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

Curriculum￿SystemBasic￿conceptsSystem￿modelingPerformance￿issuesanalysiscorrection•Time domain•Complex domain•Frequency domainChapter￿5￿￿The￿Performance￿of￿Feedback￿Control￿Systems•Test￿Input￿Signals￿•Performance￿of￿Second-Order￿Systems•The￿s-Plane￿Root￿Location￿and￿the￿Transient￿Response￿•The￿Steady-State￿Error￿of￿Feedback￿Control￿Systems•Effects￿of￿a￿Third￿Pole￿and￿a￿Zero￿on￿the￿Second-Order￿System￿Response•Performance￿Indices•Design￿Examples-Specify the measures of performanceDefine￿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￿following￿an￿input￿signal￿initiationDesign￿specifications:￿for￿a￿specified￿input￿command◦Time￿response￿indices◦Desired￿steady-state￿accuracy◦Effective￿compromise5.1￿￿￿IntroductionStep￿inputRamp￿inputParabolic￿input5.2￿￿￿Test￿Input￿SignalsUnit￿impulse￿￿￿￿￿￿The￿unit￿impulse￿function￿δ(t)￿￿has￿the￿following￿properties:￿￿￿￿￿￿￿￿￿￿￿￿The￿impulse￿input￿is￿useful￿when￿we￿consider￿the￿convolution￿integral￿ for￿ the￿ output￿y(t)￿in￿ terms￿ of￿ an￿ input￿r(t),￿which￿ is￿written￿asThe￿general￿form￿of￿standard￿test￿signals:5.2￿￿￿Test￿Input￿SignalsIf￿the￿input￿is￿a￿unit￿impulse￿function,￿we￿have￿￿￿￿The￿integral￿has￿a￿value￿only￿at￿τ￿=￿0;￿therefore,￿￿￿￿￿the￿impulse￿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￿5.2￿￿￿Test￿Input￿Signals5.2￿￿￿Test￿Input￿SignalsWith a unit step input5.3￿￿￿Performance￿of￿a￿Second-order￿System The response as a function of ζ and time is alsoshown in Figure 5.5(b) for a step input. As ζ decreases, the closed-loop roots approach the imaginary axis, and the response becomes increas-ingly oscillatory. 5.3￿￿￿Performance￿of￿a￿Second-order￿SystemWith￿an￿impulse￿function￿input￿R(s)=15.3￿￿￿Performance￿of￿a￿Second-order￿SystemThe￿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￿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￿the￿time￿response,￿and￿fv￿is￿the￿final￿value￿of￿the￿response.Standard performance measures5.3￿￿￿Performance￿of￿a￿Second-order￿SystemThe￿settling￿time,￿Ts,￿is￿defined￿as￿the￿time￿required￿for￿the￿system￿to￿settle￿within￿a￿certain￿percentage￿δ￿of￿the￿input￿amplitude.￿For￿the￿second-order￿system￿with￿closed-loop￿damping￿constant￿ζωζω„,￿Ts￿for￿which￿the￿response￿remains￿within￿2%￿of￿the￿final￿value￿is￿Hence, we will define the settling time as four time constants (that is, r = l/ζω„) of the dominant roots of the characteristic equation.5.3￿￿￿Performance￿of￿a￿Second-order￿SystemNotice:◦The￿transient￿response￿of￿the￿system￿may￿be￿described￿in￿terms￿of￿two￿factors: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￿overshoot￿and￿settling￿time.◦As￿nature￿would￿have￿it,￿these￿are￿contradictory￿requirements,￿and￿a￿compromise￿must￿be￿obtained.5.3￿￿￿Performance￿of￿a￿Second-order￿SystemLet dy(t)/dt=0, we obtain ωnβt=π.5.3￿￿￿Performance￿of￿a￿Second-order￿SystemCalculation￿of￿the￿measures5.3￿￿￿Performance￿of￿a￿Second-order￿SystemTr1 versus ξ5.3￿￿￿Performance￿of￿a￿Second-order￿SystemWhen ξ is set to 0.2When ωn is set to 5For a given ξ, the response is faster for larger ωn. The overshoot is independent of ωn.For a given ωn, the response is faster for lower ξ. The swiftness of the response, however, will be limited by the overshoot that can be accepted.5.3￿￿￿Performance￿of￿a￿Second-order￿System5.4￿￿￿Effects￿of￿a￿Third￿Pole￿and￿a￿Zero￿on￿the￿Second-order￿System￿ResponseWhen |1/γ|≥10|ζζωn|, the performance indices can be 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.Notice: the above results is only for a transfer function without finite zeros. 