5ThePerformanceofFeedbackControlSystems反馈控制系统的性能课件.ppt

1、Basic Basic conceptsconceptsSystem System modelingmodelingPerformance Performance issuesissuesanalysiscorrectionTime domainComplex domainFrequency domainTest Input Signals Test Input Signals Performance of Second-Order SystemsPerformance of Second-Order SystemsThe s-Plane Root Location and the Trans

2、ient Response The s-Plane Root Location and the Transient Response The Steady-State Error of Feedback Control SystemsThe Steady-State Error of Feedback Control SystemsEffects of a Third Pole and a Zero on the Second-Order Effects of a Third Pole and a Zero on the Second-Order System ResponseSystem R

3、esponsePerformance IndicesPerformance IndicesDesign ExamplesDesign Examples-Specify the measures of performanceDefine and measure the performance of control systems Stability Transient response：the response that disappears with time Steady-state response:the response that exists for a long time foll

4、owing an input signal initiationDesign specifications:for a specified input command Time response indices Desired steady-state accuracy Effective compromiseStep inputStep inputRamp inputRamp inputParabolic inputParabolic inputUnit impulseUnit impulse The The unit unit impulse function impulse functi

5、on (t)(t)has has the following properties:the following properties:The The impulse input is useful when we consider the convolution integral impulse input is useful when we consider the convolution integral for the for the output output y(ty(t)in terms of an input in terms of an input r(t),r(t),whic

6、h is written aswhich is written asThe general form of standard test signals:The general form of standard test signals:1/,()0()220,otherwisetftt()1,()()()t dtta g t dtg a1()()()()()ty tg trdLG s R sIf the input is a unit impulse function,we have The integral has a value only at =0;therefore,the impul

7、se response of the system G(s).The impulse response test signal can often be used for a dynamic system by subjecting the system to a large-amplitude,narrow width pulse of area A.The standard test signals are of the general form and the Laplace transform()()()ty tg td ()()y tg t1!(),()nnnr ttR ssWith

8、 a unit step input2222()()()()()()1()2generalizednnnG sKY sR sR sY sR sG sspsKss 212,1nns s 222()(2)nnnY ss ss211()1sin()1,cos,01ntny tetwhereand T h e r e s p o n s e a s a function of and time is alsoshown in Figure 5.5(b)for a step input.As decreases,the closed-loop roots approach the imaginary a

9、xis,and the response becomes increas-ingly oscillatory.With an impulse function input R(s)=1With an impulse function input R(s)=1tetysssYntnnnnnsin)(2)(222The swiftness of the response is measured by the rise time Tr and the peak time Tp.For under-damped systems with an overshoot,the 0-100%rise time

10、 Tr is a useful index.If the system is over-damped,then the peak time is not defined,and the 10-90%rise time Tr1 is normally used.The actual response matching the step input is measured by the percent overshoot and settling time Ts.The percent overshoot is defined as where Mpt is the peak value of t

11、he time response,and fv is the final value of the response.Standard performance measures.100%PtvvMfP OfThe settling time,The settling time,TsTs,is defined as the time required for the system to settle within,is defined as the time required for the system to settle within a certain percentage a certa

12、in percentage of the input amplitude.of the input amplitude.For the second-order system with closed-loop damping constant For the second-order system with closed-loop damping constant,Ts Ts for which for which the response remains within 2%of the final value is the response remains within 2%of the f

13、inal value is Hence,we will define the settling time as four time constants(that is,r=l/)of the dominant roots of the characteristic equation.nssntTTen44402.0Notice:Notice:The transient response of the system may be described in terms of two factors:The transient response of the system may be descri

14、bed in terms of two factors:The swiftness of response,as represented by the rise time and the peak time.The swiftness of response,as represented by the rise time and the peak time.The closeness of the response to the desired response,as represented by the The closeness of the response to the desired

