routhhurwitz proof stable poly

7/30/2019 RouthHurwitz Proof Stable Poly

1/13

A Simplified Proof of Hurwitz Criterion of Stability

and Generation of a Family of Stable Polynomial via

Root Locus MethodProf. N. N. Puri

ECE Department, Rutgers University

1 Problem Statement

Given:

An nth order characteristic polynomial:

Pn(s) = a0sn + a1s

n1 + + an1s + an, ai > 0 (1)

Problem:

Find the conditions under which all the roots of polynomial (1) have realparts negative. We shall refer to (1) as a Hurwitz polynomial or a stablepolynomial.

This is an age old problem whose answer was provided by now famous Routh-HurwitzCriterion.

Most textbooks do not present the proof of Hurwitz Criterion for they consider itdifficult for the undergraduate (or even the graduate) students. In what follows, we presenta very simple proof.

2 Statement of the Routh-Hurwitz Criterion

Let all the coefficients of (1) be positive (ai > 0, i = 1, , n). From these coefficients forman n n Hurwitz matrix:

Hn =

a1 a3 a5 a7 0

a0 a2 a4 a6 00 a1 a3 a5 00 a0 a2 a4 00 0 a1 a3 00 0 a0 a2 00 0 0 a1 00 0 0 a0 00 0 0 0 0

(2)

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Let all the diagonal minors (determinants) of Hn be:

1 = a1; 2 =

a1 a3a0 a2 ; 3 =

a1 a3 a5a0 a2 a40 a1 a3

; (3)

3 Hurwitz Theorem

In order that all the roots of the characteristic polynomial P(s) = a0sn + a1s

n1 + +an1s + an with all the coefficients ai > 0 have the real part negative, it is necessary andsufficient that:

All the even determinants: 2, 4, be positive (4)

OR

All the odd determinants: 1, 3, be positive (5)

Note: It can be easily shown that when all ais are positive and all odd indices determinants,1, 3, are positive, then all even indices determinants 2, 4, are automaticallypositive 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. Michialovs Theorem.

Michialov Theorem

Let

Pn(s) = a0sn + a1s

n1 + a2sn2 + + an =

ni=1

(s zi)

or

Pn(s)|s=j = Pn(j) =ni=1

|(j zi)|

ni=1

zi()

= |Pn(j)| p() (6)

Where zi are the roots of Pn(s).

p()|== = change in the argument p() as varies from to .

zi()|== =

+ if zi lies in L.H.S. if zi lies in R.H.S.

If all the roots zi of Pn(s) lie in L.H.S., then

p()|== = n (n being the order of Pn(s)) (7)

This leads us to formulate the Michialov Criterion of Stability:

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Figure 1: Figure 1

For a stable Pn(s), it is necessary and sufficient that the argument p() of thevector Pn(j) increases monotonically from 0 to n, as increases monotonicallyfrom to +.

Simply stated, the locus ofPn(j) of a stable polynomial Pn(s) traverses n quadrants inthe Pn(j) plane as 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 at1(= 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 negativevalues of .

1 = 0 < 2 < 3 < 4 < 5 < 6 < 7 < (8)

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Figure 2: Figure 2

Thus

Rn() = a0(2 22)(

2 24)(2 26)

Xn() = a1(

2

2

3)(

2

2

5)(

2

2

7) (9)

0,3,5,7, , represent zeros of Xn() (even function of)

2,4,6, , represent zeros of Rn() (odd function of)

Note that both Rn() and Xn() have simple, real roots. Thus, for an nth orderpolynomial (n even for simplicity).

Rn() = a0

n/2

i=1(

2

2

2i) = (a0n

a2n2

+ a4n4

+ an)

Xn() = a1n/2j=1

(2 22i+1) = (a1n2 a3

n4 + an1)

Pn(j) = Rn() +jXn() (10)

0 < 2 < 3 < < n

Due to the alternating property of roots ofRn() and Xn(), when Rn() is decreasing,Xn() is increasing, and vice versa. Thus let

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Figure 3: Figure 3

Xn(2i)

Rn(2i)= A2i > 0, where R

n(2i) =

d

dRn()

=2i

, i = 1, 2, ,n

2(11)

We are now in a position to utilize the results of this section to prove Hurwitz Theoremas presented in section 3 by (4) or (5).

