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Title: 
Output feedback stabilization of linear systems 

Alternate Title: 
Linear systems, Output feedback stabilization of 

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vii, 80 leaves : ill. ; 28 cm. 

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English 

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Chen, Raymond, 1949 

Copyright Date: 
1979 
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Stability ( lcsh ) Mathematics thesis Ph. D Dissertations, Academic  Mathematics  UF 

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bibliography ( marcgt ) nonfiction ( marcgt ) 
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by Raymond Chen. 

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ThesisUniversity of Florida. 

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Bibliography: leaves 7779. 

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Vita. 
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UF00099384 

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University of Florida 

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University of Florida 

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OUTPUT FEEDBACK STABILIZATION OF LINEAR SYSTEMS
By
RAYMOND CHEN
A DISSERTATION PRESENTED TO THE GRAUDATE COUNCIL OF THE
UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE
REQUIREMENTS FOR THE DEGREE OF
DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
1979
I
To My CMo thic,
ACKNOW LEDGKMEINT
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.
TABLE OF CONTENTS
Pasle
ACKNOWILEDGEMENT .......................................... iii
ABSTRACT ................................................ vi
Chapter
I. INTRODUCTION ................ ......... ..... 1
II. SnSTABILIZABLE AND RSnSTABILIZABLE IIYPERPLANES 8
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
c,B
3.1 The Convex Hulls, co(Cn ) and co(RCn )... 33
3.2 Cn Stabilizable Hlyperplanes .............. 45
IV. STABILIZABILITY OF A SYSTEM WITH DELAY .......... 48
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
LIST OF FIGURES
Page
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
OUTPUT FEEDBACK STABILIZATION OF LINEAR SYSTEMS
By
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[ < (Ba)/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
n
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 =laxn3 = 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
vi
n
of R ." ri chI mpter three, w; cliharicLcr 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
n
intersect Sn and a computer aided algorithm for determining
the Sastabilizability of a given straight line. Some dis
cussion of the directions for future research is included.
CHAPTER I
INTRODUCTION
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 xnj
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 rrab 1 amount of research in the Ii L trtre hs.
been devoted to finding a controller designed from the avail
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
problem.
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 na = det(xTABC).
Let
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.nj = 0
are real, distinct and strictly less than a}
Definition 2: For a fixed a E R, an ndimensional 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 S0stabilizable. 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(xIFGKll) = det(xIF) + kli adj(xIF)G (1.1)
Ixr
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 Snstabili
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
n
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.
n
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
included.
Given a discrete linear time invariant control system
IHxk
(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 < (8a)/2)
RC = {(al,a2 ...a) c Rn all the roots of xn + Zn a jnj 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 nj (xcni (x_)1i
i = 0,1,... ,n where x + Lj= lqij = (xc) (x)
In chapter 4, we consider the stabilization of a special
kind of system with delay:
(P) x(t) = x(t) + bx(tl) + u, y = cx(t) + dx(tl), 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 n1 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.
CHAPTER II
S STABILIZABLE AND RS STABILIZABLE HYPERPLANES
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 + : lajxnj = det(xI 
A BC).
The main purpose of this chapter is to characterize the
S stabilizability and RSnstabilizability 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,
n
Sn = ((ala2, ...a ) Rn xn + Z a.xn = 0 implies Re(x)
j=1
n
S=l 3xO
RSo = {(al,a2,...,an) Rn all the roots of xn + axn=0
j=1
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
RSnstabilizable 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
n
is S0stabilizable 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 n3
x n+ nl a xn3 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
n
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 Ddecomposition
method of Yu Naimark says that the hypersurfaces HI = ((al,a2'
E. la xn3 = 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.
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
2
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
P=0
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 Snstabilizable hyperplanes can be
characterized as those which intersect co(S ). However, if
one wishes to take the full advantage of this characterization
of Snstabilizable 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, si I'ce c(' (S ), (co(PS )
n
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.
n
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
polynomial
n n
T (x+j6) = xn + Z P.(6)xnJ
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) < n1/2c for every j = 1,2,...n. Note that
HI. (x+j6) = 0 has only stable roots x = j6, j = 1,2,...,n.
j=1
Thus, by the definition of RSg,
n
E P.(5)e. E RS (2.1)
j=13
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
1
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 k1. Fix 6. 6 > 0,
