CIRCUITSSYSTEMSSIGNALPROCESS VOL. 10, NO. 1, 1991
AN ELEMENTARY PROOF OF THE ROUTH-HURWlTZ STABILITY CRITERION* J. J. Anagnost 1 and C. A. Desoer ~
Abstract. This paper presents an elementary proof of the well-known Routh-Hurwitz stability criterion. The novelty of the proof is that it requires only elementary geometric considerations in the complex plane. This feature makes it useful for use in undergraduate control system courses. I. Introduction The determination of stability of lumped parameter, linear, time-invariant systems is one of the most fundamental problems in system theory. According to Gantmacher [2, pp. 172-173] this problem was first solved in essence by Hermite [3] in 1856, but remained unknown. In 1875 Routh also obtained conditions for stability of such systems [7]. In 1895 Hurwitz [4], unaware of Routh's work, gave another solution based on Hermite's paper. The determinantal inequalities obtained by Hurwitz are known today as the RouthHurwitz conditions, taught in virtually every undergraduate course on control theory. Unfortunately, Hurwitz's proof of the result is very complicated, involving algebraic manipulations. Indeed, the proof is so complicated that most elementary textbooks (for example, [1] and [5]) choose not to prove it at all, but rather to state it as a fact. In a recent paper, Mansour [6] proves the Routh-Hurwitz theorem in a very simple matter using the Hermite-Biehler theorem. Motivated by Mansour's proof, this paper presents a proof based on elementary geometric considerations in the complex plane. I t thus provides a clear 9eornetric insight into what makes the procedure work. It also slightly extends Mansour's work * ReceivedJuly 10, 1989;revisedDecember21, 1989.Researchsupported by HughesAircraft Company, E1 Segundo, California 90245, USA, and the National Science Foundation under Grant ECS 21818. x Electronics Research Laboratory, Department of Electrical Engineering and Computer Science, Universityof California, Berkeley,California 94720, USA.
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ANAGNOSTAND DESOER
by providing a proof of the second part of the Routh-Hurwitz criterion: the number of sign changes in the first column of the Routh table is the number of open right half-plane zeros. The idea behind the proof of the theorem is simple. It will be shown that at each step the Routh procedure (i) eliminates precisely one zero of the characteristic polynomial, (ii) preserves the position of the jco-axis zeros, and (iii) ensures that the remaining offjco-axis zeros do not cross the jco-axis. By observing the sign changes in the first column of the Routh table, it can be determined whether the eliminated zero is a zero in the open right half-plane or the open left half-plane. Thus, in n steps, precisely n zeros have been eliminated and the sign changes indicate the number of right half-plane zeros of the original polynomial. 2. Statement of the Routh-Hurwitz stability criterion Theorem 2.1 (Routh-Hurwitz). Consider an nth-order p o l y n o m i a l in s: p(s) = a o + a l s + a2 s2 + ... + a , _ is n- i + a n S , ' where a i ~ R, i = O, 1 . . . . . n and an > O, and a o ~ O.
(If a0 = 0, simply factor out an appropriate S k term and proceed.) If possible (i.e. none of the divisors are zero), construct the well-known Routh table, written in the form as shown in Table 1. We have used the notation bn_ 2 = an_ 2 -
an --an_ an- 1
3
Cn_ 4 ~ a n _ 3 - - a n - I b n _ 4 bn- 2
an bn_ 4 = an_ 4 -- --an_ an- i
5
an- 1 bn_6 Cn_ 6 -~- a n _ 5 - - b n _ 2
Then p(s) is Hurwitz (i.e., p(s) has all its zeros in the open left half-plane) if and only if each element of the first column is positive, i.e., an > 0, a,_ 1 > 0, bn-2 > 0, ..., mo > 0, no > 0. Remark on Notation. In Table 1 we have assumed (to fix notation) that n is even. The even polynomial p(s) was split into its even and odd part by p(s) = hn(s2) + sgnZ2(s2), where hn(s2) is even and of degree n, and where #,_ 2(s2) is even and of degree n - 2. Note that the coefficients of h,(s 2) are contained in the first row of the Routh table, and the coefficients of On- 2(s2) are contained in the second row of the Routh table. This explains the presence of the hn(s2) and #n- 2(s2) in the column to the left of the Routh table in Table
PROOF OF THEROUTH-HURwITZSTABILITYCRITERION
103
TabLe 1. The Routh table. hn
an
an - 2
an- 4
"""
a4
a2
all-- 2
an - x
an - 3
an- 5
"""
a3
al
bo
h._ 2
bn - 2
bn - 4
bn_ 6
""
b2
an - 4
Cn - 4
Cn - 6
Cn - 8
"""
CO
h. _.
