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A new proof of the Jury test
Automatica 35 (1999) 251—258
Brief Paper
A new proof of the Jury test L. H. Keel *, S. P. Bhattacharyya Center of Excellence in Information Systems, 330 Tenth Avenue North, Tennessee State University, Nashville, TN, USA Department of Electrical Engineering, Texas A&M University, College Station, TX, USA Received 18 July 1997; revised 23 March 1998; received in final form 24 July 1998
Abstract The problem of determining the root distribution of a real polynomial with respect to the unit circle, in terms of the coefficients of the polynomial, was solved by Jury in 1964. The calculations were presented in tabular form (Jury’s table) and were later simplified by Raible in 1974. This result is now classical and is as important in the stability analysis of digital control systems as its continuous time counterpart, the Routh Hurwitz criterion is for the stability analysis of continuous time control systems. Most texts on digital control state the Jury test but avoid giving the proof. In this paper we give a simple, insightful and new proof of the Jury test. The proof is based on the behavior of the root-loci of an associated family of polynomials that was introduced in recent literature. The proof reveals clearly the mechanism underlying the counting of the roots within and without the unit circle. 1999 Elsevier Science Ltd. All rights reserved. Keyword: Discrete systems; Polynomial; Roots; Stability test; Zero crossings
1. Introduction The stability of a discrete time linear time invariant system is determined by the root locations of the system characteristic polynomial with respect to the unit circle, the system being stable if and only if all roots lie within the unit circle. This problem was studied by Schur (1917), Cohn (1922) and Marden (1949), among others. Jury (1964), building on these and other earlier results, gave a procedure to ascertain the numbers of roots of a polynomial that lie inside and outside the unit circle. This calculation was organized in tabular form and came to be known in the control literature as Jury’s table and the Jury test. These results are the discrete time counterparts of the Routh table and Routh test, discovered nearly a century earlier and useful in the stability analysis of contin- uous time control systems.
————— *Corresponding author. Tel.:#1-615-963-7025; fax:#1-615-9637027; e-mail: [email protected]. This paper was not presented at any IFAC meeting. This paper was recommended for publication in revised form by Associate Editor C. V. Hollot under the direction of Editor Dr. R. Tempo.
An alternative to the Jury test was presented by Bistritz (1984). This test, based on constructing a sequence of polynomials of successively lower degree and determining the number of sign changes in the leading column, is quite similar in mechanism to the Routh test, and is based on properties of symmetric and antisymmetric polynomials. Raible (1974) reported a simplification of Jury’s table based on certain formulas appearing in a paper of As stro¨m (1970). Raible justified the computations in this modified table by establishing a correspondence between the signs of the first column entries of his table with those computed in Jury’s original table. In other words Raible’s proof of the modified table relies on the original proof of Jury’s table. The proof of the Jury test, given in Jury’s 1964 book (Jury, 1964), is based on Rouche´’s Theorem but is rarely presented in controls texts. Although many books on digital control contains the Jury test we have not encountered a single undergraduate textbook which contains a proof of Jury’s table (see, for example Kuo, 1992; Ogata, 1995). The stability part of the result, as opposed to the counting part, is stated in virtually every book on digital control, however also without
0005-1098/99/$—see front matter 1999 Elsevier Science Ltd. All rights reserved PII: S 0 0 0 5 - 1 0 9 8 ( 9 8 ) 0 0 1 5 2 - 6
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proof. Moreover not even a hint is given regarding the rationale or logic underlying the proof thus forcing the control engineer to accept this result on faith and without insight. In this paper we present a new, simple and hopefully insightful proof of the Jury test for root distribution with respect to the unit circle. The proof is based on elementary properties of the root loci of an associated family of polynomials. This family was first introduced in Bhattacharyya et al., (1995) and was used to give a proof of the stability part of the Jury test. However, it was not clear how the number of stable and unstable roots could be counted. In this paper we go on to show, using the same family of root loci, that roots are counted in the Jury test by sequentially sending one branch of the root locus at a time, to the origin and constructing a polynomial of lower degree by removing the root at the origin. We believe that this new proof provides valuable insight into the workings of the Jury test and is simple enough that it can be taught even to undergraduate students. The mechanism of the proof may be useful in other problems as well.
Table. Raible’s table P(z): Q(z): R(z):
p p p 2 2 p p p zL L L\ L\ q q q 2 2 q q zL\ L\ L\ L\ r r r 2 2 r zL\ L\ L\ L\ $ $ ¹(z): t t t z º(z): u u z »(z): v z p p p q "p ! p q "p ! p 2 q "p ! p L\ L\ p p L\ L\ L p L q qL qL r "q ! q r "q ! q 2 r "q ! q L\ L\ q L\ L\ q q L\ L\ L\ L\ $ t t u "t ! t u "t ! t t t u v "u ! u u
Each row of Raible’s table represents a polynomial. Thus Q(z)"q zL\#q zL\#2#q z#q L\ L\ R(z)"r zL\#r zL\#2#r z#r L\ L\ $
2. The Jury test
¹(z)"t z#t z#t
The Jury test deals with the problem of determining the root distribution with respect to the unit circle, of the polynomial
º(z)"u z#u
P(z)"pLzL#pL\zL\#pL\zL\#2#pz
»(z)"v .
#pz#p . We make the following standing assumptions throughout the paper. Assumptions (1) The polynomial P(z) has real coefficients. (2) pLO0; in other words, the polynomial is of degree n. (3) pO0; this assumption avoids the trivial case where the polynomial has a root at the origin. When this happens, one can factor out the root at the origin and proceed with a lower degree polynomial. (4) P(z) does not have a root on the unit circle. This is a reasonable assumption since we are trying to determine the root distribution with respect to the unit circle. A slight controlled perturbation of the polynomial coefficients can be designed so that any roots on the unit circle are displaced outside it, as explained later. We shall work with a simplified version of Jury’s table as given by Raible. This modified table, which we refer to as Raible’s table, is constructed as follows.
and
We assume that no entries of the first column of Raible’s table given above is zero; this is the regular case. Let i( ) ) and o( ) ) denote respectively the numbers of roots of the polynomial ( ) ) inside and outside the unit circle. Also let sign + ) , and sign + ) , denote the number > \ of positive signed elements and the number of negative signed elements in the set + ) ,. Theorem (The Jury test). In the regular case, (1) If p '0, L i(P)"sign +q ,r , , t , u , v ,, > L\ L\ 2 o(P)"sign +q ,r , , t , u , v ,. \ L\ L\ 2 (2) If p (0, L i(P)"sign +q ,r , ,t , u , v , \ L\ L\ 2 o(P)"sign +q ,r , , t , u , v ,. > L\ L\ 2 In the following sections we develop our proof of this result. Also we indicate how to proceed in the nonregular case, that is, if a leading coefficient in the first column becomes zero.
