An elementary proof of the Jury test for real polynomials

Automatica 47 (2011) 249–252

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An elementary proof of the Jury test for real polynomials✩ Younseok Choo ∗ Department of Electronic and Electrical Engineering, Hongik University, Jochiwon, Chungnam, 339-701, Republic of Korea

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Article history: Received 7 March 2010 Received in revised form 7 July 2010 Accepted 5 October 2010 Available online 23 November 2010

abstract As is well known the Jury test has been considered as the most typical way of determining the Schur stability of real polynomials and is introduced in most textbooks on digital control engineering. Its original proof is based on Rouché’s theorem. Recently Keel et al. gave a simple proof of the Jury test using Raible’s simplified table. In this paper we provide another elementary proof of the Jury test. Contrary to that of Keel et al., the proof is carried out directly for the original form of the Jury test. © 2010 Elsevier Ltd. All rights reserved.

Keywords: Discrete system Polynomial Root distribution Jury test

1. Introduction Consider a linear time-invariant discrete system with the characteristic polynomial D(z ) = dn z n + dn−1 z n−1 + · · · + d1 z + d0

(1)

where dn ̸= 0. The stability of the system is decided by the root locations of D(z ) with respect to the unit circle in the complex plane. In order for the system to be stable it is required that its characteristic polynomial be a Schur stable polynomial, i.e., D(z ) has all its roots inside the unit circle. Determining the Schur stability of a polynomial is one of the fundamental steps in many engineering problems including digital control system design. Due to its importance, many researchers have studied this topic and various techniques have been reported in the literature for checking the Schur stability of real polynomials. Among them the Jury test (Jury, 1964) provides a simple tabular form by which one can easily determine the root distribution with respect to the unit circle in the complex plane as well as the Schur stability of real polynomials. A simplified Jury tabular form was later given by Raible (1974). Over the decades the Jury test (Jury, 1964) has been considered as the most typical way of determining the Schur stability of real polynomials, and it appears in most textbooks on digital control engineering. However, as pointed out in the literature, the original ✩ The material in this paper was not presented at any conference. This paper

was recommended for publication in revised form by Associate Editor Guoxiang Gu under the direction of Editor André L. Tits. ∗ Tel.: +82 41 860 2529; fax: +82 41 862 2740. E-mail address: [email protected]. 0005-1098/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.automatica.2010.10.040

proof of the Jury test is based on Rouché’s theorem which is not introduced in control textbooks, and so some attempts have been made to give simple proofs of the Jury test (Chapellat, Mansour, & Bhattacharyya, 1990; Keel & Bhattacharyya, 1999; Mori & Kokame, 2001). The proof of Chapellat et al. (1990) is based on Rouché’s theorem and the boundary crossing theorem. Mori and Kokame (2001) studied a single-parameter characterization of the Schur stability property, from which, as in Chapellat et al. (1990), the simplified Jury stability table can be easily derived. However, in those works, only the stability part of the Jury test was proved, and no attention was given to the root distribution. On the other hand Keel and Bhattacharyya (1999) proved the root distribution part of the Jury test using Raible’s simplified table. As can been seen there, the proof is quite simple so that even undergraduate students can easily understand. The purpose of this paper is to provide another elementary proof of the Jury test. Contrary to that of Keel and Bhattacharyya (1999), the proof does not rely on Raible’s simplified table and is carried out directly for the original form of the Jury test. Although the stability part of the Jury test may be considered as a special case of the root distribution part, an independent proof is also given since it is meaningful in its own right. As can be seen in what follows, the proof of this paper is as simple as that of Keel and Bhattacharyya (1999). 2. Preliminaries In this section we present some important facts which will lead to the main results of this paper. For the nth-order polynomial D(z ) ˆ (z ) = z n D(z −1 ) and let i(·) and o(·) respectively given in (1), let D denote the number of roots of the polynomial (·) inside and outside the unit circle.

250

Y. Choo / Automatica 47 (2011) 249–252

Lemma 1 (Jury, 1964).

i.e.,

|D(z )| = |Dˆ (z )| for |z | = 1.

Dk (1)Dk (−1) = −(dk+1 − d0

The following lemma is an extension of (Theorem 1, Mori & Kokame, 2001).

or

Lemma 2.