dn - 4
dn - 6
d._ 8
"'"
do
s 12
ko lo
g2 h2 g0 ho
mo no
g-2
0
ao
1. (If we had assumed n was odd there would be a 9 . - 1 ( s 2 ) to the left of the first row, and a h,_ ~(s2) to the left of the second row.) The remainder of the notation in Table 1 is explained in Section 4. 3. Preliminary l e m m a s
We first start with a definition which makes precise the notion of net phase change. Let C denote the complex plane. Definition 3.1. Consider a polynomial p(s) and a continuous, oriented curve C c C which starts at s~ ~ C and ends at s2 ~ C. Suppose p(s) # 0 for all s ~ C. Let the curve be parametrized by the continuous function q~: [0, 1]--. C. Since p(s) # 0 for all s ~ C this means that arg(p(s)) along C is well defined mod 2n; hence, we choose arg(p(~o(0))) arbitrarily and, for all r e [0, 1], we choose arg(p(q~(r))) such that r ~ arg(p(tp(r))) is continuous. Then we define the function argnet(p(-)) ,= arg(p(~o(1))) - arg(p(tp(0))) C
= arg(p(s2) ) -- arg(p(sO). Roughly speaking, argnetc(p(. )) is simply the net phase change ofp(s) as s traverses C. For example, in Figure 2, if the plotted solid locus is p(C), then argnetc(p(-)) = 2re; the small circles in Figure 2 indicate the zeros of p(. ). The following lemma gives a relationship between the location of zeros of a polynomial and its net phase change. 3.2. Consider the polynomial p(s) = a o + a l s + a2 $2 + " . --}a._ i s " - 1 + a.s ~ where a i ~ R, i = O, 1. . . . . n and a. > O, and a o # 0 (so p(s) is o f degree n and p(O) # 0). Then p(s) has L zeros in the open left-half plane Lemma
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ANAGNOST AND DESOER
Imag
C
s - plane
Real
Figure 1. Graph of curve C. counting multiplicities, R zeros in open right half-plane counting multiplicities, and 2 K zeros jcoi, e)~ > O, on the jo)-axis with multiplicities mi, i = 1. . . . . K (i.e., there are a total of M joe-axis zeros), if and only if
(i) ptk)(jo)i) ( == (dk/dsk)p(s)l~=jo, i) = Ofor k = O, . . . , m i_ 1, i = 1 , . . . , K, but p(m')(jogi) v~ O, i = 1 , . . . , K, and p(ja~) vL O for all co e ~+\{coi: i = 1. . . . ,K}; (ii) a r g n e t c ( p ( . ) ) = (rc/2)(L- R + M ) where the oriented curve C is the jco-axis, except for indentations on the right at each jco-axis zero jco~, i.e., the curve C starts at zero and ends at +joo, as shown in Figure 1. Proof. ( =~ ) Since p(s) ?.s an nth degree polynomial, it has precisely n zeros. By assumption, precisely M are on the jo~-axis, while the remaining zeros lie in the open right half-plane, or open left half-plane. In addition, since each zero jcoi has multiplicity m i, this implies that p(k)(jooi) = 0 for k = 0, ..., m i -- 1 and ptm')(jogi) r 0. Thus, (i) is proved. To prove (ii), note that each simple real open right half-plane zero contributes -re/2 radians of phase to the net argument as s traverses C, while each simple real open left half-plane zero contributes re/2 radians of phase. Due to indentations on the right of the ira-axis zeros, each complex conjugate zero pair contributes e i t h e r + T z or - zc radians of phase depending on whether the pair resides in the closed left half-plane or open right half-plane, respectively. Since there are a total of L zeros in the open left-half plane counting multiplicities, R zeros in open right half-plane counting multiplicities, and M zeros on the ja~ axis counting multiplicities this means argnet(p(-)) = 2 (L - R + M). c
This proves (ii).
