重慶大學-過程控制-process-control-中文-翻譯-第三章

1、1Process ControlCollege of AutomationChongqing University2Dynamic ResponseOutline:輪廓nBrief Review of the Dynamic Response簡要回顧動態(tài)響應(yīng)nFirst Order Models for Processes一階模型的過程nSeconds Order Model for Processes二階模型過程nModels for Process with Dead-Time死區(qū)時間的過程模型nHigher Order Models and Approximation高階模型和近似nSp

2、ecial Features of Lead-Lag Process滯后過程的特殊特色3Dynamic Response: Brief ReviewThe First Order Model of a ProcessQ,CinC1V1Where is time constantThe general form of the 1st order model一階模型的一般形式Steady state gain穩(wěn)態(tài)增益4Dynamic Response: Brief ReviewThe Second Order Model of a ProcessWhere are time constantsTh

3、e general form of the 2st order modelQ,CinV1C1C2V22221 21222()( )ind CdCCtCdtdt )(01222tbxyadtdyadtydawith5Dynamic Response: Brief ReviewThe n-th Order Model of a ProcessWith an 0 and the zero initial condition With an 0和零初始條件)(.012112tbxyadtdyadtydadtydannnnThe corresponding Laplace transform Y(s)

4、= G(s)X(s)相應(yīng)的拉普拉斯變換0111.)(asasasabsGnnnnAny other variants?其他變量嗎?6Dynamic Response: 1st Order ProcessAnalysis of the 1st Order ProcessnStep responsetransfer functionConsider a step input, x(t) =Mu(t), and X(s) = M/sThe output is The time domain function is 7Dynamic Response: 1st Order ProcessAnalysi

5、s of the 1st Order ProcessnStep response階躍響應(yīng)Property 1性質(zhì)性質(zhì)1y increases from 0 to a new steady state MK, thus self-regulatingy增加從0到一個新的穩(wěn)態(tài)MK,從而自我調(diào)節(jié)8Dynamic Response: 1st Order ProcessAnalysis of the 1st Order ProcessnStep response Property 2性質(zhì)性質(zhì)2Steady state gain K = y/M, The larger gain K, the more s

6、ensitive is the output to the change in the input增益K越大，輸出隨輸入變化就越敏感9Dynamic Response: 1st Order ProcessAnalysis of the 1st Order ProcessnStep response Property 3At t=, the output is y =0.632MKThe formula above can be used to estimate space time 上面的公式可以用來估計空間時間10Dynamic Response: 1st Order ProcessAnal

7、ysis of the 1st Order ProcessnStep responseThe time domain function is時域功能Property 4The shorter the space time , the faster reaches the new steady state0.25, 0.5, 1, 211Dynamic Response: 1st Order ProcessAnalysis of the 1st Order ProcessnImpulse response脈沖響應(yīng)transfer function傳遞函數(shù)Consider an impulse i

8、nput, x(t) =M(t), and X(s) = M考慮一個脈沖輸入The output is Inverse transform反變換The output increases instantaneously at time t = 0, and decays exponentially to zero.輸出瞬間增加在時間t= 0，且呈指數(shù)衰減到零12Dynamic Response: 1st Order ProcessAnalysis of the 1st Order ProcessnIntegrating process: Non-self-regulating 整合過程：非自我調(diào)

9、節(jié)a0 = 0Laplace transformTime domain時域13Dynamic Response: 1st Order ProcessAnalysis of the 1st Order ProcessnIntegrating process: Non-self-regulatingpWith step response, the output is a ramp function階躍響應(yīng)，輸出的是一個斜坡函數(shù)pWith impulse response The output will not return to its original steady state 輸出不會回到原來

10、的穩(wěn)定狀態(tài) Output value is the accumulation of what is added 輸出值會累積增加nExample can bepCharging a capacity充電容量pFilling up a tank 填充的水池14Dynamic Response: 1st Order ProcessExample: Show that a storage tank with pumps at its inlet and outlet is a integrating process表明儲罐泵在其進口和出口是一個整合過程nMass balance of a conti

11、nuous flow mixed tank at constant density is :質(zhì)量守恒方程，在密度不變的情況下whereqin and q are the flow rates of the inlet and outletqin and q是進口和出口的流動速率A is the cross-section截面h is the liquid level 液位h15Dynamic Response: 1st Order ProcessExample (cont.)nAt steady state, we can define deviation variablesn在穩(wěn)定狀態(tài)下,我

