毕业设计-英语翻译原文

  • 格式:doc
  • 大小:320.06 KB
  • 文档页数:9

PID controllers: theory, design, and tuning出版作者:KarlJ Astrom ,Tore Hagglund起止页码:35-45出版日期(期刊号):1995出版单位:Library of Congress Cataloging2.5 Methods of MomentsAll average residence time was determined based on calculation of an area. All other methods discussed in Section 2.4 were based on evaluation of the step response at single points only. Such methods are quite sensitive to measurement noise. In this section we will discuss methods that are based on integrals of the step response.Area MethodsWe will first discuss a method that is based on area calculations.Static gain K and average residence time T ar are first determined as in Figure 2.6. The area A\ under the step response up to time T ar is then determined. For a system having the transfer functionsL e sTKs G -+=1)( we have10/01)1()(--=-==⎰⎰KTe dt e K dt t s A TT t T arThe time constant is thus given byKeA T 1=(2.18) The dead time is then given byKeA K A T T L ar 10-=-= (2.19) With this method parameters L and T are both determined from computations of areas. Themethod is illustrated by the following example.EXAMPLE 2.7The method based on area determination has been applied to the process model (2.7). Static gain K is first determined from the stationary values to K = 1. Area A 0is then determined to 8.0 providing the average residence time T ar = 8. Area A 1can be determined by integrating the step response up to time T ar to A 1= 1.1. From Equation (2.18), time T can be calculated to T = 3.0, and finally Equation(2.19) gives L = 5.0. To summarize, the method based on area determination gives the following three-parameter models d e ss G 0.530.311)(-+=Figure 2.15 shows the step responses of the true process model and G 3d (s), as well as the Nyquist curves of the two transfer functions.The same idea can easily be applied to a system with the transfer functionsLe sT K s G -+=2)1()( (2.20)Parameters K and residence time T ar are determined as before. In this case we haveT L T ar 2+=The area A 1 under the step response up to time T ar is then determined. For a system having transfer function (2.20) we have220//014)1()(---=--==⎰⎰K T e dt e Tt e K dt t s A TTt T t T arFigure 2.15 Step responses and Nyquist curves of the process G(s) = l/(s +1)8 (solid line) and the three-parameter model G 3d (s)(dashed line).The time constant is thus given byKe A T 421= (2.21)and the dead time isKe A K A T T L ar 22210-=-= (2.22) The following example illustrates the properties of the method.EXAMPLE 2.8The three-parameter model (2.20) has been fitted to the process model(2.7) using the method of area determination. Static gain K is determined from the stationary values to K = 1. The area A 0 is 8.0, which gives the average residence time T ar = 8.0. Furthermore the area A 1 is 1.1 and Equation (2.21) then gives T=2.0. Equation (2.22) finally gives L = 4.0 and the model becomess e e s s G 0.432)0.21(1)(-+=Figure 2.16 shows the step responses of the true process model and G3e(s), as well as the Nyquist curves of the two transfer functions.The methods based on area determination are less sensitive to high-frequency disturbances than the previous methods, where the model is determined from only a few values of the step response. On the other hand, they are more sensitive to low-frequency disturbances such as a change in static load.Figure 2.16 Step responses and Nyquist curves of the process G(s) = l/(s + I)8 (solid line) and the three-parameter modelG&,(dashed line).The Method of MomentsA drawback with the area methods is that they require a storage of the step response. Area A\ cannot be computed until area AQ is determined. Therefore, some alternative methods that are also based on integration will be considered.Let h(t) be an impulse response and G(s) the corresponding transfer function. The functions are related throughdt t h e s G st )()(0⎰∞-=Taking derivatives with respect to s givesdt t h t e s G ds s G d n st n n nn )()1()()(0)(⎰∞--== Hence,dt t h t Gnnn )()1()0(0)(⎰∞-= (2.23)The values of the transfer function and its derivatives at s = 0 can thus be determined from integrals of the impulse response.The Average Residence TimeThe impulse response is positive for systems with monotone step responses. It can be interpreted as the density function of a probability distribution if it is normalized as follows:⎰∞=)()()(dtt h t h t fThe quantity f(t)dt can then be interpreted as the probability that an impulse entering the system at time 0 will leave at time t. The average residence time is then⎰⎰⎰∞∞∞==00)()()(d tt h d tt th d t t tfT ar (2.24)Introduce)()()(t s s t g -∞=where s(t) is the unit step response. Then)()(t h dtt dg -= It follows that[]dt t g t tg dt t th ⎰⎰∞∞∞+-=0)()()(The first term of the right-hand side is zero if g(t) goes to zero at least as fast as t 1+e for large t. The average residence time can thus also be written as)())()((0∞-∞=⎰∞s dtt s s T arwhich is the definition used previously.Equation (2.23) gives a convenient way to determine parameters of different models by computing the moments. This will be illustrated by some examples.A Three-Parameter Model Consider the transfer functionsL e sTKs G -+=1)( (2.25) It follows thatdt t h G K ⎰∞==0)()0( (2.26)Taking logarithms of Equation (2.25) givessT) (1 log -sL -K log logG(s)+=Differentiating this expression gives222'''')1()()()()(1)()(sT Ts G s G s G s G sT TL s G s G +=⎪⎪⎭⎫ ⎝⎛-+--=Hence20022''20')()()0()0()()()0()0(arar ar T dtt h dt t h t T G G T dtt h dtt th