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3. Preliminary results In this section we give some preliminary technical results and in particular introduce a one-parameter family of polynomials and develop some properties of its root loci. These preliminary results are necessary for our proof of the Jury test given in the next section. Consider the polynomial P(z)"p zL#p zL\#p zL\#2#p z L L\ L\ #p z#p (1) subject to the standing assumptions stated in the previous section. Introduce the associated one parameter family of polynomials defined as
p 1 P (z) " : P(z)!j zLP , H p z L Then P (z)"P(z),
j3[0, 1].
and
(F) (G)
(6)
p 1 P< (z) " : PP(z)!j L zLPP , H p z Then
j3[0, 1].
(7)
(4)
condition " p /p "'1 is equivalent to the condition L that q and p are of opposite sign. L\ L Proof. (A) The polynomial
p 1 P (z)"P(z)!j zLP H p z L has leading coefficient
p p p !j "p 1!j L L p p L L
(8)
p 1 P< (z)"PP(z)! L zLPP "z p z
Q< (z)"z\P< (z).
p p p p p p ! zL\# p ! zL\#2# p ! L\ , L p L\ p p L L L GFFFFFFFFFFFFHFFFFFFFFFFFFI /X
1 PP(z) " : zLP , z
and
(E)
(2)
Q(z)"z\P (z). (5) Note that Q(z) obtained above corresponds exactly to the second row of Raible’s table. We also define the reverse polynomial of P(z) and its associated one parameter family as follows:
P< (z)"PP(z)
(C) (D)
(3)
1 p "z P (z)"P(z)! zLP z p L
(B) (B1) (B2)
¹his means that the numbers of roots of the family P (z) inside and outside the unit circle remain respecH tively the same, since the root loci originating inside the unit circle remain inside and those originating outside remain outside. If " p /p "'1, then L P< (z) is always of degree n for j3[0, 1] and H none of the n arcs corresponding to the root loci of the family P< (z), j3[0, 1] cross the unit circle. H i(P)"o(PP) and o(P)"i(PP). Suppose that P(z) has i roots inside the unit circle and n!i roots outside the unit circle. If " p /p "(1, then L z\P (z)"Q(z) has i!1 roots inside the unit circle and n!i roots outside the unit circle. Suppose that P(z) has i roots inside the unit circle and n!i roots outside the unit circle. If " p /p "'1, then L z\P< (z)"Q< (z) has i roots outside the unit circle and n!i!1 roots inside the unit circle. P (z) and P< (z) have the same set of roots. ¹he condition " p /p "(1 is equivalent to the condi L tion that q and p are of the same sign, and the L\ L
p pp pp p ! L zL\# p ! L L\ zL\#2# p ! L , p L\ p p GFFFFFFFFFFFFHFFFFFFFFFFFFI /K X
(9)
(10)
Proposition (A) If " p /p "(1, then L (A1) P (z) is always of degree n for j3[0, 1] and H (A2) none of the n arcs corresponding to the root loci of the family P (z), for j3[0, 1], cross the unit circle. H
and therefore when
p (1 p L this coefficient is nonzero for j3[0, 1]. Therefore P (z) is H of degree n for all j3[0, 1].
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Now suppose that z lies on the unit circle and is a zero of P (z) for some j"j 3[0, 1]. Then H
p 1 P (z )"P(z )!j zL P . H p z L Since P (z) is a real polynomial 1/z is also a zero of this H polynomial. Therefore
1 1 p 1 "P !j P (z )"0. P p zL H z z L From the above two equations we have P
p 1!j "0 p L
1 z
and
p P(z ) 1!j "0. p L Since 1!j (p /p )O0 L it follows that P(z )"P(1/z )"0. However this contradicts the assumption that P(z) has no roots on the unit circle. Therefore no root loci of the family P (z) can cross the unit circle. H (B) Proposition (B) is the dual of Proposition (A) and can be similarly proved. (C) Since PP(z)"zLP
1 , z
the reciprocal of every root of P(z) is a root of P< (z). Thus, i(P)"o(PP) and o(P)"i(PP). (D) Due to Proposition (A), the condition " p /p "(1 L ensures that no roots of
p 1 P (z)"P(z)!j zLP H p z L cross the boundary, that is, the unit circle, for j3[0, 1]. Since P (z) has one root at the origin, this root must be the terminal point of a root locus originating from a root of P(z) lying inside the unit circle.Thus, z\P (z)"Q(z) has precisely one less root, inside the unit circle than P(z), while they have the same number of roots outside the unit circle. (E) Proposition (E) is the dual of Proposition (D). It can be similarly proved.
(F) Write
p 1 p P (z)"P(z)! zLP "P(z)! PP(z), p z p L L 1 p p "PP(z)! L P(z). P< (z)"PP(z)! L zLPP z p p This implies that p P (z)"p P(z)!p PP(z) L L p P< (z)"p PP(z)!p P(z). L Consequently, !p P (z)"p P< (z). L Since p and p are nonzero constants, P (z) and P< (z) L have the same set of roots. (G) If " p /p "(1, p!p'0. Then it is obvious that L L the highest order coefficient of Q(z) becomes
p '0, L p (0, L
p '0, L p (0. L based on the Proposition
p!p '0 for " q " L L\ p (0 for L Similarly, when " p /p "'1, L p!p (0 for " q " L L\ p '0 for L Next we prove a key result above. Lemma. ¸et
P(z)"p zL#p zL\#p zL\#2#p z#p , L L\ L\ p O0, L Q(z)"q zL\#q zL\#q zL\#2#q z#q , L\ L\ L\ q O0 L\ with p!p q " L L\ p L p p !p p q " L\ L L\ p L $ p p !p p L\ . q " L p L Then,
i(Q)"i(P)!1, if sign(p )"sign(q ), L L\ o(Q)"o(P), i(Q)"o(P)!1 if sign(p )Osign(q ), L L\ o(Q)"i(P).
L.H. Keel, S.P. Bhattacharyya/Automatica 35 (1999) 251—258
Proof. Let us first consider the case when sign(p )" L sign(q ). Since L\ q "(p!p)/p , L\ L L this means that p!p'0, equivalently, L " p /p "(1. L From Proposition (D), Q(z) will have one less root inside the unit circle than P(z) while the number of roots outside the unit circle remains the same. Thus,
i(Q)"i(P)!1, if sign(p )"sign(q ), L L\ o(Q)"o(P). Now consider the case when sign(p )Osign(q ). This L L\ means that p!p(0. Equivalently, L p '1. p L Since " p /p "(1, it is clear from Proposition (D) that L i(z\P< )"i(PP)!1 and o(z\P< )"o(PP). If we consider the reverse polynomial PP(z)"zLP(1/z), we know that i(PP)"o(P) and o(PP)"i(P). Therefore, i(z\P< )" o(P)!1 and o(z\P< )"i(P). From Proposition (F), P< (z) and P (z) have the same set of roots. Therefore, z\P (z)"Q(z) has the same set of roots as z\P< (z). Thus, in this case, where a sign change occurs between p and q , we have L L\ i(Q)"o(P)!1 and o(Q)"i(P).
