A Simplified Proof of Hurwitz Criterion of Stability and Generation of a Family of Stable Polynomial via Root Locus Method Prof. N. N. Puri ECE Department, Rutgers University
1
Prob Proble lem m Stat Statem emen entt
Given: An nth order characteristic polynomial: P n (s) = a0 sn + a1 sn−1 + · · · + an−1 s + an ,
ai > 0
(1)
Problem:
Find the conditions under which all the roots of polynomial (1) have real part pa rtss nega negati tiv ve. We shal shalll refe referr to (1) (1) as a Hu Hurw rwit itzz polyn polynom omia iall or a stab stable le polynomial. This is an age old problem whose answer was provided by now famous Routh-Hurwitz Criterion. Most textbooks do not present the proof of Hurwitz Criterion for they consider it difficult for the undergraduate (or even the graduate) students. In what follows, we present a very simple proof.
2
Stat Statem emen entt of the the Routh Routh-H -Hur urwi witz tz Crit Criter erio ion n
Let all the coefficients of (1) be positive (a ( ai > 0, i = 1, 1 , · · · , n). From these coefficients form an n × n Hurwitz matrix:
H n =
a1 a0 0 0 0 0 0 0 0
a3 a2 a1 a0 0 0 0 0 0
a5 a4 a3 a2 a1 a0 0 0 0
1
a7 a6 a5 a4 a3 a2 a1 a0 0
··· ··· ··· ··· ··· ··· ··· ··· ···
0 0 0 0 0 0 0 0 0
(2)
Let all the diagonal minors (determinants) of H n be: ∆1 = a1 ; ∆2 =
3
a1 a3 a0 a2
a1 a3 a5 ; ∆3 = a0 a2 a4 ; · · · 0 a1 a3
Hurwitz Theorem
(3)
In order that all the roots of the characteristic polynomial P (s) = a0 sn + a1 sn−1 + · · · + an−1 s + an with all the coefficients ai > 0 have the real part negative, it is necessary and sufficient that: All the even determinants: ∆ 2 , ∆4 , · · · be positive
(4)
All the odd determinants: ∆ 1 , ∆3 , · · · be positive
(5)
OR
Note: It can be easily shown that when all ai ’s are positive and all odd indices determinants, ∆1 , ∆3 , · · · are positive, then all even indices determinants ∆ 2 , ∆4 , · · · are automatically positive and vice versa.
4
Preliminary Results Before Proof of the Hurwitz Theorem
In order to prove the Hurwitz Theorem we shall make use of the A.V. Michialov’s Theorem.
Michialov Theorem Let n
n−1
P n (s) = a0 s + a1 s
n−2
+ a2 s
n
+ · · · + an =
(s − z i )
i=1
or n
P n (s)|s= jω = P n ( jω) =
n
|( jω − z i )|
i=1
i=1
φz (ω) = |P n ( jω)| φ p (ω) i
(6)
Where z i are the roots of P n (s). ∞ ∆φ p (ω)|ω= ω=−∞ = ∞ ∆φz (ω)|ω= ω=−∞ = i
change in the argument φ p (ω) as ω varies from − ∞ to ∞.
+π if z i lies in L.H.S. −π if z i lies in R.H.S.