5.4￿￿￿Effects￿of￿a￿Third￿Pole￿and￿a￿Zero￿on￿the￿Second-order￿System￿ResponseSimulation￿results￿for￿Simulation￿results￿for￿ζ ζ=0.45=0.455.4￿￿￿Effects￿of￿a￿Third￿Pole￿and￿a￿Zero￿on￿the￿Second-order￿System￿ResponseWhen￿the￿transfer￿function￿has￿a￿zero,￿When￿the￿transfer￿function￿has￿a￿zero,￿5.4￿￿￿Effects￿of￿a￿Third￿Pole￿and￿a￿Zero￿on￿the￿Second-order￿System￿ResponseExample￿5.1￿￿￿Example￿5.1￿￿￿Parameter￿selectionParameter￿selectionFor￿the￿given￿system,￿select￿the￿gain￿K￿and￿the￿parameter￿p￿so￿that￿the￿time-domain￿specifications￿will￿be￿satisfied.￿◦P.O.≤5%◦Ts￿≤4sζ=0.707,￿P.O.=4.3%Ts=4/ζωn≤4,￿￿￿ζωn≥1Chose￿r12=-1±j,￿then￿P.O.=4.3%￿Ts=4sζ=0.707,￿ωn=1/ζ=1.414￿5.4￿￿￿Effects￿of￿a￿Third￿Pole￿and￿a￿Zero￿on￿the￿Second-order￿System￿ResponseExample￿5.2￿￿￿Dominant￿poles￿of￿T(s)If a>> ζωn and τ<< 1/ζωn, the pole and zero will have little effect on the step response.P.O.=55% according to Fig 5.13(a)Ts=4/3=1.33s5.4￿￿￿Effects￿of￿a￿Third￿Pole￿and￿a￿Zero￿on￿the￿Second-order￿System￿ResponseUsing a computer simulation for the actual third-order system, 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￿number￿of￿cycles￿of￿the￿damped￿sinusoid￿during￿Ts5.5￿￿￿Estimation￿of￿the￿Damping￿RatioThe frequency of the damped sinusoidal term for ζ< 1 isThe number of cycles in 1 second is ω/2π. The time constant for the exponential decay is τ = l/ζωn in seconds. The number of cycles of the damped sinusoid during one time constant is Assuming that the response decays in n visible time constants. For the second-order system, the response remains within 2% of the steady-state value after four time constants (4 τ, i.e., n = 4)Example:￿examine￿the￿response￿shown￿in￿Figure￿for￿ζ￿=￿0.4.5.5￿￿￿Estimation￿of￿the￿Damping￿RatioUse y(t) = 0 as the first minimum point and count 1.4 cycles visible (until the response settles with 2% of the final value). Then we estimate5.6￿￿The￿S-plane Root Location and the Transient ResponseEa(s), actuating signal, which is a measure of the system error.E(s)=R(s)-Y(s), the actual system error.When H(s)=1,5.7￿￿The￿Steady-state￿Error￿of￿Feedback￿Control￿SystemsN is call the type of systems. N=0, type-0 system; N=1, type-1 system; N=2, type-2 system;For a type-0 system Defineas position error constant.For N≥15.7￿￿The￿Steady-state￿Error￿of￿Feedback￿Control￿SystemsStep￿inputFor the system with N=0, the steady-state error is infinite.For the type-1 system, is defined as velocity error constant.Thus the steady-state error exists. For the type-2 system, the steady-state error is zero.Ramp￿inputRamp￿input5.7￿￿The￿Steady-state￿Error￿of￿Feedback￿Control￿SystemsFor N=0 and 1, ess=∞;For N=2Acceleration error constant5.7￿￿The￿Steady-state￿Error￿of￿Feedback￿Control￿SystemsAcceleration￿inputAcceleration￿input•For a step input When K2=0When K2>0•For a ramp input5.7￿￿The￿Steady-state￿Error￿of￿Feedback￿Control￿SystemsExample￿5.3￿Example￿5.3￿Mobile robot steering control￿ ￿In the case of the steering 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 system's damping ratio ζ and therefore a more oscillatory response to a step input. Thus, we want a compromise that provides the largest Kv based on the smallest ζ allowable.ess=05.8￿￿The￿Steady-state￿Error￿of￿of￿Nonunity￿Feedback￿SystemsA 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 5.23 (for the dc gain or steady-state calculation) .5.8￿￿The￿Steady-state￿Error￿of￿of￿Nonunity￿Feedback￿SystemsDetermine K1 and calculate the steady-state