4. Proof of the Jury test Using the preliminary results of the last section we give our new proof of the Jury test. Proof of the Theorem. The proof is based on induction. Let us first consider the case when n"1. Let º(z)"u z#u be the test polynomial and consider the corresponding Raible’s table: º(z): u u »(z): v We first consider the case when u '0. Here we need to show that i(º)"sign +v , and o(º)"sign +v ,. > \ First, if " u /u "(1, we have v '0 from the construc tion of Raible’s table. Therefore sign +v ,"1 and > sign +v ,"0. On the other hand, we know that \ i(º)"1 and o(º)"0 due to the assumption "u /u "(1. Therefore, i(º)"sign +v ,"1 and > o(º)"sign +v ,"0 which agrees with the theorem. \ Now suppose that u '0 and " u /u "'1. Then from the construction of Raible’s table, we have v (0. Therefore sign +v ,"1 and sign +v ,"0. Here we \ > also know that i(º)"0 and o(º)"1. Therefore,
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i(º)"sign +v ,"0 and o(º)"sign +v ,"1 which > \ again agrees with the theorem. We can similarly show that the theorem is true for the cases u (0, "u /u "(1, and u (0, "u /u "'1 proving that the theorem is true for n"1. Now let us assume that the theorem is true for all polynomials of degree n!1. Then we have
sign +r , , t , u , v ,"i(Q), > L\ 2 If q '0 (11) L\ sign +r , , t , u , v ,"o(Q), \ L\ 2 sign +r , , t , u , v ,"o(Q), > L\ 2 If q (0 (12) L\ sign +r , , t , u , v ,"i(Q). \ L\ 2 Here we consider the following four cases to complete the proof. (1) When p '0, " p /p "(1, then we have q '0 L L L\ from the construction of Raible’s table. From the lemma, since no sign change occurs, it follows that i(P)"i(Q)#1. Since Q(z) is of degree n!1 we have, by the induction hypothesis in Eq. (11), sign +r , , t , u , v ,"i(Q), and since q '0, we > L\ 2 L\ have sign +r , , t , u , v ,#1 > L\ 2 GFFFFHFFFFI G/ "sign +q ,r , , t , u , v ,. > L\ L\ 2 Therefore, i(P)"sign +q , r , 2 , t , u , v ,. From > L\ L\ the lemma, we have o(P)"o(Q). By the induction hypothesis in Eq. (11), since Q(z) is of degree n!1, it follows that, sign +r , , t , u , v ,"o(Q), and since \ L\ 2 q '0, we have L\ sign +r , ,t , u , v , \ L\ 2 GFFFFHFFFFI M/ "sign +q ,r , , t , u , v ,. \ L\ L\ 2 Therefore, o(P)"sign +q ,r , , t , u , v ,. \ L\ L\ 2 (2) When p '0, "p /p "'1, we have q (0 from L L L\ the construction of Raible’s table. From the lemma, since a sign change occurs, it follows that i(P)"o(Q). Since Q(z) is of degree n!1 it follows by the induction hypothesis in Eq. (12), that sign +r , ,t , > L\ 2 u , v ,"o(Q), and since q (0, we have L\ sign +r , 2, t , u , v , > L\ GFFFFHFFFFI M/ "sign +q ,r , , t , u , v ,. > L\ L\ 2 Therefore, i(P)"sign +q ,r , , t , u , v ,. From > L\ L\ 2 the lemma, o(P)"i(Q)#1. Again, by the induction hypothesis in Eq. (12), sign +r , 2 , t , u , v ,"i(Q), \ L\
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and since q
(0, we have L\ sign +r , ,t , u , v , \ L\ 2 #1 GFFFFHFFFFI G/ "sign +q ,r , , t , u , v ,. \ L\ L\ 2 Therefore, o(P)"sign +q ,r , , t , u , v ,. \ L\ L\ 2 (3) When p (0, " p /p "(1, we have q (0 from L L L\ the construction in Raible’s table. From the lemma, since no sign change occurs, i(P)"i(Q)#1 Since Q(z) is of degree n!1, we have by the induction hypothesis in Eq. (12), sign +r , , t , u , v ,"i(Q), and since \ L\ 2 q (0, we have L\ sign +r , ,t , u , v , \ L\ 2 #1 GFFFFHFFFFI G/ "sign +q ,r , , t , u , v ,. \ L\ L\ 2 Therefore, i(P)"sign +q ,r , , t , u , v ,. From \ L\ L\ 2 the lemma, o(P)"o(Q). Since Q(z) is of degree n!1, we have by the induction hypothesis in Eq. (12), sign +r , , t , u , v ,"o(Q), and since q (0, > L\ 2 L\ we have sign +r , ,t , u , v , > L\ 2 GFFFFHFFFFI M/ "sign +q ,r , , t , u , v ,. > L\ L\ 2 Therefore, o(P)"sign +q ,r , , t , u , v ,. > L\ L\ 2 (4) When p (0, " p /p "'1, we have q '0 from L L L\ the construction in Raible’s table. From the Lemma, since a sign change occurs, i(P)"o(Q). Since Q(z) is of degree n!1, we have by the induction hypothesis in Eq. (11), sign +r , , t , u , v ,"o(Q), and since \ L\ 2 q '0, we have L\ sign +r , ,t , u , v , \ L\ 2 GFFFFHFFFFI M/ "sign +q ,r , , , t , u , v ,. \ L\ L\ 2 Therefore, i(P)"sign +q ,r , , t , u , v ,. From \ L\ L\ 2 the lemma, o(P)"i(Q)#1. Since Q(z) is of degree n!1, we have by the induction hypothesis in Eq. (11), sign +r , , t , u , v ,"i(Q), and since q '0, we > L\ 2 L\ have sign +r , ,t , u , v , > L\ 2 #1 GFFFFHFFFFI G/ "sign +q ,r , , t , u , v ,. > L\ L\ 2
Therefore, o(P)"sign +q ,r , , t , u , v ,. > L\ L\ 2 Collecting the above four cases we have: (1) If p '0, then L i(P)"sign +q ,r , , t , u , v ,, > L\ L\ 2 o(P)"sign +q ,r , , t , u , v ,. \ L\ L\ 2 (2) If p (0, then L i(P)"sign +q ,r , , t , u , v ,, \ L\ L\ 2 o(P)"sign +q ,r , , t , u , v ,. > L\ L\ 2 Therefore the Theorem holds for polynomials of degree n and the proof is complete. )
5. Example Consider the test polynomial P(z)"z#0.9z!1.12z!2.504z!1.5724z!0.3774 for which Raible’s table is: Table: Raible’s table for Example P(z): 1.0000 0.9000 !1.1200 !2.5040 !1.5724 !0.3774 z Q(z): 0.8576 0.3066 !2.0650 !2.9267 !1.2327 z R(z): !0.9145 !3.9005 !5.0334 !2.4860 z S(z): 5.8437 9.7829 5.5700 z ¹(z): 0.5345 0.4581 z º(z): 0.1418 z
It is clear from the above table that sign +q , r , s , t , u , > "sign +0.8576, !0.9145, 5.8437, 0.5345, 0.1418, > "4"i(P) sign +q , r , s , t , u , \ "sign +0.8576, !0.9145, 5.8437, 0.5345, 0.1418, \ "1"o(P). In fact, the given polynomial P(z) has the following roots, 1.5,
!0.7$j0.5,
!0.5000$j0.3
which verifies the answer. Now let us examine the polynomials and their root locations at each stage of the reduction in Raible’s table. First, for P(z) , it is true that " p /p "(1 and p '0. This is equivalent to sign[p ]"sign[q ] as shown in the table. As we expect from the lemma, we have i(Q)"i(P)!1"3 and o(Q)"o(P)"1.
