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Routh–Hurwitz stability criterion

In control system theory, the Routh–Hurwitz stability criterion is a mathematical test that is a necessary and sufficient condition for the stability of a linear time-invariant (LTI) dynamical system or control system. A stable system is one whose output signal is bounded; the position, velocity or energy do not increase to infinity as time goes on. The Routh test is an efficient recursive algorithm that English mathematician Edward John Routh proposed in 1876 to determine whether all the roots of the characteristic polynomial of a linear system have negative real parts.[1] German mathematician Adolf Hurwitz independently proposed in 1895 to arrange the coefficients of the polynomial into a square matrix, called the Hurwitz matrix, and showed that the polynomial is stable if and only if the sequence of determinants of its principal submatrices are all positive.[2] The two procedures are equivalent, with the Routh test providing a more efficient way to compute the Hurwitz determinants () than computing them directly. A polynomial satisfying the Routh–Hurwitz criterion is called a Hurwitz polynomial.

The importance of the criterion is that the roots p of the characteristic equation of a linear system with negative real parts represent solutions ept of the system that are stable (bounded). Thus the criterion provides a way to determine if the equations of motion of a linear system have only stable solutions, without solving the system directly. For discrete systems, the corresponding stability test can be handled by the Schur–Cohn criterion, the Jury test and the Bistritz test. With the advent of computers, the criterion has become less widely used, as an alternative is to solve the polynomial numerically, obtaining approximations to the roots directly.

The Routh test can be derived through the use of the Euclidean algorithm and Sturm's theorem in evaluating Cauchy indices. Hurwitz derived his conditions differently.[3]

Using Euclid's algorithm

The criterion is related to Routh–Hurwitz theorem. From the statement of that theorem, we have   where:

  •   is the number of roots of the polynomial   with negative real part;
  •   is the number of roots of the polynomial   with positive real part (according to the theorem,   is supposed to have no roots lying on the imaginary line);
  • w(x) is the number of variations of the generalized Sturm chain obtained from   and   (by successive Euclidean divisions) where   for a real y.

By the fundamental theorem of algebra, each polynomial of degree n must have n roots in the complex plane (i.e., for an ƒ with no roots on the imaginary line, p + q = n). Thus, we have the condition that ƒ is a (Hurwitz) stable polynomial if and only if p − q = n (the proof is given below). Using the Routh–Hurwitz theorem, we can replace the condition on p and q by a condition on the generalized Sturm chain, which will give in turn a condition on the coefficients of ƒ.

Using matrices

Let f(z) be a complex polynomial. The process is as follows:

  1. Compute the polynomials   and   such that   where y is a real number.
  2. Compute the Sylvester matrix associated to   and  .
  3. Rearrange each row in such a way that an odd row and the following one have the same number of leading zeros.
  4. Compute each principal minor of that matrix.
  5. If at least one of the minors is negative (or zero), then the polynomial f is not stable.

Example

  • Let   (for the sake of simplicity we take real coefficients) where   (to avoid a root in zero so that we can use the Routh–Hurwitz theorem). First, we have to calculate the real polynomials   and  :
 
Next, we divide those polynomials to obtain the generalized Sturm chain:
  •   yields  
  •   yields   and the Euclidean division stops.

Notice that we had to suppose b different from zero in the first division. The generalized Sturm chain is in this case  . Putting  , the sign of   is the opposite sign of a and the sign of by is the sign of b. When we put  , the sign of the first element of the chain is again the opposite sign of a and the sign of by is the opposite sign of b. Finally, -c has always the opposite sign of c.

Suppose now that f is Hurwitz-stable. This means that   (the degree of f). By the properties of the function w, this is the same as   and  . Thus, a, b and c must have the same sign. We have thus found the necessary condition of stability for polynomials of degree 2.