Dk (1)Dk (−1) = (−1)n−k D(1)D(−1)

(k+1)

ˆ (z ) have the same number of (i) For |γ | < 1, D(z ) and D(z ) + γ D ˆ ). roots inside the unit circle, i.e., i(D) = i(D + γ D ˆ (z ) and D(z ) + γ Dˆ (z ) have the same number of (ii) For |γ | > 1, D ˆ ) = i(D + γ Dˆ ). roots inside the unit circle, i.e., i(D ˆ (z0 ) = 0 for some |γ | < 1, Proof. (i) Assume that D(z0 ) + γ D ˆ |z0 | = 1. Then γ = −D(z0 )/D(z0 ). But this is impossible since |D(z0 )| = |Dˆ (z0 )| by Lemma 1 and |γ | < 1. Hence D(z ) + γ Dˆ (z ) does not have a root on the unit circle for |γ | < 1, which in turn ˆ (z ) inside the unit circle remain implies that the roots of D(z ) + γ D ˆ (z ) = D(z ) for inside the unit circle for |γ | < 1. Since D(z ) + γ D γ = 0, we obtain (i). ˆ (z ) does not have a root (ii) Similarly we can show that D(z )+γ D on the unit circle for |γ | > 1, which again implies that the roots ˆ (z ) inside the unit circle remain inside the unit circle of D(z ) + γ D ˆ (z ) and Dˆ (z ) + D(z )/γ have the for |γ | > 1. Obviously D(z ) + γ D ˆ (z ) + D(z )/γ = Dˆ (z ) for |γ | = ∞, same set of roots. Then, since D (ii) also holds.  For the nth-order polynomial D(z ) in (1), consider a sequence of polynomials (k)

(k)

(k)

Dk (z ) = dk z k + dk−1 z k−1 + · · · + d0

(2)

recursively obtained by Dn (z ) = D(z ) 1

(3)

(k+1)

(k+1) ˆ

Dk+1 (z ))

(d Dk+1 (z ) − d0 z k+1 k = n − 1, n − 2, . . . , 1, 0. Dk (z ) =

(4)

(k+1) k+1

z

+ d(1k+1) z k + · · · + d(kk+1) z + d(kk++11)

(k)

(k+1) (k+1)

 d(k+1)  =  (kk++11) d

(k+1) (k+1)

(k+1) 

 di  (k+1)  d

− (−1)k+1 d(0k+1) )Dk+1 (1)Dk+1 (−1)

 ×

n ∏

(i)

(i)

(i)

i (i) d0

(di − d0 )(di − (−1)

 ) .

i=k+1

(0)

Then, since d0 follows. 

= (d(11) )2 − (d(00) )2 = −D1 (1)D1 (−1), the result

The following lemma is a consequence of Lemma 2, and plays a key role in proving the Jury test. Lemma 4. For any nth-order polynomial Dn (z ) and (n − 1)th-order polynomial Dn−1 (z ) related by (4), we have (n−1)

(i) If dn−1 > 0, then i(Dn−1 ) = i(Dn ) − 1 and o(Dn−1 ) = o(Dn ) (n−1)

(ii) If dn−1 < 0, then i(Dn−1 ) = o(Dn ) − 1 and o(Dn−1 ) = i(Dn ). (n)

(n)

(n−1)

Proof. (i) Let γ (n) = d0 /dn . If dn−1 (5). Since

> 0, then |γ (n) | < 1 from

(d(nn) Dn (z ) − d(0n) Dˆ n (z )) = d(nn) (Dn (z ) − γ (n) Dˆ n (z )) (n)

(n)

ˆ n ) by Lemma 2. Hence we obtain we have i(Dn ) = i(dn Dn − d0 D i(Dn−1 ) = i(Dn ) − 1 from (4). Since the degree of Dn−1 (z ) is one less than that of Dn (z ), we also have o(Dn−1 ) = o(Dn ). (n−1)

(ii) If dn−1 < 0, then |γ (n) | > 1 by (5). By Lemma 2, we have ˆ n ) = i(d(nn) Dn − d(0n) Dˆ n ) and so i(Dn−1 ) = i(Dˆ n ) − 1 from (4). i(D

ˆ n ) = o(Dn ), then i(Dn−1 ) = o(Dn ) − 1, which in turn Since i(D implies o(Dn−1 ) = i(Dn ).  A necessary and sufficient condition for the Schur stability of D(z ) can be obtained from Lemma 4. In fact Theorem 1 corresponds to the stability part of the Jury test as will be shown in what follows.