PROOF OF THE ROUTH HURWITZSTABILITYCRITERION
105
( ~ ) By assumption p(s) has precisely M / 2 pairs ofjco-axis zeros counting multiplicities, so it can be factored as K n-M p(s) = 1-I ( s2 + c~ m' 1--[ (s -- sz,), i=1
i=1
where {Szi: i = 1,..., n - M} denotes the remaining zeros of p(s). Since the curve C is indented to the right at the jco-axis zeros, we can define argnetc(P ( 9)). By computation its value is argnet(p(. )) = ~rr M + argnet ("IzIM (s _ szi)) C
\i=l
By condition (ii), the second term is equal to n(L-R)/2. Also, n - M = L + R, and M is known by (i); so conditions (i) and (ii) determine uniquely L, R, and M. [] The following two lemmas are the main results of the section. They characterize the effect of one step of the Routh-Hurwitz procedure when the leading term is even (Lemma 3.3), and when the leading term is odd (Lemma 3.4). Lemma
3.3. Consider the polynomial p(s) = a o d- als d- a2 s2 -F ... dan_ is n- 1 + ans n where ai ~ R, i = O, 1. . . . . n and an > O, and a o ~ O. Assume in addition that n is even and an-1 ~ O. Let hn(s2) and sg n_ 2(S2) be the even and odd parts of p(s), respectively, i.e., hn(s 2) : = ao + aEs2 + ... + an - 2sn - 2 + anSn, gn- 2(S2):=al + a3 s2 -]- ... + an_3 sn-4 -]- an_is n-2.
Suppose that p(s) has L zeros in the open left half-plane counting multiplicities, M jco-axis zeros counting multiplicities, and R ( = n - L - M ) zeros in the open right half-plane counting multiplicities. For any real 2, define N(S, ,~) := p(s) nt- ,~S2gn_ 2(S 2) = hn(s 2) + 2S2gn_2(S 2) + Sgn_2(S2).
Then:
(i) ja) i is a je)-axis zero of p(s) with multiplicity ml if and only if jo9 i is a jo)-axis zero of N(s, 2) with multiplicity m i f o r all 2 ~ •. (ii) Given any closed, bounded interval I c R, there exists a curve C as in Figure 1 such that N(s, 2) ~ 0 for all s e C, and f o r all 2 e I. Thus argnetc(N(., 2)) is well defined for 2 e I.
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ANAGNOSTAND DESOER
Define the interval I by I ,= [ - l a./a._ 11, [a./a._ 1 I]. Choose the curve C so that a r g n e t c ( N ( -, 2)) is well defined for 2 e I. (This can be done by part (ii).) Then: (iii) a r g n e t c ( N ( . , 2)) - a r g n e t c ( P ( . )) [ <_ n, for all ,~ ~ I. (iv) {argnetc(p(. )) - a r g n e t c ( N ( ., - a./a._ 0)} sign(a./a._ 1) = n/2. (v) N ( s, - a./ a._ a) has ( L - 1) zeros in the open left half-plane, M zeros on
the jco-axis, and R zeros in the open right half-plane, in each ease counting multiplicities if and only if a./a._ ~ > O. In addition, N(s, - a,/a._ ~) has L zeros in the open left half-plane, M zeros on the jco-axis, and R - 1 zeros in the open right half-plane, in each case counting multiplicities if and only if a./a._ ~ < O. Proofi (i) ( ~ ) T a k e 20 E ~, a r b i t r a r y , jo9, is a joocaxis z e r o of N(s, 20) with m u l t i p l i c i t y m i m e a n s t h a t (dk/dsk)N(s, 20)[~=jo,~ = 0 for k = 0, . . . , mi - 1 a n d (d"~/dsm')N( s, ),0)[~=jo, =~ 0. (dk/dsk)N(s,),o)l~=jo, = 0 for k = 0 , . . . , m i - 1 is e q u i v a l e n t to dk h~k)(--c~
+
+ 2~
{S2gn- 2(S2)} ) s=jtoi
sg._2(s 2)
=0
for
k=O ..... m i-1.
