电气学科大类Modern Control SystemsAnalysis and DesignUsing Matlab and Simulink Title: Automobile Velocity ControlName: 巫宇智Student ID: U200811997Class：电气0811电气0811 巫宇智CataloguePreface (3)The Design Introduction (4)Relative Knowledge (5)Design and Analyze (6)Compare and Conclusion (19)After design (20)Appendix (22)Reference (22)Automobile Velocity Control1.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.In 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 ameliorate 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 locus电气0811 巫宇智2.The Design Introduction:Figure 2-1 automated highway systemThe figure shows an automated highway system, and according to computing 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 twoAutomobile Velocity Controlvehicles. Our design goal is to develop a controller that can maintain the vehicles in several specification below.DS1 Zero steady-state error to a step inputDS2 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 establish settling time)3.Relative Knowledge:Controller here actually serves as a compensator, and we have some compensators for different specification and system.电气0811 巫宇智4.Design and Analysis:4.1S pecification analysisAccording to the relative knowledge above, I may consider a PI controller to compensate------------G c(s)=k p s+k I.sDs1: 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.limsGc(s)G(s)≥4 −−→ Ki>4abs→0Ds3: overshoot less than 5% to a step response.P.0≤5% −−→ξ≥0.69DS4 Settling time less than 1.5 seconds to a step input( using a 2% criterion to establish settling time)Ts≤1.5 −−→ξ∗Wn≥2.66According to DS3 and DS4, we can draw the desired regionto placeAutomobile Velocity Controlour 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:T(s)=Kp s+Kis3+10s2+(16+Kp)s+KiThe corresponding Routh array is:s3 1 Kp+ab s2a+b Kis1ba KiabKpba+-++))((电气0811 巫宇智s0KiFor stability, we have (a+b)(Kp+ab)−Kia+b>0For another 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.a=−a−b−(−Ki Kp)2<−2.66In all, the specification is equal to a PI controller with limit below.{(a+b)(Kp+ab)−Kia+b>0⋯⋯⋯⋯⋯①KiKp<−5.32+(a+b)⋯⋯⋯⋯⋯⋯⋯②Ki>4ab⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯③4.2Design process:4.2.1Controller verification:At the very beginning, we take the system with G(s)=1(s+2)(s+8)and the controller (provided by the book) withG c(s)=33+66sfor an initial discussion.Automobile Velocity ControlFigure 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)电气0811 巫宇智From figure 4-2, we can see that the overshoot is 4.75%, and the settling time is 1.04 s with zero error 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 G c(s)=33+66scompletely meets the specification .4.2.2further analysis:For next procedure, I will have some more specific discussion about the applicable range of this controller to see how much can a and b vary yet allow the system to remain stable.We don’t change the parameter of the controller, and insert the Ki=66, Kp=33 into the inequality and get this:.{ab<16.5a+b>7.32ab+33−66a+b>0If we suppose the system to be a minimum phase system, a,b>0,thus it is easy to verify the 3rd inequality. Now, we draw 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 locate 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.4.2.3Alternatives discussion：According to inequality ①②③,The range of a and b bear some relation with the inequality below:{ a +b >5.32+Ki Kp ab <Ki 4Kp +ab −Ki a +b >0Basing 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{Ki =80Kp =64Figure4-5 the range of a,b for controller(Kp=64,Ki=80)As the figure indicate,(the range between dotted lines refers tothe 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)=1(s+2)(s+8)fixed, and discuss the different compensating effect of different 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.KiKp=1.5:Figure4-6 the root locus (KiKp=1.5)Using rlocfind, we find the maximum Kp=34.8740So we choose 3 groups of parameter ([35,52.5],[30.45],[25,37.5]) to examine the reponseFigure4-7 the step response (KiKp=1.5)It’s clear that the step response preference is not satisfying with too long settling timeKiKp=2:Figure4-8 the root locus (KiKp=2)Using rlocfind, we find the maximum Kp=34.3673So we choose 3 groups of parameter to examine the response and ramp response.Figure4-9 the step response (KiKp=2)Figure4-10 the ramp response (KiKp=2)KiKp=2.5:Figure4-11 the root locus (KiKp=2.5)Using rlocfind, we find the maximum Kp=31.47Similarly, we choose 3 groups of parameter to examine the response and ramp response.Figure4-12 the step response (KiKp=2.5)Virtually, the overshoot (Kp=30, Ki=75) doesn’t meet thespecification as we expect. I guess, that may come from the effect ofzero(-2.5), thus , go back to the step response of KiKp=2, due to the elimination between zero(-2) and poles, thus the preference is within our expectation.Figure4-13 the ramp response (KiKp=2.5)pare 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 worse 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 balanceall the parameter to make a compromise, but not a single parameter.From what we are talking about, we find the 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 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 doesn’t 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。