(i)

the coefficients of Dk (z ) are given, in terms of those of Dk+1 (z ), by dk−i = dk+1 dk−i+1 − d0

)(dk(k++11)

Theorem 1. D(z ) is Schur stable if and only if

Since

ˆ k+1 (z ) = d0 D

(k+1)

di

di > 0 for i = n − 1, n − 2, . . . , 1, 0 or equivalently

|d(i i) | > |d(0i) | for i = n, n − 1, . . . , 2, 1.

i = 0, 1, 2, . . . , k.

(5)

Proof. Assume that D(z ) is Schur stable, i.e., i(D) = n. Then |dn | > |d0 |, i.e., d(nn−−11) > 0, and, by Lemma 4, i(Dn−1 ) = n − 1 and so

Lemma 3 indicates that the sign of d0 can be determined from the sign of D(1)D(−1) when some conditions are satisfied. A similar result was presented in Jury (1964) based on the Schur–Cohn determinants.

Dn−1 (z ) is Schur stable. Since Dn−1 (z ) is Schur stable, dn−2 > 0 and i(Dn−2 ) = i(D1 ) − 1 = n − 2, which implies that Dn−2 (z ) is Schur stable. Continuing the same process, we have that if D(z ) is (i) Schur stable, then di > 0 for i = n − 1, n − 2, . . . , 1, 0.

0

k−i+1

(0)

(i)

Lemma 3. (0)

(n−2)

d0 = (−1) D(1)D(−1) n

 n ∏

(i)

(i)

(i)

(di − d0 )(di − (−1)

i (i) d0

 ) .

Conversely suppose di > 0 for i = 0, 1, . . . , n − 2, n − 1. Note that i(D0 ) = o(D0 ) = 0 since D0 (z ) is a constant function. If (0) d0 > 0, then i(D1 ) = i(D0 ) + 1 = 1 and o(D1 ) = 0 by Lemma 4. (1)

Proof. It is easily seen that

Similarly if d1 > 0, then i(D2 ) = i(D1 ) + 1 = 2 and o(D2 ) = 0. Repeatedly applying Lemma 4, we have i(D) = n, o(D) = 0 and so D(z ) is Schur stable. 

ˆ k+1 (1) = Dk+1 (1) D

3. Proof of the Jury test

i=2

ˆ k+1 (−1) = (−1)k+1 Dk+1 (−1). D Then, from (4), we have (k+1)

(k+1)

−(dk(k++11)

− (−1)k+1 d(0k+1) )Dk+1 (−1)

Dk (1) = (dk+1 − d0 Dk (−1) =

)Dk+1 (1)

In this section we present a proof of the original form of the Jury test (Jury, 1964). To this end it should be noted that the kth row of the stability table in the Jury test is obtained from the coefficients of the (n − k + 1)th-order polynomial Dn−k+1 (z ) generated by (3) and (4) (see (5) also). We consider the regular cases only.

Y. Choo / Automatica 47 (2011) 249–252

Theorem 2 (Jury Test). For the nth-order real polynomial D(z ) given in (1), form the stability table: (n)

(n)

d0

(n)

d1

−dn(n−−11)

−dn(n−−21)

(n−2) dn−2

···

.. .

(1)

dn−1

−d(1n−1)

−d(0n−1)

(n−2)

··· .. .

d1

..

(n)

···

d2

d(nn)

(n−2)

d0

(6)

.

(n−2) (n−3)

(n−2) (n−3)

(0)

(n−2)

(n−2) (n−3)

(n−2) (n−3)

(0)

(n−1)

(n)

where di

(n−2)

Now assume that (11) and (12) hold for all polynomials of degree n − 1, i.e.,

o(Dn−1 ) = sign− {dn−2 , dn−2 dn−3 , . . . , dn−2 dn−3 · · · d0 }.