Routh–Hurwitz criterion for second, third and fourth-order polynomials

  • The second-degree polynomial   has both roots with negative real part (and the system with characteristic equation   is stable) if and only if both coefficients satisfy  .
  • The third-order polynomial   has all roots in the open left half-plane if and only if  ,  , and   are positive and  
  • For a fourth-order polynomial  , all the coefficients must satisfy  , and  [4] (When this is derived you do not know all coefficients should be positive, and you add  .)
  • In general the Routh stability criterion states a polynomial has all roots in the open left half-plane if and only if all first-column elements of the Routh array have the same sign.
  • All coefficients being positive (or all negative) is necessary for all roots to be located in the open left half-plane. That is why here   is fixed to 1, which is positive. When this is assumed, we can remove   from fourth-order polynomial, and conditions for fifth- and sixth-order can be simplified. For fifth-order we only need to check that   and for sixth-order we only need to check   and this is further optimised in Liénard–Chipart criterion.[5] Indeed, some coefficients being positive is not independent with principal minors being positive, like   check can be removed for third-order polynomial.

Higher-order example

A tabular method can be used to determine the stability when the roots of a higher order characteristic polynomial are difficult to obtain. For an nth-degree polynomial

  •  

the table has n + 1 rows and the following structure:

       
       
       
       
       

where the elements   and   can be computed as follows:

  •  
  •  

When completed, the number of sign changes in the first column will be the number of non-negative roots.

0.75 1.5 0 0
-3 6 0 0
3 0 0 0
6 0 0 0

In the first column, there are two sign changes (0.75 → −3, and −3 → 3), thus there are two non-negative roots where the system is unstable.

The characteristic equation of a servo system is given by:[6]

  •  
      0
    0 0
   =  0 0
  0 0 0
  0 0 0

for stability, all the elements in the first column of the Routh array must be positive. So the conditions that must be satisfied for stability of the given system as follows:[6]

 [6]

We see that if

 

then

 

Is satisfied.

  •  [7]

We have the following table :

1 11 200 0
1 1 0 0
1 20 0 0
-19 0 0 0
20 0 0 0

there are two sign changes. The system is unstable, since it has two right-half-plane poles and two left-half-plane poles. The system cannot have jω poles since a row of zeros did not appear in the Routh table.[7]

Sometimes the presence of poles on the imaginary axis creates a situation of marginal stability. In that case the coefficients of the "Routh array" in a whole row become zero and thus further solution of the polynomial for finding changes in sign is not possible. Then another approach comes into play. The row of polynomial which is just above the row containing the zeroes is called the "auxiliary polynomial".

  •  

We have the following table:

1 8 20 16
2 12 16 0
2 12 16 0
0 0 0 0

In such a case the auxiliary polynomial is   which is again equal to zero. The next step is to differentiate the above equation which yields the polynomial  . The coefficients of the row containing zero now become "8" and "24". The process of Routh array is proceeded using these values which yield two points on the imaginary axis. These two points on the imaginary axis are the prime cause of marginal stability.[8]

See also

References

  1. ^ Routh, E. J. (1877). A Treatise on the Stability of a Given State of Motion: Particularly Steady Motion. Macmillan.
  2. ^ Hurwitz, A. (1895). "Ueber die Bedingungen, unter welchen eine Gleichung nur Wurzeln mit negativen reellen Theilen besitzt". Math. Ann. 46 (2): 273–284. doi:10.1007/BF01446812. S2CID 121036103. (English translation “On the conditions under which an equation has only roots with negative real parts” by H. G. Bergmann in Selected Papers on Mathematical Trends in Control Theory R. Bellman and R. Kalaba Eds. New York: Dover, 1964 pp. 70–82.)
  3. ^ Gopal, M. (2002). Control Systems: Principles and Design, 2nd Ed. Tata McGraw-Hill Education. p. 14. ISBN 0070482896.
  4. ^ "Routh-Hurwitz Criterion". math24.net. Retrieved 2022-07-19.
  5. ^ (PDF). Archived from the original (PDF) on 2015-01-26. Retrieved 19 July 2022.
  6. ^ a b c KUMAR, Anand (2007). CONTROL SYSTEMS. PHI Learning. ISBN 9788120331976.
  7. ^ a b Nise, Norman (2015). Control Systems Engineering. Wiley. ISBN 9781118800829.
  8. ^ Saeed, Syed Hasan (2008). Automatic Control Systems. Delhi: Katson Publishers. pp. 206, 207. ISBN 978-81-906919-2-5.
  • Felix Gantmacher (J.L. Brenner translator) (1959). Applications of the Theory of Matrices, pp 177–80, New York: Interscience.
  • Pippard, A. B.; Dicke, R. H. (1986). "Response and Stability, An Introduction to the Physical Theory". American Journal of Physics. 54 (11): 1052. Bibcode:1986AmJPh..54.1052P. doi:10.1119/1.14826. Archived from the original on 2016-05-14. Retrieved 2008-05-07.
  • Richard C. Dorf, Robert H. Bishop (2001). Modern Control Systems (9th ed.). Prentice Hall. ISBN 0-13-030660-6.
  • Rahman, Q. I.; Schmeisser, G. (2002). Analytic theory of polynomials. London Mathematical Society Monographs. New Series. Vol. 26. Oxford: Oxford University Press. ISBN 0-19-853493-0. Zbl 1072.30006.
  • Weisstein, Eric W. "Routh-Hurwitz Theorem". MathWorld--A Wolfram Web Resource.
  • Stephen Barnett (1983). Polynomials and Linear Control Systems, New York: Marcel Dekker, Inc.

