重庆大学过程控制processcontrol中文翻译2.pptx

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1、1Analysis of Single Loop SystemsOutlinePID ControllersP-controller,PI controller,PD controller,PID controllerA General SISO System Transfer function with process modelSignal transmitterControllerActuator(control valves)Closed-Loop ResponseSeven examplesChoice of controllers第1页/共70页2Analysis of Singl

2、e Loop SystemsClosed-Loop Transfer FunctionsA general SISO system:The stirred-tank heater revisited 再来观察搅拌加热器模型The heat balance is represented as热平衡方程Where U is the overall heat trans.coef.热传递系数,A is the heat transfer area加热面积,is fluid density,流体密度Cp is the heat capacity,热容量V is the volume（体积）of the

3、 vessel Ti=Ti(t)is the inlet temperature 入口TH=TH(t)is steam coil temperature 蒸汽线圈的温度Q is the flow rate 流量T is the outlet temperature with the initial condition is T(0)=Ts,T 是出口温度初始条件是T(0)=Ts第2页/共70页3Analysis of Single Loop SystemsClosed-Loop Transfer FunctionsThe stirred-tank heater revisited:whereA

4、fter Laplace transformNow p is the process time constant Kd and Kp are steady state gains第3页/共70页4Analysis of Single Loop SystemsClosed-Loop Transfer FunctionsA general SISO system Temperature T is based on 1.inlet Ti,and 2.jacket temperature TH 套管温度TH第4页/共70页5Analysis of Single Loop SystemsClosed-L

5、oop Transfer FunctionsA general SISO system Steam temperature TH is chosen as the manipulated variable（被控变量）1.Because of the physical intuition,also due to 2.Having a higher process gain compared to TiReminder（提醒）:Both TH and Ti are the inputs to the process,but not necessarily the inputs to the sys

6、tem TH和Ti是过程的输入，但不一定系统输入第5页/共70页6Analysis of Single Loop SystemsClosed-Loop Transfer FunctionsA general SISO system System output is T,called the controlled variable CSytem input is R,called the set point,or the referenceThe other affecting the response are considered as load（扰动）(e.g.,the inlet temp

7、erature Ti)Reminder:Inputs means different things for process,and the closed-loop system第6页/共70页7Analysis of Single Loop SystemsClosed-Loop Transfer FunctionsA general SISO system The transfer functions for the system are Gp for the process(plant)GL for the load(disturbance扰动)Gc for the controllerGm

8、 for the measuring device（测量仪器）,with the steady state gain KmGa for the actuator（执行器）Reminder:Introducing Km on R for summing up（总结）with the same units 第7页/共70页8Analysis of Single Loop SystemsClosed-Loop Transfer FunctionsDeriving the single loop transfer functionAt the summing point:在求和点For the pro

9、cess:Substituting E into C,we have:用C取代E,有Rearranging（重排）the equation above:第8页/共70页9Analysis of Single Loop SystemsClosed-Loop Transfer FunctionsDeriving the single loop transfer functionThe characteristic polynomial（特征多项式）The term is called the forward loop transfer function（前向通道传递函数）The term belo

10、w（下面方程的专业术语）is called the open-loop functionGFL=GcGaGp第9页/共70页10Analysis of Single Loop SystemsA General SISO SystemThe system function C=GspR+GloadL1.The 1st term accounts for the set point change,called the command track function（？）2.The 2nd term account for the effect of change in disturbance Rem

11、inder:The simultaneous changes of both R and L is not taken into account in the derivation above（以上没有考虑R and L 同时改变）参考方程第10页/共70页11Analysis of Single Loop SystemsConsidering the following two casesCase 1:Servoing problem（伺服问题）Changes in set point with no disturbance(L=0)results inC=GspRThis could pe

12、rfectly track the set point（能完全反应输入）:C=RCase 2:Regulator problem（调节器问题）Changes in disturbance with a fixed（固定的）set point(R=0)C=GloadLThis could perfectly reject（除去）the disturbance:C=0Reminder:we are working with deviation variables（偏差变量）第11页/共70页12Analysis of Single Loop SystemsConsiderations of Cho

