi a] of posi Live riI-ee with l )posi Live I-uading

coefficient, there exists an f. : 0 small enough such that for

every a c (ri,Tn+ ) either h (-(c) > 0 or h( ) (-a)<0.

In the latter case, since h (k)(-a) = (n/k)Ak > 0 (we assumed

Ak > 0), one has [h(i+l) (-a) [h (k) (-a) < 0 for all a c (ln,r+E).

In the first case, one has h(i l) (--u) > 0 for all a E (n,l+c)

and h ((-a) = 0. Consequently, h (-a) < 0 for all

a c (n,n+E). Therefore, h (k)(-a) h (-a) < 0 for all

a c (Cn,n+E). By theorem 2.2.2, both cases imply that

H n Sn x ( for all a c (n,n+c).

Using (2.18), one concludes that inf{aI H n Sn -J} = n.D

With n defined as above, the assertion of theorem 2.3.1 can be
restated as follows: "A hyperplane H is S -stabilizable if

and only if a > n."

Remark: Recall that in the introduction, we noticed that if

m = n, then for any a E R there exist always parameters k.,
i = 1,2,...,n,such that all the roots of the equation

p(x) + =1kp. (x) = 0 (*) satisfy the inequality Re(x) < a.

In other words, one may "push" all the roots of (*) toward

the left side of the complex plane as far as one wishes by

the proper choice of k 's. The theorem above shows that this

is not true for the case m = n-1. In fact, the theorem

describes Lhe "limit" which can be achieved for all the

possible choices of k.'s.

Before we prove the next theorem, let us divert for a

moment to prove the following property of polynomials.


Lemma 2. 3. 2. Suppose that A is the only one (count in'j the

multiplicity) positive real root of a polynomial f whose

derivative f' has at most one positive real root. Then,

f'(x) 2 0 for all x > X.

Proof: Assume the contrary: there exists x0 > A such that

f' (x0) = 0. Then, either f"(x ) = 0 or f"(x0) > 0 or f"(x0)< 0.

We will show in the following that all these cases lead to

contradictions. Without loss of generality, we assume that f

has a positive leading coefficient.

(i) If f"(x0) = 0, then, by the assumption f' (0)=0, one

finds that x = x0 is a positive double root of f'(x) = 0.

This contradicts the hypothesis that f' has at most one posi-

tive real root.

Note that A is the only one positive real root of f(x)

and that f(x) has positive leading coefficient. Hence,

f(x) > 0 for all x > A. Therefore, there exists E > 0 small

enough such that X + E < x0 and for all x e (A,A +E)

f' (x) > 0

(ii) Suppose that f"(x0) > 0. Then, from the assumption

f'(x0) = 0, it follows that there exists 6 > 0 small enough

such that A + E < x0 6 and f' (x) < 0 for all x c (x0-6,x0)

Since f' ( + 2) > 0, and f' (x-6/2) < 0, by the intermediate

value theorem, there exists x1 where A < ./2 x, < x0 6/2

such that f'(x1) = 0. Therefore, f' contains two positive

real roots x = xl and x = x0, a contradiction.

(iii) Supposi that f"(x0) < 0. Then, ;ince f' (x0) 0,

there exists a ,. > 0 such Lhat (x) < 0 for all x ( (x0,x0+5)

Observe that f' has positive leading coefficient. Hence,

there exists 8 > x0 + 6 such that f'(x) > 0 for all x > 5.

Consequently, f'(x0 + 6/2)*f' () < 0. By the intermediate

value theorem, there exists x2 where x0 < x0 + 6/2 < x <

such that f'(x2) = 0, a contradiction. This completes the

proof. P

Theorem 2.3.3. Given a hyperplane H = {(al,a2'' ... an) RnI

A + =i1 A.a. = 0), let k = max{jl A. 0). Let h(x) =

n: A. (n)x. Let n = inf{(a H n Sn 4) If there exists a
j=0C] j a
nonnegative integer Z < k-1 such that A. < 0 for all j <

and A. S 0 for all j > 2, then n = -max{Ia h(a) = 0).

Proof: Obviously, -max{a| h(u) = 0) -maxal ilk h(j) (a) =0}=n.
Conversely, recall that h(x) = Z ( )x. The hypothesis
o nr ey The hypothesis

implies that there is exactly one variation of signs among

the coefficients of h() (x) for each 0 j < 1-1 and no

variation of sins of coefficients of h (j)(x) for j 9.

Therefore, by Descartes Rule, the polynomial h (1 has exactly

one positive real root if j 9-1 and no positive real root

if j > Z. It follows that

f -1
--max(l ( a) = 0) = max{[u I h (j) () = 0) n
j=0 j=0

Furthermore, by applying lemma 2.3.2, to h,h'; h',h",...,

h(-2)k (-1), one obtains that, for each 0 < j e-1,

max{an h(j (j ) = 0} < max{a h(j-l1)() = 0}

Consequent ly,

max{~a h )(a) = 0 max{ul h(c) = 0}


n = -max{al h(a) = 0} [

Remark: Given a hyperplane H c Rn, suppose -inf{na H n Sx 4 =r.

From the point of view of applications it is important to
actually find the points in H n S a if a > n. Unfortunately,

we do not have a general method for such a task. Following is a

special case for which we do have a method. Suppose H is a

hyperplane in Rn such that q = -max{a| h(n) = 0). Then,

h(-n) = 0. Recall that h(x) = Zn =0 j()x3. Therefore,

0oAj (n) (-n) = 0. This implies that

(( ) (-n), ( ) (-n)2 ( ) n) H
(2.2 n) n n H

Note also that

( 1)(-n), ( )(-n) .2 ) (-n)n) c RS
n n n Jn-J
since the polynomial x + jj= j)(-n)x = O has only

identical roots x = n. Hence,

S) (-n), ( ) (-n)2 ( ) (- )n) if n RSn
'1 2 n n

We remark that this method can be applied to the kind of

hyperplane described in theorem 2.3.3.

__ ~

In the fol owinq, we iI lius;t-rite I he ri;n]uts5 which we

obtained in this chapter by an example chosen from Andcerson

et al. [1].

Example: Consider a system

0 1 0 0
(E) x = 0 0 1 x + 0 u
0 13 0 1

0 5 -1
y = -1i 0 x u = Ky = [v,wjy

Then, the characteristic polynomial of the closed loop system

s + vs + (w-5v-13)s + w = 0


Let II denote the set of all the coefficients

from all possible choice of K = -v,wl, i.e.,

vectors arise

H = { (al,a2,a3) a = v, a2 = w-5v-13, a3 = w, for some

v,w r R} = {(a ,a2' a ) 5a + a2 a3 + 13 = 0).

Note that H forms a hyperplane in R3. Since A = 5 > 0 and

A3 = -1 < 0, by corollary 2.2.3, the system (E) is S0-stabili-
zable. In other words, there exists a K e R2 such that all

the roots of (2.19) have negative real parts.

Consider the polynomial h(x) defined in theorem 2.3.3.
3 2
For system (E), h(x) = -x + 3x + 15x + 13. By theorem 2.3.3,

one concludes that

inf{fal I n S3) = -max{cl h(a) = 0) C -5.91


This means Ihat rio matter h]w wi( chloor) K I v,w], t.lroc e:*:i s s

at least one root of (2.19) with real part ygrieter than or

equal to the number -max{a c RR h(ci) = 01 2 -5.91.

ft, 6

Given a < B, let C denote the set of all the points

(al,a2 ...,a n) e Rn for which the polynomial equation

x + Ejl a.x = 0 has only roots inside the circle centered

at ((ct+B)/2,0) having radius (B-a)/2. Let RCn be the set

(a ,a ...a ) E Rn all the roots of xn + a. x"- = 0

are real, distinct and strictly between a and 8).

