重大自动化过程控制processcontrol中文翻译第三章

1 Process Control College of Automation Chongqing University 2 Dynamic Response Outline: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 3 Dynamic Response: Brief Review The First Order Model of a Process Q,Cin C1 V1 Where is time constant The general form of the 1st order model一阶模型的一般形式 Steady state gain稳态增益 4 Dynamic Response: Brief Review The Second Order Model of a Process Where are time constants The general form of the 2st order model Q,Cin V1 C1 C2 V2 2 22 1 2122 2 ()( ) in d CdC CtC dtdt )( 01 2 2 2 tbxya dt dy a dt yd awith 5 Dynamic Response: Brief Review The n-th Order Model of a Process With an 0 and the zero initial condition With an 0和零初始条件 )(. 01 2 1 1 2 tbxya dt dy a dt yd a dt yd a n n n n The corresponding Laplace transform Y(s) = G(s)X(s) 相应的拉普拉斯变换 01 1 1 . )( asasasa b sG n n n n Any other variants? 其他变量吗? 6 Dynamic Response: 1st Order Process Analysis of the 1st Order Process Step response transfer function Consider a step input, x(t) =Mu(t), and X(s) = M/s The output is The time domain function is 7 Dynamic Response: 1st Order Process Analysis of the 1st Order Process Step response阶跃响应 Property 1Property 1性质性质1 1 y y increases from 0 0 to a new steady state MKMK, thus self-regulatingself-regulating y增加从0到一个新的稳态MK,从 而自我调节 8 Dynamic Response: 1st Order Process Analysis of the 1st Order Process Step response Property 2Property 2性质性质2 2 Steady state gain K = y/MK = 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 9 Dynamic Response: 1st Order Process Analysis of the 1st Order Process Step response Property 3Property 3 At t=, the output is y =0.632MK The formula above can be used to estimate space time 上面的公式可以用来估计空间时间 10 Dynamic Response: 1st Order Process Analysis of the 1st Order Process Step response The time domain function is 时域功能 Property 4Property 4 The shorter the space time , the faster reaches the new steady state 0.25, 0.5, 1, 2 11 Dynamic Response: 1st Order Process Analysis of the 1st Order Process Impulse 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 exponentially to zero. 输出瞬间增加在时间t= 0，且呈指数衰减到零 12 Dynamic Response: 1st Order Process Analysis of the 1st Order Process Integrating process: Non-self-regulating 整合过程：非自我调节 a0 = 0 Laplace transform Time domain时域 13 Dynamic Response: 1st Order Process Analysis of the 1st Order Process Integrating process: Non-self-regulating With step response, the output is a ramp function阶跃响应，输出的是一个斜坡 函数 With impulse response The 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 be Charging a capacity充电容量 Filling up a tank 填充的水池 14 Dynamic Response: 1st Order Process Example: 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 Mass 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 where qin and q are the flow rates of the inlet and outlet qin and q是进口和出口的流动速率 A is the cross-section截面 h is the liquid level 液位 h 15 Dynamic Response: 1st Order Process Example (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 qs Mass balance becomes The general solution一般解 16 Dynamic Response: 1st Order Process Example (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 The tank will overflow, while outlet slows down 容器将溢出,当出口关小 Setting qin =constant, the transfer function is The tank will be drained, while outlet speeds up 容器内液体将流干，当流出速度增加 17 Dynamic Response: 1st Order Process Example (cont.): Visualize the integrating process可视化 的积分过程 pump泵 qin q qin q Non-Self- Regulating The tank will overflow, while step-in occurs 18 Dynamic Response: 1st Order Process Other Typical 1st Order Processes E = Voltage, 电压 z = Position, K= Spring constant 弹性系数, f = friction coefficient 摩擦系数 19 Dynamic Response: 1st Order Process Other Typical 1st Order Processes An Extra Example: the charging process of an RC circuit RC电路充电过程 R C i i R V dt dQ i R dt dQ RV R V 1.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 Vc 初始值为0且VR = Vs 3. 