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Second, we observe the transition from Q(z) to R(z) . For Q(z), we see that " q /q "'1. Then let us consider its reverse polynomial QP(z)"zQ(1/z) and the corresponding reduced degree polynomial z\Q< (z). Since QP(z) satisfies the condition " q /q "(1, from the lemma, i(R)"i(z\Q< )"i(QP)!1"o(Q)!1"0, o(R)"o(z\Q< )"o(QP)"i(Q)"3. Third, let us observe the transition from R(z) to S(z). Again, we find that "r /r "'1. By the same arguments as before, we have i(S)"i(z\R< )"i(RP)!1"o(R)!1 "i(Q)!1"i(P)!2"2, o(S)"o(z\R< )"o(RP)"i(R)"o(Q)!1 "o(P)!1"0. At the fourth step we have the polynomials S(z) and ¹(z). Since " s /s "(1 is satisfied, sign[s ]"sign[t ] and we have, due to Proposition (D) i(¹)"i(S)!1"i(P)!3"1 and o(¹)"o(S)"o(P)!1. The final step can be carried out the same way since the condition " t /t "(1 is satisfied. Thus, sign[t ]" sign[u ] and i(º)"i(S)!1"i(P)!4"0 and o(º)"o(S)"o(P)!1"0. At this last step, the last root is being removed. Since º is a polynomial of degree zero i(º)"o(º)"0, we conclude that i(P)"4 and o(P)"1.
6. Singular cases The nonregular case can arise when a leading coefficient of one of the polynomials becomes zero. In this case the previous polynomial has p "p or p "!p . This L L occurs due to the presence of roots which lie on the unit circle or roots which are reciprocal with respect to each other. Such symmetries can be destroyed by a slight perturbation of the unit circle such as shrinking it slightly. This may be accomplished by the mapping z"z /(1#d) where d is a small positive real number and considering the corresponding polynomial P(z ) . This has the effect of counting unit circle roots as outside roots, which corresponds with the notion of boundedinput bounded output stability. We leave the details to the reader.
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7. Concluding remarks The new proof given here of the Jury test shows the behavior of the roots of the sequence of polynomials constructed in the test. This is done by studying the root loci of associated families of polynomials. These results complement Jury’s original proof which was based on Rouche´’s Theorem. They provide additional insight into the workings of the test. Our proof could also be more accessible to undergraduates. A similar root locus based proof can be given for Bistritz’s table (Bistritz, 1984). Finally, we point out that these results are the discrete time counterparts of recent simplifications obtained in explaining the workings of the Routh test (Ho et al. 1998).
Acknowledgements We gratefully acknowledge many helpful discussions with Professor E. I. Jury. This research was supported in part by NASA Grant NCC-5228, NSF Grants ECS9417004 and HRD-9706268.
References As stro¨m, K. J. (1970). Introduction to stochastic control theory. New York NY: Academic Press. Bhattacharyya, S. P., Chapellat, H., & Keel, L. H. (1995). Robust control: ¹he parametric approach. Upper Saddle Lake, NJ: PrenticeHall. Bistritz, Y. (1984). Zero location with respect to the unit circle of discrete-time linear system polynomials. Proc. IEEE, 72, 1131—1142. Cohn, A. (1922). U®ber die Anzahl der Wurzeln einer algebraischen Gleichung in einem Kreise. Math Z., 14, 110—148. Ho, M. T., Datta, A., & Bhattacharyya, S. P. (1998). An elementary derivation of the Routh-Hurwitz criterion. IEEE ¹rans. Automat. Control, 43, 405—409. Jury, E. I. (1964). ¹heory and application of the z-transform method. New York NY: Wiley. Kuo, B. C. (1992). Digital control systems. Philadelphia, PA: Saunders College Publishing. Marden, M. (1949). ¹he geometry of the zeroes of a polynomial in a complex variable. American Mathematical Society Surveys, (vol. 3, Ch. 10). New York NY: American Mathematical Society. Ogata, K. (1995). Discrete time control systems. Upper Saddle River, NJ: Prentice-Hall. Raible, R. H. (1974). A simplification of Jury’s tabular form. IEEE ¹rans. Automat. Control, 19, 248—250. Schur, I. (1917) U®ber Potenzreihen die im Innern des Einheitskrieses beschra¨nkt sind. J. Reine Angew. Math., 147, 205—232. Schur, I. (1917) U®ber Potenzreihen die im Innern des Einheitskrieses beschra¨nkt sind. J. Reine Angew. Math., 148, 122—145.
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L. H. Keel was born in Korea in 1953. He received the B.S. degree in Electronic Engineering from Korea University, Seoul, Korea, in 1978, and the M.S. and Ph.D. degrees in electrical engineering from Texas A&M University, College Station, Texas in 1983 and 1986, respectively. Since then he has been with the Center of Excellence in Information Systems at Tennessee State University, Nashville, TN, where he is now Professor of Electrical Engineering and Director of the Center for System Science Research. His research interests include Robust Control, System Identification, Structure and Control, and Computer-aided Control System Design. He has authored and coauthored numerous technical papers in the field of control systems and two books, including Robust Control: ¹he Parametric Approach (Upper Saddle River, NJ: Prentice-Hall, 1995).
S. P. Bhattacharyya was born in Rangoon, Burma, on 23 June, 1946, and educated at the Indian Institute of Technology, Bombay, India, 1962—1967, and Rice University, Houston, TX, 1967—1971. He established the graduate program in Automatic Control at the Federal University of Rio de Janeiro, Brazil, 1971—1980, where he was Chairman of the Department of Electrical Engineering, 1978—1980. From 1974 to 1975, he worked as a National Academy of Science Research Fellow at NASA’s Marshall Space Flight Center. In 1980, he joined Texas A&M University, College Station, TX, where he is presently Professor of Electrical Engineering, and served as Director of the Systems and Control Institute, 1990—1992. He has authored and coauthored three books and more than 150 papers in the field of control systems.