A hyperplane H c Rn is said to be stabilizable with respect

to the circle {z| Iz (B+t )/21 = (B-u)/2) if H n Cn
H is stabilizable with respect to the interval {xljc
H n RC n, In this chapter we investigate the following

two problems:

(i) What are the co(RC ) and co(Cn )?

(ii) How can one characterize the hyperplanes which are

stabilizable with respect to a given circle and the hyperplanes

which are stabilizable with respect to a given interval?

Particularly interesting from the stability point of view,

is the case of Lhe unit circle ccnte lc(d at the origin. In

fact, the question of the stabilizability of a hyperplainc

with respect to the unit circle centered at the origin arises

from the problem of direct output feedback stabilization of

a sampled data control system. The well known Schur-Cohen


criterion which characterizes C -,L consists of a system of

nonlinear inequalities. Hence, it may be a difficult task

to answer questions (i) and (ii) directly in terms of those

inequalities. Instead, by an invertible linear transformation

constructed in section 1, we shall reduce (i) and (ii) to

problems solved in chapter 2. As a main result, we prove
n n
that co(C ) and co(RC ) are both equal to the interior of
the simplex generated by the points (qil,q 2....' ,q) Rn

i = 0,1,2,...,n where q. are the coefficients of the polyno-
n-l n\ i n n" "-3
mial (x-a) (x-B) = x + Zn q n-j

3.1 The Convex Hulls, co(Cn ) and co(RCn
a,B ---------c(, P-
We will denote the standard basis of Rm by fe } jm'
j j=1

For brevity, if there is no confusion of the dimension, we

will simply write (e j=1. First, let us extend lemma 2.1.5

to the following sets

n -n+ n
S= { 0 a.e + a > 0, E a.xnj = 0 implies Re(x)<0)
j=0 3+1 0 j=0 I

nn n
n -n+1 n n-j
RS = { E a.ej+1 a > 0, all the roots x of T a.x = 0
j=0 +1 j=0

are real, distinct and x < 0}

Lemma 3.1.1. co(S) = co(RS) = Rn+ .

Proof: Obviously, by Hurwitz criterion and Lhe convexity
1n+l n4-l
property of R one has co(RS) c co(S) c R
n ~n+l n+1 th n ,n Rn
Conversely, if Z =0 a e+1 ce R + then j j(a /a )ej +.

Recall that R = co(RS) Therefore, Zn (a /a0)e can be
Realta + 0 j = 1 /a0 ]

expressed ;i: a conv:e combina'it ion of points from US In
vn 'n n
other words, there exist points i1 i (b /a S and

parameters ki > 0 satisfying Zik. 1, such that
n ~n n n ~n
Zj=1(a /a )e. = j=l[ i= k (b. ./a )] e or equivalently,

0n n+1 n+1 -n n n+
j=0 aj+l = ae + =li IIibijej+l 3. 1)

Observe that for each i, the equations xn + (b /a )xn- =0
] 1 ij 0
and a0x + j= b nx = 0 have the same roots. Hence, from
n n n n+l n ni1
S(b i/a) e e RS0, one obtains a 001 + 0 b ij+ c RS.
n nF-l
Therefore, by (3.1), one concludes that En aej+ c co(RS)
n n+l n+l n+l
for any j= 0ajej+l R This implies R 1 co(RS). U

For any fixed a < B, define

n n n+l ni,.n n-
nB = {= ae n+lc Rn+ a0 > 0 e = a.x J = 0 implies
a j jj+l j I

Ix (a+3)/21 < (l-a)/2}


n n -n+l n+l
RC = (zj a ej+1 I R la > 0, all the roots of

n axn- = 0 are real, distinct, and striclty between
j=0 a
a and 6).

Let T be the transformation on R such that for each
n -n+l n+l
Fj=0 ajie r

nl n ai C4
nP O j+l 1
ja ) = j0o

where a., j = 0,1,2,...,n are obtained from the following

polynomial expansion:

n n-j j n n-
aa. (s-c- ) (s- ,) = a sn-j
j-0 j 0

For a = 1 and B = 1, it is well known that T ,i is a

linear transformation which maps S onto Cn1 (Hermite [1],

Garden [1]). The matrix representation Q-1,l of T-1,1' due

to its application to the stability of a discrete system,

has been investigated by many researchers such as Power [1],

Fielder [1], lalijak and Moe [1], Jury and Chan [1], Duffin [1],

etc. Barnett [l] has studied the linear transformation T

induced by a more general bilinear transformation z = 'S-

and showed that the entries of the matrix representation of

T can be obtained by an interesting algorithm. For our purpose,

we need the following special case of Barnett's (i.e., 6 = 1

and y = 1).

Lemma 3.1.2. For any a < 8, T is an invertible linear
a, P
transformation on Rn+ which satisfies T (S) = Cn an.
T (RS) = RC

Proof: We divide the proof into three parts:

(i) T is linear.
n n+l n -. nl -n ~n+l
For any aj=0ajej+l and bj=0 jejl,. let j=0c jj+1
T,-(n bj j+l-
T, j=0( (a b )e+l). By the definition of Ta,, the c.,
j = 0,1,...,n, are obtained from the following polynomial


n n
n n-j n-
c.s = ( (a (s-)b ) (s-)) (3.2)
j=0 j=0

n n+l n n+1 n -n+1
If E a 3+ and .j bjl are the images of an e
j=0 l j-0 j nd j0 j j+l
a n n+l
and Zj0 b ej+1 under T respectively, then the right hand

side of (3.2) equals (a.+b )s Thus, c = aj + b.,

j = 0,1,2, .. n. Hence,

n n+l n n+l n -n+l
T ( (a.tb.)e ) = T ( a e ) + T ( ( b.en+ )
j,S j 3 j+1 aB j=0 ] j+1 a, j=0 ] n

(ii) T a is invertible.
To show that T6 is invertible, since TB is linear,
a,)? a,)?
it suffices to show that T is onto. In other words, for
a n ~ -n+l an 1n+1
any j=0 a ej+l we need to find Zj=0 a ij+ such that

n n
Z a.(s-a)n- (s-B) = a.s" (3.3)
j=0 3 j=0

Substitute (6z-a)/(z-l) for s in (3.3) and multiply both sides

by (z-l)n. Then, (3.3) becomes

n n
S a.z = a. (z-a)n (z-1) (3.4)
j=0 3 j=0 3

In view of this, one may simply choose a.'s to be the coeffi-

cients of the polynomial expansion of the right hand side of

(3.4). This proves that T is onto. Hence, T is


(iii) T (S) C and T (RS) = RC

Consider the 1-1 transformation

s --> z = -P


which maps the interior of Lhe circle {s| I=-(itiF)/2 (B,-r)/2)

onto the open left half plane {z I Re(z) < 0}. In particular,

(3.5) maps the interval (sla < s < B) onto the negative real

axis. Hence, for any nth degree polynomial p(z), the state-

ment: "all the roots of p(z) = 0 are in {z Re(z) < 0}" is

equivalent to the statement: "all the roots of p(s)

(s-B)"p(S-') = 0 are in {sI Is-(a+B)/21 < (--a)/2)." Further-

more, {zl p(z) = 0) c {z Re(z) < 0} if and only if

{s p(s) = 0} c {sj a < s < B). Observe that for any p(z)
n n
n ax"-, (s-n = a(s-a)n(s-) = a.sn-
j=0 ] j=0 j=0
n +n+1
Recall that this was how the image of E a e+ under T
j=0 j+1l a,u

was defined. Hence, T ,, determines a one to one correspon-

dence between the Hurwitz polynomials and polynomials whose

roots lie inside the circle {sI Is-(a+p)/21 = (B-a)/2).