5 Proof of Hurwitz Theorem

Let us consider f(), an odd function of . From (10) and (11), using partial fractions,

f() =Xn()

Rn()=

Xn()

a0n/2

i=1(2 22i)

=n/2i=1

Xn(2i)

Rn(2i( 2i))+

Xn(22i)

Rn(22i( + 2i))

=n/2i=1

A2i

1

2i+

1

+ 2i

(12)

Now consider the function f(z) which is an odd function of z and expand it in powersof z1:

f(z) =n/2i=1

A2i

1

(z 2i)+

1

(z+ 2i)

=

c1z

+c3z3

+ =n1i=0

c2k+1z2k+1

(13)

The coefficients c2k+1 are completed via contour integration of (13), yielding

c2k+1 =1

2j

c

f(z)z2kdz, where contour c includes all poles of f(z).

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From Residue theorem

c2k+1 =n/2i=1

A2i(2i)

2k + (2i)2k

> 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 definitematrix which is related to matrix Hn.

To extract the properties of c2k+1 (k = 0, 1, , n/2), let us consider a function 2(z),

and f(z):

2(z) = (x0 + zx1 + + zn1xn1)

2 =n1p=0

n1q=0

xpxqzp+q (15)

f(z) =n1k=0

c2k+1z2k+1

=n/2i=1

A2i

1

z i+

1

z+ i

(16)

Let us evaluate a scalar function V,

V =1

2j

c

f(z)2(z)dz, c containing all the poles of f(z) (17)

From (15) and (17)

V =1

2j

c

f(z)c2k+1

n1p=0

n1q=0

zp+q

z2k1

dz = n1

p=0

n1q=0

cp,qxpxq (18)

where cp,q = 0 when p + q = even,

cp+q+1 when p + q = oddAlso

V =1

2j

c

n/2i=1

A2i

1

z i+

1

z+ i

2(z)dz =

n/2i=1

A2i

2(i) + 2(i)

(19)

Equation (19) implies that V is a positive definite function, which implies that matrixC associated with V in (18) is a positive definite matrix given by

V = xTC x, C =

c1 c3 cn1

c3 c3 c5 ...

......

......

cn1 c2n1

a positive definite matrix (20)

Let us relate this matrix C to coefficients a0, a1, , an:From (10), (12) and (13):

f() =Xn()

Rn()=

c1

+c33

+ +c2n1n1

=a1

n1 a3n3 + a5

n5

a0n a2n2 + a4n4 + an(21)

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Cross multiplication and collecting terms of similar powers of ,

c1a0 = a1 (22)

c3a0 c1a2 = a3

c5a0 c3a2 + c1a4 = a5...

..

.cn1a0 cn3a2 + c1an2 = an1

......

c2m1a0 c2m3a2 + c2m5a4 + (1)ma2m = 0, m =

n

2+ 1,

an+1 = an+2 = = a2n = 0

all c2k+1 > 0

Equation (22) can be written in matrix form:

a1 a3 a5 a7 0

a0 a2 a4 a6 00 a1 a3 a5 00 a0 a2 a4 0...

......

......

...0 0 0 0 0 an

=

c1 c3 c5 c7 c2n11 0 0 0 00 c1 c3 c5 00 1 0 0 0...

......

......

...1 0 0 0 0 0

a0 a2 a4 a6 an0 a0 a2 a4 an20 0 a0 a2 an40 0 0 a0 an6...

......

......

...0 0 0 0 0 a0

(23)

Since the determinant of product of matrices is equal to the product of individualdeterminant:

a1 a3 a5 a7 0

a0 a2 a4 a6 00 a1 a3 a5 00 a0 a2 a4 0...

......

......

...0 0 0 0 0 an

= an0

c1 0 c3 0 c2n1

0 c3 0 c5 c2n3c3 0 c5 0 c2n5...

......

......

......

......

......

...c2n1 0 0 0 0 0

(24)

Since the matrix C is positive definite, the determinants of Hn and all its diagonalminors are positive. This completes the proof of the Hurwitz Criterion.