Brief Paper
A new proof of the Jury test L. H. Keel *, S. P. Bhattacharyya Center of Excellence in Information Systems, 330 Tenth Avenue North, Tennessee State University, Nashville, TN, USA Department of Electrical Engineering, Texas A&M University, College Station, TX, USA Received 18 July 1997; revised 23 March 1998; received in final form 24 July 1998
Abstract The problem of determining the root distribution of a real polynomial with respect to the unit circle, in terms of the coefficients of the polynomial, was solved by Jury in 1964. The calculations were presented in tabular form (Jury’s table) and were later simplified by Raible in 1974. This result is now classical and is as important in the stability analysis of digital control systems as its continuous time counterpart, the Routh Hurwitz criterion is for the stability analysis of continuous time control systems. Most texts on digital control state the Jury test but avoid giving the proof. In this paper we give a simple, insightful and new proof of the Jury test. The proof is based on the behavior of the root-loci of an associated family of polynomials that was introduced in recent literature. The proof reveals clearly the mechanism underlying the counting of the roots within and without the unit circle. 1999 Elsevier Science Ltd. All rights reserved. Keyword: Discrete systems; Polynomial; Roots; Stability test; Zero crossings
1. Introduction The stability of a discrete time linear time invariant system is determined by the root locations of the system characteristic polynomial with respect to the unit circle, the system being stable if and only if all roots lie within the unit circle. This problem was studied by Schur (1917), Cohn (1922) and Marden (1949), among others. Jury (1964), building on these and other earlier results, gave a procedure to ascertain the numbers of roots of a polynomial that lie inside and outside the unit circle. This calculation was organized in tabular form and came to be known in the control literature as Jury’s table and the Jury test. These results are the discrete time counterparts of the Routh table and Routh test, discovered nearly a century earlier and useful in the stability analysis of contin- uous time control systems.
————— *Corresponding author. Tel.:#1-615-963-7025; fax:#1-615-9637027; e-mail: [email protected]. This paper was not presented at any IFAC meeting. This paper was recommended for publication in revised form by Associate Editor C. V. Hollot under the direction of Editor Dr. R. Tempo.
An alternative to the Jury test was presented by Bistritz (1984). This test, based on constructing a sequence of polynomials of successively lower degree and determining the number of sign changes in the leading column, is quite similar in mechanism to the Routh test, and is based on properties of symmetric and antisymmetric polynomials. Raible (1974) reported a simplification of Jury’s table based on certain formulas appearing in a paper of As stro¨m (1970). Raible justified the computations in this modified table by establishing a correspondence between the signs of the first column entries of his table with those computed in Jury’s original table. In other words Raible’s proof of the modified table relies on the original proof of Jury’s table. The proof of the Jury test, given in Jury’s 1964 book (Jury, 1964), is based on Rouche´’s Theorem but is rarely presented in controls texts. Although many books on digital control contains the Jury test we have not encountered a single undergraduate textbook which contains a proof of Jury’s table (see, for example Kuo, 1992; Ogata, 1995). The stability part of the result, as opposed to the counting part, is stated in virtually every book on digital control, however also without
0005-1098/99/$—see front matter 1999 Elsevier Science Ltd. All rights reserved PII: S 0 0 0 5 - 1 0 9 8 ( 9 8 ) 0 0 1 5 2 - 6
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proof. Moreover not even a hint is given regarding the rationale or logic underlying the proof thus forcing the control engineer to accept this result on faith and without insight. In this paper we present a new, simple and hopefully insightful proof of the Jury test for root distribution with respect to the unit circle. The proof is based on elementary properties of the root loci of an associated family of polynomials. This family was first introduced in Bhattacharyya et al., (1995) and was used to give a proof of the stability part of the Jury test. However, it was not clear how the number of stable and unstable roots could be counted. In this paper we go on to show, using the same family of root loci, that roots are counted in the Jury test by sequentially sending one branch of the root locus at a time, to the origin and constructing a polynomial of lower degree by removing the root at the origin. We believe that this new proof provides valuable insight into the workings of the Jury test and is simple enough that it can be taught even to undergraduate students. The mechanism of the proof may be useful in other problems as well.
Table. Raible’s table P(z): Q(z): R(z):
p p p 2 2 p p p zL L L\ L\ q q q 2 2 q q zL\ L\ L\ L\ r r r 2 2 r zL\ L\ L\ L\ $ $ ¹(z): t t t z º(z): u u z »(z): v z p p p q "p ! p q "p ! p 2 q "p ! p L\ L\ p p L\ L\ L p L q qL qL r "q ! q r "q ! q 2 r "q ! q L\ L\ q L\ L\ q q L\ L\ L\ L\ $ t t u "t ! t u "t ! t t t u v "u ! u u
Each row of Raible’s table represents a polynomial. Thus Q(z)"q zL\#q zL\#2#q z#q L\ L\ R(z)"r zL\#r zL\#2#r z#r L\ L\ $
2. The Jury test
¹(z)"t z#t z#t
The Jury test deals with the problem of determining the root distribution with respect to the unit circle, of the polynomial
º(z)"u z#u
P(z)"pLzL#pL\zL\#pL\zL\#2#pz
»(z)"v .
#pz#p . We make the following standing assumptions throughout the paper. Assumptions (1) The polynomial P(z) has real coefficients. (2) pLO0; in other words, the polynomial is of degree n. (3) pO0; this assumption avoids the trivial case where the polynomial has a root at the origin. When this happens, one can factor out the root at the origin and proceed with a lower degree polynomial. (4) P(z) does not have a root on the unit circle. This is a reasonable assumption since we are trying to determine the root distribution with respect to the unit circle. A slight controlled perturbation of the polynomial coefficients can be designed so that any roots on the unit circle are displaced outside it, as explained later. We shall work with a simplified version of Jury’s table as given by Raible. This modified table, which we refer to as Raible’s table, is constructed as follows.
and
We assume that no entries of the first column of Raible’s table given above is zero; this is the regular case. Let i( ) ) and o( ) ) denote respectively the numbers of roots of the polynomial ( ) ) inside and outside the unit circle. Also let sign + ) , and sign + ) , denote the number > \ of positive signed elements and the number of negative signed elements in the set + ) ,. Theorem (The Jury test). In the regular case, (1) If p '0, L i(P)"sign +q ,r , , t , u , v ,, > L\ L\ 2 o(P)"sign +q ,r , , t , u , v ,. \ L\ L\ 2 (2) If p (0, L i(P)"sign +q ,r , ,t , u , v , \ L\ L\ 2 o(P)"sign +q ,r , , t , u , v ,. > L\ L\ 2 In the following sections we develop our proof of this result. Also we indicate how to proceed in the nonregular case, that is, if a leading coefficient in the first column becomes zero.