External links

  • A MATLAB script implementing the Routh-Hurwitz test
  • Online implementation of the Routh-Hurwitz Criterion

routh, hurwitz, stability, criterion, this, article, needs, additional, citations, verification, please, help, improve, this, article, adding, citations, reliable, sources, unsourced, material, challenged, removed, find, sources, news, newspapers, books, schol. This article needs additional citations for verification Please help improve this article by adding citations to reliable sources Unsourced material may be challenged and removed Find sources Routh Hurwitz stability criterion news newspapers books scholar JSTOR April 2009 Learn how and when to remove this template message In control system theory the Routh Hurwitz stability criterion is a mathematical test that is a necessary and sufficient condition for the stability of a linear time invariant LTI dynamical system or control system A stable system is one whose output signal is bounded the position velocity or energy do not increase to infinity as time goes on The Routh test is an efficient recursive algorithm that English mathematician Edward John Routh proposed in 1876 to determine whether all the roots of the characteristic polynomial of a linear system have negative real parts 1 German mathematician Adolf Hurwitz independently proposed in 1895 to arrange the coefficients of the polynomial into a square matrix called the Hurwitz matrix and showed that the polynomial is stable if and only if the sequence of determinants of its principal submatrices are all positive 2 The two procedures are equivalent with the Routh test providing a more efficient way to compute the Hurwitz determinants D i displaystyle Delta i than computing them directly A polynomial satisfying the Routh Hurwitz criterion is called a Hurwitz polynomial The importance of the criterion is that the roots p of the characteristic equation of a linear system with negative real parts represent solutions ept of the system that are stable bounded Thus the criterion provides a way to determine if the equations of motion of a linear system have only stable solutions without solving the system directly For discrete systems the corresponding stability test can be handled by the Schur Cohn criterion the Jury test and the Bistritz test With the advent of computers the criterion has become less widely used as an alternative is to solve the polynomial numerically obtaining approximations to the roots directly The Routh test can be derived through the use of the Euclidean algorithm and Sturm s theorem in evaluating Cauchy indices Hurwitz derived his conditions differently 3 Contents 1 Using Euclid s algorithm 2 Using matrices 2 1 Example 2 2 Routh Hurwitz criterion for second third and fourth order polynomials 2 3 Higher order example 3 See also 4 References 5 External linksUsing Euclid s algorithm EditThe criterion is related to Routh Hurwitz theorem From the statement of that theorem we have p q w w displaystyle p q w infty w infty where p displaystyle p is the number of roots of the polynomial f z displaystyle f z with negative real part q displaystyle q is the number of roots of the polynomial f z displaystyle f z with positive real part according to the theorem f displaystyle f is supposed to have no roots lying on the imaginary line w x is the number of variations of the generalized Sturm chain obtained from P 0 y displaystyle P 0 y and P 1 y displaystyle P 1 y by successive Euclidean divisions where f i y P 0 y i P 1 y displaystyle f iy P 0 y iP 1 y for a real y By the fundamental theorem of algebra each polynomial