If all the roots z i of P n (s) lie in L.H.S., then ∞ ∆φ p (ω)|ω= ω=−∞ = nπ
(n being the order of P n (s))
This leads us to formulate the Michialov Criterion of Stability: 2
(7)
Figure 1: Figure 1
For a stable P n (s), it is necessary and sufficient that the argument φ p (ω) of the vector P n ( jω) increases monotonically from 0 to nπ , as ω increases monotonically from −∞ to +∞. Simply stated, the locus of P n ( jω) of a stable polynomial P n (s) traverses n quadrants in the P n ( jω) plane as ω varies from 0 to ∞ and then goes off to infinity in the nth quadrant. Figure 2 shows that the locus of a 7th order stable polynomial. Note that it starts at ω1 (= 0) with a positive value and travels through 7 quadrants successively and then goes off to ∞. Also note that mirror image of the locus about real axis occurs for negative values of ω. ω1 = 0 < ω 2 < ω3 < ω 4 < ω5 < ω6 < ω 7 < ∞
3
(8)
Figure 2: Figure 2 Thus Rn (ω) = a0 (ω 2 − ω22 )(ω 2 − ω42 )(ω 2 − ω62 ) X n (ω) = a1 ω(ω 2 − ω32 )(ω2 − ω52 )(ω 2 − ω72 )
0, ±ω3 , ±ω5 , ±ω7 , · · · , ±ω2 , ±ω4 , ±ω6 , · · · ,
(9)
represent zeros of X n (ω) (even function of ω) represent zeros of Rn (ω) (odd function of ω)
Note that both Rn (ω) and X n (ω) have simple, real roots. Thus, for an nth order polynomial (n even for simplicity). n/2
Rn (ω) = a0
2 (ω 2 − ω2i ) = (a0 ω n − a2 ω n−2 + a4 ωn−4 − · · · + an )
i=1 n/2
X n (ω) = a1 ω
2 (ω 2 − ω2i+1 ) = −ω(a1 ω n−2 − a3 ω n−4 + · · · − an−1 )
j=1
P n ( jω) = Rn (ω) + jX n (ω) 0 < ω2 < ω3 < · · · < ω n
(10)
Due to the alternating property of roots of Rn (ω) and X n (ω), when Rn (ω) is decreasing, X n (ω) is increasing, and vice versa. Thus let 4
Figure 3: Figure 3
d −X n (±ω2i ) n = 0, where (±ω ) = (ω) = 1, 2, A > R R , i · · · , 2i 2i n n dω 2 Rn (±ω2i ) ω=±ω2
(11)
i
We are now in a position to utilize the results of this section to prove Hurwitz Theorem as presented in section 3 by (4) or (5).
5
Proof of Hurwitz Theorem
Let us consider f (ω), an odd function of ω. From (10) and (11), using partial fractions, −X n (ω) = f (ω) = Rn (ω) a0 n/2
=
A2i
i=1
n/2
−X n (ω) n/2 2 i=1 (ω
2 ) − ω2i
1 1 + ω − ω2i ω + ω2i
=
i=1
−X n (ω2i ) −X n (−2ω2i ) + Rn (ω2i (ω − ω2i )) Rn (−2ω2i (ω + ω2i ))
(12)
Now consider the function f (z ) which is an odd function of z and expand it in powers of z −1 : n/2
f (z ) =
A2i
i=1
n−1 1 1 c1 c3 c2k+1 + = + 3 + ··· = 2k+1 (z − ω2i ) (z + ω2i ) z z i=0 z
The coefficients c2k+1 are completed via contour integration of (13), yielding c2k+1
1 = 2πj
c
f (z )z 2k dz,
where contour c includes all poles of f (z ). 5
(13)
From Residue theorem n/2
c2k+1 =
2k
2k
A2i (ω2i ) + (−ω2i )
i=1
> 0,
k = 0, 1, · · · , n − 1
(14)
This contains all the necessary and sufficient information about the stability of P (s). Genius of both Hurwitz and Routh lies in translating these conditions to the parameters, a0 , a1 , · · ·, an . In fact the coefficients c2k+1 (k = 0, 1, · · · , n/2) from a positive definite matrix which is related to matrix H n . To extract the properties of c2k+1 (k = 0, 1, · · · , n/2), let us consider a function θ2 (z ), and f (z ): 2
n−1
θ (z ) = (x0 + zx 1 + · · · + z
2
xn−1 ) =
n−1 n−1
x p xq z p+q
(15)
p=0 q=0 n−1
c2k+1 = f (z ) = 2k+1 z k=0
n/2
A2i
i=1
1 1 + z − ωi z + ωi
(16)
Let us evaluate a scalar function V , 1 V = 2πj
c
f (z )θ2 (z )dz,
c containing all the poles of f (z )
(17)
From (15) and (17) 1 V = 2πj
c
n−1 n−1
f (z )c2k+1
p=0 q=0
where c p,q =
p+q
z z 2k−1
n−1 n−1
dz =
(18)
c p,q x p xq
p=0 q=0
0 when p + q = even, c p+q+1 when p + q = odd
Also 1 V = 2πj
n/2
c i=1
A2i
n/2 1 1 2 + θ (z )dz = A2i θ2 (ωi ) + θ2 (−ωi ) z − ωi z + ωi i=1
(19)
Equation (19) implies that V is a positive definite function, which implies that matrix C associated with V in (18) is a positive definite matrix given by
T
V = x C x,
C =
c1 c3 c3 .. . cn−1
.. .