T L G G T -=-==+=-=⎰⎰⎰⎰∞∞∞∞(2.27)Gain K is thus given by Equation (2.26) and average residence time T ar and time constant T by Equation (2.27). The dead time L can then be computed toT T L ar -=It has thus been shown that the parameters of the model can be obtained from the first two moments of the impulse response. We illustrate the procedure with an example.EXAMPLE 2.9Consider the process model8)1(1)(+=s s G The first two derivatives with respect to s become109')1(72)(;)1(8)(+-=+-=s s G s s G Hence G(0) = 1, G'(0) = - 8 , and G"(0) = 72. Equations (2.26) and (2.27) now give86472812=-===T T K arWe thus find 8.222≈=T and 2.5228≈-=L . This result can be compared with the previous methods in Examples 2.2 and 2.7.Another Three-Parameter ModelThe method of moments will now be applied to determine the parameters of the transfer functionsLe sT K s G -+=2)1()( We havesT) (1 log 2 -sL -K log G(s) log +=Hence()222''''12)()()()(12)()(sT Ts G s G s G s G sT TL s G s G +=⎪⎪⎭⎫ ⎝⎛-+--=Hence2022''200'21)(2)(21)0(2)0()()(2)0()0()()0(ar ar ar T dtt h dtt h tT G G T dtt h dtt th T L G G T dtt h G K -=-==+=-===⎰⎰⎰⎰⎰∞∞∞∞∞(2.28)We illustrate the method with an example.EXAMPLE 2.10Consider the process model (2.7). It follows from the previous example that G(0) = 1, G'(0) = 8, and G"(0) = 72. We thus find K = 1,T ar = 8, T = 2 and L = 4. This is the same model as the one obtained in Example 2.8.Other Input SignalsFrom a practical point of view it is a drawback to have methods that require special input signals. The method of moments can be applied to any signal provided that the system is initially at rest.Let U(s) and Y (s) be the Laplace transforms of an arbitrary input and the correspondingoutput, respectively. Taking derivatives we getetcs U s G s U s G s U s G s Y s U s G s U s G s Y s U s G s Y )()()()(2)()()()()()()()()()()('''''''''''++=+==Hence,()()()()()()()()()()()()()()()ectU G U G U G Y U G U G Y U G Y 0000200000000000'''''''''''++=+== (2.29)The transfer function G(0) and its derivatives can thus be calculated from experiments with arbitrary inputs by calculating the following moments of the input and output()()()()()()()()dt t y t Y dtt u t Un n n n n n ⎰⎰∞∞-=-=001010and using Equation (2.29).By using these formulas it is possible to calculate G^(0) for any signals for which the moments()dt t u t u n n ⎰∞=and()dt t y ty nn ⎰∞=exist. This means that the signals must decay sufficiently fast.A typical case where the method can be used is when an experiment is performed in a closed loop with a pulse-like perturbation signal on the process input.Weighted MomentsThe method just discussed cannot be used if the signals do not go to zero or, equivalently, to a priori known mean values that can be subtracted in the calculations of moments, because the moments will then be infinite. There is, however, a simple modification that can be used in this case. It follows from the definition of the Laplace transform that()()()()()dt t y t e ds s Y d s Yn st nn n n ⎰∞--==01 The weighted moments()()()()a Y dt t y e t y n nar n n 10-==-∞⎰will exist provided that y(t) does not grow faster than e at for large t. By computing y n and the analogously defined moment u n , we can compute Y (n )(a)and U (n )(a)., and thus also G (n )(a).A Three-Parameter ModelConsider a system with the transfer function()sL e sTKs G -+=1 (2.30) We havesT) (1 log 2 -sL -K log G(s) log +=Hence()()()()()()()222''''11sT Ts G s G s G s G sT TL s G s G +=⎪⎪⎭⎫ ⎝⎛-+--=Thus we get()()()()()22'''221a a G a G a G a G aT T =⎪⎪⎭⎫ ⎝⎛-=+ (2.31) Hence,()()a G G L aaT -∂∂=∂-='1 (2.32)The average residence time thus becomes()()aa G G T L T ar ∂-∂+∂∂-=+=12' Furthermore the static gain is given by()()aL e a G aT K +=1 (2.33)The formulas are illustrated by an example.EXAMPLE 2.11Consider a system with the transfer function()()811+=s s G We have()()()()()()10''9'8172;18;11a a G a a G a a G +=+-=+=Computing the derivatives at the origin from the first terms in the Taylor series expansion gives()()()()()()()()()10109'9981101817218019118110a a a a a G a aa a a G ++-=+-+-≈++=+++≈The estimate of the average residence time becomes()()()()()()2'91011018911101800ˆa a a a a a G G T ar +++=+++≈-= From these expressions it follows that a must be small in order to give reasonably goodapproximations. To discuss the values of a, it is reasonable to normalize and consider aT ar . In this case, T ar = 8. With aT ar = 1 we get G(0) = 0.74, G'(0) = -5.54, and t ar = 7.53. With aT ar = 0.5 we get G(0) = 0.91, G'(0) = -7.1, and f ar = 7.83, giving errors in the range of 10%. With aT ar =0.2 we get ()()-7.81,0G 0.98,0G ==, and 96.7ˆ=arT . It follows from Equation (2.31) that()()aa a +=+-+=122164172a 22 It follows from Equations (2.32) and (2.33) that()()()()()()a a e aa K aL aT +--++=+-=-+=1/22872211111228221122The average residence time becomes()()()2221222122218ˆaa aL T T ar-+-+-+=+=With aT ar = 1, 0.5, and 0.2, we get the estimates f ar = 8.26, t ar = 8.06, and f ar - 8.01, respectively. This method of estimating the average residence time gives slightly better results than the extrapolation method.The example shows that we can obtain reasonable estimates of the model parameters and the average residence time by using weighted moments. It also seems reasonable to choose parameter a so that aT ar is in the range of 0.2 to 1. The best results are obtained for a small value of a. There is, however, an advantage in using larger values of a because there is then a less risk for disturbances to enter the system.。