自動(dòng)控制理論甲2第八次課件英文版

1、Principle of Automatic ControlCHAPTER 5Control-System Characteristics (Ch.6)Jy= Outline of this chapterroduction! Rouths stability criterion! Mathematical and physical form! Feedback system types! Steady-se error coefficients! Use of steady-se error coefficients! Nonunity-feedback system! Summary ro

2、ductionroductionOpen- and closed-loop transfer functions have certainbasic characteristicst permit transient and steady-seyses of the feedback-controlled system.Five factors of prime importance in feedback-controlsystems areStability Steady-se errorControllability ObservabilityParameter sensitivity

3、roductionroduction Rouths stability criterion provides a means for determining stability without evaluating the roots of the equation. The steady-se error are obtained from the open-loop transfer function for unity-feedbacksystems(or equivalent unity-feedback systems), yielding the figures of merit

4、() and a ready means for classifing systems. Based on se-space mDetermined by systems characteristicequation, many methods can be used.roductionroduction# The RouthHurwitz method isroduced as a useful tool forassessing system stability. The technique allows us to computethe number of roots of the ch

5、aracteristic equationhe righthalf-plane without actually computing the values of the roots.# Thus we can determine stability without the addedcompuional burden of determining characteristic rootslocations. This gives us a design method for determining valuesof certain system parameters stability.t w

6、ill lead to closed-loop# For stable systems we willroduce the notion of relativestability, which allows us to characterize the degree ofstability. Degree of stability is asso are.ted with how far left they roductionThe Concept of StabilityThe concept of stability can be illustrated by a cone placed

7、on a plane horizontal surface.A nesary andsufficient condition for a feedback system to be stable ist all the poles of the system transfer function have negative real parts. roductionThe Concept of StabilityThe following are physical exles of a stable,unstable and critically stable systemUnstableSta

8、bleNeutral orCritically stable roductionThe Concept of StabilityXR R P7RY pR5Y Rr0 8R0S 6X MPY uyV 4_a Y 6o R iRX4)XRI R* XR -dJdZXX ?YR Y=4 yV X X#c ?Y Y=_a 6X roduction roductionWays of Determining StabilityCompute poles from equating characteristic polynomial (denominator of closed-loop transfer

9、function) to zeroWithout explicitly computing the polesRouth(-Hurwitz) CriterionRoot LocusNyquist Criterion by systems characteristic equation. From open-loop transfer function Feneral nonlinear systems (which of course includeslinear systems as a spel case) we can use LyapunovTheory to determine st

10、abilityWe will begin our study of stability of linear systems by presenting the Routh-Hurwitz Stability Criterion roductionWays of Determining Stability E Routh Hurwitz Kfb vf=vSR H_XRX vbvRfb vCU 80,RvS R H_,vq8R0 Nyquist o0ARfb vRX,RX .,vR fby=QP +eR kfb fb _ kfb6 xh qfb vkRXRX Rouths stability cr

11、iterion(See P176-182)! Rouths stability criterion! Exles! Application of Rouths criterion! Relative stabilityy= Rouths stability criterionThe Rouths stability criterion Consider the nth-order system with closed-loop transfer functionRy G Hwhere To find the poles of the closed-loop system, we need to

12、 solve the equation sn1! b s b 0n Q s n101n s ,!, Ifis written in factored form, we have (are theroots ofs )(s bn s 1)(2 )!n 0 Multiplying the factors together we findt Rouths stability criterionThe Rouths stability criterion In other words, for a nth-degree equation we obtain If all the roots are i

13、n leflf of s-plane (LHP) : All coefficients of polynomial musve same sign All coefficients must be nonzero(except b0).$ ExThese are nesary but not sufficient conditions for stabilityles: s2 5 0 not stable, violates sec ond conditions3 s2 2s 8 0 nothing can be said, conditions are met(s) b sn b sn1al