c3 · · · cn−1 ··· c5 · · · .. .. .. . . . · · · c2n−1
a positive definite matrix
(20)
Let us relate this matrix C to coefficients a0 , a1 , · · · , an : From (10), (12) and (13): −X n (ω) c1 c3 c2n−1 a1 ω n−1 − a3 ω n−3 + a5 ω n−5 − · · · = + 3 + · · · + n−1 = f (ω) = Rn (ω) ω ω ω a0 ω n − a2 ω n−2 + a4 ωn−4 − · · · + an 6
(21)
Cross multiplication and collecting terms of similar powers of ω, c1 a0 = a1 c3 a0 − c1 a2 = −a3 c5 a0 − c3 a2 + c1 a4 = a5 .. .. . . cn−1 a0 − cn−3 a2 + · · · − c1 an−2 = −an−1 .. .. . . c2m−1 a0 − c2m−3 a2 + c2m−5 a4 − · · · + (−1)m a2m = 0,
(22)
m=
n + 1, · · · 2
an+1 = an+2 = · · · = a2n = 0 all c2k+1 > 0 Equation (22) can be written in matrix form: a1 a0 0 0 .. .
a3 a2 a1 a0 .. .
a5 a4 a3 a2 .. .
a7 a6 a5 a4 .. .
··· ··· ··· ··· .. .
0 0 0 0 .. .
· · · an · · · an−2 · · · an−4 = · · · an−6 .. .. . . 0 0 0 0 0 an 0 0 a0 (23) Since the determinant of product of matrices is equal to the product of individual determinant:
c1 −c3 c5 −c7 1 0 0 0 0 c1 −c3 c5 0 1 0 0 .. .. .. .. . . . . 1 0 0 0
a1 a0 0 0 .. .
a3 a2 a1 a0 .. .
a5 a4 a3 a2 .. .
a7 a6 a5 a4 .. .
··· ··· ··· ··· .. .
0 0 0 0 .. .
0
0
0
0
0
an
= an0
c1 0 c3 .. . .. . c2n−1
· · · c2n−1 0 ··· 0 ··· 0 ··· .. .. . . 0 0
a0 a2 a4 0 a0 a2 0 0 a0 0 0 0 .. .. .. . . . 0 0 0
0 c3 0 · · · c2n−1 c3 0 c5 · · · c2n−3 0 c5 0 · · · c2n−5 .. .. .. .. .. . . . . . .. .. .. .. .. . . . . . 0 0 0 0 0
a6 a4 a2 a0 .. .
(24)
Since the matrix C is positive definite, the determinants of H n and all its diagonal minors are positive. This completes the proof of the Hurwitz Criterion.
7
6
Generation of a Family of Stable Polynomials via Root Locus Method (Order Reduction Algorithm)
Consider an nth order stable polynomial: P n (s) = a0 sn + a1 sn−1 + · · · + an Starting with this polynomial, we shall generate a family of stable polynomial P i (s) (i = n − 1, n − 2, · · · , 1), such that: P i (s) = a0 (i)si + a1 (i)si−1 + · · · + ai (i) = Ri (s) + X i (s) ak (i)|i=n = ak , k = 0, 1, · · · , n Ri (s) = Even degree part of P i (s) X i (s) = Odd degree part of P i (s)
(25)
Let P i (s) be the starting polynomial. We shall generate P i−1 (s) as following: Case 1, i = even integer: Let P i (s) =
a0 (i)si + a2 (i)si−2 + · · · + ai (i)
i−2
+s a1 (i)s
i−4
+ a3 (i)s
+ · · · + ai−1 (i)
= Ri (s) + X i (s) i
2
=
i
+1
a2k (i)si−2k +
k=0
2
a2k−1 (i)si−2k+1
k=1
where Ri (s) is even degree and X i (s) is odd degree. Consider the Root Locus of the following feedback system with variable ki , where ki ≤ 0: + -
-
ki Gi (s) =
ski X i (s) Ri (s)+X i (s)
=
ski X i (s) P i (s)
-
ki ≤ 0
− 6
Then the closed loop characteristic polynomial is: P i (ki , s) = P i (s) + ki sX i (s)
(26)
and Root Locus Equation is: P i (ki , s) = P i (s) + ki sX i (s) = 0 or
1 sX i (s) = − = Gi (s) P i (s) ki 8
(27)
