matlab期末大作业Word文档下载推荐.docx

1、Compare and Conclusion 19After design 20Appendix 22Reference 221.Preface:With the high pace of human civilization development, the car has been a common tools for people. However, some problems also arise in such tendency. Among many problems, the velocity control seems to a significant challenge. I

2、n a automated highway system, using the velocity control system to maintain the speed of the car can effectively reduce the potential danger of driving a car and also will bring much convenience to drivers.This article aims at the discussion about velocity control system and the compensator to ameli

3、orate the preference of the plant, thus meets the complicated demands from people. The discussion is based on the simulation of MATLAB.Key word: PI controller, root locus2.The Design Introduction:Figure 2-1 automated highway systemThe figure shows an automated highway system, and according to comput

4、ing and simulation, a velocity control system for maintaining the velocity if the two automobiles is developed as below. Figure 2-2 velocity control systemThe input, R（s）, is the desired relative velocity between the two vehicles. Our design goal is to develop a controller that can maintain the vehi

5、cles in several specification below. DS1 Zero steady-state error to a step input DS2 Steady-state error due to a ramp input of less than 25% of the input magnitude.DS3 Percent overshoot less than 5% to a step input.DS4 Settling time less than 1.5 seconds to a step input（ using a 2% criterion to esta

6、blish settling time）3.Relative Knowledge:Controller here actually serves as a compensator, and we have some compensators for different specification and system.Table 3-1 CompensatorscompensatorAppilicable conditionAdjust steady errorAdd the type of system, thus adjust the steady errorPI controller:

7、adjust steady error and add an zeroIncrease the phase marginIncrease the phase margin, keep the steady error unchangedPID controller: balance all the preference（steady preference, as well as dynamic preference）4.Design and Analysis:4.1Specification analysisAccording to the relative knowledge above,

8、I may consider a PI controller to compensate-.Ds1: zero steady error to step response:To introduce an integral part to add the system type is enough.Ds2: Steady-state error due to a ramp input of less than 25% of the input magnitude.Ds3: overshoot less than 5% to a step response. According to DS3 an

9、d DS4, we can draw the desired region to place our close-loop poles.（as the shadow indicate） Figure 4-1 Desired region for locating the dominant polesAfter adding the controller, the system transfer function become:The corresponding Routh array is: 1 Kp+ab a+b Ki 0 Ki For stability, we have For anot

10、her consideration, we need to put the break point of root locus to the shadow area in Figure 4-1 to ensure the dominant poles placed on the left of s=-2.66 line.In all, the specification is equal to a PI controller with limit below.4.2 Design process:4.2.1Controller verification:At the very beginnin

11、g, we take the system with G（s）= and the controller （provided by the book） with for an initial discussion. Figure 4-2 step response（ a=2,b=8,Kp=33,Ki=66） Figure 4-3 ramp response（ a=2,b=8,Kp=33,Ki=66）From figure 4-2, we can see that the overshoot is 4.75%, and the settling time is 1.04 s with zero e

12、rror to the step input.From figure 4-3, it is clear that the ramp steady-state error is a little less than 25%.Thus, the controller with completely meets the specification .4.2.2further analysis:For next procedure, I will have some more specific discussion about the applicable range of this controll

13、er to see how much can a and b vary yet allow the system to remain stable.We dont change the parameter of the controller, and insert the Ki=66, Kp=33 into the inequality and get this:If we suppose the system to be a minimum phase system, a,b0, thus it is easy to verify the 3rd inequality. Now, we dr

14、aw to see the range of a and b. Figure4-4 the range of a,b for controller（Kp=33,Ki=66）Actually, the shade area can not completely meets the specification, for the constraint conditions represented in the 3 inequality is not enough, we need to draw the root locus for a certain system（a and b） to loca

15、te the actual limit for controller.However, this task is rather difficult, in a way, the 4 variables （a,b,Ki,Kp） all vary in terms of others change . Thus we can approximately locate the range of a and b from the figure above.4.2.3 Alternatives discussion：According to inequality ,The range of a and

16、b bear some relation with the inequality below:Basing our assumption on the range in the previous discussion, we can easily see that in order to increase the range, we can increase Ki and decrease the ratio of Ki to Kp.Thus, I adjust the parameter to Figure4-5 the range of a,b for controller（Kp=64,K

17、i=80）As the figure indicate, （the range between dotted lines refers to the previous controller, while the range between red lines refers to the new alternatives）, the range increase as we expect.Next step, we may keep the system of G（s）= fixed, and discuss the different compensating effect of differ