Similarly, T determines a 1-1 correspondence between the

polynomials which have only distinct negative real roots and

the polynomials of which all the roots are distinct real and

strictly between a and 6. If the leading coefficient a0 of a

Hurwitz polynomial is negative, then aj < 0, j = 1,2,...,n.
m n n n+1
Hence, a0 = j=0 aj < 0 which means that T (LE=0a ej+l
n n-nl -n -n
E a j+ C ,. Therefore, T (S) = co(C ) and
jj l + ,E rSJ
r, (RS) = RIC ,

Following a similar technique as; Iarrnr-LLtt 11, we compute

explicitly the matrix representation Q, of Ta, relative

to the usual standard basis of Rn

Note that Q is an ntl by n+l matrix such that

( a ej+ 1) = (a0'al,...'1 )Qa,1".

lHence, from the definition of T ,, for uach fixed i, the row

(qi0,il' ...' in) of Q ,B represents the coefficients of the


(s- )n-i(s- )i (3.6)

By a simple combinatorial argument, one may expand (3.6) as

n i n-i-kk
(s n-ic s-B) = E [(- )) () (n- )aj-k k] (3.7)
j=0 k=0 j-k

Hence, for each 0 < i < n and 0 < j < n, one obtains

q i (-1 ) ()( _n )aj- Bk (3.8)
k=C k -j-k

In particular, from (3.7), it is easy to see that qi0 = 1,

i = 0,1,2,...,n.

We remark that the determinant of Q is equal to
(a-B) (see appendix II).
Now consider a transformation T on R defined by

n ~ n+l1 n *n+l
f=0an -+n+1
a,B j=O aej+

where the a., j = 0,1,2,...,n are obtained from the following

polynomial expansion:

n n
7 a. (z-a)n (z-1) = T a.zn-
j=0 j=0 I

By the transformation z -> s and with the same

technique used in establishing lemma 3.1.2, one can show
n+l which
that T oB is a linear invertible transformation on R which

satisfies the relations T ,( ) S and T (Cn)

Furthermore, one finds that the entries qij of the matrix

representation Qa, of T are

qij = (-l) ( (j-k)ak-k n-i-j

We remark that one may use the algorithm proposed by Barnett [1]

to generate the entries of matrix QC.

For the unit circle case, i.e., a = -1 and 8 1, it is

interesting to note that the matrices Q1,1 and 0-I, which

are associated with T_1,l and T_-,1' respectively, are equal.

Duffin [1] and Jury and Chan [1] have computed the matrix

explicitly and have proved the following relations between the

entries of Q-1,1

qij = qi-l,j qi-l,j qij- i i, j (3.10)

-Q I, Q_- = (-2)n(n+l)

Obviously, (3.10) can be easily programmed on a computer to

generate the matrix Q-1, 1

Now, recall that if T is an invertible affine transfor-

mation on R, then for any set S c R, co(T(S)) = co(T(S)).

Using this, we obtain the following proposition.

Proposition 3.1.3. For any fixed a < i and every n, the sets

co(C ) and co(RC ) are both equal to the polygonal convex

set in Rn+ formed by the intersection of the following half

n j- -
2 [(-l) ( ( ),( j k n-i-j la > 0 j = 0,1,2,...,n.
i=0 k=0

Proof: Recall that T (C ) = S aid T (,C = S

From lemma 3.1.1, it follows that

(co(C ) co(T0 c B ') = co(S) = R+1


-n ~n n+l
T (co(RC )) =co(T (RC )) = co(RS) = R+
a,B a(cx a ,B a, +

n n+1 n n
Hence, E ae+1 E co(C co(RC, ) if and only if
j=0 aj j+l a c ,(
T n -n+l
T (= a ej+) (a0,a ,.., a )QB > 0 or equivalently,

S j n-i i ) j-k n-i-j+ka > 0 for each
i = 0,1,2,...,n.j-k k

Observe that, in terms of the usual basis vector of

Rn+ the positive orthant R n+ can be expressed as
n n+l
{ E=0 kiei+1 ki > 0). It follows that
i 0i n+l,

co(Cn) = co(T (S)) = T (co(S)) = T ,R(Rn+

n >n+l
= i TaX ei+l) ki > 0)
i= ,q

= { E ki(qi0', il .. in)' ki > 0

where (qi0',qjl... qin) is the ith row (0 i

Q ., This gives us another characterize ti on

co(RC ).
C ,

Sn) of matrix

of co(C ) and

Proposition 3.1.4. For any fixed a < 6 and n, co(cn ) and
ScoRcn e a t n co
co(R n, ) are equal to the interior of the convex cone
C,% r

11.n I 11 4 1
(;i., iraitcd by thi vector' j qi. wi h origin as

vertex, where fur each i, the ij, j = 0,1,..., n are Lhe
n-i i
coefficients of the polynomial expansion (x-n) (x-S) =

En q.x-3
j=0 niji
For M c Rn+ and a c R, let M denote the subset of M

defined by (a0,al,...,an) E M a = a). Note that n,

Cn RCn and RC are related by the following relations:

(C ) = -11 x c
e,S 1 a
a 1c,

(RCn ) = {1)x RCn
a,S 1 aF

The next lemma shows the relation between their convex hulls.

This will enable us to answer the first question which we

raised in the beginning of this chapter.

Lemma 3.1.5. (i) (co(Cn = { x co( ) and (co(RC ))

{1) x co(RCn )

(ii) co(Cn,) = co(RC ).

Proof: (i) Every point from (co(Cn ))l, is of the form

ki (bi0,bil ...bin
where (b. b ... .b) e C and the k.'s satisfy the
i l1 in a, 1
relations k. > 0, .k. = 1 and Ekib = 1. Since, for each
(b i0,bi ... ,b i ) c b 0 > 0, it follows that
0 1 in aR i0

Ek (bi0,bil ... bin) = {l) k ibi0(bil/bi0 bi2/bi0 ...
i i
bin/bi0) c {/1 b co(Cn

The last inclusion follows from the definition of co(C" ,)
C ,

Conversely, if (l,al,a2 ... ,a ) r {1) x co(C ) then
r, e
(l,ala2 ...,an) = {1} x { k (bil, b ,..., bi ]
k (lbil,...,b ) where (b ,b ...,b ) c C and

E.k. = 1 with k. > 0. Since for each i, (l,b l,b i ,.. ,b. ) c
a i i 1 12 in
~n ,n
C 6; this implies that (l,al .. ., ) e (co(C ) Therefore,
0 t n nt,
(co(n )) = {1} x co(Cn ). The same proof holds for

(co(RC )) = (1) x co(RC ).

(ii) Since co(C ) = co(RC ), by (i), it follows that

c o 'n -n C
{l) x co(C ) = (co(C )) = (co(RC )) = {1} x co(RC

This implies that co(Cn ) = co(RC" .

Combining the previous lemma and proposition 3.1.3, we

obtain the description of the convex hulls of C,6 and RC a,

Proposition 3.1.6. Co(CN ) and co(RC ) are both equal to

n j n-i
((al,a2, .. ) (-1) ( ) Bn-3 + z [ (-1 ) I ( )
1 i=l k=0

i j-k n- +k]a.a > 0, j 0,1,2,...,n .

Proof: Apply proposition 3.1.3 with a0 = 1.

Consider the matrix Q = (qij) associated with T ,B.

Recall that qi. equals one for each i, 0 < i s n. From

proposition 3.1.4, it follows that

l} x co(Cn ) (co(C ))
E k ( 'qil-,
i= ki( l'qi2,'''q in) i > 0}1
S {1 { ) ki(qil'qi2' .. ri-) Iki > 0, k = 1i .
i=0 1


co(Cn ) -= 0ki(qil,q' *. ,qin) ki > 0 ik = 1}.
1 1

Proposition 3.1.7. co(Cn ) and co(RC ) are both equal to

the open simplex generated by the n+1 points (qil,q.'i2 ..,qin)E

Rn, i = 0,1,2,...,n where qij are the coefficients of the
3=1 i ; explicitly,
polynomials (x-a) -1(x-S) = xn I1 qijxn3; explicitly,
qij (-)j (n-i i ()j-kk
q = k =0 (j-k k

In particular, if a = -1 and 1 = 1, then one obtains

the following conclusions:

Corollary 3.1.8. co(Cn ) and co(RCn ) are both equal to
-1,- -1,1
the open simplex generated by the n+l points (qil',i2 .... q

c Rn, i = 0,1,2,...,n where qij are the coefficients of the
n-i i n .n n-
polynomials (x+l) (x-l) = x + Lj..q ijx explicitly,
Sk n-i i
q.. = (-1) ( )
13 k=0 k k

Example: Suppose n = 2. Then, co(C ) is the shaded
region in the figure 3.1, and is bounded by the following

three lines:

1 + al + a2 = 0, 1 a2 = 0, 1 a1 + a2 = 0

In this case, the convex hull of C is actually equal to
C itself. Since
1 2 1
-1,1 1 0 -1
1 -2 1


Figure 3.1

by corollary 3.1.8, this implies that co(C2 ) is the
triangle in R2 having the points (2,1), (0,-1) 'd (-2,1)

as vertices.

