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1、1Dynamic ResponseOutline:轮廓Brief Review of the Dynamic Response简要回顾动态响应First Order Models for Processes一阶模型的过程Seconds Order Model for Processes二阶模型过程Models for Process with Dead-Time死区时间的过程模型Higher Order Models and Approximation高阶模型和近似Special Features of Lead-Lag Process滞后过程的特殊特色第1页/共70页2Dynamic Res
2、ponse:Brief ReviewThe First Order Model of a ProcessQ,CinC1V1Where is time constantThe general form of the 1st order model一阶模型的一般形式Steady state gain稳态增益第2页/共70页3Dynamic Response:Brief ReviewThe Second Order Model of a ProcessWhere are time constantsThe general form of the 2st order modelQ,CinV1C1C2V
3、2with第3页/共70页4Dynamic Response:Brief ReviewThe n-th Order Model of a ProcessWith an 0 and the zero initial condition With an 0和零初始条件The corresponding Laplace transform Y(s)=G(s)X(s)相应的拉普拉斯变换Any other variants?其他变量吗?第4页/共70页5Dynamic Response:1st Order ProcessAnalysis of the 1st Order ProcessStep resp
4、onsetransfer functionConsider a step input,x(t)=Mu(t),and X(s)=M/sThe output is The time domain function is 第5页/共70页6Dynamic Response:1st Order ProcessAnalysis of the 1st Order ProcessStep response阶跃响应Property 1性质性质1y increases from 0 to a new steady state MK,thus self-regulatingy增加从0到一个新的稳态MK,从而自我调
5、节第6页/共70页7Dynamic Response:1st Order ProcessAnalysis of the 1st Order ProcessStep response Property 2性质性质2Steady state gain K=y/M,The larger gain K,the more sensitive is the output to the change in the input增益K越大，输出随输入变化就越敏感第7页/共70页8Dynamic Response:1st Order ProcessAnalysis of the 1st Order Process
6、Step response Property 3At t=,the output is y=0.632MKThe formula above can be used to estimate space time 上面的公式可以用来估计空间时间第8页/共70页9Dynamic Response:1st Order ProcessAnalysis of the 1st Order ProcessStep responseThe time domain function is时域功能Property 4The shorter the space time,the faster reaches the
7、 new steady state0.25,0.5,1,2第9页/共70页10Dynamic Response:1st Order ProcessAnalysis of the 1st Order ProcessImpulse response脉冲响应transfer function传递函数Consider an impulse input,x(t)=M(t),and X(s)=M考虑一个脉冲输入The output is Inverse transform反变换The output increases instantaneously at time t=0,and decays expon
8、entially to zero.输出瞬间增加在时间t=0，且呈指数衰减到零第10页/共70页11Dynamic Response:1st Order ProcessAnalysis of the 1st Order ProcessIntegrating process:Non-self-regulating 整合过程：非自我调节a0=0Laplace transformTime domain时域第11页/共70页12Dynamic Response:1st Order ProcessAnalysis of the 1st Order ProcessIntegrating process:No
9、n-self-regulatingWith step response,the output is a ramp function阶跃响应，输出的是一个斜坡函数With impulse responseThe output will not return to its original steady state输出不会回到原来的稳定状态Output value is the accumulation of what is added输出值会累积增加Example can beCharging a capacity充电容量Filling up a tank 填充的水池第12页/共70页13Dyn
10、amic Response:1st Order ProcessExample:Show that a storage tank with pumps at its inlet and outlet is a integrating process表明储罐泵在其进口和出口是一个整合过程nMass balance of a continuous flow mixed tank at constant density is:质量守恒方程，在密度不变的情况下whereqin and q are the flow rates of the inlet and outletqin and q是进口和出口的
11、流动速率A is the cross-section截面h is the liquid level 液位h第13页/共70页14Dynamic Response:1st Order ProcessExample(cont.)At steady state,we can define deviation variables在稳定状态下,我们可以定义偏差变量h=h hs,qin=qin qs,and q=q qsnMass balance becomesnThe general solution一般解第14页/共70页15Dynamic Response:1st Order ProcessExam
12、ple(cont.)The transfer function Step input in either qin or q Leading to a ramp response,thus no steady state阶跃输入qin或q，导致斜坡响应，因此，没有稳定的状态nThe tank will overflow,while outlet slows downn容器将溢出,当出口关小Setting qin=constant,the transfer function is nThe tank will be drained,while outlet speeds upn容器内液体将流干，当