(0)

dn−2−i

> 0, then |d(11) | > |d0 | and D1 (z ) is Schur stable, i.e., i(D1 ) = 1 and o(D1 ) = 0. If (0) (1) (1) d0 < 0, then |d0 | > |d1 | and D1 (z ) has a root outside the unit circle, i.e., i(D1 ) = 0 and o(D1 ) = 1. Hence we have i(D1 ) = (0) (0) sign+ {d0 } and o(D1 ) = sign− {d0 }. (n−2)

d0

d0

(n−1) dn−1−i

(0)

We prove (11) and (12) by induction. If d0

(1)

i(Dn−1 ) = sign+ {dn−2 , dn−2 dn−3 , . . . , dn−2 dn−3 · · · d0 }

(1)

d1

251

If dn−1 > 0, then i(Dn−1 ) = i(Dn ) − 1 and o(Dn−1 ) = o(Dn ) by Lemma 4. On the other hand

= di and

(n−1)

0

(n−1)

i = 0, 1

(n−1)

(n−1)

(n−1) (n−2)

(n−1) (n−2)

(0)

sign+ {dn−1 , dn−1 dn−2 , . . . , dn−1 dn−2 · · · d0 }

= sign− {dn(n−−22) , dn(n−−22) dn(n−−33) , . . . , dn(n−−22) dn(n−−33) · · · d(00) } = o(Dn−1 ) = i(Dn ) = i(D)

(n)

(n−1)

P2 = P1 (|dn−1 | − |d0 (2)

(n−1)

(n−1) (n−2)

(n−1) (n−2)

(0)

sign− {dn−1 , dn−1 dn−2 , . . . , dn−1 dn−2 · · · d0 }

|)

= sign+ {dn(n−−22) , dn(n−−22) dn(n−−33) , . . . , dn(n−−22) dn(n−−33) · · · d(00) } + 1 = i(Dn−1 ) + 1 = o(Dn ) = o(D).

(2)

Pn−1 = Pn−2 (|d2 | − |d0 |)

Hence the proof is completed.



(1)

Pn = Pn−1 (|d1 | − |d0 |). (k)

Assume that dk Then

̸= 0, k = n − 1, n − 2, . . . , 0 (i.e., regular case).

(−1)n D(1)D(−1) > 0

(7)

|d(i i) | > |d(0i) |,

(8)

i = n, n − 1, . . . , 2.

(ii) (Root distribution) The number of roots of D(z ) inside the unit circle and outside the unit circle are respectively given by i(D) = sign− {P1 , P2 , P3 , . . . , Pn }

(9)

o(D) = sign+ {P1 , P2 , P3 , . . . , Pn }

(10)

where sign+ {·} and sign− {·} respectively denote the number of positive and negative signed elements in the set {·}. (0)

> 0 if and only if (−1) D(1)D(−1) > 0. Since d0 = (d1 ) −(d0 ) , the (1) (1) constraint (7) can be replaced by |d1 | > |d0 |. Then the stability Proof. (i) If (8) is satisfied, then Lemma 3 implies that d0 (0)

n

(1) 2

(1) 2

part is the same as Theorem 1. (ii) To prove the root distribution part, note that (n−1) (n−2) sign{−dn−1 dn−2

· · · d(nn−−kk) },

(n−1) (n−2)

(0)

i(D) = sign+ {dn−1 , dn−1 dn−2 , . . . , dn−1 dn−2 · · · d0 } (n−1)

(n−1) (n−2)

(n−1) (n−2)

D(1) < 0,

(ii) The root distribution part of the theorem indicates that D(z ) is Schur stable if and only if P1 < 0,

Pi > 0,

i = 2, 3, . . . , n

or equivalently

|dn | > |d0 | |d(i i) | > |d(0i) |,

i = n − 1, n − 2, . . . , 1.

Hence, as noted in Jury (1964), the stability part is a special case of the root distribution part in the Jury test. (iii) (Singular cases) While computing the elements of the stability table (6), we may encounter a row with its first element equal to zero. The Jury test is not directly applicable to such singular cases, and some modifications are required (Jury, 1964). For example, consider the (n − k + 1)th row whose elements (k) dk−i , i = 0, 1, . . . , k, are obtained from the coefficients of the (k)

k = 1 , 2 , . . . , n.

