重大自动化过程控制_process_control_中文_翻译_第三章

1ProcessControlCollegeofAutomationChongqingUniversity2DynamicResponseOutline:轮廓BriefReviewoftheDynamicResponse简要回顾动态响应FirstOrderModelsforProcesses一阶模型的过程SecondsOrderModelforProcesses二阶模型过程ModelsforProcesswithDead-Time死区时间的过程模型HigherOrderModelsandApproximation高阶模型和近似SpecialFeaturesofLead-LagProcess滞后过程的特殊特色3DynamicResponse:BriefReviewTheFirstOrderModelofaProcessQ,CinC1V1WhereistimeconstantThegeneralformofthe1stordermodel一阶模型的一般形式Steadystategain稳态增益4DynamicResponse:BriefReviewTheSecondOrderModelofaProcessWherearetimeconstantsThegeneralformofthe2stordermodelQ,CinV1C1C2V22'''22121222()()indCdCCtCdtdt)(01222tbxyadtdyadtydawith5DynamicResponse:BriefReviewThen-thOrderModelofaProcessWithan0andthezeroinitialconditionWithan0和零初始条件)(...012112tbxyadtdyadtydadtydannnnThecorrespondingLaplacetransformY(s)=G(s)X(s)相应的拉普拉斯变换0111...)(asasasabsGnnnnAnyothervariants?其他变量吗?6DynamicResponse:1stOrderProcessAnalysisofthe1stOrderProcessStepresponsetransferfunctionConsiderastepinput,x(t)=Mu(t),andX(s)=M/sTheoutputisThetimedomainfunctionis7DynamicResponse:1stOrderProcessAnalysisofthe1stOrderProcessStepresponse阶跃响应Property1性质1yincreasesfrom0toanewsteadystateMK,thusself-regulatingy增加从0到一个新的稳态MK,从而自我调节8DynamicResponse:1stOrderProcessAnalysisofthe1stOrderProcessStepresponseProperty2性质2SteadystategainK=y/M,ThelargergainK,themoresensitiveistheoutputtothechangeintheinput增益K越大，输出随输入变化就越敏感9DynamicResponse:1stOrderProcessAnalysisofthe1stOrderProcessStepresponseProperty3Att=τ,theoutputisy=0.632MKTheformulaabovecanbeusedtoestimatespacetimeτ上面的公式可以用来估计空间时间τ10DynamicResponse:1stOrderProcessAnalysisofthe1stOrderProcessStepresponseThetimedomainfunctionis时域功能Property4Theshorterthespacetimeτ,thefasterreachesthenewsteadystate0.25,0.5,1,211DynamicResponse:1stOrderProcessAnalysisofthe1stOrderProcessImpulseresponse脉冲响应transferfunction传递函数Consideranimpulseinput,x(t)=M(t),andX(s)=M考虑一个脉冲输入TheoutputisInversetransform反变换Theoutputincreasesinstantaneouslyattimet=0,anddecaysexponentiallytozero.输出瞬间增加在时间t=0，且呈指数衰减到零12DynamicResponse:1stOrderProcessAnalysisofthe1stOrderProcessIntegratingprocess:Non-self-regulating整合过程：非自我调节a0=0LaplacetransformTimedomain时域13DynamicResponse:1stOrderProcessAnalysisofthe1stOrderProcessIntegratingprocess:Non-self-regulatingWithstepresponse,theoutputisarampfunction阶跃响应，输出的是一个斜坡函数WithimpulseresponseTheoutputwillnotreturntoitsoriginalsteadystate输出不会回到原来的稳定状态Outputvalueistheaccumulationofwhatisadded输出值会累积增加ExamplecanbeChargingacapacity充电容量Fillingupatank填充的水池14DynamicResponse:1stOrderProcessExample:Showthatastoragetankwithpumpsatitsinletandoutletisaintegratingprocess表明储罐泵在其进口和出口是一个整合过程Massbalanceofacontinuousflowmixedtankatconstantdensityis:质量守恒方程，在密度不变的情况下whereqinandqaretheflowratesoftheinletandoutletqinandq是进口和出口的流动速率Aisthecross-section截面histheliquidlevel液位h15DynamicResponse:1stOrderProcessExample(cont.)Atsteadystate,wecandefinedeviationvariables在稳定状态下,我们可以定义偏差变量h’=h–hs,q’in=qin–qs,andq’=q–qsMassbalancebecomesThegeneralsolution一般解16DynamicResponse:1stOrderProcessExample(cont.)ThetransferfunctionStepinputineitherq’inorq’Leadingtoarampresponse,thusnosteadystate阶跃输入q‘in或q’，导致斜坡响应，因此，没有稳定的状态Thetankwilloverflow,whileoutletslowsdown容器将溢出,当出口关小Settingq’in=constant,thetransferfunctionisThetankwillbedrained,whileoutletspeedsup容器内液体将流干，当流出速度增加17DynamicResponse:1stOrderProcessExample(cont.):Visualizetheintegratingprocess可视化的积分过程pump泵qinqqinqNon-Self-RegulatingThetankwilloverflow,whilestep-inoccurs18DynamicResponse:1stOrderProcessOtherTypical1stOrderProcessesE=Voltage,电压z=Position,K’=Springconstant弹性系数,f=frictioncoefficient摩擦系数19DynamicResponse:1stOrderProcessOtherTypical1stOrderProcessesAnExtraExample:thechargingprocessofanRCcircuitRC电路充电过程RCiiRVdtdQiRdtdQRVRV1.Thesourcesendcurrentthroughtheresistorandchargethecapacity电压通过电阻向电容充电2.TheVcisinitiallyzeroandVR=VsVc初始值为0且VR=Vs3.Asthecapacitorcharges,VcincreasesandVRdecreasesandthecurrentidecreases为电容器充电，Vc的增加，VR减小，电流i降低4.Inthesteadystate,Vc=Vs,andi=0,thatis,thecapacitorischargedtoVs在稳定状态下，Vc=Vs，I=0，即，电容被充电至VsRcVVVccVVdtdVRCcdVdQCcCdVdQdtdVRCVcR20DynamicResponse:1stOrderProcessOtherTypical1stOrderProcessesAnExtraExample(cont.):RCiiVccVVdtdVRCwhere=RCistimeconstantK=1issteadystategainx(t)=Vsistheinput)1(/tsceVVCharging:Discharging:/tsceVV21DynamicResponse:1stOrderProcessOtherTypical1stOrderProcessesAnExtraExample(cont.):RCiiV)1(/tsceVVCharging:Discharging:/tsceVV22DynamicResponse:2ndOrderProcessAnalysisofthe2ndOrderProcessThegeneralform一般形式where)(01222tbxyadtdyadtydaBeingrearranged，重新整理with)0,0(thusforastableprocess在一个稳定的过程23DynamicResponse:2ndOrderProcessAnalysisofthe2ndOrderProcessCorrespondingLaplaceTransform相应的拉普拉斯变换Where说明：dampingratio阻尼比naturalperiodofoscillation自然振荡周期naturalfrequency固有频率24DynamicResponse:2ndOrderProcessAnalysisofthe2ndOrderProcessCharacteristicpolynomial特征多项式Thepolesare极点是：Noticingagainthatastableprocessrequires再一次注意到一个稳定的过程需要)00(i.e.,25DynamicResponse:2ndOrderProcessThreeCas