Design and Simulation of Active Suspension System by Using Matlab

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k k1 1 1 1x1− x2= Y1 = −(+ )Y1 − ( + )Y1−m1m2m1m2(11)b2k 21 1− (W− x2)− (W − x2)+ Fa( + )m2m2m1m2Since we cannot use second derivatives in the state-spacerepresentation, we integrate this equation to get Y1 :b b bY1 1 21 = −(+ )Y1− (W − x2) +m1m2m2(12)k1k1k 21 1+ ∫ ( −(+ )Y1− (W − x2)+ Fa( + ))dtm1m2m2m1m2No derivatives of the input appear in this equation, and Y 1is expressed in terms of states and inputs only, except forthe integral. Let's call the integral Y 2 . The state equation forY 2 is:k k k1 1Y (1 1)Y22 = − + 1 − (W − x2)+ Fa( + )m1m2m2m1m(13)2Assuming that x 2 = x 1 -Y 1 and substituting the equation (13)into the equation (12) we get state-space for Y 1 :b1b1b2Y 1 = −(+ )Y1− (W − x1+ Y1)+ Y2m1m2m(14)2Then we shall substitute the derivative of Y 1 into theequation (9) of the derivative of x 1 :b1b2b1b1b1b2k1x1= − x1+ ( ( + + ) − )Y1m1m2m1m1m2m2m1(15)b1b2Fab1+ W + − Y2m1m2m1m1The state variables are x 1 , x 1 , Y 1 and Y 2. The matrix fromthe above equations (13), (14) and (15) is:⎡ 0 100 ⎤⎢ b b b b b b k bx1 201(1 1 2)1 1 ⎥⎡ 1 ⎤ ⎢−+ + − −m m m m m m m m⎥⎢ x⎥ 1 2 1 1 2 2 1 11⎢⎥⎢ ⎥ = ⎢ b2b b bY0 (1 1 2) 1⎥⎢ 1 ⎥− + +⎢ mm m m⎥⎢21 2 2Y⎥ ⎢k 2k k k⎥⎣ 2 ⎦ ⎢ 0 − (1+1+2) 0 ⎥⎢⎣m2m1m2m2⎥⎦⎡ 0 0 ⎤⎢ 1 b bx1 2 ⎥⎡ 1 ⎤ ⎢m m m⎥⎢x⎥ 1 1 21⎢⎥b⎡Fa⎤* ⎢ ⎥ + ⎢02 ⎥(16)⎢Y− ⎢1 ⎥mW⎥⎢⎥2⎣ ⎦⎢ ⎥Y ⎢2 1 1 k⎥⎣ ⎦ ⎢ + −2 ⎥⎢⎣m1m2m2⎥⎦⎡ x1⎤⎢ ⎥⎡ ⎤[ ] ⎢x1 ⎥FaY = 0 0 1 0 + [ 0 0] ⎢Y⎥⎢ ⎥ (17)1 ⎣W⎦⎢ ⎥⎣Y2⎦For the majority of problems the state-space method ismuch simpler than the Laplace Transform method. We canput the above state-space equations (16) and (17) intoMatlab by defining the four matrices of the standard statespaceequation:X = AX + BW; Y = CX + DW(18)Thus, we have created a new m-file, formed on the basis ofstate-space equation. By adding the commands'step(A,B,C,D,1)' and 'step(A,0.1*B,C,D,2)' and running theprogram, we obtain a completely identical situation as inthe Figures 2. and 3.DESIGNING PID CONTROLLERThe ultimate goal of the system's synthesis lies in thechoice of the structure and determining the value ofadjustable system parametres so that the system is providedwith predetermined characteristics in relation to thetransitional and stationary work regime. The transferfunction for PID controller is:2K K s K s KKIK sD + P + IP + + D =(19)sswhere K P is the proportional gain, K I an integral gain andK D is the derivative gain. We shall be needing all of thethree gains for our controller. To start with, by iteration wehave come to the following supposition: K P =190000,K I =715000 and K D =582000. We enter this into Matlab byadding the following command into the m-file:numc=[KD,KP,KI];denc=[1 0].Let's simulate the system's response (the distance betweenX 1 -X 2 ) to unit step road disturbance. From Figure 4. we canfind the transfer function from the road disturbance W tothe output (X 1 -X 2 ):⎡ numf numc⎢ W −⎣ denf denc⎡ numf numc⎢ W −⎣ denf denc( X1−X2)⎤ nump2 ⎥⎦ denp( X − X ) = ( X − X )⎤2 ⎥⎦denpnump( X − X ) = ( X − X )numf ⎛ denp numc⎞W = ⎜ + ⎟( X1− X 2 )denf ⎝ nump denc⎠nump* numf* denc=W denf11( denp* denc+nump* numc)1122(20)As nump=denf, we can simplify the transfer function fromthe equation (20) and enter it into the m-file in thefollowing way:numa=conv(numf,denc);dena=polyadd(conv(denp,denc),conv(nump,numc)).Now we have created the close-loop transfer function inMatlab, which will represent the plant, road disturbance andas well as the controller. Let's see what the close-loop stepresponse for this system looks like before we begin thecontrol process. To simulate 0.1 m high step as ourdisturbance we should multiply 'numa' by 0.1. Let's add thefollowing to our m-file:t=0:0.05:5;step(0.1*numa,dena,t).4