In fact, one may generalize proposition 3.1.7 to the

following theorem.

Theorem 3.1.9. Given a < B, let 1, be the open interval

(a,8). Let Da be an open disk on the complex plane with a

and B as the end points of a diameter,i.e., {z IJz-(6+a)/2[ <

(B-a)/2}. Let F be a set of complex numbers such that

I ca F c D Let (F) = {(al,a2 .. .,a ) e Rnl x +

Z na.x- = 0 implies x c F). Then, co(n()) = co(C ,)

co(RC ).

Proof: The hypothesis on F implies that

n n
RCn e (F) c Cn


co(RC ) c co(T(r)) c co(C ))

By the relation co(RCn) = co(Cn ), one concludes that

co(n(()) = co(C ) = co(C .

3.2 C -stabilizable llyperplanes
Now, we consider the problem of characterizing hyperplanes

which are stabilizable with respect to a given circle in

the complex plane.

In the following, for any hyplerpl]ane 11= { (a ,a2 ... n)
Rn 1
SR +ni, A a = 0), we denote by H the set {(aoal, .
a) c Rn+ Z Aa = 0). Note that H is a hyperplane in RnH1

Lemma 3.2.1. Suppose H is a hyperplane in Rn. Then, for

any fixed a < B, the property H n C" n 4 is equivalent to
H n C

Proof: (=). If (al,a2 ...,a) H n Cn, then by the
2 ,' then b the
definitions of H and C respectively, (l,a ...'an) c H and
(n n
(1 al,...,a ) c Hence, (1,al,a2,. ,an) c n C

( -) If (a0'"a,,I an) E H n Cu,, then (al/a0 a2 /a 0
1 1
a /a0) H n Cn,6
n 0a

Using this lemma and proposition 3.1.4, we obtain the

following criterion for determining the stabilizability of a

hyperplane with respect to a given circle.

Theorem 3.2.2. Given a < B, let Q, be the matrix associated

with the transformation T ,8 (see (3.8)). Suppose H = {(al,a2,

...,an) Rn A0 + g =Aa= = 0) is a hyperplane in Rn. Let

Ai, i = 0,1,...,n,be the numbers obtained from (AO,A ..A n

Q,B(AO,Al"... An) explicitly, A = [=0 (-) =O -i ) j-k

6 kA. Then, the following statements are equivalent.

(1) H n C ,

(2) H n RCn
X ,
(3) At least two of the numbers Ai, i = 0,1,2,...,n,are

of strictly opposite signs.

Proof: (1) (3). The proof follows from a sequence of

equivalences below. First, observe that, by the last lemma,

H n C x S is equivalent to H n ,x 4 Since is
n+1 n
an open connected set in R the property H n C ,
,n n-
holds if and only if H n co(Cn ) f Recall that co(Cn

is equal to the open cone generated by the row vectors

(qio0',qil .. in), i = 0,1,2,...,n of the matrix Q(,B. Hence,
H n co(C ,) i means that there exist parameters k > 0,
n n ~n+1
i = 0,l,...,n, such that Zn [l kq l
S0,,.On,such that n i= kiqij ej+l E H or equivalently,

n n n n
z A ( kq ij) = E ( E qijAj)ki = 0
j=0 3 i=0 1 i=0 j=0 13

for some k. > 0, i = 0,1,2,...,n. By lemma 2.2.1, this is

true if and only if at least two of the numbers A = E nqijA.,

i = 0,1,...,n are of strictly opposite signs.

(1) (2). Since Cn and RC are open connected
X B (1, P

sets,the property H n Cn x ( holds if and only if
a, P

H n co(Cn ) Also, H n RCn if and only if

H n co(RCn ) n 4. Recall that co(C ) = co(RCn ) Hence,

H1 n Cn if and only if H n RC, -L
Qc pP


In this chapter we wish to illustrate the applicability

of the "convex hull technique" which was used in the last two

chapters, to the study of the stabilizability of the following

delayed control system.

(0) x(t) = ax(t) + bx(t-l) + u, y = cx(t) + dx(t-l), u = ky

where a, b, c, d, k are real numbers.

Our purpose is to determine the existence of a real

number k such that the corresponding characteristic equation

of (D)

S+ + + kc) + (b + kd)e- 0 (4.1)

has only roots in the left half complex plane. Note that
-2 -+ 2 x
with a, b, c, d fixed and c + d z 0, the set of all the

coefficients of (4.1) which arise from all possible choices

of k E R forms a straight line

I = ((a,b) ad bc (ad + bc) 0) (4.2)

Now, consider an exponential polynomial equation of

the form

S+ a + be- = 0 (4.3)
A a + be = 0 (4.3)


Let VD denote the set { (a,b) e R2 i f a be = 0 implies

ReA < a). A hyperplane H is said to be 0 -stabiliz;ile if

H n Da < ;. In particular, when a = 0 we call D0 the domain

of stability of (4.3). Thus, a system (V) is stabilizable

if and only if H n 90 x* As in the previous chapters, we

will study two main problems related to D : (i) the convex

hull of a and (ii) the characterization of hyperplanes

(straight lines) which intersect c .

4.1 Convex Hull of D
It is known that for any (a,b) e R equation (4.3) has

only a finite number of roots with positive real parts

(J. Hale [I], Lemma 20.1). This kind of roots will be called

unstable roots in the following. In terms of the D-decomposi-

tion method by Yu Naimark [1] one may partition the coefficient

space R into infinitely many regions. We indicate in

figure 4.1 these regions and the number of unstable roots of

the equation when the coefficients belong to each region.

The boundary of these regions is the curve a = -- cos y
sin y
b = Y where nr < y < (n+l)n with n = 0,1,2,... This
sin y
figure actually suggests most of the techniques used in the

proofs of the theorems in this chapter.

Hayes [l] proved the following analytic characterization

of V0'

Theorem 4.1.1 [Hayes] All the roots of A + a + be- = 0,

where a and b are real, have negative real parts if and only if

(i) a 1 > 0






C~t6 :=o

Figure 4.1

(ii) a b > 0

(iii) /a2+r2 > b where if a x 0, r is the root of the

equation r = -a tan r in (0,n) and if a = 0,

r = n/2. O

Hence, DO is the set of all points (a,b) which satisfy con-

ditions (i), (ii) and (iii) of the Hayes theorem. By the

continuity of the roots with respect to the coefficients of

exponential polynomials, one obtains that V0 is open in R .


Lemma 4.1.2. DV is connected.

Proof: It suffices to show that for any pair of points

(al,b1) and (a2,b2) form DO there is a path P c P0 which

connects them.