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， V =Vs ，I = 0，即，电容被充电至Vs Rc VVV c c VV dt dV RC c dV dQ C c CdVdQ dt dV RCV c R 20 Dynamic Response: 1st Order Process Other Typical 1st Order Processes An Extra Example (cont.): R C i i V c c VV dt dV RC where = RC is time constant K = 1 is steady state gain x(t) = Vs is the input )1( /t sc eVV Charging: Discharging: /t sc eVV 21 Dynamic Response: 1st Order Process Other Typical 1st Order Processes An Extra Example (cont.): R C i i V )1( /t sc eVV Charging: Discharging: /t sc eVV 22 Dynamic Response: 2nd Order Process Analysis of the 2nd Order Process The general form 一般形式 where )( 01 2 2 2 tbxya dt dy a dt yd a Being rearranged，重新整理 with )0, 0 (thus for a stable process 在一个稳定的过程 23 Dynamic Response: 2nd Order Process Analysis of the 2nd Order Process Corresponding 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 natural frequency固有频率 24 Dynamic Response: 2nd Order Process Analysis of the 2nd Order Process Characteristic 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 )00 (i.e., 25 Dynamic Response: 2nd Order Process Three Cases of the Poles Case 1. overdamped process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 21 and 1 21 and or, In the case of having real poles, we have在实极点的情况下，我们有 26 Dynamic Response: 2nd Order Process Three Cases of the Poles Case 1. overdamped process过阻尼过程 How about the forms of transfer function in term of ? 传递函数有哪种形式按照 /1 and /1 21 1 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 27 Dynamic Response: 2nd Order Process Three Cases of the Poles三种极点 Case 1. overdamped process过阻尼 1 28 Dynamic Response: 2nd Order Process Three Cases of the Poles Case 2. critically damped process临界阻尼过程 repeating poles are重极点 /1 1 The coefficient can be considered as time constant 该系数 可视为时间常数 2 1)s( K Step response 21 in term of依据 This is the fastest response without oscillatory behaviour 是不伴有震荡的最快速响应 29 Dynamic Response: 2nd Order Process Three Cases of the Poles Case 3. underdamped process欠阻尼 10 Step response Being 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 2 1 j based on 基于 30 Dynamic Response: 2nd Order Process Key Features 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 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 31 Dynamic Response: 2nd Order Process Key Features of Underdamped Process 1. Overshoot超调 The overshoot increases as becomes smaller The OS becomes zero as approaches 1 The time to reach the peak value is Peak Time峰值时间Tp The time to hit the final value of y(t) is Rise Time上升时间 tr 32 Dynamic Response: 2nd Order Process Key Features of Underdamped Process 2. Frequency and Period周期 Noting that（注意）T=2 Tp The unit of the frequency is radian/time频率的单位是弧度/时间 The relationship between frequency and period 33 Dynamic Response: 2nd Order Process Key Features of Underdamped Process 3. Settling time调节时间 Ts The dominant factor forcing the oscillation to decay to zero is 震荡衰减到0的主导因素是： )/(t e in To settle with 5% of the final value is Ts=3/(/) 5%误差带所需要的调节时间 T= To settle with 2% of the final value is Ts=4/(/) n 1 34 Dynamic Response: 2nd Order Process Key Features of Underdamped Process 4. Decay Ratio 衰减率 OS为超调（overshoot） The decay ratio is the square of the overshoot Both quantities are functions of only这两个量只是 函数（调节时间和衰减率） 35 Dynamic Response: 2nd Order Process Other Typical 2nd Order Processes E = Voltage, z = Position, K= Spring constant, f = Friction Coefficient M = Mass h = force 36 Dynamic Response Processes 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 exponential function指数函数 37 Dynamic Response Processes with Dead Time A Simple Example 38 Dynamic Response Processes with Dead Time The 1nd and 2nd order models have the s-domain function S域函数 Td是延迟时间 and Dealing with the exponential functions处理的指数函数 Estimation 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） 39 Dynamic Response Processes with Dead Time The 1nd order Pad approximation The 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 The numerator has a positive zero, making the process unstable分子有一 个积极的零,使过程不稳定 The 2nd order Pad approximation Having two negative poles and at least one zero 有两个负极点和至少一个零点 40 Dynamic Response Processes with Dead Time Example: 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 approximation Observation: the approximation is acceptable at larger times compared with the original transfer function. 