time is very short, but that the response does not meet the5% overshoot requirement. We have solved this problemmultiplying K D, K P and K I by 2. Rerunning this m-file weshould get the following plot:Figure 5. Closed-Loop Response with PID ControllerFrom the graph, the percent overshoot is about 9,5%, whichis 4,5 % larger than the requirement, but the settling time issatisfactory - 3 sec. To choose the proper gain that yieldsreasonable output from the beginning, we start withchoosing a pole and two zeros for PID controller. A pole ofthis controller must be at zero and one of the zeros has to bevery close to the pole at the origin, at 1. The other zero, wewill put further from the first zero, at 3, actually we canadjust the second-zero's position to get the system to fulfillthe requirement. Let's add the following command in the m-file, so we can adjust the second-zero's location and choosethe gain to have a rough idea what gain you should use forK D, K P and K I .z1=1; z2=3; p1=0;numc=conv([1 z1],[1 z2]); denc=[1 p1];num2=conv(nump,numc);den2=conv(denp,denc);rlocus(num2,den2);[K,p]=rlocfind(num2,den2)We should see the close-loop poles and zeros on the s-planelike this and we can choose the gain and dominant poles onthe graph by ourselves:Figure 7. Closed-Loop Response with PID ControllerNow let's see if the percent overshoot and settling timemeet the initial requirements. The overshoot is about 5%,and the settling time is approximately 2,5s. For thisproblem, it turns out that the PID design method adequatelycontrols the system. This can be seen by looking at the rootlocus plot, for the task can be achieved by simply changingonly the gains of a PID controller.CONTROLLER DESIGN BY ROOT LOCUSMETHODThe necessary condition for the stability of a stationarycontinuous linear system with concentrated parametres isthat all the roots of its characteristic equation. have negativereal parts or, which is the same, lie in the left semi-plain ofthe complex s variable. Root locus design offers thepossibility of adjusting the poles, zeros and transferfunction gain, so that, with closed feedback, the system getsthe desired dynamic characteristics [4]. Let us first see whatare the poles of the open-loop automatic control system:R roots(denp)=-23.7792+35.1432i-0.1098+5.2504i-23.7792-35.1432i-0.1098-5.2504iFigure 6. Root Locus with PID ControllerTherefore, the dominant poles are roots -0.1098±5.2504i,which are located close to the complex axis with a smalldamping ratio, and which have a dominant influence on theparametres of the transitional regime. The main idea of rootlocus design is to estimate the response of the closed-loopfrom the open-loop locus plot. By adding zeros and/or polesto the original system (adding a compensator), the rootlocus and thus the closed-loop response will be modified.Let's first view the root locus for the plant.Now that we have the close-loop transfer function,controlling the system is simply a matter of changing theK D, K p and K I variables. Figure 5. tells us that the settling5
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Page 3: When we want to consider input W(s)
Page 7 and 8: FREQUENCY DESIGN METHODSince the ph

Design and Simulation of Active Suspension System by Using Matlab

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