StabilityAsymptotically StableSlide 3Slide 4Routh CriteriaExampleSlide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Stability•BIBO stability:Def: A system is BIBO-stable if any bounded input produces bounded output. otherwise it’s not BIBO-stable.))(( resp. impulse)( where finite )(stable-BIBO :Thm10sHthdtthLcancelled pole/zero allafter plane half leftopen in the )( of poles"" all stable-BIBO sHAsymptotically StableA system is asymptotically stable if for any arbitrary initial conditions, all variables in the system converge to 0 as t→∞ when input=0.0part real have seigenvalue all a.s. is systemA Thm: If a system is A.S. then it is BIBO-stableBut BIBO-stable does not guarantee A.S.in generalIf system is C.C. & C.O., then no pole/zero cancellation In this case, BIBO stable  Asymp Stable315411321253116420111 : : : )(Let nnnnnnnnnnnnnnnnnnnnnnnnnsaaaaaaaaaasaaasaaaasasasasasdRouth TableRouth Stability criterion:1) d(s) is A.S. iff 1st col have same sign2) the # of sign changes in 1st col = # of roots in right half plane2nd order: stable iff all coeff same sign3rd order: stable iff a) all coeff same sign b) b*c>a*dRouth CriteriaRegular case: 1st col all non zero(1) A.S. 1st col. all same sign(2) #sign changes in 1st col. = #roots with Re(.)>0Special case 1: one whole row=0Solution: 1) use prev. row to form aux. eq. A(s)=0 2) get 3) use coeff of in 0-row 4) continue )(sAdsdExample)1(444)(44:0:1414:1616:142814:781:47884)(222123452345sssAssssssssssssd :row prev. ←whole row=0stable marginally.originally col. 1st in 0 have did wesince A.S.Not ButR.H.P in roots no col. in change sign No :atedifferenti14:8:1414:218)(0122ssssssdssdA103229.15.147884)().()1(444)()()()(234522-jjssssssdsdjssssAsAsdA(s) :roots has Indeed of roots are these& :roots has :example prev in :e.g. original of roots all are of root The :Fact 7 )474)(1()(044447747448744744788412322232324234352323452sssssds-sss-sssss-sssss-sssssssss0114444)()(12100121121122)(12332443452345:s:s:sssdssdAsAsss:s:s:sssssssd :row From e.g.0)Re( withroots nocol. 1st in change sign No :row From1:2:2)(1)(0122sssdssdAssAsunstable. is roots. double are &0))fence.(Re( the on roots areThey at root double at root double of roots are of roots But)()()()1(12)().(12)(22222424sdjsjsjsjsssssAsdsssA 3: 32: 3 : 3 0: 2 1:3 2 1:322)( e.g.continue & 0by "0" replace :solution0row but whole 0col1st in # one :2 case Special012234234s-ssssssssssdreplace0)Re( have two these :roots has :Verify0)Re( i.e. RHP, in roots 2col. 1st in changes sign 2 0 assume wesince2928140570902.009057.00)(032.j.jsdUseful case: parameter in d(s)How to use: 1) form table as usual2) set 1st col. >03) solve for parameter range for A.S.2’) set one in 1st col=03’) solve for parameter that leads to M.S. or leads to sustained oscillationExamples+3s(s+2)(s+1) KpppppplcpKsKsssKssssdsKssssKsGK3)2(3)3()1)(2()()3()1)(2()3()(23.. char.poly :Solstability for of range find:Q+003)2(30303:33)2(3:33:21:0123pppppppppKKK KKsKKsKsKs col. 1st : A.S.For :table Routh=6037)1(0342046341)2(32020034)2(3)(22223kkkkkkkkkkkskksssd 2) 1) :need we: A.S.Forprod. outer two mid of prod 2) 0coeff all 1) criteria? Routh order 3rd remember e.g.A.S.for need weall over also. and but or 528.01372013737113737137)1(2kkkkkkkkk>0.5)137(3434340)(43)(,1374)2(3221kkjssAkssssdkkkk:freq oscinoscillatio sustained to leads :row From0row AndM.S. is this At :get we set