15、 response,as represented by the overshoot and settling time.overshoot and settling time.As nature would have it,these are contradictory requirements,and a As nature would have it,these are contradictory requirements,and a compromise must be promise must be obtained.22122222122()()(2)2()sin2nnnnnnntn

16、nnnndy tLsY ssdts ssssdy tLetdtss tedttdyssssYntnnnnnsin)(2)(222tedttdyntnnsin)(Let dy(t)/dt=0,we obtain nt=.%100.21eOP1()1sin()ntny tet 222()(2)nnnY ss ss21pnT2/11PtMe Calculation of the measuresCalculation of the measures%100.21eOP21pnT8.03.0,60.016.2 1nrTTr1 versus When is set to 0.2When n is set

17、 to 5For a given,the response is faster for larger n.The overshoot is independent of n.For a given n,the response is faster for lower.T h e s w i f t n e s s o f t h e response,however,will be limited by the overshoot that can be accepted.)1)(12(1)(2ssssTWhen|1/|10|n|,the performance indices can be

18、represented by the ones of the second order system.In the case,the poles of the second order system are called dominant poles of the system.%100.21eOPnsT44 Notice:the above results is only for a transfer function without finite zeros.2222)(/()(nnssasasTnFor the given system,select the gain For the g

19、iven system,select the gain K K and the parameter and the parameter p p so that the time-domain so that the time-domain specifications will be satisfied.specifications will be satisfied.P.O.5%P.O.5%Ts 4sTs 4s=0.707,P.O.=4.3%=0.707,P.O.=4.3%Ts=4/Ts=4/n n4,4,n n11Chose r12=-1Chose r12=-1 j,then P.O.=4

20、.3%Ts=4sj,then P.O.=4.3%Ts=4s=0.707,=0.707,n n=1/=1/=1.414=1.414 222)(22ssKpssKsTIf a n n and 1/n n,the pole and zero will have little effect on the step response.P.O.=55%according to Fig 5.13(a)Ts=4/3=1.33s)1)(2()(/()()()(222sssasasTsRsYnnnUsing a computer simulation for the actual third-order syst

21、em,we find that the percent overshoot is equal to 38%and the settling time is 1.6 seconds.Thus,the effect of the third pole of T(s)is to dampen the overshoot and increase the settling time(hence the real pole cannot be neglected).According to the percent overshoot P.O.According to the percent oversh

22、oot P.O.According to the number of cycles of the damped sinusoid during According to the number of cycles of the damped sinusoid during TsTs%100.21eOP1()1sin()ntny tet The frequency of the damped sinusoidal term for 0For a ramp inputMobile robot steering controlsKKsG/)(211In the case of the steering

23、 control system,we want to increase the gain factor KK2 in order to increase Kv and reduce the steady-state error.However,an increase in KK2 results in an attendant decrease in the systems damping ratio and therefore a more oscillatory response to a step input.Thus,we want a compromise that provides

24、 the largest Kv based on the smallest allowable.ess=0A nonunity feedback systemA speed control system:K1 and K2 account for the conversion of one set of units to another set of units.The equivalent block diagram with K1=K2.A unity feedback system.If K1=K2,the system is transformed to that of Fig 5.2

25、3(for the dc gain or steady-state calculation).Determine K1 and calculate the steady-state error for a unit step input.Solution:Select K1=K2=2Example 5.4 Example 5.4 Steady-state errorSteady-state erroror 5.9%of the magnitude of the step input.Assume we cannot insert a gain K1 following R(s).The act

26、ual error is E(s)=1(s)-T(s)R(s).Try to determine an appropriate gain K so that the steady-state error to a step input in minimized.To achieve zero steady-state error,we require thatThus K=4 will yield a zero steady-state error.Example 5.5 Feedback SystemExample 5.5 Feedback SystemSolution:A performa

27、nce index is a quantitative A performance index is a quantitative measure of the performance of a system measure of the performance of a system and is chosen so that emphasis is given and is chosen so that emphasis is given to the important system specifications.to the important system specification