18、ent PI parameter.When carefully checking the controller, we may find that the controller actually add a zero（ -Ki/Kp） , an integral part and a gain part, so we can only change the zero and draw the locus root and examine the step response and ramp response.: Figure4-6 the root locus （）Using rlocfind

19、, we find the maximum Kp=34.8740So we choose 3 groups of parameter （35,52.5,30.45,25,37.5） to examine the reponse Figure4-7 the step response （Its clear that the step response preference is not satisfying with too long settling time Figure4-8 the root locus （Using rlocfind, we find the maximum Kp=34

20、.3673So we choose 3 groups of parameter to examine the response and ramp response. Figure4-9 the step response （ Figure4-10 the ramp response （ Figure4-11 the root locus （Using rlocfind, we find the maximum Kp=31.47Similarly, we choose 3 groups of parameter to examine the response and ramp response.

21、 Figure4-12 the step response （ Virtually, the overshoot （Kp=30, Ki=75） doesnt meet the specification as we expect. I guess, that may come from the effect of zero（-2.5）, thus , go back to the step response of, due to the elimination between zero（-2） and poles, thus the preference is within our expec

22、tation. Figure4-13 the ramp response （5.Compare and ConclusionMainly from the step response and ramp response, it can be concluded that, in a certain ratio of Ki to Kp, the larger Kp brings smaller ramp response error, as well as larger range of applicable system. Nevertheless, the larger Kp means w

23、orse step response preference（including overshoot and settling time）.This contradiction is rather common in control system. In all, to get the most satisfying preference, we need to balance all the parameter to make a compromise, but not a single parameter.From what we are talking about, we find the

24、 controller provided by the book（Kp=33, Ki=66） may be one of the best controller in comparison to some degree, with satisfying step response and ramp response preference, as well as a wider range for the variation of a and b, further, it use a zero（s=-2） to transfer the 3rd order system to 2nd order

25、 system, in doing so, we may eliminate some unexpected influence from the zero.The controller verified above （in Figure4-9 and Figure 4-10） with Kp=34, Kp=68 may be a little better, but only a little, and it doesnt leave some margin.6.After Design这是一次艰难，且漫长的大作业，连续一个星期，每天忙到晚上3点，总算完成了这个设计，至少我自己是很满意的。其

26、实与其说是大作业，不如说就是一次课程设计。运用所学的自控知识，加上matlab操作知识，去探究了一下用根轨迹法去研究校正的问题。这次选题很多，有超前滞后校正，有状态反馈校正，但这两种在前不久的自控实验中都已经做过，所以这一次挑战一下自己，选择这个根轨迹法来做。一开始觉得这个选题并不难，而且书上也要代码等，但真正做起来，发现很有点棘手。由于题目中要求研究某种控制器对某些系统的校正能力，相当于PI控制器中的Kp，Ki和系统中的两个极点全是变化的。 一段时间琢磨和不断仿真试验后，我决定换个角度去思考，分别控制系统不变和控制器不变，去研究控制器的控制范围，以及各种控制器对一个特定系统的矫正效果，最后在

27、通过比较分析论证。题目要求找到一个适用范围更广的控制器，但经过我不断摸索，证明出课后题中所给控制器已经是最优解了，再优的也只是提高一点点的极限值，虽然我没有找到更优解了，但这个过程中我充分了解到设计和探究的步骤，不管结果是否正确，我在这个探究过程中收获颇多。7.hold offclgn=1; d=1 10 16;tau=2.5;nc=1 tau; dc=1 0;num,den=series（n,d,nc,dc）;rlocus（num,den）hold onplot（-2.66 -2.66,-20 20）;z=0.69;plot（0 -20*z,0 20*sqrt（1-z2）,0 -20*z,0

28、 -20*sqrt（1-z2）gridrlocfind（num,den）% draw the root locus%Appendix：a=0:0.01:20;b=6.57-a;c=0:0.10:20d=20./cplot（a,b,c,d）%plot the inequality%Kp=30;Ki=75;nc=Kp Ki;numa,dena=series（n,d,nc,dc）;g=tf（numa,dena）f=feedback（g,1）;h=f*tf（1,1,0）;step（h）, gridhold on;t=0:100;plot（t,t,r-）;% ramp response%na,da=cl

29、oop（numa,dena）;step（na,da）, grid%step response%8.Reference：1）Robert H. Bishop，Modern Control System Analysis And Design Using MATLAB And Simulink，Beijing：Tsinghua University Press，2003，122）胡寿松，自动控制原理（第五版），北京：科学出版社，2007，63）胡寿松，自动控制原理简明教程（第四版简明版），北京：科学出版社，2003, 44）张德丰，Matlab Simulink 建模与仿真，北京，电子工业出版社，2009,3