The root locus starts (ki = 0) at the poles of Gi (s) (the roots of P i (s)), which are in the stable half of the s plane. Note that only the even part of P i (ki , s) changes with ki , the odd part being the same as P i (s). Thus none of the roots of P i (ki , s) can cross the imaginary axis in the s plane. As −ki tends to ∞, the branches terminate on the zeros of Gi (s) (roots of sX i (s)) which are simple roots of multiplicity one. However a very interesting thing happens. At −ki =
a0 (i) = −ki∗ a1 (i)
(28)
One of the roots of P i (ki , s) moves to −∞ in the s plane. All the other roots of P i (ki , s) are still in the left hand side of the s plane. As −ki is increasing beyond −ki∗ , the real root reappears but this time in the R.H.S. of the s plane, making the polynomial P i (ki , s) unstable. Thus the order of the polynomial P i (s) can be reduced by one provided ki is chosen as ki∗ = − aa01 (i) . This yields as lower order stable polynomial. (i) P i−1 (s) = P i (ki , s)|k =k i
∗
i
a0 (i) = Ri (s) − sX i (s) + X i (s) a1 (i)
(29)
Figure 5 shows that the typical Root Locus for i = 4 (even) 6 ?
-
6
The presence of the roots of R4 (s) act as a “barrier wall” which is not crossed by Root Locus branches. Case 2, i = odd integer Feedback System P i (ki , s) when i = odd integers is modified as: + -
-
ki Gi (s) =
ski Ri (s) Ri (s)+X i (s)
− 6
9
=
ski Ri (s) P i (s)
-
ki ≤ 0
i
2
P i (s) =
i
a2k−1 (i)si−2k+1 +
k=1
2
a2k (i)si−2k
(30)
k=0
a0 (i) P i−1 (s) = P i (ki , s)|k =k = Ri (s) + X i (s) − sRi (s) a1 (i) a0 (i) where − ki∗ = a1 (i) ∗
i
i
(31)
Figure 7 shows the typical Root Locus plot for i=5 (odd integer) 6 ? ?
-
6 6
The main point to be noted the introduction of algorithm (29) (for i even) or (31) (for i odd) transforms an ith order polynomial to (i − 1)th order. Its (i − 1) roots still lie on the left hand side of s plane. The last ith root moves to −∞ for −ki = aa01(i) 0. (i) > From equation (29) i = even integer
i
2
P i−1 (s) =
k=0
or
i
a2k (i)si−2k −
a0 (i) 2 a2k−1 (i)si−2k + X i (s) a1 (i) k=1
i
2
P i−1 (s) =
k=0
i
2 a0 (i) i−2k i−2k + a2k (i) − a2k−1 (i)s s a2k−1 (i)si−2k+1 a1 (i) k=0
Thus a0 (i) i a2k+1 (i), k = 1, 2, · · · , 2 a1 (i) i a2k−1 (i − 1) = a2k−1 (i), k = 1, 2, · · · , 2 a2k (i − 1) = a2k (i) −
Similarly from (31) 10
(32)
i = odd integer
a2k (i − 1) = a2k−1 (i) k = 1, 2, · · · , a2k−1 (i − 1) = a2k (i) −
7
i+1 2
a0 (i) i a2k+1 (i), k = 1, 2, · · · , 2 a1 (i)
(33)
Generation of a Family of Hurwitz Polynomials via Root Locus Method (Order Augmentation Algorithm)
Consider an nth-order Hurwitz Polynomial: Case 1: n is even P n (s) = Rn (s) + X n (s), (n = 2, 4, 6, · · ·) Rn (s) = a0 (n)sn + a2 (n)sn−2 + a4 (n)sn−4 + · · · + an (n) X n (s) = a1 (n)sn−1 + a3 (n)sn−3 + a5 (n)sn−5 + · · · + an−1 (n)s
(34)