Since coi is real, e q u a t i n g the i m a g i n a r y a n d real p o r t i o n s of this e x p r e s s i o n to z e r o yields h.(k)(--co i2) + 20((dk/dsk)(sEgn_ 2(S2)})ls=j~o, ---- 0 a n d
( Td~sk {sg,_ =(s 2) }) ~=~, =
0
for
k = 0 . . . . , mi - 1.
T h i s l a t t e r e x p r e s s i o n implies t h a t ~,"{k)_2~t_ ~,~"2sJ= 0; u s i n g this in the f o r m e r e x p r e s s i o n s h o w s t h a t h ~ ) ( - c o 2) = 0, k = 0 , . . . , m i - 1. T h u s
p(k)(jo)i) = h~)(--a)~) +
{sg,_ a(s 2)
= O,
k = 0 . . . . , m i - 1.
~=jr i F r o m (d"Tdsm')N(s, 2o)[,=j~ , ~ 0 a n d ~0,-2~-~"(*) ~ ,.,2~ = 0, k = 0, ... , mi - 1, we f u r t h e r c o n c l u d e t h a t (d"/ds=')p(s)[,=j~,, g= O. H e n c e , ja), is a ja)-axis z e r o of p(s) with m u l t i p l i c i t y m,. ( ~ ) jo9 i is a jo)-axis z e r o of p(s) w i t h m u l t i p l i c i t y m~ m e a n s t h a t p(k)(jr = h.(k)(--09 i2) + ((dk/dsk){sg._ 2(S2)})I~=j~, = 0, k = 0 . . . . , mi - 1. E q u a t i n g the real a n d i m a g i n a r y p a r t s of p(k)(jooi) = 0 yields h ~ ) ( - m ~ ) = 0 a n d v."(k)-ztr--,~i~"2~J= 0 for k = 0, . . . , m~ - 1. This in t u r n implies t h a t 20((dk/dsk){S2gn_2(S2)})ls=jo , = 0 for a n y ) , o e ~ a n d for k = 0 . . . . , m i - 1 . Thus,
dk ~N(s, 2o)l~=j~,= p~(/aJ,) + 2o( ~dk {s~g.-=(s~)}) +
( ~d~sk {sg._=(s2) }) ~j~, =
0
s~ jta,i for
k = 0,...,mi
-- 1.
PROOF OF THE ROUTH-HuRWITZ STABILITYCRITERION
107
It can be shown by similar reasoning that (dm'/dsm')N(s,2o)ts=jo,, # 0. This proves (i). (ii) This statement merely asserts the existence of a curve C which ensures that argnetc(N(., 2)) is well defined for all 2 in the closed, bounded interval I. Since the details are not relevant to the rest of the proof, they are left to the Appendix. (iii) For simplicity, first assume that p(s) has no j09-axis zeros. For this case we take curve C to be the positive j09-axis. Note that N(j09, 2) ~ 0 on •xI. Since sZg,_ z(s 2) is an even function of s, -20920,_ 2(-092) only contributes to the real part of N(j09, 2). In particular, points a and b in Figure 2 are fixed points: indeed, if 091 satisfies P(J091) = b, then 9,-2(-09~) = 0, which means that N(j091, 2) = b for all 2 e I. Next, order the zeros of 092g n_ 2(--092) by 0 = 09o < 091 ~ 092 < ... _< 09k < ~ . First consider 09 ~ (0, 091)- Since there are no zeros of 099,_2(-09 2) in this interval, this implies that sign{Im(N(j09, 2))} is a constant on (0, 091) x L By the choice of curve C, N(j09, 2 ) # 0 for all 09 e ~+ and for all 2 ~ I, so argnetto ' i~j(N(., 2)) is well defined. In particular, argnetto ' j~,,1(N(., 2)) is well defined which implies by definition that argnetto" j,o,j(N(., 2)) = arg(N(j09 1, 2)) - arg(N(j0, 2)). But arg(N(j09 1, 2)) = arg(p(j091) ) and arg(N(j0, 2)) = arg(p(j0)), since 0 and 091 correspond to the fixed points a and b in Figure 2. Thus, the net argument change between zero and 091 is independent of 2. The same reasoning applies for 09 ~ (091,092), for (092, 09a), etc., up to (09k- 1,09k)" Hence, argnet(N(., 2)) = argnet(p(. )). [0, j~k] [0, jo~k] Thus, the only difference in phase occurs for (09k, ~ ) . Since there are no zeros of cog._ 2(-092) in this interval, this again implies that sign{Im(N(j09, 2))} is a constant (see Figure 2). This in turn implies that largnett/,o~,j~ol(N(., 2 ) ) -
NOr ~.)~~. ~ 0
I Imag ~) = N(jco, 0)
:/ !:.