13、osing VariablesChoice of controlled variables（受控变量）Those that are dictated by the problem 那些取决于这个问题Those that are not self-regulating 那些无法自动调节Those that may exceed equipment or process constraints那些可能超过设备或工艺约束Those that my interact with other variable那些受其他变量干扰第12页/共70页13Analysis of Single Loop Syste

14、msConsiderations of Choosing VariablesChoice of manipulated variables（操纵变量）Those that have a direct effect on the process（直接影响过程）,especially the output variable.Those that have a large steady state gain.(Good sensitivity)(高敏感度)Those that have no dead time(usually not satisfied).（无延迟）Those that have

15、no interaction with other control loops.（与其他控制环没有相互作用）Reminder:the remaining（剩下的）ones are taken as（被当作）load（干扰）variables in a SISO system after choosing the controlled and manipulated variables第13页/共70页14Analysis of Single Loop SystemsSynthesis of a Single Loop SystemThe stirred-tank heater revisite

16、d.Each transfer function with its element现实中：许多因素在传递环中影响的安全性、可靠性、准确性、动态性和成本。工程师需要综合考虑第14页/共70页15Analysis of Single Loop SystemsSynthesis of a Single Loop SystemThe stirred-tank heater revisited.Each transfer function with its element第15页/共70页16Analysis of Single Loop SystemsSynthesis of a Single Loo

17、p SystemThe stirred-tank heater revisitedPhysical unit conversion among transfer function传递函数的物理单位转化第16页/共70页17Analysis of Single Loop SystemsSynthesis（综合）of a Single Loop SystemProcess model(i.e.,process identification识别)Transfer functions are usually derived based on either1.Scientific or engineer

18、ing principles,or 2.Making a set point experiment and fitting the data to a model制造了一组实验 拟合数据模型The stirred-tank heater has,Assuming（假设）Kp=0.85 oC/oCp=20 min第17页/共70页18Analysis of Single Loop SystemsSynthesis of a Single Loop SystemSignal transmitter（信号发射器）The transducer is placed far downstream,or w

19、ell insulated.The response is slow,then该传感器尽可能地放在下游,或好绝缘的地方,响应较慢：where Km is the measurement gain of the transducer 传感器的量度增益 m is the time constant of the device td is accounts for transport lag 传输延迟 第18页/共70页19Analysis of Single Loop SystemsSynthesis of a Single Loop SystemSignal transmitter What y

20、ou see is not always what is occurring because of the transducer第19页/共70页20Analysis of Single Loop SystemsSynthesis综合 of a Single Loop SystemSignal transmitter发射器A thermocouple（热电偶）typically has the resolution分辨率 of 0.05 mv/0CA RTD（电阻式温度检测器）has better resolution and response time Reminder:this is to

21、o small a change in output for 12-bit A-D convertersThe transducer output should have 0-1 v or 4-20 mAHere we simplify简化 Km as It is slope of a calibration curve斜坡校正曲线,obtaining得到 in practice by1.The least square最小二乘 fitting,if the curve is linear2.The tangent切线 at operating point操作点,if the curve is

22、 nonlinear第20页/共70页21Analysis of Single Loop SystemsSynthesis of a Single Loop SystemControllerThe amplified放大的 signal is sent to发送到 the controller 1.Being physically a computer or a little black box2.Being implemented实现likely by a PID algorithms算法The controller usually has a front panel 面板1.with ph

23、ysical units（单位）such as oC(relative scales of 0-100%相对尺度)2.Needs to convert the change into electrical signal电信号The reason that Km is part of the controller in the block diagram方框图第21页/共70页22Analysis of Single Loop SystemsSynthesis of a Single Loop SystemActuator/control valve 控制阀Driving the range o

24、f the manipulated variable操纵变量的范围Responding faster than the processInterface接口 between controller and actuatorChosing a fail-close valve for safety1.Pneumatic气动阀 valve with 3-15 psi2.A current-to-pressure(I/P)transducer 传感器Reminder:Emergency power or spring-loaded valve is not proper for safey issue

25、应急电源或弹簧阀是不正确的安全问题第22页/共70页23Analysis of Single Loop SystemsSynthesis of a Single Loop SystemActuator/control valve Three assumptions（假设）for a valve:1.The proper amplifier or transducer interfacing the controller output with the valve适当的放大器或传感器接口作用于输出阀门2.The valve is of linear characteristics3.The ch