13、流出速度增加第15页/共70页16Dynamic Response:1st Order ProcessExample(cont.):Visualize the integrating process可视化的积分过程 pump泵qinqqinqNon-Self-RegulatingnThe tank will overflow,while step-in occurs第16页/共70页17Dynamic Response:1st Order ProcessOther Typical 1st Order ProcessesE=Voltage,电压z=Position,K=Spring consta
14、nt弹性系数,f=friction coefficient摩擦系数第17页/共70页18Dynamic Response:1st Order ProcessOther Typical 1st Order ProcessesAn Extra Example:the charging process of an RC circuit RC电路充电过程RCiiV1.The source send current through the resistor and charge the capacity 电压通过电阻向电容充电 2.The Vc is initially zero and VR=Vs V
15、c 初始值为0且VR=Vs3.As the capacitor charges,Vc increases and VR decreases and the current i decreases 为电容器充电，Vc的增加，VR减小，电流i降低4.In the steady state,Vc=Vs,and i=0,that is,the capacitor is charged to Vs 在稳定状态下，Vc=Vs，I=0，即，电容被充电至Vs 第18页/共70页19Dynamic Response:1st Order ProcessOther Typical 1st Order Process
16、esAn Extra Example(cont.):RCiiVwhere =RC is time constant K=1 is steady state gain x(t)=Vs is the input Charging:Discharging:第19页/共70页20Dynamic Response:1st Order ProcessOther Typical 1st Order ProcessesAn Extra Example(cont.):RCiiVCharging:Discharging:第20页/共70页21Dynamic Response:2nd Order Process A
17、nalysis of the 2nd Order Process nThe general form 一般形式wherenBeing rearranged，重新整理with for a stable process在一个稳定的过程第21页/共70页22Dynamic Response:2nd Order Process Analysis of the 2nd Order ProcessnCorresponding Laplace Transform相应的拉普拉斯变换Where 说明：damping ratio阻尼比 natural period of oscillation自然振荡周期 nat
18、ural frequency固有频率第22页/共70页23Dynamic Response:2nd Order Process Analysis of the 2nd Order ProcessnCharacteristic polynomial特征多项式The poles are 极点是：Noticing again that a stable process requires再一次注意到一个稳定的过程需要第23页/共70页24Dynamic Response:2nd Order Process Three Cases of the PolesnCase 1.overdamped proce
19、ss过阻尼过程In term of the two time constants依据两个时间常数Time constant can be derived below被推倒or,In the case of having real poles,we have在实极点的情况下，我们有第24页/共70页25Dynamic Response:2nd Order Process Three Cases of the PolesnCase 1.overdamped process过阻尼过程How about the forms of transfer function in term of?传递函数有哪种
20、形式按照Step response in term of the time constant阶跃响应下的时间常数 Response is sluggish compared with underdamped or critically damped processes响应比较缓慢与欠阻尼或临界阻尼的进程相比第25页/共70页26Dynamic Response:2nd Order Process Three Cases of the Poles三种极点nCase 1.overdamped process过阻尼第26页/共70页27Dynamic Response:2nd Order Proce
21、ss Three Cases of the PolesnCase 2.critically damped process临界阻尼过程repeating poles are重极点 The coefficient can be considered as time constant该系数 可视为时间常数Step responsein term of依据 This is the fastest response without oscillatory behaviour是不伴有震荡的最快速响应 第27页/共70页28Dynamic Response:2nd Order Process Three C
22、ases of the PolesnCase 3.underdamped process欠阻尼Step responseBeing rearranged as被调整为 The real part determines the exponential decay,thus the amount of can be considered as the time constant 决定了指数衰减，从而 可视为时间常数Two conjugate poles are两个共轭极点是based on基于 第28页/共70页29Dynamic Response:2nd Order Process Key Fe
23、atures of Underdamped Process欠阻尼过程的主要特点(2)making control system design specifications with respect to the dynamic response 制作动态响应的控制系统设计规范(1)fitting experimental data in the measurements of natural period and damping factor,把测量自然周期和阻尼因子拟合过后的实验数据Features Derived from the figure for:图的特征第29页/共70页30Dyn
24、amic Response:2nd Order Process Key Features of Underdamped Process1.Overshoot超调nThe overshoot increases as becomes smallernThe OS becomes zero as approaches 1nThe time to reach the peak value is Peak Time峰值时间TpnThe time to hit the final value of y(t)is Rise Time上升时间 tr 第30页/共70页31Dynamic Response:2
25、nd Order Process Key Features of Underdamped Process2.Frequency and Period周期nNoting that（注意）T=2 Tp nThe unit of the frequency is radian/time频率的单位是弧度/时间 The relationship between frequency and period第31页/共70页32Dynamic Response:2nd Order Process Key Features of Underdamped Process3.Settling time调节时间 Ts