of degree n must have n roots in the complex plane i e for an ƒ with no roots on the imaginary line p q n Thus we have the condition that ƒ is a Hurwitz stable polynomial if and only if p q n the proof is given below Using the Routh Hurwitz theorem we can replace the condition on p and q by a condition on the generalized Sturm chain which will give in turn a condition on the coefficients of ƒ Using matrices EditLet f z be a complex polynomial The process is as follows Compute the polynomials P 0 y displaystyle P 0 y and P 1 y displaystyle P 1 y such that f i y P 0 y i P 1 y displaystyle f iy P 0 y iP 1 y where y is a real number Compute the Sylvester matrix associated to P 0 y displaystyle P 0 y and P 1 y displaystyle P 1 y Rearrange each row in such a way that an odd row and the following one have the same number of leading zeros Compute each principal minor of that matrix If at least one of the minors is negative or zero then the polynomial f is not stable Example Edit Let f z a z 2 b z c displaystyle f z az 2 bz c for the sake of simplicity we take real coefficients where c 0 displaystyle c neq 0 to avoid a root in zero so that we can use the Routh Hurwitz theorem First we have to calculate the real polynomials P 0 y displaystyle P 0 y and P 1 y displaystyle P 1 y f i y a y 2 i b y c P 0 y i P 1 y a y 2 c i b y displaystyle f iy ay 2 iby c P 0 y iP 1 y ay 2 c i by dd Next we divide those polynomials to obtain the generalized Sturm chain P 0 y a b y P 1 y c displaystyle P 0 y a b y P 1 y c yields P 2 y c displaystyle P 2 y c P 1 y b c y P 2 y displaystyle P 1 y b c y P 2 y yields P 3 y 0 displaystyle P 3 y 0 and the Euclidean division stops Notice that we had to suppose b different from zero in the first division The generalized Sturm chain is in this case P 0 y P 1 y P 2 y c a y 2 b y c displaystyle P 0 y P 1 y P 2 y c ay 2 by c Putting y displaystyle y infty the sign of c a y 2 displaystyle c ay 2 is the opposite sign of a and the sign of by is the sign of b When we put y displaystyle y infty the sign of the first element of the chain is again the opposite sign of a and the sign of by is the opposite sign of b Finally c has always the opposite sign of c Suppose now that f is Hurwitz stable This means that w w 2 displaystyle w infty w infty 2 the degree of f By the properties of the function w this is the same as w 2 displaystyle w infty 2 and w 0 displaystyle w infty 0 Thus a b and c must have the same sign We have thus found the necessary condition of stability for polynomials of degree 2 Routh Hurwitz criterion for second third and fourth order polynomials Edit The second degree polynomial P s s 2 a 1 s a 0 displaystyle P s s 2 a 1 s a 0 has both roots with negative real part and the system with characteristic equation P s 0 displaystyle P s 0 is stable if and only if both coefficients satisfy a i gt 0 displaystyle a i gt 0 The third order polynomial P s s 3 a 2 s 2 a 1 s a 0 displaystyle P s s 3 a 2 s 2 a 1 s a 0 has all roots in the open left half plane if and only if a 2 displaystyle a 2 a 1 displaystyle a 1 and a 0 displaystyle a 0 are positive and a 2 a 1 gt a 0 displaystyle a 2 a 1 gt a 0 For a fourth order polynomial P s s 4 a 3 s 3 a 2 s 2 a 1 s a 0 0 displaystyle P s s 4 a 3 s 3 a 2 s 2 a 1 s a 0 0 all the coefficients must satisfy a i gt 0 displaystyle a i gt 0 and a 3 a 2 a 1 gt a 1 2 a 3 2 a 0 displaystyle a 3 a 2 a 1 gt a 1 2 a 3 2 a 0 4 When this is derived you do not know all coefficients should be positive and you add a 3 a 2 gt a 1 displaystyle a 3 a 2 gt a 1 In general the Routh stability criterion states a polynomial has all