14、l roots nnb sn2 products of all roots taken 2 aimenb sn3 products of all roots taken 3 aimen! b (1)n product of all n rootsnn !sn1 ( ! sn2 ! sn31)n ! 0n1n13213 21421 nCharacteristic polynomialQ s s b sn bn1 ! s bnn101)( G)( )(closed(GH)( sRouths stability criterionThe Rouths stability criterionStep1

15、: Order the coefficients of the characteristic$polynomialo the array s b snbn1! s b nn101Step 2: Complete the array as followsbn bn3 1n 1c 1bbn1bnc Continued until the rest of the cs are all equal to zero. 1 6bn7cbn Rouths stability criterionThe Rouths stability criterion5!Step 3: The number of root

16、s of withsitive realpart is equal to the number of changes in sign of thecolumn of the Rourray.There must be no changes in signhecolumn for astable system. This is a ne for stability.sary and sufficient condition d c1c2 11111d 12c1d 13c1!Continue until no more d terms are present. The rest of the ro

17、ws are formedhis way down to s0 row.Noticet the rest of the s1 and s0 rows contain only one term each as Routhian array in P177.$The Routh Criterion gives nesary and sufficient conditions for stability. Here is how it works:snsn1 31sn2sn3 21!s1 s0bn42!n5 !3 !j1k1!snsn1bnbn 6 !bn1 bn3!Rouths stabilit

18、y criterionThe Rouths stability criterion bn1! s b ns b sn1n01 % The Routh criterion sestthe number of roots of Q(s) with itive real parts is equal to thenumber of changes in sign of the column of the Routhianarray.There are two roots of Q(s) with%itive real parts.! Rouths stability criterionThe Rou

19、ths stability criterionQ s s5 s 10s72 2152 240Exle 1(P178)152240s s 8 Incolumn there are two%changes of sign, Q(s) has two roots he righlf s- plane (RHP),system is unstable.% In fact, the roots of Q(s) are: c bn bn3 1bn11 c 1 1 88bn1 240c 1 0bn s1 32 4Change sign second time Change sign time Rouths

20、arrayss ssc bn bn3 11bbn1n12bnn3c 1bnnnc 1 bnnn!Change sign second time Change sign time sn sn1 sn2sn3 21!s1 s0bn42n5 !3 !j1 k1Rouths array / Routhian arrayCharacteristic polynomial, Q(s)Rouths stability criterionThe Rouths stability criterion$Three distinct cases may happen:(1) No elementhecolumn i

21、s zerocolumn but there is(2) There is a zero inan elementhe same rowt is not zero(3) There is a row whose elements are all equalto zero Rouths stability criterioncolumn is zeroCase One: No elementheExle 2 Second Order SystemThe characteristic polynomial of a 2nd order system isThe Rouths array is th

22、enwhere A nesary and sufficient condition for stability of 2nd order systems ist all coefficients have the same sign.c a0() 2 1111a10 a0s2 aa02s1 s0a10c10 sa2s a2 01Rouths stability criterioncolumn is zeroCase One: No elementheExle 3 Third Order System3rd order system isThe characteristic polynomial

23、 of aThe Rourray is thenwhereRouths stability criterioncolumn is zeroCase One: No elementheExle 4 Unstable System-3rd systemThe characteristic polynomial is thenThe Rourray is then Two changes of sign he column imps two roots in righlf of s-plane. This is an unstable system. From a2a3 24 1 22 0 , it

24、 is an unstable system. s3 21s 2s1s 0124 220240(s) (s )()(3)s 2 s 8)(s )3 ss2s 24If the roots of a third order system are 1 17 s3then the system is clearly unstable (two unstable poles).A nesary and sufficient condition for stability of a 3rd order system ist all coefficients beitive anda2a3 0c 11 a

25、2 a2 a3 ,a2d 11c1c10 a0s313s202s1 s0c10d10(s) a s3 a s2 a s a a 032103Rouths stability criterioncolumn is zeroCase One: No elementheExle 5 Design ExleThe Routhian array is thenAnswer: Rouths stability criterioncolumn is zeroCase One: No elementheExle 6 Consider the system with characteristic polynom