26、nThe dominant factor forcing the oscillation to decay to zero is 震荡衰减到0的主导因素是：innTo settle with 5%of the final value is Ts=3/(/)n5%误差带所需要的调节时间 T=nTo settle with 2%of the final value is Ts=4/(/)第32页/共70页33Dynamic Response:2nd Order Process Key Features of Underdamped Process4.Decay Ratio 衰减率OS为超调（ove
27、rshoot）nThe decay ratio is the square of the overshoot nBoth quantities are functions of only这两个量只是函数（调节时间和衰减率）第33页/共70页34Dynamic Response:2nd Order Process Other Typical 2nd Order ProcessesE=Voltage,z=Position,K=Spring constant,f=Friction CoefficientM=Massh=force第34页/共70页35Dynamic ResponseProcesses
28、 with Dead Time过程控制的延迟时间The time delay between the input and output in a process输入与输出的时间延迟Being also called dead time or transport lag传输延迟The Laplace transform of a time delay is an exponentialfunction指数函数第35页/共70页36Dynamic ResponseProcesses with Dead TimeA Simple Example第36页/共70页37Dynamic ResponseP
29、rocesses with Dead TimenThe 1nd and 2nd order models have the s-domain function S域函数nTd是延迟时间andnDealing with the exponential functions处理的指数函数nEstimation with Taylor series expansion泰勒级数展开估计Estimation with Pad approximation(higher accuracy)Pad逼近估计（精度更高）第37页/共70页38Dynamic ResponseProcesses with Dead T
30、imenThe 1nd order Pad approximation nThe Denominator introduces a negative pole,probably impacting the characteristic polynomial of the original process介绍了负极分母,可能影响特征多项式的原工艺nThe numerator has a positive zero,making the process unstable分子有一个积极的零,使过程不稳定nThe 2nd order Pad approximation nHaving two nega
31、tive poles and at least one zeron有两个负极点和至少一个零点第38页/共70页39Dynamic ResponseProcesses with Dead TimenExample:Using the 1nd order Pad approximation帕德近似to plot the step response of the 1st process with dead time 使用的一介帕德近似逼近延迟时间绘制的第一过程的阶跃响应Pad approximationObservation:the approximation is acceptable at la
32、rger times compared with the original transfer function.逼近的函数和原函数相比可以接受 第39页/共70页40Dynamic ResponseProcesses with Dead TimeExample(cont.)Generating the required plot 生成需要的图形（MATLAB）Pad approximation第40页/共70页41Dynamic ResponseProcesses with Dead TimeThe response of the dead time process第41页/共70页42Dyn
33、amic ResponseProcesses with Dead TimeTwo plants have different intermediate variables but have the same input-output behavior!两个工厂有不同的中间变量,但有相同的投入产出的行为!第42页/共70页43Dynamic ResponseProcesses with Dead TimeTwo plants have different intermediate variables but have the same input-output behavior!第43页/共70
34、页44Dynamic ResponseHigher Order ProcessAll linearized higher order system can be broken down into the 1st and 2nd order units所有线性化高阶系统可以分成一阶和二阶单位The complex process like two interacting tanks can be formulated in coupled differential equations复杂的过程，像两个相互作用的容器能制定耦合微分方程All these problems are considere
35、d linear所有这些问题都能被线性化第44页/共70页45Dynamic ResponseHigher Order ProcessA series of well-mixed vessels where the volumetric flow rate,and the respective volumes are constant一系列混合容器，其中体积流速，和各自的容量是恒定的第45页/共70页46Dynamic ResponseHigher Order ProcessA series of well-mixed vessels(cont.)混合容器The steady state ga
36、in is unity in the process在过程中稳态增益不变The more tanks in the series,the more sluggish is the response of the overall process 容器越多，整个响应过程的滞后越长Processes that are products of the 1st order functions are called as multicapacity processes 多容量过程If all of space time（空间时间关系）are equal,第46页/共70页47Dynamic Respons
37、eHigher Order ProcessExample:showing how the unit step response Cn(t)becomes more sluggish as n increases显示单位阶跃响应Cn(t)随著n增加变得更加缓慢第47页/共70页48Dynamic ResponseHigher Order ProcessExample(cont.)the Matlab code for the plot绘制图形的MATLAB代码The response is obviously slower,as n incresesThe curves can be appro