Figure 2. Graph of co ~
N(jco, 2), 0 <
f.0 =0
co < ~ (n even).
][08
ANAGNOST AND DESOER
argnettj~,~,j~l(p(-))l
2eI.
Since
argnetEo, io,~l(N(. ,
2))=
l argnet(N(., 2)) - argnet(p(. ))[ < rc [ 0 , joo]
[ 0 , joo]
for all 2 e I. This proves (iii) for the case where p(s) has no j~o-axis zeros. (iv) N o t e that by the definition of I that - a,/a,_ 1 ~ I. O r d e r the zeros of ~oO,- 2 ( - c02) as before, and use arguments identical to that of (iii) to obtain argnet(p(. )) - argnet(N(., - a , / a , _ 1)) [o, joo]
[o, joo]
= argnet (p(.)) - argnet (N(., - a,/a n_ 1))[jtok, joo]
[.]ok, jc~]
Since cok is a fixed point (i.e., independent of 2), we then obtain = arg (p(j~o)) - arg (N(jco, - a , / a , _ 1)). oj---~ oo
r
r
Since we are taking the limit as c~ ---} ~ , we only need to consider the leading term of each polynomial. Performing this operation, and using the properties of arg, we obtain in succession = arg (a,(jrn)") r ---}oD
arg (a,_ 1(jog)"- l) go--* oo
= arg [a,(jog)"/a._ l(jo)) n- 1] = arg [a,j~o/a,_ 1]. a~ ---~oo
Thus, if a,/a,_ 1 > 0, the net a r g u m e n t difference is re/2, and if a,/a._ 1 < O, then the net a r g u m e n t difference is -re/2. This proves (iv) for the case where p(s) has no jog-axis zeros. If p(s) has jo>axis zeros, then (i) shows that N(s, 2) has the same j~o-axis zeros with the same multiplicities. This means that the only difference in a r g u m e n t can come from the non-j~o-axis zeros. If we extract thej~o-axis zeros by Pl(S) = p(s)/I-I~/2 (S 2 "~ f-D2), then pl(s) has no jco-axis zeros, so we can apply the arguments above. F o r example, to prove (iii) we k n o w from above that [argnet(Ni(., 2)) -- argnet(pl(-))1 < re, [0, joo]
[0, j ~ ]
where Ni(s, 2) .-= N(s, 2 ) / F I ~ / 2 (s 2 + ~02) for all 2 e I. If we choose a c o n t o u r C like Figure 1, indented on the right of the jco-axis zeros, we then obtain [argnet(N(., 2)) - argnet(p(. ))1 _< rc C
C
for all 2 e 1, which proves (iii). Statement (iv) is proved similarly.
PROOFOFTHEROUTH-HURWlTZSTABILITYCRITERION
109
(v) The net argument difference between p(s) and N(s, - a n / a . _ l ) as s traverses C is sign(a./an_ 1)z~/2, by applying (iv) above. Applying both logical implications of Lemma 3.2 shows that N(s, - a , / a . _ 1) and p(s) have the same number of zeros on the jog-axis, and a difference of at most one in the number of open right half-plane or open left half-plane zeros, depending on the sign of a,/a,_ 1. This proves (v). [] In the case that n is odd (rather than even as in the statement of Lemma 3.3), we have the corresponding results to Lemma 3.3 Lemma 3.4. Consider the polynomial p(s) = a o -4- als + a2s 2 + ... + a,_ Is"- 1 + a,s" where ag ~ g~, i = O, i , . . . , n and a, > O, and ao ~ O. Assume n is odd. Let h, _ l(s z) and s9. _ l(s z) be the even and odd parts of p(s), respectively, i.e., gn_l(S2) := a I --[-a3 s2 -k ... q- an_ 2Sn- 3 -[- ansn-l, h,_ l(s 2) .'= a 0 + azs 2 + + a,_ 3s"- 3 + a. _ is"- 1. Assume a,_ 1 ~ O. Suppose that p(s) has L zeros in the open left half-plane counting multiplicities, M fio-axis zeros counting multiplicities, and R ( = n - L - M ) zeros in the open right half-plane counting multiplicities. For any real 2, define No(s, 2) .'= p(s) + 2sh,_ l(s 2) = s o , _ l ( s 2) + 2sh._l(s 2) + h,_l(s2). Then, statements (i)-(v) of Lemma 3.3 hold, with No(s, 2) replacing N(s, 2).