12、們可以定義偏差變量h=h hs, qin =qin qs, and q =q qsnMass balance becomesnThe general solution一般解16Dynamic Response: 1st Order ProcessExample (cont.)nThe transfer function nStep input in either qin or q Leading to a ramp response, thus no steady state階躍輸入qin或q，導(dǎo)致斜坡響應(yīng)，因此，沒有穩(wěn)定的狀態(tài)nThe tank will overflow, while ou

13、tlet slows downn容器將溢出,當出口關(guān)小Setting qin =constant, the transfer function is nThe tank will be drained, while outlet speeds upn容器內(nèi)液體將流干，當流出速度增加17Dynamic Response: 1st Order ProcessExample (cont.): Visualize the integrating process可視化的積分過程 pump泵qinqqinqNon-Self-RegulatingnThe tank will overflow, while

14、step-in occurs18Dynamic Response: 1st Order ProcessOther Typical 1st Order ProcessesE = Voltage, 電壓z = Position,K= Spring constant彈性系數(shù), f = friction coefficient摩擦系數(shù)19Dynamic Response: 1st Order ProcessOther Typical 1st Order ProcessesnAn Extra Example: the charging process of an RC circuit RC電路充電過程R

15、CiiRVdtdQiRdtdQRVRV1.The source send current through the resistor and charge the capacity 電壓通過電阻向電容充電 2. The Vc is initially zero and VR = Vs Vc 初始值為0且VR = Vs3. As the capacitor charges, Vc increases and VR decreases and the current i decreases 為電容器充電，Vc的增加， VR減小，電流i降低4. In the steady state, Vc=Vs,

16、and i=0, that is, the capacitor is charged to Vs 在穩(wěn)定狀態(tài)下， Vc=Vs ，I = 0，即，電容被充電至Vs RcVVVccVVdtdVRCcdVdQC cCdVdQ dtdVRCVcR20Dynamic Response: 1st Order ProcessOther Typical 1st Order ProcessesnAn Extra Example (cont.): RCiiVccVVdtdVRCwhere = RC is time constant K = 1 is steady state gain x(t) = Vs is t

17、he input )1(/tsceVVCharging: Discharging: /tsceVV21Dynamic Response: 1st Order ProcessOther Typical 1st Order ProcessesnAn Extra Example (cont.):RCiiV)1(/tsceVVCharging: Discharging: /tsceVV22Dynamic Response: 2nd Order Process Analysis of the 2nd Order Process nThe general form 一般形式where)(01222tbxy

18、adtdyadtydanBeing rearranged，重新整理with )0, 0 (thus for a stable process在一個穩(wěn)定的過程23Dynamic Response: 2nd Order Process Analysis of the 2nd Order ProcessnCorresponding Laplace Transform相應(yīng)的拉普拉斯變換Where 說明：damping ratio阻尼比 natural period of oscillation自然振蕩周期 natural frequency固有頻率24Dynamic Response: 2nd Ord

19、er Process Analysis of the 2nd Order ProcessnCharacteristic polynomial特征多項式The poles are 極點是：Noticing again that a stable process requires再一次注意到一個穩(wěn)定的過程需要)00 (i.e., 25Dynamic Response: 2nd Order Process Three Cases of the PolesnCase 1. overdamped process過阻尼過程In term of the two time constants依據(jù)兩個時間常數(shù)T

20、ime constant can be derived below被推倒21 and 121 and or,In the case of having real poles, we have在實極點的情況下，我們有26Dynamic Response: 2nd Order Process Three Cases of the PolesnCase 1. overdamped process過阻尼過程How about the forms of transfer function in term of ? 傳遞函數(shù)有哪種形式按照 /1 and /121 1Step response in ter

21、m of the time constant階躍響應(yīng)下的時間常數(shù) Response is sluggish compared with underdamped or critically damped processes響應(yīng)比較緩慢與欠阻尼或臨界阻尼的進程相比27Dynamic Response: 2nd Order Process Three Cases of the Poles三種極點nCase 1. overdamped process過阻尼 128Dynamic Response: 2nd Order Process Three Cases of the PolesnCase 2. c

22、ritically damped process臨界阻尼過程repeating poles are重極點 /1 1The coefficient can be considered as time constant該系數(shù) 可視為時間常數(shù)21)s(KStep response21 in term of依據(jù) This is the fastest response without oscillatory behaviour是不伴有震蕩的最快速響應(yīng) 29Dynamic Response: 2nd Order Process Three Cases of the PolesnCase 3. under

23、damped process欠阻尼 10Step responseBeing rearranged as被調(diào)整為 The real part determines the exponential decay, thus the amount of can be considered as the time constant 決定了指數(shù)衰減，從而 可視為時間常數(shù)Two conjugate poles are兩個共軛極點是21jbased on基于 30Dynamic Response: 2nd Order Process Key Features of Underdamped Process欠阻