roots in the open left half plane if and only if all first column elements of the Routh array have the same sign All coefficients being positive or all negative is necessary for all roots to be located in the open left half plane That is why here a n displaystyle a n is fixed to 1 which is positive When this is assumed we can remove a 3 a 2 gt a 1 displaystyle a 3 a 2 gt a 1 from fourth order polynomial and conditions for fifth and sixth order can be simplified For fifth order we only need to check that D 2 gt 0 D 4 gt 0 displaystyle Delta 2 gt 0 Delta 4 gt 0 and for sixth order we only need to check D 3 gt 0 D 5 gt 0 displaystyle Delta 3 gt 0 Delta 5 gt 0 and this is further optimised in Lienard Chipart criterion 5 Indeed some coefficients being positive is not independent with principal minors being positive like a 2 gt 0 displaystyle a 2 gt 0 check can be removed for third order polynomial Higher order example Edit A tabular method can be used to determine the stability when the roots of a higher order characteristic polynomial are difficult to obtain For an nth degree polynomial D s a n s n a n 1 s n 1 a 1 s a 0 displaystyle D s a n s n a n 1 s n 1 cdots a 1 s a 0 the table has n 1 rows and the following structure a n displaystyle a n a n 2 displaystyle a n 2 a n 4 displaystyle a n 4 displaystyle dots a n 1 displaystyle a n 1 a n 3 displaystyle a n 3 a n 5 displaystyle a n 5 displaystyle dots b 1 displaystyle b 1 b 2 displaystyle b 2 b 3 displaystyle b 3 displaystyle dots c 1 displaystyle c 1 c 2 displaystyle c 2 c 3 displaystyle c 3 displaystyle dots displaystyle vdots displaystyle vdots displaystyle vdots displaystyle ddots where the elements b i displaystyle b i and c i displaystyle c i can be computed as follows b i a n 1 a n 2 i a n a n 2 i 1 a n 1 displaystyle b i frac a n 1 times a n 2i a n times a n 2i 1 a n 1 c i b 1 a n 2 i 1 a n 1 b i 1 b 1 displaystyle c i frac b 1 times a n 2i 1 a n 1 times b i 1 b 1 When completed the number of sign changes in the first column will be the number of non negative roots 0 75 1 5 0 0 3 6 0 03 0 0 06 0 0 0In the first column there are two sign changes 0 75 3 and 3 3 thus there are two non negative roots where the system is unstable The characteristic equation of a servo system is given by 6 b 0 s 4 b 1 s 3 b 2 s 2 b 3 s b 4 0 displaystyle b 0 s 4 b 1 s 3 b 2 s 2 b 3 s b 4 0 b 0 displaystyle b 0 b 2 displaystyle b 2 b 4 displaystyle b 4 0b 1 displaystyle b 1 b 3 displaystyle b 3 0 0b 1 b 2 b 0 b 3 b 1 displaystyle b 1 b 2 b 0 b 3 over b 1 b 1 b 4 b 0 0 b 1 displaystyle b 1 b 4 b 0 times 0 over b 1 b 4 displaystyle b 4 0 0 b 1 b 2 b 0 b 3 b 3 b 1 2 b 4 b 1 b 2 b 0 b 3 displaystyle b 1 b 2 b 0 b 3 b 3 b 1 2 b 4 over b 1 b 2 b 0 b 3 0 0 0b 4 displaystyle b 4 0 0 0for stability all the elements in the first column of the Routh array must be positive So the conditions that must be satisfied for stability of the given system as follows 6 b 1 gt 0 b 1 b 2 b 0 b 3 gt 0 b 1 b 2 b 0 b 3 b 3 b 1 2 b 4 gt 0 b 4 gt 0 displaystyle b 1 gt 0 b 1 b 2 b 0 b 3 gt 0 b 1 b 2 b 0 b 3 b 3 b 1 2 b 4 gt 0 b 4 gt 0 6 We see that if b 1 b 2 b 0 b 3 b 3 b 1 2 b 4 0 displaystyle b 1 b 2 b 0 b 3 b 3 b 1 2 b 4 geq 0 thenb 1 b 2 b 0 b 3 gt 0 displaystyle b 1 b 2 b 0 b 3 gt 0 Is satisfied s 4 6 s 3 11 s 2 6 s 200 0 displaystyle s 4 6s 3 11s 2 6s 200 0 7 We have the following table 1 11 200 01 1 0 01 20 0 0 19 0 0 020 0 0 0there are two sign changes The system is unstable since it has two right half plane poles and two left half plane poles The system cannot have jw poles since a row of zeros did not appear in the Routh table 7 Sometimes the