Consider the half line L = {(a,l) a > -1}. We assert

that L c D. Since (a,l) c L implies a + 1 > 0, the condi-

tions (i) and (ii) of theorem 4.1.1 are satisfied. For

condition (iii), suppose first that a x 0. Then, by the

relations r = -a tan r and 0 < r < n, one derives

/+r = / r2cot2r+r2 r > 1 = b
,dTl =sin r

If a = 0, then

/a7 = /0+( )2 > 1 = b

In both cases (iii) holds. Hence, L c 0D. Now, for any

fixed (albl) and (a2,b2), we construct the path P as

P = P1 u P2 u P3, where

P1 I (il'b) I= 1 t (bl-1) 0 t 1 )

P2 = { (,1) a a a +t(a2-al) 0 t 1) L c P0

P3 = ((a2b) lb =14t(b2-1), 0 < t s ]

Note that if (aOb0) E V, then for any b with -a0 < b < bO,

the point (a0,b) satisfies (i), (ii), and (iii) of theorem

4.1.1. By this property one obtains P1 c 0. Similarly,

P3 c P0. Hence, P c D0. O

We remark that in general the stability domain of an

exponential polynomial is not necessarily connected. For

example, the stability domain of x(t) 4- ax(t-l) + bx(t) = 0

is a disconnected open set in R bounded by the straight

lines b = (-1) 2-a + [ 2 n l k = 0,1,2...,n. (See

figure 4.2.) This suggests an interesting problem (not

treated here): "What types of exponential polynomials have

connected stability domains?"

In the following we are going to show that

co(V0) = {(a,b)e R2 a+l>0 ab, a+b>0, a-b+2>0 (4.4)

The right hand side will be denoted by M in this chapter.

This is a polygonal convex cone in R To prove (4.4),

since co(00) and M are open convex sets, it suffices to

show that their closures are equal, namely, colT = M.

First, we need the following lemmas.

Lemma 4.1.3. Let V0 and M be defined as above. Then DO c M.


Figure 4.2

Proof: Obscive that, by the IlHaycs thtorenm, any point (a,b)

chosen from V0 satisfies a + 1 > 0 and a + b > 0. Therefore,

one needs only to show that conditions (i), (ii) and (iii)

of Hayes theorem imply a b + 2 > 0.

Consider a + 2 / +r where r = a tan r with 0 < r < 7

when a x 0, and r = T if a = 0. Suppose a s 0. Then, by the

substitution of a = r cot r, one obtains

a + 2 /a2+r2 = -r cot r + 2 /r2cot r+r

2(l+cos r) (tan r r > 0
sin r 2 2

The last inequality holds, since +cos r as well as
i yn sin r
r r
tan are positive for all 0 < r < r.

Suppose now a = 0, in which case r = 7/2. Then, one has

a +2- /a2r2 2 > 0

In both cases, the inequality a + 2 a2+r2 > 0 holds.

By (iii), this implies that a + 2 b > 0. The proof is

completed. D

Lemma 4.1.4. The point 9 = (-1,1) belongs to coTDT.

Proof: Let L be the half line defined in lemma 4.1.2. Recall

that L1 c 00. This lemma follows readily from the fact that

for any E > 0,

(-1 + ) c L N )

For the convenience of the proof of the next lemma, we

adopt the following equivalent representation of M;

M ={(a,b) I a+]10, b = ka + (k+1) for su me -l'k l) (4.5)

Lemma 4.1.5. Mr c o( 0).

Proof: Let (a,,b0) be an arbitrary point chosen from M.

Then, by (4.5) b0 = k0a0 + (k+1l) for some -1 < k0 < 1. We

claim that for this fixed k0 there exists an a > max{0,a0]

such that the half line L defined as

L = {(a,b) a>a, b = k0a + (k0+1))

lies in DO. Observe that if n > max{0,a0), then for any

(a,b) E L one has a + 1 > 0; moreover,

a + b = a + k0a + (k0+l) = (k0+l) (a+l) > 0

since -1 < k0 < 1 and a0 + 1 > 0. Therefore, conditions (i)

and (ii) of theorem 4.1.1 are satisfied for all (a,b) rL ,

a being any number greater than max{0,a0 .

For condition (iii), it suffices to show that for som

a > max(0,a0), one has a + r b > 0 for all (a,b) c L ,

where r = -a tan r and 0 < r < P. Since a = -r cot r and

0 < r < n, it follows that

222 22 2
a +r -b = (-r cot r) +r L-k0r cot r + (k +1) 1
2 2 2
= r csc r [kgr cot r (k0+1) ]
2 2 sin r 2
= csc r{l [k0cos r (k04l) ] }
sin r 12
Note that 1 [kocos r (k1il) ---r is continuous in a

neighborhood of r = a; moreover, one has

sin 11 2 2
1 k cos n (k0+1) 1- ] = 1 k > 0

Hence, there exists an c, with 0 < E < T/2, such that for all

r satisfying a E < r < in,

a2+r2-b2 = r csc2r{ [k cos r (k+l)sin r2 >0 (4.6)

Now, let a = 2 max{0,a, -(n-c)cot(n-E)}. Note that -r cot r

is an increasing function of r in the interval 0 < r < a.

Hence, for every a > a, the corresponding r obtained from

-r cot r = a satisfies the inequalities n < r < T. Con-

sequently, (4.6) holds for all (a,b) r L This completes

the proof of the claim.

Now, by this claim, one may represent (a0,b0) as a con-

vex combination of the point (-1,1) and some (a,b) e L c D,0


a-a a-a
(a0'b 0) (-1,1) + (1 + )(a,b).

Therefore, (a0,b0) c co(0). Hence, one concludes that

M c co(00).

Proposition 4.1.6. co( 0) = {(a,b) E R2 a+l>0, a+b>0,


Proof: By lemma 4.1.3 and the convexity of M, one obtains

coTPT c coTHM = M. On the other hand, lemma 4.1.5 implies

that M c FcoD ). Therefore, M = coo(). Since for any m

dimensional convex set Y c Rm, interior (Y) = interior (Y)

(see e.g., Rockafellar [1] p. 46), also since co(t0) and

M ire open convex ;ets in R2, one conci]licus that co (V ) = M

{(a,b) a + 1 > 0, a + b > 0, a b + 2 > 0). []

4.2 V -Stabilizable Hyp erplanos

Now, let H be a hyperplane in R2 defined by Aa + Bb + C =0.

Since D0 is an open connected set, it follows that H n 00x

if and only if H n co(00) x 4. Observe that, equivalently,

the open convex cone co(00) can be expressed as:

co(D0) = {(-l+kl+k2, 1+kl-k2) jk > 0, k2 > 0}

Hence, II n co(00) -x means that there exist k > 0 and

k2 > 0 such that (-lk+k2, 1+kl-k2) c H or equivalently,

A(-l+kl+k2) + B(l+kl-k2) + c = (AfB)k1 + (A-B)k2 + (B-A-C) = 0

from some kI > 0 and k2 > 0. By lemma 2.2.1, this is true if

and only if at least two of the numbers A + B, A B, B A + C

are of strictly opposite signs.

Summarizing the analysis above, we obtain the following


Theorem 4.2.1. Suppose H is a hyperplane in R2 defined by

Aa + Bb + C = 0. Then, H n D0 4. if and only if at least

two of the numbers A B, A + B, B A + C are of strictly

opposite signs. []

Expressed in terms of the original parameters a, b, c,

d, this theorem can be restated as:

Theorem 4.2.2. The system (D) is D -stabilizable if and only

if at least two of the numbers d-c, d+c, -c-d-ad-bc are of

strictly opposite signs.

Proof: Reca] from (4.2) that the hyperplane coi-rr';ponding

to (D) is defined by da cb (ad+bc) = 0. Honce, the

A, B, C in the previous theorem are d, -c and -(ad+bc),

respectively. O

One of the most interesting questions in the stabiliza-

tion of a delayed control system is the following: "Can a

system with delay be stabilized by a feedback without delay?"

In general, this cannot be done as the example at the end of

this chapter shows. However, for a system like (P), the

theorem above tells us that (V) may indeed be stabilized by

a feedback without delay. In fact, under this kind of feed-

back, one has d = 0 and consequently, d c = -c and d + c = c

are of strictly opposite signs.