Proof. The proofs of (i) and (ii) are identical to the analogous results of Lemma 3.3, and are thus omitted. The proofs of (iii)-(v) are also nearly identical to their counterparts in Lemma 3.3. The key point is to note that jo~2h,_~(-co 2) contributes only to the imaginary part of No(it0, 2 ). This means that points a and b of Figure 3 are fixed points, i.e., if co1 satisfies
N0~o,7.),7.r ~~'~-~"~'a
Imag l
Real p(jco)= N(jco,0) Figure 3. Graph of co ~ N(jo9, 2), 0 _< o9 < go (n odd).
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ANAGNOST AND DESOER
p(jogi) = a, then h . _ l ( - o ) 2) --0, which means that No(Je)l, 2 ) = b for all 2 ~ I. The details are left to the reader []
4. Proof of Theorem 2.1 (Routh-Hurwitz) Let us first emphasize some notation. To fix notation, assume n is even. Let h,_w(S 2) be the even polynomial of degree n - w whose coefficients lie in the row 2w. (See Table 1.) Let g,_ w(s2) be the even polynomial of degree n - w whose coefficients lie in the row 2w + 1. (Again see Table 1.) To construct the Routh table, perform the calculations indicated in Section 2. This corresponds to finding a 2,_w ~ R, or a/~._ ~ e R, such that
h._w(S 2) = hn_w+ 2(s 2) + ~n_wS2gn_w(S2),
(4.1)
O.- ~(s2) = 9 . - ~ + 2(s2) + /~.- wh,- ~ +2(s2),
(4.2)
where the leading terms of h._ w(S2) and 0n- w(s2), respectively, are canceled. If this procedure cannot be performed (i.e., the leading term of 9._w(s 2) or h._ w(s2) is zero), then a zero is in the first column of the Routh table. The standard procedure given in elementary textbooks is to replace the zero by e > 0, and proceed. See Section 5 for some of the implications of this.
Proof of Theorem 2.1 (Routh-Hurwitz). ( =~ ) If p(s) is Hurwitz, then each of its zeros is in the open half-plane. Consider the first step of the Routh procedure. By Lemma 3.3(v), N(s, - a,/a n_ l) = h.(s2) - a,/a,_ ls2g,_ 2(s 2) + sg._2(s 2) has the same number of zeros in the open left half-plane as p(s) except for the eliminated zero. Since all the zeros of p(s) are in the open left half-plane, the eliminated zero must also be in the left half-plane. Thus, N(s, - a . / a . _ 1) = SOn-2(S2) + h.-2(s2) has precisely n - 1 zeros in the open left half-plane, and a./a._ 1 is positive. By using Lemma 3.4, in the next step we have that h._2(s 2) + SOn_g(S2) has precisely n - 2 zeros in the open left-half plane, and a./a._ ~ is positive. After n steps, n zeros have been eliminated and each element in the first column is positive. ( ~ ) If each element in the first column is positive, then use of Lemma 3.3(v) and Lemma 3.4(v) show that precisely n zeros in the open left halfplane have been eliminated. Thus, p(s) is Hurwitz. [] Remark 4.1. In the case that n is odd, the proof of the Routh-Hurwitz theorem is nearly identical. Simply write p(s) = sg._ l(s 2) + h,_ l(s2), where so,_ i(s 2) and h,_ i(s 2) are the odd and even parts of p(s), respectively. Apply Lemmas 3.4 and 3.3 appropriately in a manner similar to that in the proof of the even case. The details are left to the reader.