24、尼過程的主要特點(2) making control system design specifications with respect to the dynamic response 制作動態(tài)響應(yīng)的控制系統(tǒng)設(shè)計規(guī)范(1) fitting experimental data in the measurements of natural period and damping factor,把測量自然周期和阻尼因子擬合過后的實驗數(shù)據(jù)Features Derived from the figure for: 圖的特征31Dynamic Response: 2nd Order Process Key

25、Features of Underdamped Process1. Overshoot超調(diào)nThe overshoot increases as becomes smallernThe OS becomes zero as approaches 1nThe time to reach the peak value is Peak Time峰值時間TpnThe time to hit the final value of y(t) is Rise Time上升時間 tr 32Dynamic Response: 2nd Order Process Key Features of Underdamp

26、ed Process2. Frequency and Period周期nNoting that（注意）T=2 Tp nThe unit of the frequency is radian/time頻率的單位是弧度/時間 The relationship between frequency and period33Dynamic Response: 2nd Order Process Key Features of Underdamped Process3. Settling time調(diào)節(jié)時間 TsnThe dominant factor forcing the oscillation to

27、decay to zero is 震蕩衰減到0的主導(dǎo)因素是： )/(teinnTo settle with 5% of the final value is Ts=3/(/)n5%誤差帶所需要的調(diào)節(jié)時間 T=nTo settle with 2% of the final value is Ts=4/(/)n134Dynamic Response: 2nd Order Process Key Features of Underdamped Process4. Decay Ratio 衰減率OS為超調(diào)（overshoot）nThe decay ratio is the square of the

28、overshoot nBoth quantities are functions of only這兩個量只是函數(shù)（調(diào)節(jié)時間和衰減率）35Dynamic Response: 2nd Order Process Other Typical 2nd Order ProcessesE = Voltage, z = Position,K= Spring constant, f = Friction CoefficientM = Massh = force36Dynamic ResponseProcesses with Dead Time過程控制的延遲時間The time delay between th

29、e input and output in a process輸入與輸出的時間延遲nBeing also called dead time or transport lag傳輸延遲nThe Laplace transform of a time delay is an exponentialfunction指數(shù)函數(shù)37Dynamic ResponseProcesses with Dead TimenA Simple Example38Dynamic ResponseProcesses with Dead TimenThe 1nd and 2nd order models have the s-

30、domain function S域函數(shù)nTd是延遲時間andnDealing with the exponential functions處理的指數(shù)函數(shù)nEstimation with Taylor series expansion泰勒級數(shù)展開估計Estimation with Pad approximation (higher accuracy)Pad逼近估計（精度更高）39Dynamic ResponseProcesses with Dead TimenThe 1nd order Pad approximation nThe Denominator introduces a negati

31、ve pole, probably impacting the characteristic polynomial of the original process介紹了負極分母,可能影響特征多項式的原工藝nThe numerator has a positive zero, making the process unstable分子有一個積極的零,使過程不穩(wěn)定nThe 2nd order Pad approximation nHaving two negative poles and at least one zeron有兩個負極點和至少一個零點40Dynamic ResponseProces

32、ses 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遲時間繪制的第一過程的階躍響應(yīng)Pad approximationObservation: the approximation is acceptable at larger times compared with the original transfer function.逼近的函數(shù)和原函數(shù)相比可以接受 41Dyn

33、amic ResponseProcesses with Dead TimenExample (cont.) Generating the required plot 生成需要的圖形（MATLAB）Pad approximation42Dynamic ResponseProcesses with Dead TimenThe response of the dead time process43Dynamic ResponseProcesses with Dead TimenTwo plants have different intermediate variables but have the

34、same input-output behavior!兩個工廠有不同的中間變量,但有相同的投入產(chǎn)出的行為!44Dynamic ResponseProcesses with Dead TimenTwo plants have different intermediate variables but have the same input-output behavior!45Dynamic ResponseHigher Order ProcessnAll linearized higher order system can be broken down into the 1st and 2nd o

35、rder units所有線性化高階系統(tǒng)可以分成一階和二階單位nThe complex process like two interacting tanks can be formulated in coupled differential equations復(fù)雜的過程，像兩個相互作用的容器能制定耦合微分方程nAll these problems are considered linearn所有這些問題都能被線性化46Dynamic ResponseHigher Order ProcessnA series of well-mixed vessels where the volumetric f

36、low rate, and the respective volumes are constant一系列混合容器，其中體積流速，和各自的容量是恒定的n1nnncctcdd47Dynamic ResponseHigher Order ProcessnA series of well-mixed vessels (cont.)混合容器pThe steady state gain is unity in the process在過程中穩(wěn)態(tài)增益不變pThe more tanks in the series, the more sluggish is the response of the overal