Given a e R, recall that D is the set of all coefficients

(a,b) such that the real parts of all the roots of A + a + be-

= 0 are smaller than a. Consider a substitution of X by n + a

in equation (4.3).

+ a + be1 = q + (a+a) t be-'e-n

The substitution A = n + a induces an affine transformation
T of the coefficient space R of equation (4.3). This

transformation is defined by

T (a,b) (x,y) = (a,b) o-" + (a,0)

1 0 -a
Since det = e 0, T is invertible. Its inverse

is defined by
is defined by

I 0
T (x,y) = (a,b) = (x,y) + (-a,0)
(t 0 c U

Note that 1 = a + n implies that the condition Re(A) < a is

equivalent to Re(rl)<0. Hence, T maps DP onto VO, i.e.,

T, ( ) = 1 0

Suppuo, c I nL H is a hIypurpl] ne

C = 0. Then HI n V X z if and only

T (H) n T (V ) = T (ii) n DO'

Observe that

-1 l
T (H) = {(x,y) T (x,y) c H}
S1( 0
= ((x,y) [(x,y) 0 ea +

in RH defined by Aa 4 Bb +

if ; x Ta(H n aV ) =

(-a,0) ] + C = 0}

= (x,y) Ax + Be y + C An = 0}

Therefore, by theorem 4.1.1 we have


the following generaliza-

Theorem 4.2.3. Suppose that H is the hyperplane in R2

defined by Aa + Bb + C = 0. Then H n 1X X if and only if

at least two of the numbers A Be A + Be and Be + C A -

An are of strictly opposite signs. E

Recall -hat for polynomials of degree n, the number

inf{a]H n S" .' is always bounded from below for any

hyperplane H. Indeed, theorem 2.3.1 describes this property.

However the next theorem shows that this is not the case for

the V 's.

Theorem 4.2.4. Suppose H is the hyperplane in R2 defined by

Aa + Bb + C 0. If A = 0, then inf{faH n ) = -e

If A = 0,then

inf{all rnD x} = inf{aeRj (A-Be) (A+Bea) (Be+C-A-Aa) = 0) (4.7)

Proof: If A = 0, then A b Bea = Bec and A Bea -Be

These numbers are always of opposite signs for any a E R.

Hence, inf{ae i n a 4) = --. Suppose A x 0. Without loss

of generality, assume A > 0. Let denote the number on the

right hand side of (4.7). Note that, for any P < n, the

numbers A Be, A + Be and Be + C A AB all have the

same sign as A. Hence, ln inf{fa(H n Da 4}. On the other

hand, we have one of the following four cases:

(1) If n satisfies A Be' = 0, then there exists an

E > 0 such that (A-Bea)A < 0 and (A+Bea)A > 0

for all a such that n < a < n + E. Therefore

(A-Be ) (A+Be~ ) > 0 (4.8)

for all a such that n < a < n + E.

(2) Suppose that n satisfies A + Be0 = 0. Then, as in

(1), one may find E > 0 small enough such that

(4.8) holds.

(3) Suppose n is a double root of Be + C A Aa = 0.

This implies that Be A = 0. This case is then

reduced to case (1).

(4) If n is a root of multiplicity one of Bea + C A -

Aa = 0, then one may find E > 0 such that

(Bc"/ + C An) A < 0 and (A Be") A > 0 for

all a such that n < a < n c. Therefore,

(Be" t C A An)(A Be ) < 0 for all a such that

n < a < n + e. From the discussion above, one

concludes that min[rXIH n Da ,} < n.

II nce, If l ill r1 PI / i). L]

Remark: The stability domain of an arbitrary mixed type of

exponential polynomial H(x,ex) is one of the most interesting

subsets of the coefficient space of H(x,eX). Most of the

properties of the stability domains of polynomials are not

necessarily true for stability domains of exponential polyno-

mials. For example, the stability domain of )2 + (aA + b)e- =0
is the set bounded by b = 0 and curve a = y sin y, b = y cos y

where 0 < y (see El'sgol'ts [1]) Illustrated by figure

4.3, we see that this is a bounded convex set. Recall,

however, that the stability domain of any polynomial of

degree greater than or equal to three is an unbounded noncon-

vex set. In chapter one, we also mentioned that the stability

domain of any polynomial is connected. However, the stability

domain of the equation 2 + aAe + b = 0 (see figure 4.2)

is disconnected.

It seems that a "nice" characterization of the stability

domains of exponential polynomials is essential for under-

standing these topological properties mentioned above, or,

even the stability domain itself. The most general charac-

terization available is a theorem proved by Pontryagin [1]

(see next section). However, the difficulty in applying



Figure 4.3

Pontryagin's theorem arises from the fact that, generally,

it is not easy to determine whether a transcendental equation

has all the roots real or not. Bellmanand Cooke's book [11

contains characterizations of stability domains for some

special exponential polynomials. Much still remains to be

done for a better understanding of the stability domain of

general exponential polynomials.

4.3 An Example

In this section, we give an example of a control system

with delay which can not be stabilized by feedbacks without

delay. First, let us review a theorem proved by Pontryagin


Definition: Let h(z,w) = Ea z wn with m,n nonnegative
integers. A term a z w is called the principal term of
h(z,w) if as 0 and, if for each other term a with
rs mn
a z 0, we have either r > m, s > n or r = m, s > n or
r > m, s = n.

Theorem 4.3.1. [Pontryagin] Let H(z) = h(z,ez), where

h(z,t) is polynomial with a principal term. The function

H(iy) is now separated into real and imaginary parts i.e.,

we set H(iy) F(y) + iG(y). If all the roots of the

function H(z) lie to the left hand side of the imaginary axis,

then the zeros of the functions F(y) and G(y) are real, inter-

lacing and

G'(y)F(y) G(y)F'(y) > 0


for each y 1 R. Moreover, in order that all the roots of the

function lie to the left of the imaginary axis, it is sufficient

that one of the following conditions be satisfied:

(1) All the zeros of the functions F(y) and G(y) are

real and interlacing and the inequality (4.9) is

satisfied for at least one value of y;

(2) All the zeros of the function F(y) are real and for

each zero y = yo of F, condition (4.9) is satisfied

i.e., F' (y )G(y0) < 0;

(3) All the zeros of the function G(y) are real and for

each zero y = yo of G the inequality (4.9) is

satisfied i.e., G'(yQ)F(y0) > 0.

Lemma 4.3.2. If the exponential polynomial

h(X) = e (2 + pX + q) + (rX + m)


has only stable roots, then (q + m)(p + q + r) > 0.

Proof: Observe that, by substituting iy for 1, one obtains

h(iy) = F(y) + iG(y) where

F(y) = (q y2)cos y + m py sin y

G(y) = (q y )sin y + py cos y + ry

From Pontryagin's Theorem and the hypothesis that h(A) has only

stable roots, it follows that G'(y)F(y) G(y)F'(y) > 0 for

all y E R. In particular, if y = 0, then G'(0)F(0) G(0)F'(0)=

(q + m)(p + q + r) > 0. Now, consider the following control


Lx(L) ot ) (t) 1 -2 x (t-1)
(E) + +
x2(t) -al -a x2(t) 1 -2 x2(t-1)

x, (t)
+ u u = [fl 2f

where a.,fi, i = 1,2 are real numbers. Note that there is

no delay term in feedback control. The characteristic

equation of (E) is

A2 + A(a-f2) + (al -f ) + e [X-(a2- 2 + i + 2al-2f) ] = 0

Compare this equation with (4.10). Then

(q + m)(q + p + r) = [(al-f) (a2-f2 + 1 + 2a1-2f) ][al-f +

a2-f2 + 1] = (a]-f) + (a2-f2) + 12 < 0

for any real numbers fl and f2. Hence, the system (E) is

always unstable for any feedback without delay.