i = 1,2,...,n. For every c > 0, choose q0 > 0 large enough
such that for all j, 1 j < ki
r
kT Pj(q06162' '6n < Vn (2.3)
0
and
r Pk(0'61 .2' 6 < n (2.4)
k k'
q 0
as well as
k
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 k1 for all j k1 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)
q0
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 < ki
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 )xn3
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)
=1
belongs to I'. I
belongs to S tl ce, using Ihe relrition i S c o(lr; ), we
obtain
+ 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
0
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
k1
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 < {(k1) + + (nk) }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
q0
Now we state the main lemma.
Lemma 2.1.5. For every positive intcqer n, co(RS) = co (
0 0
n
R+.
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,
then
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.xnj (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
proof.
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 (yn) (2.12)
j=0 j'
Thus, by substituting y = x + a, (2.12) lecomlis
n n (nj)
Z a.(x+) = p(x+) = E ( x
j=0 3 j=0 (nj)
with a0 = 1. In view of (2.11), it follows tI it, for e.ach j,
1 5 j < n
lrjsn
(nj) ji (2.
a. = P =_ ) 1 ( ) (2.13)
3 Tnl) 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,
~ ni j nii a I +
a. = E ( )a [Aa' + (l )a'.'] = A[ 2 (n 1) 'a ] +
S i=0 ji i=O i 1
i=0
Hence, T is affine. Represented in matrix form,
T n(a 2 n)(n)
T (ala2, ...an = (al'a2 ... an + (( ) (2) ( )
(. ni 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
invertible.
(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 nj
every number vx such that xg + nxj=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)
ni ji
where Q ( ij) is an n x n matrix with qij (j) (a)
11 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 niji
i [( ). la > 0
i= 0 1i
i = 1,2,...,n
Proof: Let T be the transformation defined in lenma 2.1.6.
a
The theorem follows readily from the discussion preceding
n
lemma 2.1.6 and (2.13). The same argument holds for RSP. U
n
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(T1(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 jk (a) Jrklejrkr 0] (2.14)
n
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 (ni) ()
Note that the entries of the row vector ((n) (u) (n) () ...,
(n)(a)n) represent the coefficients of the polynomial (xc)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) =
naU
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 hyperplane Ii = { (a,i ,. ,a ) r' I
nn
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.
n
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 Snstabili
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.
jln nj j 0 j 
Then, the following statements are equivalenLt.
(i) II is RSnstabi]izable.
(ii) H is Snstabilizable.
a
(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
(Ly)
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
y
H n co(Sn) 4. Also, H n RSn 4 if and only if H n co(RSn)
1 a a
nn
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 ni) 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= 1i 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 S0stabilizable 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.
1
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 Snstabilizable
hyperplane not only can one find points (al,a2,...,an) e H
such that x" + =lajxn3 = 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.xnJ = 0 implies
x e r}. Then, the property H n 1n() T I is equivalent to
H n RSn d.
I'roof Ob; rrv 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
an
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 nj nj
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.
C(
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 < kl such that h i)() = 0. Also, since h (x)
i s a polynomi a] of posi Live riIee with l )posi Live Iuading
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
n
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.,
1
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 = n1. 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 (x06,x0)
Since f' ( + 2) > 0, and f' (x6/2) < 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 < k1 such that A. < 0 for all j <
and A. S 0 for all j > 2, then n = max{Ia h(a) = 0).
3
Proof: Obviously, max{a h(u) = 0) maxal ilk h(j) (a) =0}=n.
j=0
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 < 11 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 91 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 e1,
max{an h(j (j ) = 0} < max{a h(jl1)() = 0}
Consequent ly,
.l1
max{~a h )(a) = 0 max{ul h(c) = 0}
j=0
Hence,
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
n
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 JnJ
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;trite 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 + (w5v13)s + w = 0
(2.19)
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 = w5v13, 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 S0stabili
l12
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
C1
__~_
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.
CHAPTER III
C ,STABILIZABLE HYPERPLANLS
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 (Ba)/2. Let RCn be the set
a,
(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 = (Bu)/2) if H n Cn
Ca,B
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 SchurCohen
32
n
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
cBc
RQn
the simplex generated by the points (qil,q 2....' ,q) Rn
i = 0,1,2,...,n where q. are the coefficients of the polyno
nl n\ i n n" "3
mial (xa) (xB) = x + Zn q nj
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 nj
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 n4l
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
0
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 nFl
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 < (la)/2}