(n−1) (n−2)

(−1)n D(−1) > 0 for dn > 0 (−1)n D(−1) < 0 for dn < 0.

polynomial Dk (z ). If dk

Then (9) and (10) are respectively equivalent to (n−1)

Remarks. (i) According to the sign of dn , the condition (7) can be written as (Jury, 1964) D(1) > 0,

(i) (Stability) D(z ) is Schur stable if and only if

sign{Pk } =

(0)

If dn−1 < 0, then i(Dn−1 ) = o(Dn ) − 1, o(Dn−1 ) = i(Dn ) by Lemma 4, and

P1 = |d0 | − |d(nn) |

(1)

(n−1) (n−2)

= sign− {dn(n−−22) , dn(n−−22) dn(n−−33) , . . . , dn(n−−22) dn(n−−33) · · · d(00) } + 1 = o(Dn−1 ) = o(Dn ) = o(D).

and let

.. .

(n−1) (n−2)

sign− {dn−1 , dn−1 dn−2 , . . . , dn−1 dn−2 · · · d0 }

1

(n−1)

(0)

and

.. .

  d(2) d(2)   2 (1) i  , d1−i =  (2) (2)  d d2−i  0   d(1) d(0)   1 (0) 01  d0 =  (0) (1)  d d 

(n−1) (n−2)

= sign+ {dn(n−−22) , dn(n−−22) dn(n−−33) , . . . , dn(n−−22) dn(n−−33) · · · d(00) } + 1 = i(Dn−1 ) + 1 = i(Dn ) = i(D)

n−1−i

0

(n−1) (n−2)

sign+ {dn−1 , dn−1 dn−2 , . . . , dn−1 dn−2 · · · d0 }

  d(n) d(n)   n i  =  (n) , i = 0, 1, 2, . . . , n − 1 (n)  d dn−i  0   d(n−1) d(n−1)   n −1  i =  (n−1) , i = 0, 1, . . . , n − 2 (n−1)  d d

(0)

o(D) = sign− {dn−1 , dn−1 dn−2 , . . . , dn−1 dn−2 · · · d0 }.

(11) (12)

is zero but remaining elements are (k+1)

not all zero, then the elements dk+1−i , i = 0, 1, . . . , k + 1, of the (n − k)th row are replaced by the coefficients of a new (k + 1)th-order polynomial having the same number of roots inside and on the unit circle as Dk+1 (z ). Then the same multiplication rule is used for computing remaining elements (k) of the stability table. Similarly, if dk−i = 0 for all 0 ≤ i ≤ k,

252

Y. Choo / Automatica 47 (2011) 249–252

the elements of the above row are replaced by the coefficients of a new (k + 1)th-order polynomial having the same number of roots inside the unit circle as Dk+1 (z ). For details and other methods, readers are referred to (Jury, 1964). 4. Conclusions In this paper we presented an elementary proof of the Jury test. Contrary to that of Keel and Bhattacharyya (1999), the proof was carried out directly for the original form of the Jury test. Although the stability part of the Jury test may be considered as a special case of the root distribution part, its proof was also given independently since it is meaningful in its own right. It is believed that the work of this paper may help control engineers to clearly understand the Jury test.

Acknowledgement This research was supported by the 2010 Hongik University Research Fund. References Chapellat, H., Mansour, M., & Bhattacharyya, S. P. (1990). Elementary proofs of some classical stability criteria. IEEE Transactions on Education, 33, 232–239. Jury, E. I. (1964). Theory and application of the z-transform method. New York, NY: Wiley. Keel, L. H., & Bhattacharyya, S. P. (1999). A new proof of the Jury test. Automatica, 35, 251–258. Mori, T., & Kokame, H. (2001). Single-parameter characterizations of Schur stability. IEICE Transactions on Fundamentals of Electronics, Communications and Computer Sciences, E84-A, 2061–2064. Raible, R. H. (1974). A simplification of Jury’s tabular form. IEEE Transactions on Automatic Control, 19, 248–250.