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6 Generation of a Family of Stable Polynomials via

Root Locus Method (Order Reduction Algorithm)

Consider an nth order stable polynomial:

Pn(s) = a0sn + a1s

n1 + + an

Starting with this polynomial, we shall generate a family of stable polynomial Pi(s)(i = n 1, n 2, , 1), such that:

Pi(s) = a0(i)si + a1(i)s

i1 + + ai(i) = Ri(s) + Xi(s) (25)

ak(i)|i=n = ak, k = 0, 1, , n

Ri(s) = Even degree part of Pi(s)

Xi(s) = Odd degree part of Pi(s)

Let Pi(s) be the starting polynomial. We shall generate Pi1(s) as following:

Case 1, i = even integer: Let

Pi(s) =

a0(i)si + a2(i)s

i2 + + ai(i)

+s

a1(i)si2 + a3(i)s

i4 + + ai1(i)

= Ri(s) + Xi(s)

=

i

2+1

k=0

a2k(i)si2k +

i

2k=1

a2k1(i)si2k+1

where Ri(s) is even degree and Xi(s) is odd degree.

Consider the Root Locus of the following feedback system with variable ki, whereki 0:

- - kiGi(s) = skiXi(s)Ri(s)+Xi(s) = skiXi(s)Pi(s)+

6

ki 0-

Then the closed loop characteristic polynomial is:

Pi(ki, s) = Pi(s) + kisXi(s) (26)

and Root Locus Equation is:

Pi(ki, s) = Pi(s) + kisXi(s) = 0

orsXi(s)

Pi(s)=

1

ki= Gi(s) (27)

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The root locus starts (ki = 0) at the poles of Gi(s) (the roots of Pi(s)), which are inthe stable half of the s plane. Note that only the even part of Pi(ki, s) changes withki, the odd part being the same as Pi(s). Thus none of the roots ofPi(ki, s) can crossthe imaginary axis in the s plane. As ki tends to , the branches terminate on thezeros of Gi(s) (roots of sXi(s)) which are simple roots of multiplicity one. Howevera very interesting thing happens. At

ki = a0(i)a1(i)

= ki (28)

One of the roots of Pi(ki, s) moves to in the s plane. All the other roots ofPi(ki, s) are still in the left hand side of the s plane. As ki is increasing beyondki , the real root reappears but this time in the R.H.S. of the s plane, making thepolynomial Pi(ki, s) unstable.

Thus the order of the polynomial Pi(s) can be reduced by one provided ki is chosen

as ki = a0(i)a1(i)

. This yields as lower order stable polynomial.

Pi1(s) = Pi(ki, s)|ki=ki=

Ri(s) a0(i)

a1(i)sXi(s)

+ Xi(s) (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 byRoot Locus branches.

Case 2, i = odd integerFeedback System Pi(ki, s) when i = odd integers is modified as:

- - kiGi(s) = skiRi(s)Ri(s)+Xi(s) = skiRi(s)Pi(s)+

6

ki 0-

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Pi(s) =

i

2k=1

a2k1(i)si2k+1 +

i

2k=0

a2k(i)si2k (30)

Pi1(s) = Pi(ki, s)|ki=ki= Ri(s) +

Xi(s)

a0(i)

a1(i)sRi(s)

(31)

where ki =a0(i)

a1(i)

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) (fori 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 =a0(i)

a1(i) > 0.

From equation (29)

i = even integer

Pi1(s) =

i

2k=0

a2k(i)si2k

a0(i)

a1(i)

i

2k=1

a2k1(i)si2k

+ Xi(s)

or

Pi1(s) =

i

2

k=0

a2k(i) a0(i)

a1(i)a2k1(i)s

i2k

si2k +

i

2

k=0a2k1(i)s

i2k+1

Thus

a2k(i 1) = a2k(i) a0(i)

a1(i)a2k+1(i), k = 1, 2, ,

i

2

a2k1(i 1) = a2k1(i), k = 1, 2, ,i

2(32)

Similarly from (31)

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i = odd integer

a2k(i 1) = a2k1(i) k = 1, 2, ,i + 1

2

a2k1(i 1) = a2k(i) a0(i)

a1(i)a2k+1(i), k = 1, 2, ,

i

2(33)

7 Generation of a Family of Hurwitz Polynomials via

Root Locus Method (Order Augmentation Algo-

rithm)

Consider an nth-order Hurwitz Polynomial:

Case 1: n is even

Pn(s) = Rn(s) + Xn(s), (n = 2, 4, 6, ) (34)

Rn(s) = a0(n)sn

+ a2(n)sn2

+ a4(n)sn4

+ + an(n)Xn(s) = a1(n)s

n1 + a3(n)sn3 + a5(n)s

n5 + + an1(n)s

Assume Pn(s) is Hurwitz, we are required to generate a new Hurwitz polynomialPn+1(s) of one degree higher, using the polynomial Pn(s).