PROOF OF THE ROUTH-HURWlTZ STABILITY CRITERION
111
5. The second part of the Routh-Hurwitz theorem Based on Lemma 3.3 we have the second part of the Routh-Hurwitz criterion. Theorem 5.1. Consider an nth-order polynomial in s: p ( s ) = a o + a l s + a2 s2 + . . . + a n _ i S n - 1 + ans n,
where a~ ~ ff~, i = O, 1 , . . . , n and a, > 0 and ao ~ O. As before, to f i x notation assume that n is even. Suppose when calculatin9 the Routh table that no element in the first column is zero. 7hen the number of sign chanoes in the first column of the Routh table is the number of open right half-plane zeros of p(s).
Proof. At each step the algorithm (i) eliminates precisely one zero of p(s), (ii) preserves the position of the jr,-axis zeros, and (iii) ensures that the remaining offjco-axis zeros do not cross the j~o-axis. By Lemma 3.3(v) and Lemma 3.4(v) the eliminated zero is in the open left half-plane if the ratio of the associated leading coefficients is positive, whereas the eliminated zero is in the open right half-plane if the ratio of the associated leading coefficients is negative. Thus, the number of sign changes in the first column indicates the number of open fight-half plane zeros of p(s) that were eliminated. [] Remark 5.2. If a zero does appear in the first column during the Routh procedure, care must be exercised in ascertaining the zero positions of the original polynomial. By adding an e > 0 to a column, the position of the zeros of the original polynomial are being perturbed (since the zeros of a polynomial are continuous functions of their coefficients provided a, remains bounded away from zero). Attempting to deduce properties of the zeros of the original polynomial based on the properties of the perturbed polynomial can often lead to erroneous conclusions as the following example shows. Example. Let p(s) = (s z + a)(s z + b) = s 4 + (a + b)s 2 --l- ab, where a, b ~ R. The Routh table for this example is 1
e
a+b
ab
0
0
a+b
ab
0
~ab a+b
0
ab
0
-
0
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ANAGNOST AND DESOER
If a > 0 and b > 0, then there are two sign changes in the first column since e > 0. This leads to the "conclusion" that there are two zeros in closed right half-plane. Note that adding an e > 0 merely pushes the jo-axis zeros of the p(s) off the jo-axis. Much more insidious examples can be constructed that make it very difficult to tell the position of the zeros of the original polynomial. (See, for example Example 4, p. 184 of [2]). However, we do have the following proposition.
Proposition 5.3. Suppose that during construction of the Routh table a zero in the first column is encountered. Then (i) I f there are one or more nonzero elements in the same row, then p(s) has at least one zero in the open right half-plane. (ii) I f the row is zero, then (a) p(s) has at least one pair of jco-axis zeros, or (b) p(s) contains a factor of the form (s + go)(S - go)for some ao E ~ or (c) p(s) contains a factor of the form (s + ao + floJ) (s + c% - floJ) (s - ao + floJ) (s - ~o - floj) for some ~o, floe ~. Proof. (i) Since there is a zero in the first column in the Routh table, the Routh-Hurwitz theorem (Theorem 2.1) shows that p(s) has at least one zero in the closed right half-plane. Without loss of generality, assume that the zero is in the second element of the first column, i.e., sg,_ 2(s2) has a leading coefficient of at most order n - 3. (Since the row is nonzero by supposition, sg._ 2(s2) is at least of order 1.) Suppose that the only zeros of p(s) in the closed right half-plane are rio-axis zeros, say M counting multiplicities. Then extract the rio-axis zero pairs from p(s) by M/2
M/2
M/2
Pl(s)"=P(S)/ 1-[ (s2 + co~) = hn(s2)/I-[ (s2 + co/2) + sg~-2(sg)/1-I (s2 + c0/2)9 i=1
i=1
i=l
(Note that since sg~_2(s z) is nonzero, M < n.) By supposition, pl(s) is Hurwitz and thus has every coefficient strictly positive. However, h,(s2)/1-[~=/~2 (s 2 + co~) is of order n-M, while sg~_2(s2)/1-I~/12 (s 2 + co~) is of order at most n - M - 3 (but at least 1.) Thus, pl(s) has its n - M - 1 coefficent equal to zero, which contradicts the fact that pl(s) is Hurwitz. This proves (i). (ii) Encountering a zero row during construction of the Routh table means that at some step hn_w+2(s
2) =
--,,~n_wS2On_w(S2),
or an - w +
2($2)
=
- - Idn -
whn- w + 2(S2)"
Hence the polynomial hn_w+2(s 2) + sg~_~,(s 2) or gg,_w+2(s 2) + h~_w+2(s 2) equals ( 1 - 2n_,~s)sg~_w(S 2) or (1-#._wS)h,_w+2(s2), respectively. Since
PROOF OF THE ROUTH-HURWITZ STABILITYCRITERION
113
g,_w(s 2) and hn_w+2(s2) have real coefficients, this means that g._w(s 2) or h,_w + 2(s 2) have zeros of the type stated in the proposition. Working our way back up the Routh table, note that p(s) can be written as linear combinations ofg,_w(S 2) and hn_w+2(S2) o r gn_w+2(s2) and h,_w+ 2(s2). (Use (4.1) and (4.2) and the Routh table, Table 1.) Thus, p(s) also has the stated property. []
Acknowledgments The authors would like to thank R. J. Minnichelli and an anonymous reviewer for their meticulous reading and comments on a first version of this work.