37、l process 容器越多，整個響應(yīng)過程的滯后越長pProcesses that are products of the 1st order functions are called as multicapacity processes 多容量過程pIf all of space time （空間時間關(guān)系）are equal, n.2148Dynamic ResponseHigher Order ProcessExample: showing how the unit step response Cn(t) becomes more sluggish as n increases顯示單位階躍

38、響應(yīng)Cn(t)隨著n增加變得更加緩慢49Dynamic ResponseHigher Order ProcessnExample (cont.) the Matlab code for the plot繪制圖形的MATLAB代碼The response is obviously slower, as n incresesThe curves can be approximated by the 1st order model with dead time 這些特征曲線可近似為滯后的一階模型350Dynamic ResponseApproximation of Higher Order Proc

39、essnHigher modelspBeing factored into the form partial functions考慮部分函數(shù)的形式pTime constants have a large enough difference時間常數(shù)有很大的的差異nThe reduced-order model approximationpThrowing away the small time scale terms p扔掉小時間關(guān)系pRetaining the ones with dominant poles (larger time constants)p固定主導(dǎo)極點(大時間常數(shù))51Dyn

40、amic ResponseApproximation of Higher Order ProcessnUsing the first order function with dead time pThis approximation can not be adequate合適 to many practicespIt is however the way one designs controller with empirical tuning methods 然而它是一個與經(jīng)驗調(diào)諧設(shè)計控制器的方法) 1() 1)(1(21sssKXYnniidsttsKeXYd11 with ) 1(52Dy

41、namic ResponseApproximation of Higher Order ProcesspUsing the second order function with dead time利用二階時滯函數(shù)pThis model provides a better approximation 提供更好的近視pHow to determine the dominant poles?主極點) 1() 1)(1(21sssKXYnniidsttssKeXYd2 , 121 with ) 1)(1(53Dynamic ResponseApproximation of Higher Order P

42、rocessnExample: Find the simplest（最簡單的） approximation of the transfer functionnSolutionnThe dominant pole is at -1/3nCorresponding to the largest time constant 3相應(yīng)的最大時間常數(shù)3nThe dead time = 0.1+0.5+1.0 =1.654Dynamic ResponseApproximation of Higher Order ProcessnExample (cont.): the Matlab code for the

43、 plot of the step response of original and approximation functionsPad approximationniidsttsKeXYd11 with ) 1(55Dynamic ResponseTwo More Examples of Higher Order ProcessnExample 1: Stirred Tank Heatern攪拌槽內(nèi)的加熱器The heat balance is represented as熱平衡方程為：Where U is the overall heat trans. coef.熱傳遞系數(shù), A is

44、the heat transfer area加熱面積, is fluid density, 流體密度Cp is the heat capacity, 熱容量V is the volume of the vessel Ti = Ti(t) is the inlet temperature 入口TH =TH(t) is steam coil temperature 蒸汽線圈的溫度Q is the flow rate 流量T is the outlet temperature with the initial condition is T(0) = Ts, T 是出口溫度初始條件是T(0)= Ts5

45、6Dynamic ResponseTwo More Examples of Higher Order ProcessnExample 1 (cont.) Stirred Tank Heater攪拌槽內(nèi)的加熱器Rearranging the equation方程整理Defining , which leads to At the steady state, Defining the deviation variables定義偏差變量 57Dynamic ResponseTwo More Examples of Higher Order ProcessnExample 1 (cont.) Stir

46、red Tank HeaterDefining the deviation variables偏差變量We have the steady state equationIn the form of deviation variables以偏差變量的形式58Dynamic ResponseTwo More Examples of Higher Order ProcessnExample 1 (cont.) Stirred Tank HeaterOmitting the apostrophe without confusion省略撇號不會出現(xiàn)混淆The Laplace transformwithT

47、he final form in s domain（域） is59Dynamic ResponseTwo More Examples of Higher Order ProcessnExample 1 (cont.) Stirred Tank Heater原函數(shù)：Another form of the transfer functionwhereAfter Laplace transformNow p is the process time constant Kd and Kp are steady state gains60Dynamic ResponseTwo More Examples

48、of Higher Order ProcessnExample 1 (cont.): An Experiment實驗Keeping the inlet（入口） temperature constantIncreasing the steam temperature by 10oC增加蒸汽溫度10攝氏度The final form and its inverse function參考方程：61Dynamic ResponseTwo More Examples of Higher Order ProcessExample 1 (cont.): An ExperimentPlotting the time functionAs time progresses隨著