and
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).
n+l
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 nj j n n
aa. (sc ) (s ,) = a snj
j0 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 Q1,l of T1,1' 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.
n
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
expansion
n n
n nj 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 j0 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.
0,)?
To show that T6 is invertible, since TB is linear,
a,)? a,)?
it suffices to show that T is onto. In other words, for
n,)B
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.(sa)n (sB) = a.s" (3.3)
j=0 3 j=0
Substitute (6za)/(zl) for s in (3.3) and multiply both sides
by (zl)n. Then, (3.3) becomes
n n
S a.z = a. (za)n (z1) (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
invertible.
(iii) T (S) C and T (RS) = RC
Consider the 11 transformation
s > z = P
(3.5)
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)
(sB)"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", (sn = a(sa)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 = (Ba)/2).
Similarly, T determines a 11 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 nnl n n
E a j+ C ,. Therefore, T (S) = co(C ) and
jj l + ,E rSJ
n
r, (RS) = RIC ,
Following a similar technique as; IarrnrLLtt 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
n,S
( a ej+ 1) = (a0'al,...'1 )Qa,1".
j=0
lHence, from the definition of T ,, for uach fixed i, the row
(qi0,il' ...' in) of Q ,B represents the coefficients of the
polynomial
(s )ni(s )i (3.6)
By a simple combinatorial argument, one may expand (3.6) as
n i nikk
(s nic sB) = E [( )) () (n )ajk k] (3.7)
j=0 k=0 jk
Hence, for each 0 < i < n and 0 < j < n, one obtains
q i (1 ) ()( _n )aj Bk (3.8)
k=C k jk
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
n(n+l)
(aB) (see appendix II).
n+l
Now consider a transformation T on R defined by
nB
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. (za)n (z1) = 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) ( (jk)akk nij
k=O0
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 0I, 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 Q1,1
qij = qil,j qil,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 Q1, 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
spaces:
n j 
2 [(l) ( ( ),( j k nij 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
also,
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 ni i ) jk nij+ka > 0 for each
i = 0,1,2,...,n.jk 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 >n+l
= i TaX ei+l) ki > 0)
i= ,q
n
= { E ki(qi0', il .. in)' ki > 0
i=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
at,
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
ni i
coefficients of the polynomial expansion (xn) (xS) =
En q.x3
j=0 niji
For M c Rn+ and a c R, let M denote the subset of M
a
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 )
nS
(ii) co(Cn,) = co(RC ).
Proof: (i) Every point from (co(Cn ))l, is of the form
ki (bi0,bil ...bin
1
n
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
n
(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 ]
1
i
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 ni
((al,a2, .. ) (1) ( ) Bn3 + z [ (1 ) I ( )
1 i=l k=0
i jk 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
n
S {1 { ) ki(qil'qi2' .. ri) Iki > 0, k = 1i .
i=0 1
IHtruCO,
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 (xa) 1(xS) = xn I1 qijxn3; explicitly,
qij ()j (ni i ()jkk
q = k =0 (jk 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
ni i n .n n
polynomials (x+l) (xl) = x + Lj..q ijx explicitly,
Sk ni i
q.. = (1) ( )
13 k=0 k k
Example: Suppose n = 2. Then, co(C ) is the shaded
1,1
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
1,1
C itself. Since
1,i
1 2 1
1,1 1 0 1
1 2 1
2,1)
Figure 3.1
by corollary 3.1.8, this implies that co(C2 ) is the
1,i
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[ <
(Ba)/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
Consequently,
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
\t,
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, .
i=1
a) c Rn+ Z Aa = 0). Note that H is a hyperplane in RnH1
i=l
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
o,0
En
H n C
Ca,
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
n
( ) 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 ) jk
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,
n
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
1
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
CHAPTER IV
STABILIZABILITY OF A SYSTEM WITH DELAY
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(tl) + u, y = cx(t) + dx(tl), 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
2
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 Ddecomposi
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
P=4L
P=i
?=3
p==0
62
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 .
Also,
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 (bl1) 0 t 1 )
P2 = { (,1) a a a +t(a2al) 0 t 1) L c P0
P3 = ((a2b) lb =14t(b21), 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(tl) + bx(t) = 0
is a disconnected open set in R bounded by the straight
lines b = (1) 2a + [ 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, ab+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.
P=2
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 Lk0r 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+r2b2 = r csc2r{ [k cos r (k+l)sin r2 >0 (4.6)
Now, let a = 2 max{0,a, (nc)cot(nE)}. 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
i.e.,
aa aa
(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,
ab+2>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+klk2) jk > 0, k2 > 0}
Hence, II n co(00) x means that there exist k > 0 and
k2 > 0 such that (lk+k2, 1+klk2) c H or equivalently,
A(l+kl+k2) + B(l+klk2) + c = (AfB)k1 + (AB)k2 + (BAC) = 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.
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 dc, d+c, cdadbc are of