26、ialQ(s s6 s2s9s3 5 2 12 20The Rourray is then after dividing by 3 after dividing by 4 after multiplying by 7% Incolumn there are two changes of sign, Q(s) has tworootshe righlf s- plane(RHP), system unstable. P178s sss sTheorem 1(P178): The coefficients of any row may be multipd or divided by aitive

27、 number without changing the signs of thecolumn.P178when 0, 8 0K0,8, thesystemis stable.a2a3 2K For a 3rd system, if the system is a stable systems3 41s 2s1s 02K8 K02K0Consider the third order system with the following characteristic polynomial (ss4K . Find for which values of K the system is stable

28、.Rouths stability criterioncolumnCase Two: A zero coefficienthe If element of a row is zero, two methodshe book could be used. Methhod 1 Substitude s=1/xhe original equationLet s=1/x, and rearrange the characteristic polynomial Q(s)Exle 7The Routhian array isThe new Routhian array isQ(x) 5x4 2x3 2x2

29、 x 1 0 Rouths stability criterioncolumnCase Two: A zero coefficienthe If element of a row is zero, two methodshe book could be used. Methhod 2 Multiply the original polynomial by the factor(s+1)Multiply(s+1), and rearrange the polynomial Q(s)Q s s4 s5Exle 8The Routhian array isQ(s ss3s37 54 2The new

30、 Routhian array is P180Same result is obtained as method 1. Two changes in sign.Unstable system with two poles in righlf of s-planes5 31s4 42s3 92s2 s1 s07510 101110s4 21s3 21s2 505(P179)Theorem2: The number of roots x withitive real parts will be the same as the number of s roots withitive real par

31、tsNote: This method does not work when the coefficients of Q(s) and of Q(x) are identical Two changes in sign. !Unstable system with two poles in righlf of s-planex4x3x2x52 1211 25s4 21s3 21s2 505Q(s) 1 1 2 2 5 0 x4x3x2xQ s s4 s5Rouths stability criterionCase Three: A zero row s2s )( ), ( j )(s j),s

32、2 s s 2)(s nnnn If the ith row of the Routhian array is zero, we form the following auxiliary(f/R) polynomial from the previous(nonzero) row (!ml) i1 i1 i3U (ss!2 3 where i are the coefficients of the previous row. Order(kM) of the polynomial indicates number of symmetrical roots We then replace the

33、 ith row by the coefficients of the derivative of the auxiliary polynomial with respect to s and complete the array to obtain info about other roots, besides the symmetrical roots. Rouths stability criterionCase Three: A zero rowExle 9 case 3The Routhian array is thenThen use the auxiliary polynomia

34、l of the row preceding the row of zeros to go on. Roots ofK=8U(s) are also roots of s2 2( j2)(s j)2Ks028 s2U s3 41s 2 82s1 00s 0 08All roots stable except the pair on imaginary axis(s=2j)dU 4sdss3 41s 2 82s1 04s 0 08s3 41s 2s1s 02K8 K02K0Go back to the Design Exle 5 and make K=8 to get the Routhian

35、array presented next ( (s s4 8)Theorem 3(P170): A zero row. If all elements in a row are zero then the polynomial has real or complex roots symmetric(=U) to origin(1P) of the formRouths stability criterionCase Three: A zero rowExle 10 (P180) case 30The Routhian array isThen the auxiliary equation is

36、s0 (s j3)(s j3 U Rouths stability criterionExlesEx. 11. A characteristic equation of a system is known as s4 s3 s2 s Using Rouths criterion to determine the stability of the system.Solution: All coefficients of the polynomial areitive andnonzero, it meets nesary conditions for stabilityThe Rourray i

37、s then Actually, the characteristic equation can be factored s+2) (s+3) (s2 +s+1) = 0All roots have negative real part, so the system is stable.s4s3s2 s1 s0112 6611 061/455 / 6106 All are itive , thesystem is stable.There are no changes of signhecolumn, there are no roots withimaginary axis(s=3j).it