38、ximated by the 1st order model with dead time 这些特征曲线可近似为滞后的一阶模型第48页/共70页49Dynamic ResponseApproximation of Higher Order ProcessHigher modelsBeing factored into the form partial functions考虑部分函数的形式Time constants have a large enough difference时间常数有很大的的差异The reduced-order model approximationThrowing awa
39、y the small time scale terms 扔掉小时间关系Retaining the ones with dominant poles(larger time constants)固定主导极点(大时间常数)第49页/共70页50Dynamic ResponseApproximation of Higher Order ProcessUsing the first order function with dead time This approximation can not be adequate合适 to many practicesIt is however the way
40、one designs controller with empirical tuning methods 然而它是一个与经验调谐设计控制器的方法第50页/共70页51Dynamic ResponseApproximation of Higher Order ProcessUsing the second order function with dead time利用二阶时滞函数This model provides a better approximation 提供更好的近视How to determine the dominant poles?主极点第51页/共70页52Dynamic Re
41、sponseApproximation of Higher Order ProcessExample:Find the simplest（最简单的）approximation of the transfer functionnSolutionnThe dominant pole is at-1/3nCorresponding to the largest time constant 3相应的最大时间常数3nThe dead time=0.1+0.5+1.0 =1.6第52页/共70页53Dynamic ResponseApproximation of Higher Order ProcessE
42、xample(cont.):the Matlab code for the plot of the step response of original and approximation functionsPad approximation第53页/共70页54Dynamic ResponseTwo More Examples of Higher Order ProcessExample 1:Stirred Tank Heater搅拌槽内的加热器The heat balance is represented as热平衡方程为：Where U is the overall heat trans.
43、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 of the 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)
44、=Ts第54页/共70页55Dynamic ResponseTwo More Examples of Higher Order ProcessExample 1(cont.)Stirred Tank Heater搅拌槽内的加热器Rearranging the equation方程整理Defining,which leads to At the steady state,Defining the deviation variables定义偏差变量 第55页/共70页56Dynamic ResponseTwo More Examples of Higher Order ProcessExample
45、 1(cont.)Stirred Tank HeaterDefining the deviation variables偏差变量We have the steady state equationIn the form of deviation variables以偏差变量的形式第56页/共70页57Dynamic ResponseTwo More Examples of Higher Order ProcessExample 1(cont.)Stirred Tank HeaterOmitting the apostrophe without confusion省略撇号不会出现混淆The Lap
46、lace transformwithThe final form in s domain（域）is第57页/共70页58Dynamic ResponseTwo More Examples of Higher Order ProcessExample 1(cont.)Stirred Tank Heater 原函数：Another form of the transfer functionwhereAfter Laplace transformNow p is the process time constant Kd and Kp are steady state gains第58页/共70页59
47、Dynamic ResponseTwo More Examples of Higher Order ProcessExample 1(cont.):An Experiment实验Keeping the inlet（入口）temperature constantIncreasing the steam temperature by 10oC增加蒸汽温度10摄氏度The final form and its inverse function参考方程：第59页/共70页60Dynamic ResponseTwo More Examples of Higher Order ProcessExample
48、 1(cont.):An ExperimentPlotting the time functionAs time progresses随着时间的推移nThe exponential term decays awayn指数项衰减了The temperature approaches the new value MKp(self regulating)温度接近新值MKp(自我调节)nThe normalized（正常的）response is 63.2%at the process time p第60页/共70页61Dynamic ResponseTwo More Examples of High
49、er Order ProcessExample 2:Interacting tanks-in-series相互作用的水池One valve connecting two tanks 阀门一The flow between them depends on the difference in the hydrostatic heads他们之间流动取决于之间的差异静压头Mass balance of the 1st tankMass balance of the 2st tankWhere R1 and R2 are the resistance（阻力）of the flow through val
50、ve 1 and 2q0h1h2Valve 1Valve 2第61页/共70页62Dynamic ResponseTwo More Examples of Higher Order ProcessExample 2(cont.)Interacting tanks-in-series相互作用的水池Rearranging in the deviation variables(without aphostrophe)重排偏差变量where第62页/共70页63CramerRuleDynamic ResponseTwo More Examples of Higher Order ProcessExam