26、ange in steam flow rate蒸汽流量 can be translated to changes in heating coil temperature第23页/共70页24Analysis of Single Loop SystemsSynthesis of a Single Loop SystemActuator/control valveThe valve under the assumption for the coil has Valve gain,Kv=0.6 oC/mv,and Valve time constant,v=0.2 minThe actuator c

27、ontroller function would appear出现 as The system characteristic polynomial is nowGp前面有表达式第24页/共70页25Analysis of Single Loop SystemsSynthesis of a Single Loop SystemPut all together第25页/共70页26Analysis of Single Loop SystemsSynthesis of a Single Loop SystemThe control performence 控制性能良好第26页/共70页27Analy

28、sis of Single Loop SystemsSynthesis of a Single Loop SystemThe control performence 说明：oscillatory震荡的第27页/共70页28Analysis of Single Loop Systems Closed-Loop System ResponseOutline of deriving transfer function传递函数的推导1.Draw the control loop 画控制回路,and derive the closed-loop transfer functions.推导闭环传递函数2.

29、Pick either the servo or the regulator problem.伺服或调节器问题3.With the given choices of Gc,Gp,Ga and Gm,(a)plug插入 their transfer functions into the closed-loop equation.(b)The characteristic polynomial should fall out（发生）nicely.特征多项式可以计算出来第28页/共70页29Analysis of Single Loop Systems Closed-Loop System Resp

30、onseOutline of deriving transfer function传递函数的推导 4.Rearrange the expressions such that we can (a)redefine the parameters as time constants,重新排列表达式,这样我们可以重新定义时间常数 (b)steady state gains for the closed-loop system.第29页/共70页30Analysis of Single Loop SystemsClosed-Loop System ResponseLearning how to set

31、up提出 problems properly.Better understanding for why a system may have a faster response,why a system may become underdamped欠阻尼,and when there is an offset补偿,how controller settings may affect the closed-loop system response.如何设置控制器可以影响闭环系统响应第30页/共70页31Analysis of Single Loop SystemsClosed-Loop Syste

32、m ResponseAssumptions to the derivation of推导 transfer functionSimplifying简化 the system1.Setting Gm=1,Ga=1 2.Using the servoing伺服体统 problem with L=0,The resulting system is represented as Having the characteristic polynomial特征多项式第31页/共70页32Analysis of Single Loop SystemsClosed-Loop System ResponseExa

33、mple 1:Derive the transfer function for a 1st order process with p-control导出P控制器的一介传递函数,find the value of controlled variable at steady state after a unit step change in set point找到控制变量的值在稳定状态后，在设置点的单位阶跃变化.Solution:Gc=Kc in p-control,and in the 1st order processFrom the simplified transfer function

34、将Kc和Gp带入简化的传递函数第32页/共70页33Analysis of Single Loop SystemsClosed-Loop System ResponseExample 1(cont.):Rearranging the equationWhere is the close loop steady state gainis the close loop time constantThe probable location of the closed-loop pole闭环极点 is第33页/共70页34Analysis of Single Loop SystemsClosed-Lo

35、op System ResponseExample 1(cont.):the differential equation微分方程The first order process一阶过程In the unity单一 feedback loop with Gc=Kc,we have x=Kc(r-y)Therefore,The Laplace transform 第34页/共70页35Analysis of Single Loop SystemsClosed-Loop System ResponseExample 1(cont.):Observation 观察 At condition K1,the

36、 controlled variable will change in magnitude less than a given change in set point,due to控制变量将在强度低于给定的设置点的变化,由于 C=R/(s+1)At condition 第35页/共70页36Analysis of Single Loop SystemsClosed-Loop System ResponseExample 1(cont.):Observation The steady state error稳态增益误差(offset)Considering a change in set poi

37、nt（给定输入）R=M/s The eventual最终的 change in the controlled variable is The offset稳态误差 is defined to as ess=r-c(),thus The offset ess decreases as Kc increases参考方程：C=K R/(s+1)第36页/共70页37Analysis of Single Loop SystemsClosed-Loop System ResponseExample 1(cont.):Observation The steady state error(offset)fr