28、s.ISE:ISE:IAE:IAE:ITAE:ITAE:ITSE:ITSE:TdtteISE02)(TdtteIAE0|)(|TdttetITAE0|)(|TdttteITSE02)(TdtttytrtefI0),(),(),(The calculation of the integral squared error(ISE).121)(2sssTFor a step inputExample 5.6 Example 5.6 Performance criteriaPerformance criteriaThe performance index ITAE provides the best

29、selectivity of the performance indices.The value of the damping ratio selected on the basis of ITAE is 0.7.For a second-order system,this results in a swift response to a step with a 4.6%overshoot.S e l e c t K3 t o minimize the effect of the disturbance D(s).Example 5.7 Example 5.7 Space telescope

30、control systemSpace telescope control systemWith K1=0.5,K1K2Kp=2.5 and a unit step disturbanceThen the natural frequency of the vehicle is 2.5/20.25/snfcyclesK3=3.2 and =0.5.(ISE)K3=4.2 and =0.665.(IAE)Complex systems with high-order transfer functions Complex systems with high-order transfer functi

31、ons lower-order approximate modellower-order approximate model Method 1:delete a certain insignificant pole,in the meanwhile Method 1:delete a certain insignificant pole,in the meanwhile retain the steady-state response.retain the steady-state response.Example:Example:)2(30/)()130)(2(30/)30)(2()(ssK

32、sGsssKsssKsG)()()()(ssMsLsH Criteria:Select ci and di in such a way that L(s)has a frequency response very close to that of H(s)q=1,2,in which the poles are in the left-hand s-plane and m n.where p g n,K without change.Poles:S=-1,-2,-3 -1.029,-1.555Choose K and K1 so that:(1)The percent overshoot of

33、 the output to a step command r(t)10%;(2)The steady-state error to a ramp command is minimized;(3)The effect of a step disturbance is reduced.(1)Select K and K1 to meet P.O.10%for R(s)=A/s.Set D(s)=0.When When =0.6,P.O.=9.5%.=0.6,P.O.=9.5%.(2)Examine the steady-state error for a ramp input.(2)Examin

34、e the steady-state error for a ramp input.(3)Reduce the effect of a step disturbance.(3)Reduce the effect of a step disturbance.The steady-state error due to a unit step disturbance is equal to-1/K.The steady-state error due to a unit step disturbance is equal to-1/K.The transient response of the er

35、ror due to the step disturbance input can be The transient response of the error due to the step disturbance input can be reduced by increasing K.reduced by increasing K.(4)In summary,we need large K,large K/K1 and(4)In summary,we need large K,large K/K1 and =0.6.=0.6.2.12.12.16.022112KKKKKKKKnnnSel

36、ect K=25,K1=6,K/K1=4.17;Select K=100,K1=12,K/K1=8.33.Realistically,we must limit K so that the systems operation remains linear.K=100,eK=100,essss=B/8.33=0.12B=B/8.33=0.12BGoal:Goal:Achieve the fastest response Achieve the fastest response to a step input r(t);to a step input r(t);Limit the overshoo

37、t and Limit the overshoot and oscillatory nature of the oscillatory nature of the response;response;Reduce the effect of a Reduce the effect of a disturbance on the output disturbance on the output position of the read head.position of the read head.Neglect the effect of the coil inductance.Compromi

38、se:Ka=40Be aware of key test signals used in controls and of the resulting transient response characteristics of second-order systems to test signal inputs.Recognize the direct relationship between the pole locations of second-order systems and the transient response.Be familiar with the design form

39、ulas that relate the second-order pole locations to percent overshoot,settling time,rise time,and time to peak.Be aware of the impact of a zero and a third pole on the second-order system response.Gain a sense of optimal control as measured with performance indices.AssignmentsAssignmentsSkills CheckSkills CheckE5.4 E5.8 E5.9 E5.18E5.4 E5.8 E5.9 E5.18