Appendix. Proof of Lemma 3.3(ii) The idea is to choose a curve C as in Figure 1 with a sufficiently small indentation about each jco-axis zero so that (ii) holds. Take any bounded interval I c ~. From (i) of Lemma 3.3, for any 2 6 I, N(s, 2) has the same jco-axis zeros with the same multiplicities as p(s), say a total of M. Therefore, only n - M zeros of N(s, 2) depend on 2. Let {zi(2), i = 1. . . . . n - M} be such that N(zi(2), 2) = 0. At 2 = - a J a , _ 1, note that the degree of N(s, 2) drops by precisely one (since a,_ 1 r 0). Thus, precisely one member of {zi(2 ), i = 1. . . . , n - M}, say zs(2 ), goes to infinity as 2 ~ - an/a,_ 1, and it tends to infinity along the real axis, as an asymptotic expansion shows. This means that there is a closed interval I j c I with [zs(2) l < ~ for all 2 61 s, satisfying min
min [zs(A) - job I) =
IE{1 . . . . . M} 3. e I
min
rain [Zs(2) - jr
).
i~{1, ...,M}.~el.r
This latter term has an achievable positive minimum, since the locus Zs(Is) is closed and bounded, and never crosses the imaginary axis. Call the minimum distance Rj. So now consider z(I):= {zi(2), i = 1. . . . , n - M , i S J, 2 ~ I } c C, the locus of all off-imaginary axis zeros of N(s, 2)(except for zs(2)) as 2 varies over I. Since I is closed and bounded, the locus is a closed and bounded set. Therefore, mint ~ ~1. . . . . M~(Iz ( I ) - joJi]) is a finite, positive constant, say R. Let R * = min(Rj, R ) > 0. Using this value of R*, we can then make the radius of the indentation about each jo~-axis zero jab equal to R*/2, say. This allows construction of the desired contour C, which proves (ii). []
References [1] Doff, R. C., Modern Control Systems, 3rd edn., Addison-Wesley, Reading, MA, 1980. [2] Gantmacher, F. R., The Theory of Matrices, Vol. 2, Chelsea, New York, 1959.
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[3] Hermite, C., Sur le nombre des racines d'une ~quation alg6brique comprise entre des limites donn6es, d. Reine Angew. Math., 51, 39-51, 1856. I-4] Hurwitz, A., On the conditions under which an equation has only roots with negative real parts, Math. Ann., 46, 273-284, 1895. Also in Selected Papers on Mathematical Trends in Control Theory, Dover, New York, 1964, pp. 70-82. I-5] Kuo, B. C., Automatic Control Systems, 4th edn., Prentice-Hall Englewood Cliffs, NJ, 1982. [6] Mansour, M., A Simple Proof of the Routh-Hurwitz Criterion, Report #88-04, Institute of Automatic Control & Industrial Electronics, Swiss Federal Institute of Technology, December, 1988. I7] Routh, E. J., A Treatise on the Stability of a Given State of Motion, London, Macmillan, 1877.