presence of poles on the imaginary axis creates a situation of marginal stability In that case the coefficients of the Routh array in a whole row become zero and thus further solution of the polynomial for finding changes in sign is not possible Then another approach comes into play The row of polynomial which is just above the row containing the zeroes is called the auxiliary polynomial s 6 2 s 5 8 s 4 12 s 3 20 s 2 16 s 16 0 displaystyle s 6 2s 5 8s 4 12s 3 20s 2 16s 16 0 We have the following table 1 8 20 162 12 16 02 12 16 00 0 0 0In such a case the auxiliary polynomial is A s 2 s 4 12 s 2 16 displaystyle A s 2s 4 12s 2 16 which is again equal to zero The next step is to differentiate the above equation which yields the polynomial B s 8 s 3 24 s 1 displaystyle B s 8s 3 24s 1 The coefficients of the row containing zero now become 8 and 24 The process of Routh array is proceeded using these values which yield two points on the imaginary axis These two points on the imaginary axis are the prime cause of marginal stability 8 See also EditControl engineering Derivation of the Routh array Nyquist stability criterion Routh Hurwitz theorem Root locus Transfer function Lienard Chipart criterion variant requiring fewer computations Kharitonov s theorem variant for unknown coefficients bounded within intervals Jury stability criterion analog for discrete time LTI systems Bistritz stability criterion analog for discrete time LTI systems References Edit Routh E J 1877 A Treatise on the Stability of a Given State of Motion Particularly Steady Motion Macmillan Hurwitz A 1895 Ueber die Bedingungen unter welchen eine Gleichung nur Wurzeln mit negativen reellen Theilen besitzt Math Ann 46 2 273 284 doi 10 1007 BF01446812 S2CID 121036103 English translation On the conditions under which an equation has only roots with negative real parts by H G Bergmann in Selected Papers on Mathematical Trends in Control Theory R Bellman and R Kalaba Eds New York Dover 1964 pp 70 82 Gopal M 2002 Control Systems Principles and Design 2nd Ed Tata McGraw Hill Education p 14 ISBN 0070482896 Routh Hurwitz Criterion math24 net Retrieved 2022 07 19 Stability Analysis Tools PDF Archived from the original PDF on 2015 01 26 Retrieved 19 July 2022 a b c KUMAR Anand 2007 CONTROL SYSTEMS PHI Learning ISBN 9788120331976 a b Nise Norman 2015 Control Systems Engineering Wiley ISBN 9781118800829 Saeed Syed Hasan 2008 Automatic Control Systems Delhi Katson Publishers pp 206 207 ISBN 978 81 906919 2 5 Felix Gantmacher J L Brenner translator 1959 Applications of the Theory of Matrices pp 177 80 New York Interscience Pippard A B Dicke R H 1986 Response and Stability An Introduction to the Physical Theory American Journal of Physics 54 11 1052 Bibcode 1986AmJPh 54 1052P doi 10 1119 1 14826 Archived from the original on 2016 05 14 Retrieved 2008 05 07 Richard C Dorf Robert H Bishop 2001 Modern Control Systems 9th ed Prentice Hall ISBN 0 13 030660 6 Rahman Q I Schmeisser G 2002 Analytic theory of polynomials London Mathematical Society Monographs New Series Vol 26 Oxford Oxford University Press ISBN 0 19 853493 0 Zbl 1072 30006 Weisstein Eric W Routh Hurwitz Theorem MathWorld A Wolfram Web Resource Stephen Barnett 1983 Polynomials and Linear Control Systems New York Marcel Dekker Inc External links EditA MATLAB script implementing the Routh Hurwitz test Online implementation of the Routh Hurwitz Criterion Retrieved from https en wikipedia org w index php title Routh Hurwitz stability criterion amp oldid 1123311569, wikipedia, wiki, book, books, library,

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