strictly opposite signs.
Proof: Reca] from (4.2) that the hyperplane coirr';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'en
The substitution A = n + a induces an affine transformation
2
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
1
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
tion.
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 (ABe) (A+Bea) (Be+CAAa) = 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 (ABea)A < 0 and (A+Bea)A > 0
for all a such that n < a < n + E. Therefore
(ABe ) (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
2
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
62
A=
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
[1].
Definition: Let h(z,w) = Ea z wn with m,n nonnegative
mn
m,n
rs
integers. A term a z w is called the principal term of
rs
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
mn
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
(4.9)
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)
(4.10)
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
system:
Lx(L) ot ) (t) 1 2 x (t1)
(E) + +
x2(t) al a x2(t) 1 2 x2(t1)
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(af2) + (al f ) + e [X(a2 2 + i + 2al2f) ] = 0
Compare this equation with (4.10). Then
(q + m)(q + p + r) = [(alf) (a2f2 + 1 + 2a12f) ][alf +
2
a2f2 + 1] = (a]f) + (a2f2) + 12 < 0
for any real numbers fl and f2. Hence, the system (E) is
always unstable for any feedback without delay.
CIIAPlTE'R V
CONCLUDING RItEMARKS
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 pstabilizability 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 RouthHurwitz criterion
and the transformation T constructed in section 2.1 to deter
n
mine if a point belongs to S Wie now know how to characterize
S stabilizability of affine sets with dimension equal to n,
nl, and 0. For affine sets with dimensions between n2 and
1 inclusive, no complete answer is known yet. Observe that
in general, for an affine set of dimension lower than nl,
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 nl, 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 Sostabi
lizability for the case of straight lines in terms of Nyquist
S66
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 n1 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
algorithm.
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
direction.
Proposition 5.1.1. Suppose f(x) and g(x) are monic polynomials
of degree n and n1 respectively. Then, for any E > 0, A < 0,
there exists a K e R such that for every k > K, n1 of the roots
of f(x) + kg(x) = 0 are within the cneighborhoods of the zeros
of g(x) and the remaining root is smaller than A.
Proof: Let A1 ,2',.' n1 be
(counting the multiplicity).
centered at A. with radius r.
1 1
does not contain any A. A..
S=1
2
2M = inf jg(z) Choose K' >
1
the nl roots of g(x) = 0
For each i, let Ci be a circle
< E such that the Interior(Ci)
1
Let M = sup If(z) and
ztCi
0, such that M1 < M2K' for each
1 1
i 1,2,...n1. 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,...,n1. 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)) = .
Xfo
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
n
of SO: Every straight line in R with a direction (rl,r2.
nI
rn) where (r2/rl, r3/r ,...,r /r) e SO with r1 x 0 inter
n
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.)xn3 = 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 <
km+1
(iv) Use LienardChipart criterion to test each f + k.g,
i = 0,1,... ,n. If t.her 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)
n
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
1
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) (k1)
(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 LienardChipart criterion is the
one with k. f (1,5).
1
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 ni 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.
APPENDIX I
Given < R rca]] 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
I
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.xnj = 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 + xn3 = 0 and xn + na xn3 = 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') + (lt) (x ,x", ,x")
Note for each t c [0,1] the set {tx' + (lt)x', tx2 + (lt)x",
.. ,tx' + (lt)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! + (lt)x'') = t Re(x!) + (lt)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
Ct
APPIENDTX II
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 ) ni(s )
n
Sq nj
j=0
(T~ .1)
ni i
Now, let f(x,y) = x y Then,
f(a+s,+t s) = (sa)n s ) (II.
In terms of Taylor's theorem, one has
1 n 3 nj n1
f(a+h, b+k) ( + k ) (x n ) xa
(n7 i2iY x=a
=0 y=b
By substituting a = a, b B and h = k = s into the above
equation one obtains
n
Sni 1 3 nj ni 1 n
(sa) n (sB) = .Y [ (n(jT + ) (x y )x:_s
j0 (nj) rx +y x=)
y=8
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 ni 1
ij (nj ax + y x y =
y=F
(II.
2)
(11.3)
4)
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 nj, ni i
m ( + ) (x y )
ij (nj)' (y n
We claim that dct (fn (x, y) ) = (yx) n (11+1)/2 To prove this,
~n
we use mathematical induction on n. Let M = (nm .) be the
n
matrix obtained from M (x,y) by starting from the (n+l)th
row of Hn (x,y), subtracting successively the (jl)th row
from the jth 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 nj ni i n3 ni+l i
mij = ( j) [ ( + ) (x y )( + ) (x y ]
1,j (nj). ox !y ax ay
1 / _x\
(nj x + [(x y )(yx)i
1 + nj ni i ni ni il
,{ (yx)( + ) (x y ) + (x y )
(nj). DX dy
1 + ( nj ni i1
(n j (yx) ( + ) (x y )
Sn1
= (yx)mil. j
n
Also, m.i = 0 for i > 2, since .il = 1 for each i = 1,2,...,
n+l. Hence, by factoring (yx) out of 2nd, 3rd, and (n+l)th
row of M and applying induction hypothesis, one obtains
76
det(Mn(x,y) ) = (yx)ndet(Mnl(x,y)) = (yx) n (yx) (n1)n/2
= (yx)n (n+1)/2
Now, observe that Mn (a,B) = Qn Therefore,
det(Q" ) = dcet(M (x,y)) = (aB) (n+l)/2
y=B
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BIOGRAPHICAL SKETCH
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

