Systems Analysis and Control

1 Systems Analysis and ControlMatthew M. PeetArizona State UniversityLecture 10: Routh-Hurwitz Stability CriterionOverviewIn this Lecture, you will learn:The Routh-Hurwitz Stability Criterion: Determine whether a system is stable. An easy way to make sure feedback isn t destabilizing Construct the Routh TableM. PeetLecture 10: Control Systems2 / 28A Stability TestWe know that for a system with Transferfunction G(s) =n(s)d(s)Input-Output Stability implies that all roots ofd(s)are in the Left Half-PlaneIAll have negative real (s)Re(s)CRHPQ uestion:How do we determine if all roots ofd(s)have negative real part?

2 Example: G(s) =s2+s+ 1s4+ 2s3+ 3s2+s+ 1M. PeetLecture 10: Control Systems3 / 28A Stability TestAnother VariationDetermining stability is not that hard (Matlab).Now suppose we add feedback:Controller:Static Gain: K(s) =kClosed Loop Transfer Function: y(s) = G(s) K(s)1 + G(s) K(s) u(s)x1x2mcmwuClosed Loop Transfer Function:k(s2+s+ 1)s4+ 2s3+ (3 +k)s2+ (1 +k)s+ (1 +k)We know that increasing the gain reduces steady-state error. But how high can we go?What is the maximum value ofkfor which we have stability?

3 M. PeetLecture 10: Control Systems4 / 28A Stability TestSuppose we are given a polynomial denominatord(s) =sn+an 1sn 1+ +a0 Fact:n(s)d(s)is unstable if any roots ofd(s)have negative real :How to determine if any roots ofa(s)have negative real partSimple CaseAll Real Roots. Suppose all the roots ofd(s)had negative real (s) = (s p1)(s p2) (s pn)Observe what happens as we expand out the roots:d(s) = (s p1)(s p2)(s p3)(s p4) (s pn)= (s2 (p1+p2)s+p1p2)(s p3)(s p4) (s pn)= (s3 (p1+p2+p3)s2+ (p1p2+p2p3+p1p3)s p1p2p3)(s p4) (s pn)= =sn (p1+p2+ +pn)sn 1+ (p1p2+p1p3+ )sn 2 (p1p2p3+p1p2p4+ )sn 3+ + ( 1)np1p2 pnM.

4 PeetLecture 10: Control Systems5 / 28A Stability TestSo if we writed(s) =sn+an 1sn 1+ +a0we getan 1= (p1+p2+ +pn)an 2= (p1p2+ )an 3= (p1p2p3+ )Critical Point:Ifd(s)is stable, all thepiare negative. an 1= (p1+p2+ +pn)>0 an 2= (p1p2+ )>0 an 3= (p1p2p3+ )>0 Conclusion:All the coefficients ofd(s)are positive!!! Also true if thepiare complexIHarder to show. If any coefficient is negative,d(s)is unstable. Note!If allaiare positive, that proves PeetLecture 10: Control Systems6 / 28 Example: Suspension ProblemController:Static Gain: K(s) =kClosed Loop Transfer Function:k(s2+s+ 1)s4+ 2s3+ (3 +k)s2+ (1 +k)s+ (1 +k)x1x2mcmwuExamine the denominator:d(s) =s4+ 2s3+ (3 +k)s2+ (1 +k)s+ (1 +k)All coefficients are positive for all positivek >0 Conclusion:We don t know anything PeetLecture 10: Control Systems7 / 28 Example.

5 Another ExampleConsider the very simple transfer function G(s) =1s3+s2+s+ 2 The coefficients ofd(s) =s3+s2+s+ 2are all , the roots ofd(s)are at p1= p2,3=.177 IPositive Real Part - UnstableM. PeetLecture 10: Control Systems8 / 28 Routh s MethodIntroduced in 1874 Generalizes the previous method Introduces additional combinationsof coefficients Based on Sturm s is the idea of the Routh Table Step 1:Write the polynomial asd(s) =ansn+an 1sn 1+ +a1s+a0M. PeetLecture 10: Control Systems9 / 28 Routh s MethodStep 2 Write the coefficients in 2 rows First row starts withan Second row starts withan 1 Other coefficients alternate between rows Both rows should be same lengthIContinue until no coefficients are leftIAdd zero as last coefficient if necessaryM.