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3. Preliminary results In this section we give some preliminary technical results and in particular introduce a one-parameter family of polynomials and develop some properties of its root loci. These preliminary results are necessary for our proof of the Jury test given in the next section. Consider the polynomial P(z)"p zL#p zL\#p zL\#2#p z L L\ L\ #p z#p (1) subject to the standing assumptions stated in the previous section. Introduce the associated one parameter family of polynomials defined as
p 1 P (z) " : P(z)!j zLP , H p z L Then P (z)"P(z),
j3[0, 1].
and
(F) (G)
(6)
p 1 P< (z) " : PP(z)!j L zLPP , H p z Then
j3[0, 1].
(7)
(4)
condition " p /p "'1 is equivalent to the condition L that q and p are of opposite sign. L\ L Proof. (A) The polynomial
p 1 P (z)"P(z)!j zLP H p z L has leading coefficient
p p p !j "p 1!j L L p p L L
(8)
p 1 P< (z)"PP(z)! L zLPP "z p z
Q< (z)"z\P< (z).
p p p p p p ! zL\# p ! zL\#2# p ! L\ , L p L\ p p L L L GFFFFFFFFFFFFHFFFFFFFFFFFFI /X
1 PP(z) " : zLP , z
and
(E)
(2)
Q(z)"z\P (z). (5) Note that Q(z) obtained above corresponds exactly to the second row of Raible’s table. We also define the reverse polynomial of P(z) and its associated one parameter family as follows:
P< (z)"PP(z)
(C) (D)
(3)
1 p "z P (z)"P(z)! zLP z p L
(B) (B1) (B2)
¹his means that the numbers of roots of the family P (z) inside and outside the unit circle remain respecH tively the same, since the root loci originating inside the unit circle remain inside and those originating outside remain outside. If " p /p "'1, then L P< (z) is always of degree n for j3[0, 1] and H none of the n arcs corresponding to the root loci of the family P< (z), j3[0, 1] cross the unit circle. H i(P)"o(PP) and o(P)"i(PP). Suppose that P(z) has i roots inside the unit circle and n!i roots outside the unit circle. If " p /p "(1, then L z\P (z)"Q(z) has i!1 roots inside the unit circle and n!i roots outside the unit circle. Suppose that P(z) has i roots inside the unit circle and n!i roots outside the unit circle. If " p /p "'1, then L z\P< (z)"Q< (z) has i roots outside the unit circle and n!i!1 roots inside the unit circle. P (z) and P< (z) have the same set of roots. ¹he condition " p /p "(1 is equivalent to the condi L tion that q and p are of the same sign, and the L\ L
p pp pp p ! L zL\# p ! L L\ zL\#2# p ! L , p L\ p p GFFFFFFFFFFFFHFFFFFFFFFFFFI /K X
(9)
(10)
Proposition (A) If " p /p "(1, then L (A1) P (z) is always of degree n for j3[0, 1] and H (A2) none of the n arcs corresponding to the root loci of the family P (z), for j3[0, 1], cross the unit circle. H
and therefore when
p (1 p L this coefficient is nonzero for j3[0, 1]. Therefore P (z) is H of degree n for all j3[0, 1].
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Now suppose that z lies on the unit circle and is a zero of P (z) for some j"j 3[0, 1]. Then H
p 1 P (z )"P(z )!j zL P . H p z L Since P (z) is a real polynomial 1/z is also a zero of this H polynomial. Therefore
1 1 p 1 "P !j P (z )"0. P p zL H z z L From the above two equations we have P
p 1!j "0 p L
1 z
and
p P(z ) 1!j "0. p L Since 1!j (p /p )O0 L it follows that P(z )"P(1/z )"0. However this contradicts the assumption that P(z) has no roots on the unit circle. Therefore no root loci of the family P (z) can cross the unit circle. H (B) Proposition (B) is the dual of Proposition (A) and can be similarly proved. (C) Since PP(z)"zLP
1 , z
the reciprocal of every root of P(z) is a root of P< (z). Thus, i(P)"o(PP) and o(P)"i(PP). (D) Due to Proposition (A), the condition " p /p "(1 L ensures that no roots of
p 1 P (z)"P(z)!j zLP H p z L cross the boundary, that is, the unit circle, for j3[0, 1]. Since P (z) has one root at the origin, this root must be the terminal point of a root locus originating from a root of P(z) lying inside the unit circle.Thus, z\P (z)"Q(z) has precisely one less root, inside the unit circle than P(z), while they have the same number of roots outside the unit circle. (E) Proposition (E) is the dual of Proposition (D). It can be similarly proved.
(F) Write
p 1 p P (z)"P(z)! zLP "P(z)! PP(z), p z p L L 1 p p "PP(z)! L P(z). P< (z)"PP(z)! L zLPP z p p This implies that p P (z)"p P(z)!p PP(z) L L p P< (z)"p PP(z)!p P(z). L Consequently, !p P (z)"p P< (z). L Since p and p are nonzero constants, P (z) and P< (z) L have the same set of roots. (G) If " p /p "(1, p!p'0. Then it is obvious that L L the highest order coefficient of Q(z) becomes
p '0, L p (0, L
p '0, L p (0. L based on the Proposition
p!p '0 for " q " L L\ p (0 for L Similarly, when " p /p "'1, L p!p (0 for " q " L L\ p '0 for L Next we prove a key result above. Lemma. ¸et
P(z)"p zL#p zL\#p zL\#2#p z#p , L L\ L\ p O0, L Q(z)"q zL\#q zL\#q zL\#2#q z#q , L\ L\ L\ q O0 L\ with p!p q " L L\ p L p p !p p q " L\ L L\ p L $ p p !p p L\ . q " L p L Then,
i(Q)"i(P)!1, if sign(p )"sign(q ), L L\ o(Q)"o(P), i(Q)"o(P)!1 if sign(p )Osign(q ), L L\ o(Q)"i(P).
L.H. Keel, S.P. Bhattacharyya/Automatica 35 (1999) 251—258
Proof. Let us first consider the case when sign(p )" L sign(q ). Since L\ q "(p!p)/p , L\ L L this means that p!p'0, equivalently, L " p /p "(1. L From Proposition (D), Q(z) will have one less root inside the unit circle than P(z) while the number of roots outside the unit circle remains the same. Thus,
i(Q)"i(P)!1, if sign(p )"sign(q ), L L\ o(Q)"o(P). Now consider the case when sign(p )Osign(q ). This L L\ means that p!p(0. Equivalently, L p '1. p L Since " p /p "(1, it is clear from Proposition (D) that L i(z\P< )"i(PP)!1 and o(z\P< )"o(PP). If we consider the reverse polynomial PP(z)"zLP(1/z), we know that i(PP)"o(P) and o(PP)"i(P). Therefore, i(z\P< )" o(P)!1 and o(z\P< )"i(P). From Proposition (F), P< (z) and P (z) have the same set of roots. Therefore, z\P (z)"Q(z) has the same set of roots as z\P< (z). Thus, in this case, where a sign change occurs between p and q , we have L L\ i(Q)"o(P)!1 and o(Q)"i(P).