Let us analyze the Root Locus for the following closed loop system:

- -KnRn(s)sPn(s)

= Kn(a0(n)sn+a2(n)sn

2++an(n))s(a0(n)sn+a1(n)sn1++an(n))

+ 6

-

The closed loop characteristic polynomial for this system is:

Pn+1(Kn, s) = sPn(s) + KnRn(s) (35)

For small values ofKn, all the roots ofPn+1(Kn, s) are in L.H.S. just as the polynomialPn(s). Since the imaginary part of Pn+1(Kn, s) is the same as Pn(s), thus for n even

Pn(s) = a0(n)sn + a1(n)s

n1 + + an(n)

Pn+1(Kn, s) = sa0(n)s

n + a1(n)sn1 + + an(n)

+Kn

a0(n)s

n + a2(n)sn1 + + an(n)

(36)

Let

Pn+1(s) = a0(n + 1)sn+1 + a1(n + 1)s

n + + an+1(n + 1)

Kn =an+1(n + 1)

an

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Thus the coefficients of the new augmentation polynomial Pn+1(s) are obtained fromPn(s) as:

a2j(n + 1) = a2j(n)

a2j+1 = a2j(n) + Kna2j(n)

Kn =an+1(n + 1)

an, j = 0, 1, , n/2

Conditions of stability in terms of (n + 1)th polynomial are:a2j+1(n) = a2j(n + 1)

an+1(n+1)an(n)

a2j(n) > 0

a2j(n) > 0

j = 1, 2, , n/2

Case 2: When n is an odd integer

Pn(s) = Xn(s) + Rn(s)

=a0(n)s

n + a1(n)sn1 + + an(n)

+a1(n)s

n1 + a3(n)sn3 + + an1(n)

- -KnRn(s)sPn(s)

= a1(n)sn1+a3(n)sn

3++an(n)s(a0(n)sn+a1(n)sn1++an(n))

+ 6

-

Pn+1(Kn, s) = sPn(s) + KnRn(s)

= a0(n)sn+1 + a1(n)s

n + a2(n)sn1 + a3(n)s

n2 + + an(n)s

+Kna1(n)sn1 + Kna3(n)s

n3 + + Knan(n)

Let Knan(n) = an+1(n + 1) or Kn =an+1(n+1)

an(n)

Pn+1(Kn, s)|Kn=an+1(n+1)

an(n)

= Pn+1(s)

WherePn+1(s) = a0(n + 1)s

n+1 + a1(n + 1)sn + + an+1(n + 1)

The coefficients of the new polynomial Pn+1(s) are

a0(n + 1) = a0(n)

a2j(n + 1) = a2j(n) +an+1(n + 1)

an(n)a2j1(n)

a2j+1(n + 1) = a2j+1(n) j = 1, 2, ,n + 1

2

The conditions of stability, in terms of (n + 1)th polynomial are:a2j(n) = a2j(n + 1)

an+1(n+1)an(n)

a2j1(n) > 0

a2j+1(n) > 0

j = 1, 2, ,

n + 1

2

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References

[1] Aizerman, M.A. , Theory of Automatic Control, Pergamon Press, pp. 184185,1963.

[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 Algebraand Its Applications, pp. 105110, Vol. 372, 2003.

[4] Puri, N.N. and Weygandt, C.N. , Second Method of Liapunov and Rouths CanonicalForm, Journal of Franklin Institute, pp. 365384, Vol. 276, No. 5, Nov. 1963.

[5] Hurwitz, A. , Uber die Bedingungen, unter Welchen eine Gleichung nur Wurzeln mitnegativen reellen Teilen besitzt, Math. Ann., pp. 273284, Vol. 46, 1985.

routhhurwitz proof stable poly

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