6 PeetLecture 10: Control Systems10 / 28 Routh s MethodStep 3 Complete the third row. Call the new entriesb1, ,bkIThe third row will be the same length as the first twob1= det a4a2a3a1 a3b2= det a4a0a30 a3b3= det a40a30 a3 The denominator is the first entry from the previous row. The numerator is the determinant of the entries from the previous tworows:IThe first columnIThe next column following the coefficientbk= det anan 2kan 1an 2k 1 an 1 IIf a coefficient doesn t exist, PeetLecture 10: Control Systems11 / 28 Routh s MethodStep 4 Treat each following row in the same way as the third row There should ben+ 1rows total, including the first PeetLecture 10.

7 Control Systems12 / 28 Routh s MethodStep 4 Now examine the first columnTheorem number of sign changes in the first column of the Routh table equals thenumber of roots of the polynomial in the Closed Right Half-Plane (CRHP).Note:Any row can be multiplied by any positive constant without changing PeetLecture 10: Control Systems13 / 28 Routh s MethodNumerical ExampleSuppose we have a stable transfer function G(s) =1(s+ 2)(s+ 3)(s+ 5)To improve performance, we close the loop with a gain of 1000 Controller: K(s) = 1000 The Closed-Loop Transfer Function is1000s3+ 10s2+ 31s+ 1030 Question:Have we destabilized the system?

8 M. PeetLecture 10: Control Systems14 / 28 Routh s MethodNumerical Example We divide the second row by 10 There aretwosign changes:1 72and 72 103 ITwo poles in the isDestabilizing!M. PeetLecture 10: Control Systems15 / 28 Another Numerical ExampleRecall the suspension Problem with feedback:Closed Loop Transfer Function:k(s2+s+ 1)s4+ 2s3+ (3 +k)s2+ (1 +k)s+ (1 +k)Question:Can feedback destabilize thesuspension system? Is it stable for anyk >0???x1x2mcmwuLets start the Routh Table:s413 +k1 +ks321 +k0s2b1b2b3We need to find PeetLecture 10: Control Systems16 / 28 Another Numerical ExampleStart by calculating the coefficients in the first row:b1= det 1 3 +k2 1 +k 2=12(5 +k)andb2= det 1 1 +k2 0 2= 1 +kwhich givess413 +k1 +ks321 +k0s212(5 +k)1+k0sc100So far, so good.

9 Now calculate the next PeetLecture 10: Control Systems17 / 28 Another Numerical ExampleThe coefficients for the next row arec1= det 21 +k12(5 +k) 1 +k 12(5 +k)=k2+ 2k+ 15 +kandc2= +k1 +ks321 +k0s212(5 +k) 1 +k0sk2+2k+15+k001d100 Again, the first column is all positive for anyk >0 Now calculate the final PeetLecture 10: Control Systems18 / 28 Another Numerical ExampleThere is only one non-zero coefficient in the last det 12(5 +k) 1 +kk2+2k+15+k0 k2+2k+15+k=k+ 1s413 +k1 +ks321 +k0s212(5 +k)1+k0sk2+2k+15+k0011 +k00 Conclusion:No matter whatk >0is, the first column is always positive.

10 No sign changes for anyk. Stable for anyk. We ll find out why later the suspension PeetLecture 10: Control Systems19 / 28 Stability of QuadraticsWhat about a simple second-order system?1s2+bs+cWe know the poles are atp1,2= b b2 4c024681012 2 1 ResponseTime (sec)AmplitudeCalculate det 1cb0 b= bcb=cThe Routh table iss21c0sb001c0 0 Thus a quadratic is stable if and only if both coefficients are PeetLecture 10: Control Systems20 / 28 Stability of 3rd order systemsNow consider a third order system:1s3+as2+bs+c det 1ba c a= c aba=b ca det a cb ca0 b ca=cThe Routh table iss31b0s2ac0sb ca0 01c0 0So for 3rd order, stability is equivalent to.