Con ; id -r a ;y t;L ( i, ') wi th pa ram t- i in K. In

chapter 2 and chiptur 3 we discussed respectively the S -

stabilizability and C p-stabilizability of a system (n,K)n

for which Xv (K) forms a hyperplane. Suppose XT (K) contains

only one point. Then, T (K) n Sn n is equivalent to

X> (K) c Sn. One can use the well known Routh-Hurwitz criterion

and the transformation T constructed in section 2.1 to deter-
mine if a point belongs to S Wie now know how to characterize

S -stabilizability of affine sets with dimension equal to n,

n-l, and 0. For affine sets with dimensions between n-2 and

1 inclusive, no complete answer is known yet. Observe that

in general, for an affine set of dimension lower than n-l,

the fact that this affine set intersects the convex hull of

certain set S does not always imply that the intersection of

that affine set and S is nonempty. Hence, the "convex hull

technique" gives less information for affine sets with

dimension lower than n-l, since only necessary conditions are

obtained. However, because of their simplicity when used on

a computer, those necessary conditions can be used as a quick

check to eliminate systems which are "obviously" not S -stabi-

lizable. Skoog [11 obtained some partial results of So-stabi-

lizability for the case of straight lines in terms of Nyquist


plot. In an inspiring paper, Anderson, Bose and Jury []

showed that Output Foltidback Stabilization problem can be

handled by algorithms from decision algubra (see Jacobson [1]).

This can be a very fruitful direction to pursue. It is also

interesting to point out that the case of a hyperplane, which

is the most complicated among all affine sets with dimension

less than n-1 when one uses decision algebra, turns out to be

the easiest for our approach, but the case of a line which is

the most difficult for us is the easiest by decision algebra


In the following, we will present a partial result and a

computer aided algorithm concerning lines which intersect

S They should only be considered as an attempt in this


Proposition 5.1.1. Suppose f(x) and g(x) are monic polynomials

of degree n and n-1 respectively. Then, for any E > 0, A < 0,

there exists a K e R such that for every k > K, n-1 of the roots

of f(x) + kg(x) = 0 are within the c-neighborhoods of the zeros

of g(x) and the remaining root is smaller than A.

Proof: Let A1 ,2',.' n-1 be

(counting the multiplicity).

centered at A. with radius r.
1 1
does not contain any A. A..
2M = inf jg(z) Choose K' >

the n-l roots of g(x) = 0

For each i, let Ci be a circle

< E such that the Interior(Ci)
Let M = sup If(z) and
0, such that M1 < M2K' for each
1 1

i 1,2,...n-1. Hence, if(z) < Ikg(z) for any k > K',

z C. and i ,2, ,n- Therefore, by Rouch6 theorem,

f(x) + kg(x) = 0 has as many zeros in Tnterior(C.) as g(x) = 0

does for any k > K' and i = 1,2,...,n-1. Now, consider

k = -f(x)/g(x) Since (-f(x)/g(x)) = -1 p(x)/g (x) with

deq p(x) < dog q (x) there exist Al c R such that -f(x)/g(x)

is strictly ducLie' sing in the interval (-(,/,Al). Also, since

deg f dcg q = 1, it follows that there exists some A2 such

that k = -f(x)/g(x) > 0 for all x < A2 and lim (-f(x)/g(x)) = .
Hence, k = -f(x)/g(x) is invertible in (-o,o) with a = min{Al,

A2',A 1. This implies that for every k > -f(a)/g(a), the x

obtained from k = -f(x)/g(x) satisfies x a< < A, or equiva-

lently, f(x) + kg(x) has a zero smaller than A for every

k > -f(c)/g(C) Now the proposition follows readily if one

chooses K = max{K', -f(cu)/g((x) .

This proposition shows an interesting geometric property
of SO: Every straight line in R with a direction (rl,r2.

rn) where (r2/rl, r3/r ,...,r /r) e SO with r1 x 0 inter-
sects S0. Furthermore, an infinite portion of the line is

contained in SO

For straight lines which do not have such kind of

"direction" we propose the following algorithm.

Consider f(x) = x + na x and g(x) = Er.x n
n n
Let f(x) + kq(x) = + n (a+kr.)xn-3 = p(x,k).

(i) Form the Hurwitz matrix H(k) of p(x,k) treated as

a polynomial of x.

(ii) Solve the real roots of det(HI(k)) = 0. (For

digital computer, this should not be a difficult task). Let

the distinct real roots be ki < hk2 < ,..., < km.

(iii) Cljiuo se arbitrary points k. c (k.,k i ), i = 0,1,2 ...,m

satisfying = k0 < k0 < k < < k < < ... < k < km <

(iv) Use Lienard-Chipart criterion to test each f + k.g,

i = 0,1,... ,n. If t.he-r exists a 0 < i n m such that

f(x) k g(x) = 0 has only stable roots, then for all

k L (k ,ki+ ) the f(x) + kg(x) = 0 will have only stable roots.

Otherwise, the straight line {(al + kr ...,an + krn) k E R)
does not intersect SO.

The proof of this algorithm follows from the next lemma.

Lemma 5.1.2. Let f(x) and g(x) be two polynomials of degrees

n and e respectively with 2 < n. Let k., i = 1,2, ..,m be
the distinct real roots of det(H(k)) = 0 where H(k) is the

Hurwitz matrix of f(x) + kg(x). Then, for each fixed i,

all the polynomial equations f(x) + kg(x) = 0 with k c (ki,k.+)

will have the same number of unstable roots.

Proof: Let i be fixed. For each k c (k.,ki+l), let pj(k),

j = 1,2,...,n denote the roots of f(x) + kg(x) = 0. Recall

that, by Orlando's formula (Gantmacher Ill vol. 2, p. 196),
n(n+1l)/2 n
det(H(k)) = (-1) 1 j Ai (k) r
there exist k' and k" both in (ki,ki+ ) and k' < k", such

that f(x) + k'g(x) and f(x) + k"g(x) have different number

of unstable roots, then by the continuity property of roots

with respect to the coefficients, there exists k with

ki < k' < k < k" < ki+1 such that the equation f(x) + kg(x) = 0

has either a pair of conjugate of purely imaginary roots or a

zero root. Hence, det(l(k)) = 0. This contrdicts the fact

that between k1 and ki+1, there is no reil root of det(I1(k)) -0.

Therefore, for every k c (ki,ki+1), the polynomials f + kg

have the same number of unstable roots. n

By this Icmin,i, one nses that it is enough to test only

owner point from e(c.ch ilntlirvia (ki,kitl) to deti:lmine whether

all the roots of f(x) + kg(x) = 0 are stable or not for any

k E (kiki+l). The major advantage of this algorithm over the

conventional root locus method is that it involves only solving

real roots of an nth degree polynomial. This is a well devel-

oped area in numerical analysis. One may program that easily

on a digital computer. The other advantage is that it also

gives the numerical ranges of k for which f + kg has only

stable roots. The major drawback of this algorithm is that

it is very sensitive to double real roots. A small round off

error may cause the loss of a double real roots of det(H(k)) =0.

In other words, one may lose one of the partition points k..

However, this can be prevented by predetermining the number

of distinct real roots by Sturm's theorem (see Gantmacher [11

vol. 2, p. 175). Note this algorithm also works for any poly-

nomial of the form E.n f.(k)xn- when f. are polynomials of k.

Example: Suppose f(x) = x + x x + 5 and g(x) = x + 3x 1.

We wish to find the range of k such that f + kg = 0 has only

stable roots.

(i) det(il(k)) = 3(-k+5) (k+2) (k-1)

(ii) The three roots k = 5, k = -2, k = 1 separate the

real line into four intervals (-C ,-2) (-2,1), (1,5), (5,-).

( i) Pick: one point k. from each interval above and

substitute th.m into H(k) It is easy to check that the

only H(k.) which satisfies Lienard-Chipart criterion is the

one with k. f (1,5).
lence, one concludes that every polynomial equation

f t kg = 0 with k L (1,5) has only stable roots.

For further research, this author feels that the study

of the S -stabilizability of affine sets with dimension

strictly lower than n-i deserves an immediate attention.