Assume P n (s) is Hurwitz, we are required to generate a new Hurwitz polynomial P n+1 (s) of one degree higher, using the polynomial P n (s). Let us analyze the Root Locus for the following closed loop system: + -
-
K n Rn (s) sP n (s)
=
-
K n (a0 (n)sn +a2 (n)sn 2 +···+an (n)) s(a0 (n)sn +a1 (n)sn 1 +···+an (n)) −
−
− 6
The closed loop characteristic polynomial for this system is: P n+1 (K n , s) = sP n (s) + K n Rn (s)
(35)
For small values of K n , all the roots of P n+1 (K n , s) are in L.H.S. just as the polynomial P n (s). Since the imaginary part of P n+1 (K n , s) is the same as P n (s), thus for n even P n (s) = a0 (n)sn + a1 (n)sn−1 + · · · + an (n)
n
n−1
P n+1 (K n , s) = s a0 (n)s + a1 (n)s
+ · · · + an (n)
+K n a0 (n)sn + a2 (n)sn−1 + · · · + an (n)
Let P n+1 (s) = a0 (n + 1)sn+1 + a1 (n + 1)sn + · · · + an+1 (n + 1) an+1 (n + 1) K n = an 11
(36)
Thus the coefficients of the new augmentation polynomial P n+1 (s) are obtained from P n (s) as: a2 j (n + 1) = a2 j (n) a2 j+1 = a2 j (n) + K n a2 j (n) an+1 (n + 1) K n = , j = 0, 1, · · · , n/2 an Conditions of stability in terms of (n + 1)th polynomial are:
a2 j+1 (n) = a2 j (n + 1) −
an+1(n+1) a2 j (n) an (n)
> 0 a2 j (n) > 0
j = 1, 2, · · · , n/2
Case 2: When n is an odd integer P n (s) = X n (s) + Rn (s) =
a0 (n)sn + a1 (n)sn−1 + · · · + an (n) + a1 (n)sn−1 + a3 (n)sn−3 + · · · + an−1 (n) + -
-
K n Rn (s) sP n (s)
=
a1 (n)sn 1 +a3 (n)sn 3 +···+an (n) s(a0 (n)sn +a1 (n)sn 1 +···+an (n)) −
−
-
−
− 6
P n+1 (K n , s) = sP n (s) + K n Rn (s) = a0 (n)sn+1 + a1 (n)sn + a2 (n)sn−1 + a3 (n)sn−2 + · · · + an (n)s +K n a1 (n)sn−1 + K n a3 (n)sn−3 + · · · + K n an (n) Let K n an (n) = an+1 (n + 1) or K n =
an+1(n+1) an (n)
P n+1 (K n , s)|K = n
an+1 (n+1)
= P n+1 (s)
an (n)
Where P n+1 (s) = a0 (n + 1)sn+1 + a1 (n + 1)sn + · · · + an+1 (n + 1) The coefficients of the new polynomial P n+1 (s) are a0 (n + 1) = a0 (n) an+1 (n + 1) a2 j −1 (n) an (n) n+1 a2 j+1 (n + 1) = a2 j+1 (n) j = 1, 2, · · · , 2 a2 j (n + 1) = a2 j (n) +
The conditions of stability, in terms of (n + 1)th polynomial are:
a2 j (n) = a2 j (n + 1) −
an+1 (n+1) (n) an (n) a2 j −1
> 0 a2 j+1 (n) > 0
12
j = 1, 2, · · · ,
n+1 2
References [1] Aizerman, M.A. , “Theory of Automatic Control”, Pergamon Press, pp. 184–185, 1963. [2] Gantmacher, F.R. , “The Theory of Matrix”, Vol. II, Chelsea Publishing Company, N.Y., 1960. [3] Holz, Olga “Hermite – Biehler, Routh – Hurwitz and Total Positivity”, Linear Algebra and Its Applications , pp. 105–110, Vol. 372, 2003. [4] Puri, N.N. and Weygandt, C.N. , “Second Method of Liapunov and Routh’s Canonical Form”, Journal of Franklin Institute , pp. 365–384, Vol. 276, No. 5, Nov. 1963. ¨ [5] Hurwitz, A. , “Uber die Bedingungen, unter Welchen eine Gleichung nur Wurzeln mit negativen reellen Teilen besitzt”, Math. Ann., pp. 273–284, Vol. 46, 1985.
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