error for a unit step input.Solution:Select K1=K2=25.8￿￿The￿Steady-state￿Error￿of￿of￿Nonunity￿Feedback￿SystemsExample￿5.4￿￿￿￿Steady-state￿erroror 5.9% of the magnitude of the step input.Assume we cannot insert a gain K1 following R(s). The actual 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.5.8￿￿The￿Steady-state￿Error￿of￿of￿Nonunity￿Feedback￿SystemsExample￿5.5￿￿Feedback￿SystemSolution:A￿performance￿index￿is￿a￿quantitative￿measure￿of￿the￿performance￿ of￿ a￿ system￿ and￿ is￿chosen￿so￿that￿emphasis￿is￿given￿to￿the￿important￿system￿specifications.ISE:￿IAE:￿￿ITAE:ITSE:￿5.9￿￿￿￿Performance￿IndicesThe calculation of the integral squared error (ISE).For a step input5.9￿￿￿￿Performance￿IndicesExample￿5.6￿￿￿Performance￿criteriaThe performance index ITAE provides the best selectivity of the performance indices. The value of the damping ratio ζ 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.Select K3 to minimize the effect of the disturbance D(s).5.9￿￿￿￿Performance￿IndicesExample 5.7 Example 5.7 Space telescope control systemSpace telescope control systemWith K1=0.5, K1K2Kp=2.5 and a unit step disturbance5.9￿￿￿￿Performance￿IndicesThen the natural frequency of the vehicle is K3 = 3.2 and ζ= 0.5. (ISE)K3 = 4.2 and ζ= 0.665. (IAE)Complex￿systems￿with￿high-order￿transfer￿functions￿￿lower-order￿approximate￿model◦Method￿1:￿delete￿a￿certain￿insignificant￿pole,￿in￿the￿meanwhile￿retain￿the￿steady-state￿response.Example:￿5.10￿￿￿The￿Simplification￿of￿Linear￿Systems-￿Method￿2:￿frequency￿response￿methods-￿Method￿2:￿frequency￿response￿methods 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,…… 5.10￿￿￿The￿Simplification￿of￿Linear￿Systemsin which the poles are in the left-hand s-plane and m < n.where p ≤ g < n, K without change.Example￿5.9￿￿￿￿Example￿5.9￿￿￿￿A￿simplified￿modelA￿simplified￿model5.10￿￿￿The￿Simplification￿of￿Linear￿SystemsPoles: S=-1, -2, -3 →→→ -1.029, -1.5555.10￿￿￿The￿Simplification￿of￿Linear￿SystemsChoose K and K1 so that:(1) The percent overshoot of 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.5.11 Design Example: Hubble Telescope Pointing Control(1) Select K and K1 to meet P.O. ≤10% for R(s)=A/s. Set D(s)=0.When￿ζ=0.6,￿P.O.=9.5%.￿5.11 Design Example: Hubble Telescope Pointing Control(2)￿Examine￿the￿steady-state￿error￿for￿a￿ramp￿input.(3)￿Reduce￿the￿effect￿of￿a￿step￿disturbance.◦The￿steady-state￿error￿due￿to￿a￿unit￿step￿disturbance￿is￿equal￿to￿-1/K.◦The￿transient￿response￿of￿the￿error￿due￿to￿the￿step￿disturbance￿input￿can￿be￿reduced￿by￿increasing￿K.(4)￿In￿summary,￿we￿need￿large￿K,￿large￿K/K1￿and￿ζ=0.6.￿￿￿￿￿￿￿￿￿Select 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 system’s operation remains linear.5.11 Design Example: Hubble Telescope Pointing Control5.11 Design Example: Hubble Telescope Pointing ControlK=100,￿￿ess=B/8.33=0.12B5.12 Sequential Design Example: Disk Drive Read SystemGoal:◦Achieve￿the￿fastest￿response￿ to￿ a￿ step￿ input￿r(t);◦Limit￿ the￿ overshoot￿ and￿oscillatory￿ nature￿ of￿ the￿response;◦Reduce￿ the￿ effect￿ of￿ a￿disturbance￿on￿the￿output￿position￿of￿the￿read￿head.5.12 Sequential Design Example: Disk Drive Read SystemNeglect the effect of the coil inductance.5.12 Sequential Design Example: Disk Drive Read SystemCompromise: Ka=405.12 Sequential Design Example: Disk Drive Read SystemBe￿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￿formulas￿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.AssignmentsSkills￿CheckE5.4￿￿￿￿￿￿E5.8￿￿￿￿￿￿￿E5.9￿￿￿￿￿E5.185.13￿￿￿￿Summary。