38、ive real parts except the pair ondU 2sdss s91s91s22 180s219s12s09Consider the systems characteristic equationQ(s s4 s1118 18 Rouths stability criterionExlesEx. 12. A characteristic equation of a system is known as s5 s4 s3 s2 s 0Using Rouths criterion to determine the stability of the system.Solutio

39、n: All coefficients of the polynomial areitive andnonzero, it meets nesary conditions for stabilityThe Rourray is then21s ss 95(after multiplying by 3)1115(after multiplying by 5/2)s s s174(0after multiplying by 11)1500 Rouths stability criterionExlesEx. 13. A characteristic equation of a system is

40、known as s3 s2 s Use Rouths criterion to determine the stability of the system.Solution: For a three-order systems3s2116s1s00 U2160dU1 20s dsThere are a pair of imaginary roots.S1,2=j4Oscillation frequency nAuxiliary polynomialThe system is unstable.1 10 0There are 2 changes of signs: , the system i

41、s unstable. There are 2 roots with itive real parts. Thesystem is unstable.Rouths stability criterionApplication of Rouths criterionExle 14 (P181) - Design Exle Consider the 4th-order system with closed-loop transfer function Ry G The value of K is an adjustable parameter inHthe system. It is import

42、ant to know the rangeof values of K for which the system is stable.0K0 Rouths stability criterionApplication of Rouths criterion(supplement)! 63jX RX _jXRX X i8XR XR s = z - 1 KRf= QP 0Rf= f=tv D R/ vR/ T X % Vai TD: s4Ks3 Ks280 KK(80 )(25 ) 98Ks(80 K)0s14K281.,712 432000 0K2RT a2a3 0 K 0 30 - K 0 K

43、Rd K 2RT T 0 T - 5 0; !T 5 8kWYVKP101 Rouths stability criterionRelative StabilityK)(02s17)Exle-19. LcRCUX ;SlNRKRN ,Vs$ 2 D9RK d ,f=!(s)= s3 7s2 17s Ks3 s2 s1 s017? K17K a a 02377 K 77 K 1 1197/j 17; n 17SlNS1 K119= 0 , K = 119Nn? s2 + K = 0n=?10T 250 0;T 25T 52=a1*a2-a0*a30 2a1*a1*a4/a3XRTRd T s41

44、T100s32100s2T 5100s110T 2500s0T 5100Rouths stability criterionRelative StabilityK)(02Exle-19. LcRCUs17)X ;SlNRKRN ,Vs$ 2 D9RK d s = z 2! (z)= (z -2)3+ 7(z -2)2 + 17(z2)+ K= z3+ z2+ z+ K 14- K 14 a2 a3 0z3 z2R 11K-141z1 z0-K+150-14+K K+15 0 K 0 K 15 AQVC7jQVC0)3nWY dAQg.G;# SWYRVC EN*WYVRKR4+91 G (s)

45、 01 G (s) 0abs3 4s2 ( K 5)s K 0s3 3.5s2 3.5s 12.25 0 20s j1.876H!=_LK3 AQV (jQIBVAQ$V (jQ5%VG (s) 11.25b(s 0.5)(s 1)(s 2)G (s) K (s 1)as(s 1)(s 5)(1K 14 K 15 ,Vs$ 2 D9 Stability of se equation m We have learnt the stability criterion for a control system, however, is it useful to a system which repr

46、esented by ase space m? Yes. It is because of the characteristic equation for a systembeing changeleo matter what mis described.Se equationIts characteristic equation isWhich is as same ast of matrix A, so the roots of thecharacteristic equation of course are same.Thus, the nesary and sufficient con

47、dition for a stablesystem represented by se equation ist all eigenvalues ofmatrix A having negative real part. Rouths criterion therefore can be used. Mathematical and physical form(See P182-183)y= s 0 x!AxBu tMathematical and physical formMathematical and physical formIn various systems the control