4. Proof of the Jury test Using the preliminary results of the last section we give our new proof of the Jury test. Proof of the Theorem. The proof is based on induction. Let us first consider the case when n"1. Let º(z)"u z#u be the test polynomial and consider the corresponding Raible’s table: º(z): u u »(z): v We first consider the case when u '0. Here we need to show that i(º)"sign +v , and o(º)"sign +v ,. > \ First, if " u /u "(1, we have v '0 from the construc tion of Raible’s table. Therefore sign +v ,"1 and > sign +v ,"0. On the other hand, we know that \ i(º)"1 and o(º)"0 due to the assumption "u /u "(1. Therefore, i(º)"sign +v ,"1 and > o(º)"sign +v ,"0 which agrees with the theorem. \ Now suppose that u '0 and " u /u "'1. Then from the construction of Raible’s table, we have v (0. Therefore sign +v ,"1 and sign +v ,"0. Here we \ > also know that i(º)"0 and o(º)"1. Therefore,
255
i(º)"sign +v ,"0 and o(º)"sign +v ,"1 which > \ again agrees with the theorem. We can similarly show that the theorem is true for the cases u (0, "u /u "(1, and u (0, "u /u "'1 proving that the theorem is true for n"1. Now let us assume that the theorem is true for all polynomials of degree n!1. Then we have
sign +r , , t , u , v ,"i(Q), > L\ 2 If q '0 (11) L\ sign +r , , t , u , v ,"o(Q), \ L\ 2 sign +r , , t , u , v ,"o(Q), > L\ 2 If q (0 (12) L\ sign +r , , t , u , v ,"i(Q). \ L\ 2 Here we consider the following four cases to complete the proof. (1) When p '0, " p /p "(1, then we have q '0 L L L\ from the construction of Raible’s table. From the lemma, since no sign change occurs, it follows that i(P)"i(Q)#1. Since Q(z) is of degree n!1 we have, by the induction hypothesis in Eq. (11), sign +r , , t , u , v ,"i(Q), and since q '0, we > L\ 2 L\ have sign +r , , t , u , v ,#1 > L\ 2 GFFFFHFFFFI G/ "sign +q ,r , , t , u , v ,. > L\ L\ 2 Therefore, i(P)"sign +q , r , 2 , t , u , v ,. From > L\ L\ the lemma, we have o(P)"o(Q). By the induction hypothesis in Eq. (11), since Q(z) is of degree n!1, it follows that, sign +r , , t , u , v ,"o(Q), and since \ L\ 2 q '0, we have L\ sign +r , ,t , u , v , \ L\ 2 GFFFFHFFFFI M/ "sign +q ,r , , t , u , v ,. \ L\ L\ 2 Therefore, o(P)"sign +q ,r , , t , u , v ,. \ L\ L\ 2 (2) When p '0, "p /p "'1, we have q (0 from L L L\ the construction of Raible’s table. From the lemma, since a sign change occurs, it follows that i(P)"o(Q). Since Q(z) is of degree n!1 it follows by the induction hypothesis in Eq. (12), that sign +r , ,t , > L\ 2 u , v ,"o(Q), and since q (0, we have L\ sign +r , 2, t , u , v , > L\ GFFFFHFFFFI M/ "sign +q ,r , , t , u , v ,. > L\ L\ 2 Therefore, i(P)"sign +q ,r , , t , u , v ,. From > L\ L\ 2 the lemma, o(P)"i(Q)#1. Again, by the induction hypothesis in Eq. (12), sign +r , 2 , t , u , v ,"i(Q), \ L\
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and since q
(0, we have L\ sign +r , ,t , u , v , \ L\ 2 #1 GFFFFHFFFFI G/ "sign +q ,r , , t , u , v ,. \ L\ L\ 2 Therefore, o(P)"sign +q ,r , , t , u , v ,. \ L\ L\ 2 (3) When p (0, " p /p "(1, we have q (0 from L L L\ the construction in Raible’s table. From the lemma, since no sign change occurs, i(P)"i(Q)#1 Since Q(z) is of degree n!1, we have by the induction hypothesis in Eq. (12), sign +r , , t , u , v ,"i(Q), and since \ L\ 2 q (0, we have L\ sign +r , ,t , u , v , \ L\ 2 #1 GFFFFHFFFFI G/ "sign +q ,r , , t , u , v ,. \ L\ L\ 2 Therefore, i(P)"sign +q ,r , , t , u , v ,. From \ L\ L\ 2 the lemma, o(P)"o(Q). Since Q(z) is of degree n!1, we have by the induction hypothesis in Eq. (12), sign +r , , t , u , v ,"o(Q), and since q (0, > L\ 2 L\ we have sign +r , ,t , u , v , > L\ 2 GFFFFHFFFFI M/ "sign +q ,r , , t , u , v ,. > L\ L\ 2 Therefore, o(P)"sign +q ,r , , t , u , v ,. > L\ L\ 2 (4) When p (0, " p /p "'1, we have q '0 from L L L\ the construction in Raible’s table. From the Lemma, since a sign change occurs, i(P)"o(Q). Since Q(z) is of degree n!1, we have by the induction hypothesis in Eq. (11), sign +r , , t , u , v ,"o(Q), and since \ L\ 2 q '0, we have L\ sign +r , ,t , u , v , \ L\ 2 GFFFFHFFFFI M/ "sign +q ,r , , , t , u , v ,. \ L\ L\ 2 Therefore, i(P)"sign +q ,r , , t , u , v ,. From \ L\ L\ 2 the lemma, o(P)"i(Q)#1. Since Q(z) is of degree n!1, we have by the induction hypothesis in Eq. (11), sign +r , , t , u , v ,"i(Q), and since q '0, we > L\ 2 L\ have sign +r , ,t , u , v , > L\ 2 #1 GFFFFHFFFFI G/ "sign +q ,r , , t , u , v ,. > L\ L\ 2
Therefore, o(P)"sign +q ,r , , t , u , v ,. > L\ L\ 2 Collecting the above four cases we have: (1) If p '0, then L i(P)"sign +q ,r , , t , u , v ,, > L\ L\ 2 o(P)"sign +q ,r , , t , u , v ,. \ L\ L\ 2 (2) If p (0, then L i(P)"sign +q ,r , , t , u , v ,, \ L\ L\ 2 o(P)"sign +q ,r , , t , u , v ,. > L\ L\ 2 Therefore the Theorem holds for polynomials of degree n and the proof is complete. )
5. Example Consider the test polynomial P(z)"z#0.9z!1.12z!2.504z!1.5724z!0.3774 for which Raible’s table is: Table: Raible’s table for Example P(z): 1.0000 0.9000 !1.1200 !2.5040 !1.5724 !0.3774 z Q(z): 0.8576 0.3066 !2.0650 !2.9267 !1.2327 z R(z): !0.9145 !3.9005 !5.0334 !2.4860 z S(z): 5.8437 9.7829 5.5700 z ¹(z): 0.5345 0.4581 z º(z): 0.1418 z
It is clear from the above table that sign +q , r , s , t , u , > "sign +0.8576, !0.9145, 5.8437, 0.5345, 0.1418, > "4"i(P) sign +q , r , s , t , u , \ "sign +0.8576, !0.9145, 5.8437, 0.5345, 0.1418, \ "1"o(P). In fact, the given polynomial P(z) has the following roots, 1.5,
!0.7$j0.5,
!0.5000$j0.3
which verifies the answer. Now let us examine the polynomials and their root locations at each stage of the reduction in Raible’s table. First, for P(z) , it is true that " p /p "(1 and p '0. This is equivalent to sign[p ]"sign[q ] as shown in the table. As we expect from the lemma, we have i(Q)"i(P)!1"3 and o(Q)"o(P)"1.