38、om another perspective表示法 It is the difference between controlled variable and set pointSuppose R=1/s,Where is called position error constant静态位置误差系数静态位置误差系数The same result as before第37页/共70页38Analysis of Single Loop SystemsClosed-Loop System ResponseExample 2:Derive the transfer function for a 2nd

39、order overdamped process with p-control.导出二阶P控制的欠阻尼传递函数 The second order process has time constants 2 and 4 min,and process gain 1.0 units（单位）,What proportional gain would provide us with a system with damping ratio阻尼比 of 0.7?什么比例增益会给我们提供阻尼比为0.7的系统第38页/共70页39Analysis of Single Loop SystemsClosed-Loo

40、p System ResponseExample 2:Derive the transfer function for a 2nd order overdamped process with p-control.Sulution:In p-control Gc=Kc,In the 2nd order process,According to the transfer function of the closed-loop第39页/共70页40Analysis of Single Loop SystemsClosed-Loop System ResponseExample 2(cont.):Th

41、e transfer function can easily be obtained获得Defining new quantitiesWhere the steady state gain is the system natural time period is the damping ratio阻尼比 is 第40页/共70页41Analysis of Single Loop SystemsClosed-Loop System ResponseExample 2(cont.):ObservationsUsing ,and=0.7,Kp=1,1=2,2=4We have Kc=1.291.Th

42、e entire整个的 system may be underdamped欠阻尼,even though the open-loop process is overdamped过阻尼第41页/共70页42Analysis of Single Loop SystemsClosed-Loop System ResponseExample 2(cont.):Observations2.The characteristic polynomial（特征多项式）may have either real roots,or complex roots（复根）,depending on chosing Kc3.

43、This 2nd order system has the same steady state error as the one in the 1st order system(previous example)这二阶系统和一阶系统(前面的例子)具有相同的稳态误差4.This 2nd order process with p-control has a unique decay ratio衰减比 for controller design criterion控制器的设计准则due to 因为第42页/共70页43Analysis of Single Loop SystemsClosed-Loo

44、p System ResponseExample 3:Derive the transfer function of a first order process with PI control.What is the offset偏移量 in this system?SolutionPI controller has The 1st order process is第43页/共70页44Analysis of Single Loop SystemsClosed-Loop System ResponseExample 3(cont.):Observations1.The system is of

45、 the 2nd order due to the integral action积分作用 2.The system steady state gain is unity一致,thus zero offset零点偏移3.Any 2nd order system may exhibit underdamped behavior,due to 因为Rewritting the transfer functionWhere the system natural time period the damping ratio阻尼比参考方程第44页/共70页45Analysis of Single Loop

46、 SystemsClosed-Loop System ResponseExample 4:Derive the transfer function of a 1st order process with PD control.Solution:The PD control has The 1st process is第45页/共70页46Analysis of Single Loop SystemsClosed-Loop System ResponseExample 4(cont.):Observations1.The system remains the 1st order系统仍是一阶2.I

47、t becomes a lead-lag element 超前滞后环节Rewritting System steady state gainSystem time constantsame as the 1st order process with p control第46页/共70页47Analysis of Single Loop SystemsClosed-Loop System ResponseExample 5:Derive the transfer function of an integrating process with p-control.What is the offse

48、t偏移量 in this system?Solution:The integrating process has Gp=1/AsThe p-control Gc=KcObservation:1.The system has unity gain单位增益 with zero offset零点偏移2.We usually dont need PI control because the process can introduce the integral action by itself under such circumstances在这样的情况下我们通常不需要PI控制因为本身具有积分行为第47

49、页/共70页48Analysis of Single Loop SystemsClosed-Loop System ResponseExample 6:Provide the illustrative说明 closed-loop time response using利用 data of example 3 in MatlabSolution:第48页/共70页49Analysis of Single Loop SystemsClosed-Loop System ResponseExample 7:design a servo-controller for a mixed system给一个混

50、合系统设计了一个伺服控制器1.Injecting a blue dye into a stream of water for making denim蓝色染料注入到牛仔布2.Blending the injected dye into the pipe flow by a mixer用搅拌器混合染料进入管道3.Detecting the dye concentration by a photodetector用光电探测器探测染料的浓度 4.Transmitting the analog output to a controller传输模拟输出到一个控制器5.Sending a signal t