48、led variable, labeled C, shown in Fig. 6.1may be the physical form ofition, speed, flow, temperature, etc.Once the blockshe diagram are related to transfer functions, thesystem can beyzed witho variable may be.Generally, the importantare what the physical controlledties are theseveral derivatives of

49、c(cRO For any specific control system, each of these “mathematical” functions has a definite “physical” meaning.CR G HFig.6.1 Mathematical and physical formMathematical and physical formOften the input signal to a system has an irregular form, asshown in Fig.6.2. We cannot get the system response ea

50、sily.For LTI system, the signal form shown in Fig.6.2 may beconsidered to be com input signals: step, red of three basic forms of known types of and parabola.For a given LTI system, it canbeyzed separay for eachof these types of inputs.Fig. 6.2 rtExlescontrol system for a missileition-control system

51、temperature profile of pros controlFor exle, if c isition, then Dc is velocity and D2c is acceleration.Outline of this chapterroduction! Rouths stability criterion! Mathematical and physical form! Feedback system types! Steady-se error coefficients! Use of steady-se error coefficients! Nonunity-feed

52、back system! Summary stability, accuracy, controllability, observability, sensitivity Feedback system types(See P183-191)steady se errorfeedback system typesysis of system typesy= Characterizing feedback control system ?Feedback system typesSteady se errorControlled variable Overshoot Steady se e!rr

53、o!r Transient Se Steady Se Time Settling time Feedback system typesSteady se error Control systems are inherently dynamic, their performanceis usually specifiederms of both the transient responseand the steady se responseF (s)ControllerE (s) t!R (s)C(s) For a stable system, transient response will d

54、isappear with time, i.e. H(s)( )ss)(t( t)ss=0KCpfFeedback system typesSteady se error e( ) t c t ,System error is defined ashoweverc(t) usually cannot be obtained directly, it is measured by equipment to produce a feedback signal z(t), thus errormay be e( ) t z t If the system is unity feedback, z(t

55、)=c(t).Steady se error is defined asIf the system is unity feedback z(t)=c(t), then Feedback system typesSteady se errorAs shownhe Fig., the Laplace form of system output isThe Laplace form of system error isF (s)Controller E (s)R (s)C(s) Z (s)H(S) E (s) H s C(s)1)(f ( H s)( G s)( )H (s)1G s )(H (s)

56、Produced by disturbanceCpProduced byset-poinputf)(G s )( )(f)(1 G s )(H s1 G s )(H (s)(ss lim ( c t)tR ECGZH)(ss lim ( z t)tFeedback system typesSteady se errorF (s)f E (s)E (s) H s C(s)R (s)C(s) f ( H s) )( 1)(pC G s)( )H (s)1G s )(H (s)Z (s) H(S) lim sGf (s) )(e lim s)(esfs0 1 GsrG s)( )H (s)s )(H

57、 ss0pcDisturbanteady se errorSet-posteady se errorSo, the system steady se error isFeedback system typesFeedback system types (XV,? For an unity feedback system as shown in Fig.6.4, which is called “tracker”. The open loop transfer function for this system is G(s)=C(s)/E(s)Generally, G(s) has the fo

58、llowing mathematical form: Standard form CER G Fig. 6.4 a1, a2,. b1, b2,. are constant coefficients Km is gain constant of the transfer function m=0,1,2,.denotes the transfer function type K 1( b s b 2 !sw ) )(m21wmssm1( a sa 2 !su )21uForward G(s) witity gainG(s) Km (1 T1s)(1 T2 s)!sm (1 T s)(1 T s

59、)!abwhere:b,. T1, T2,. are constant coefficients, Km is gain constant of transfer function.Steady se error is determined both with inputs r(t) and system m.Motiviion of system “Type” For unity feedback system with typical input,how to have steady se error estimate quickly?how to design the system wi

60、th error elimination when input is known? sssresfBy final value theoremFeedback system typesFeedback system types (XV,? A system “type” designation is based upon the value of the exponent m of s in Eq.(6.21)open-loop G(s).Once a physical system has been expressedType “0” system, when m=0 Type “1” sy