Besides the engineering application, this study also provides

a better understanding of the set Sn itself. After this, the

investigation of S -stabilizability of systems with multiple

control would be the next goal. For nonlinear control systems,

a problem, initially proposed by Letov [1], of determining the

minimal number state measurementsneeded for stabilizing the

system by a feedback control was considered by Casti and

Letov [11.

In a practical situation, it is not enough just to know

that the system is stabilizable by a constant gain feedback.

The numerical scheme for finding a feedback gain matrix is

also important. The papers by Miller, Cochran and Howze El];

Luus [1], NcBrinn and Roy 11], as well as Sirisona and Choi [1]

represent some of the research in this direction.


Given < R r-ca]] that

Sn = {(a1 ,a ) a R" xn + a.xI- = 0 implies
a U 2 n ji1 ]

Ro(x) < j

In this appendix we wish to show that Sn is connected. Since,

in Rn, connectedness is equivalent to path connectedness, it

suffices to show that for every a' = (al,a, ... ,a) S n

a" = (a ,a ...,a) Sn there exists a path p: r0,1] + Rn

such that p(10,11) c Sn with p(0) = a' and p(1) = a". Now
consider a mapping T from Cn into Cn where C is the set of

all complex numbers, such that for each (x,,x2,...,x n) Cn,

1 (Xl'2, ..X. x ) = (ala2 .... an)

where a.'s satisfy the relation {x xn + n a.xn-j = 0} =
I j=1 j
{xlx 2...,x n. Note that 7 is continuous. Also, if

{x1,x,2....,x is symmetric with respect to the real axis

i.e., {xlx2 .. = {xl,x2 ...,xn ) where x. represents the

conjugate of x., then (xl,x2, ...,x ) E Rn.

Now, for the a' and a" given above, arrange the roots of

x + xn-3 = 0 and xn + na xn-3 = 0 such that
1=1 I j1 j

{xl n = a xn-) = {x', ...,x mx m+ ,....,x }

x j a" 1 2-}. 2 ( r+l" ..x

where x i = 2mrl,...,n and x', j = 2r+l,...,n are real roots;

, i = 1,2,...,2m, and x', j 1,2...,2 are complex roots
2j ]
such that x, i 1,2, .. ,in ind x" = x j1, j 1,2

...,r Note that ii(x', .,x ) (a ,. ,a') and i (x ',x

., ") = (a ,a ...,a") Consider the path p: [0,1] Cn

defined by

p(t) = t(x{',x ,...,x') + (l-t) (x ,x", ,x")

Note for each t c [0,1] the set {tx' + (l-t)x', tx2 + (l-t)x",

.. ,tx' + (l-t)x" is symmetric with respect to the real
n n
axis. Hence, ; (p(L0,1])) c Rn. Also, since a' c Sn and

a" Sn by the definition of Sn, one has Re(x') < a and

Re(x ) < a for every i = 1,2,... It follows that for any

t e [0,1] and for every i = 1,2,.... n,

Re(tx! + (l-t)x'') = t Re(x!) + (l-t)Re(x') <

Therefore, r(p([0,1])) c S". Note that r(p(0)) = a" and

7(p(l)) = a'. Hence, l(p) is a path in Sn which connects

a' and a". This proves Sn is a connected set in Rn


Po" r a, t l -:I< II l t I 0 L (li] ) b I thle' lit l ix r L.pr l ;t' ntLa-

tion of T1B lielincd in section 3.1. Roca ll that for each

fixed i, the row (qi0 ,il ,...,in) of Q a, represents the

coefficients of

(s- ) n-i(s- )

Sq n-j

(T~ .1)

n-i i
Now, let f(x,y) = x y Then,

f(-a+s,-+t s) = (s-a)n- s- ) (II.

In terms of Taylor's theorem, one has

1 n 3 n-j n-1
f(a+h, b+k) ( + k )- (x n- ) xa
(n-7 i2i-Y x=a
=0 y=b

By substituting a = -a, b -B and h = k = s into the above

equation one obtains

Sn-i 1 3 n-j n-i 1 n-
(s-a) n (s-B) = .Y [ (n(jT + ) (x y )x:_s
j0 (n-j) rx +y x=-)

By comparing the right hand sides of (II.1) and (II.3), one

derives another expression for the entries of Q namely,
Ll p

1 3 3 n- j n-i 1
ij (nj ax + y x y =-





To show that for each fixed n, det(n ) = (-B)(n+l)/2

let .ln(x,y) = (mn .) be the (n+1) x (n+1) matrix with

n 1 n-j, n-i i
m ( + ) (x y )
ij (n-j)' (y n

We claim that dct (fn (x, y) ) = (y-x) n (11+1)/2 To prove this,
we use mathematical induction on n. Let M = (nm .) be the

matrix obtained from M (x,y) by starting from the (n+l)-th

row of Hn (x,y), subtracting successively the (j-l)-th row

from the j-th row, j = n+l,n,...,2. These are elementary

row operations which do not change the determinant of n(x,y)

Therefore, det(n (x,y)) = det(U). Furthermore, for any i 2 2

j 2,

n 1 n-j n-i i n-3 n-i+l i-
mij = ( j) [ ( + ) (x y )-(- + ) (x y ]
1,j (n-j). ox !y ax ay

1 / _x\
(n-j x + [(x y )(y-x)i

1 + n-j n-i i- ni n-i i-l
,{ (y-x)( + ) (x y ) + (x y )
(n-j). DX dy

1 + ( n-j n-i i-1
(n j (y-x) ( + ) (x y )

= (y-x)mi-l. j-

Also, m.i = 0 for i > 2, since .il = 1 for each i = 1,2,...,

n+l. Hence, by factoring (y-x) out of 2nd, 3rd, and (n+l)-th

row of M and applying induction hypothesis, one obtains


det(Mn(x,y) ) = (y-x)ndet(Mn-l(x,y)) = (y-x) n (y-x) (n-1)n/2

= (y-x)n (n+1)/2

Now, observe that Mn (-a,-B) = Qn Therefore,

det(Q" ) = dcet(M (x,y)) = (a-B) (n+l)/2


B. O. D. Anderson and N. K. Bose and E. I. Jury:
[11 Outout Feedback Stabilization and Related Problems -
Solution Via Decision Methods, IEEE Trans. Automatics
Control, AC-20(1), pp. 53-66, 1975.

S. Barnett:
[ll Some Applications of Matrices to Location of Zeros of
Polynomials, Int. J. Control, 17(4), pp. 823-831,

R. Bellman and K. L. Cooke:
[13 Differential-Difference Equations, Academic Press,
New York, 1963.

J. M. Brasch Jr. and J. B. Pearson:
[ll Pole Placement Using Dynamic Compensation, IEEE Trans.
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Raymond Chen was born on August 19, 1949, in Tainan,

Taiwan, Republic of China. In 1971, he obtained his B.S.E.

degree in Mathematics from National Cheng Kung University in

Taiwan. After two years of military service in the Navy, he

entered the University of Florida and received his Master of

Science degree in Mathematics in 1974. He is married to the

former Chen Shi Chuan in 1975. He is a member of the American

Mathematical Society, Society of Industrial and Applied

Mathematics, and Phi Kapa Phi.

I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.

V. M. Popov, Chairman
Professor of Mathematics

I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.

L. S. Block
Associate Professor of Mathematics

I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.

J. K. Brooks
Professor of Mathematics

I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.

A. K. Varma
Professor of Mathematics

I certify that I have read this study and that in my
opinion it conforms to acceptable standards of scholarly
presentation and is fully adequate, in scope and quality,
as a dissertation for the degree of Doctor of Philosophy.

T. E. Bullock
Professor of Electrical Engineering

This dissertation was submitted to the Department of Mathematics
in the College of Arts and Sciences and to the Graduate Council,
and was accepted as partial fulfillment of the requirements
for the degree of Doctor of Philosophy.

June, 1979

Dean, Graduate School

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