逼近的函数和原函数相比可以接受 41 Dynamic Response Processes with Dead Time Example (cont.) Generating the required plot 生成需要的图形（MATLAB） Pad approximation 42 Dynamic Response Processes with Dead Time The response of the dead time process 43 Dynamic Response Processes with Dead Time Two 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! 44 Dynamic Response Processes with Dead Time Two plants have different intermediate variables but have the same input-output behavior! 45 Dynamic Response Higher Order Process All 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 considered linear 所有这些问题都能被线性化 46 Dynamic Response Higher Order Process A 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 n1n n n cc t c d d 47 Dynamic Response Higher Order Process A series of well-mixed vessels (cont.)混合容器 The steady state gain 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, n . 21 48 Dynamic Response Higher Order Process Example: 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 49 Dynamic Response Higher Order Process Example (cont.) the Matlab code for the plot绘制图形的MATLAB代码 The response is obviously slower, as n increses The curves can be approximated 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 3 50 Dynamic Response Approximation of Higher Order Process Higher models Being 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 approximation Throwing away 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) 51 Dynamic Response Approximation of Higher Order Process Using the first order function with dead time This approximation can not be adequate合适 to many practices It is however the way 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 ) 1() 1)(1( 21 sss K X Y n n i id st t s Ke X Y d 1 1 with ) 1( 52 Dynamic Response Approximation of Higher Order Process Using 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 ) 1() 1)(1( 21 sss K X Y n n i id st t ss Ke X Y d 2 , 1 21 with ) 1)(1( 53 Dynamic Response Approximation of Higher Order Process Example: Find the simplest（最简单的） approximation of the transfer function Solution The dominant pole is at -1/3 Corresponding to the largest time constant 3相应的最大时间常数3 The dead time = 0.1+0.5+1.0 =1.6 54 Dynamic Response Approximation of Higher Order Process Example (cont.): the Matlab code for the plot of the step response of original and approximation functions Pad approximation n i id st t s Ke X Y d 1 1 with ) 1( 55 Dynamic Response Two More Examples of Higher Order Process Example 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 U is the overall heat trans. coef. 热传递系数, A A is the heat transfer area加热面积, is fluid density, 流体密度 C Cp p is the heat capacity, 热容量 V 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)= Ts 56 Dynamic Response Two More Examples of Higher Order Process Example 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 57 Dynamic Response Two More Examples of Higher Order Process Example 1 (cont.) Stirred Tank Heater Defining the deviation variables偏差变量 We have the steady state equation In the form of deviation variables以偏差变量的形式 58 Dynamic Response Two More Examples of Higher Order Process Example 1 (cont.) Stirred Tank Heater Omitting the apostrophe without confusion省略撇号不会出现混淆 The Laplace transform with The final form in s domain（域） is 59 Dynamic Response Two More Examples of Higher Order Process Example 1 (cont.) Stirred Tank Heater原函数： Another form of the transfer function where After Laplace transform Now p is the process time constant Kd and Kp are steady state gains 60 Dynamic Response Two More Examples of Higher Order Process Example 1 (cont.): An Experiment实验 Keeping the inlet（入口） temperature constant Increasing 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： 61 Dynamic Response Two More Examples of Higher Order Process Example 1 (cont.): An Experiment Plotting the time function As