L.H. Keel, S.P. Bhattacharyya/Automatica 35 (1999) 251—258
Second, we observe the transition from Q(z) to R(z) . For Q(z), we see that " q /q "'1. Then let us consider its reverse polynomial QP(z)"zQ(1/z) and the corresponding reduced degree polynomial z\Q< (z). Since QP(z) satisfies the condition " q /q "(1, from the lemma, i(R)"i(z\Q< )"i(QP)!1"o(Q)!1"0, o(R)"o(z\Q< )"o(QP)"i(Q)"3. Third, let us observe the transition from R(z) to S(z). Again, we find that "r /r "'1. By the same arguments as before, we have i(S)"i(z\R< )"i(RP)!1"o(R)!1 "i(Q)!1"i(P)!2"2, o(S)"o(z\R< )"o(RP)"i(R)"o(Q)!1 "o(P)!1"0. At the fourth step we have the polynomials S(z) and ¹(z). Since " s /s "(1 is satisfied, sign[s ]"sign[t ] and we have, due to Proposition (D) i(¹)"i(S)!1"i(P)!3"1 and o(¹)"o(S)"o(P)!1. The final step can be carried out the same way since the condition " t /t "(1 is satisfied. Thus, sign[t ]" sign[u ] and i(º)"i(S)!1"i(P)!4"0 and o(º)"o(S)"o(P)!1"0. At this last step, the last root is being removed. Since º is a polynomial of degree zero i(º)"o(º)"0, we conclude that i(P)"4 and o(P)"1.
6. Singular cases The nonregular case can arise when a leading coefficient of one of the polynomials becomes zero. In this case the previous polynomial has p "p or p "!p . This L L occurs due to the presence of roots which lie on the unit circle or roots which are reciprocal with respect to each other. Such symmetries can be destroyed by a slight perturbation of the unit circle such as shrinking it slightly. This may be accomplished by the mapping z"z /(1#d) where d is a small positive real number and considering the corresponding polynomial P(z ) . This has the effect of counting unit circle roots as outside roots, which corresponds with the notion of boundedinput bounded output stability. We leave the details to the reader.
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7. Concluding remarks The new proof given here of the Jury test shows the behavior of the roots of the sequence of polynomials constructed in the test. This is done by studying the root loci of associated families of polynomials. These results complement Jury’s original proof which was based on Rouche´’s Theorem. They provide additional insight into the workings of the test. Our proof could also be more accessible to undergraduates. A similar root locus based proof can be given for Bistritz’s table (Bistritz, 1984). Finally, we point out that these results are the discrete time counterparts of recent simplifications obtained in explaining the workings of the Routh test (Ho et al. 1998).
Acknowledgements We gratefully acknowledge many helpful discussions with Professor E. I. Jury. This research was supported in part by NASA Grant NCC-5228, NSF Grants ECS9417004 and HRD-9706268.
References As stro¨m, K. J. (1970). Introduction to stochastic control theory. New York NY: Academic Press. Bhattacharyya, S. P., Chapellat, H., & Keel, L. H. (1995). Robust control: ¹he parametric approach. Upper Saddle Lake, NJ: PrenticeHall. Bistritz, Y. (1984). Zero location with respect to the unit circle of discrete-time linear system polynomials. Proc. IEEE, 72, 1131—1142. Cohn, A. (1922). U®ber die Anzahl der Wurzeln einer algebraischen Gleichung in einem Kreise. Math Z., 14, 110—148. Ho, M. T., Datta, A., & Bhattacharyya, S. P. (1998). An elementary derivation of the Routh-Hurwitz criterion. IEEE ¹rans. Automat. Control, 43, 405—409. Jury, E. I. (1964). ¹heory and application of the z-transform method. New York NY: Wiley. Kuo, B. C. (1992). Digital control systems. Philadelphia, PA: Saunders College Publishing. Marden, M. (1949). ¹he geometry of the zeroes of a polynomial in a complex variable. American Mathematical Society Surveys, (vol. 3, Ch. 10). New York NY: American Mathematical Society. Ogata, K. (1995). Discrete time control systems. Upper Saddle River, NJ: Prentice-Hall. Raible, R. H. (1974). A simplification of Jury’s tabular form. IEEE ¹rans. Automat. Control, 19, 248—250. Schur, I. (1917) U®ber Potenzreihen die im Innern des Einheitskrieses beschra¨nkt sind. J. Reine Angew. Math., 147, 205—232. Schur, I. (1917) U®ber Potenzreihen die im Innern des Einheitskrieses beschra¨nkt sind. J. Reine Angew. Math., 148, 122—145.
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L. H. Keel was born in Korea in 1953. He received the B.S. degree in Electronic Engineering from Korea University, Seoul, Korea, in 1978, and the M.S. and Ph.D. degrees in electrical engineering from Texas A&M University, College Station, Texas in 1983 and 1986, respectively. Since then he has been with the Center of Excellence in Information Systems at Tennessee State University, Nashville, TN, where he is now Professor of Electrical Engineering and Director of the Center for System Science Research. His research interests include Robust Control, System Identification, Structure and Control, and Computer-aided Control System Design. He has authored and coauthored numerous technical papers in the field of control systems and two books, including Robust Control: ¹he Parametric Approach (Upper Saddle River, NJ: Prentice-Hall, 1995).
S. P. Bhattacharyya was born in Rangoon, Burma, on 23 June, 1946, and educated at the Indian Institute of Technology, Bombay, India, 1962—1967, and Rice University, Houston, TX, 1967—1971. He established the graduate program in Automatic Control at the Federal University of Rio de Janeiro, Brazil, 1971—1980, where he was Chairman of the Department of Electrical Engineering, 1978—1980. From 1974 to 1975, he worked as a National Academy of Science Research Fellow at NASA’s Marshall Space Flight Center. In 1980, he joined Texas A&M University, College Station, TX, where he is presently Professor of Electrical Engineering, and served as Director of the Systems and Control Institute, 1990—1992. He has authored and coauthored three books and more than 150 papers in the field of control systems.