Inverted pendulum倒立摆的matlab建模

ECE 451控制工程Inverted pendulum09/29/2013Introduction:Inverted pendulum is a typical fast，multi-varaibles，nonlinear，unstable system， ithasssicationfiddcontrontofotheinvertedpullum.assumetheininputisastepsignal thegrationalaccelerationg=9.8m/s 2andlinearize1 .匹配模型化Mmass of the car0.5 kg公斤平衡重0.2公斤coefficientoffrictionforccart 0.1 n/m/seclengthtopendulumcenterofmass 0.3 mimassmomentofinertiaofthependulum0. 006 kg.m 2Fforce applied to the cartxcoordinate of cart positionpendulumanglefromvertical (down )从水平方向和垂直方向N=md2dt2(x lsin)Force analysisconsiderthehorizontaldirectioncartforce，we get the equation:Mx=F-bx-Nconsidertthehorizontaldirectionpendulumforce，we get the equation:N=mx mlcos-ml2sin到To get rid of P and N，we get this equation:-Plsin-Nlcos=I邮件待办事宜，引导至与n，toobtainasecondmotionequation :mll2mglmsin=-mlxcosutorepresenttthecontrolledobjectwiththeinputforcef，线性化two motion equationsapplylaplacetransformtotheequationabovethetransferfunctionofangleandpositionLet v=xputtheequationaboveintothesecondequationWe get the transfer function状态空间执行：Solve the algebraic equation，obtain solution as follows:finallywethesystemstatespaceequations .2.PID控制器设计很快就设计好了KD (s ) isthetransferfunctionofthecontroller .g (s ) isthetransferfunctionofthecontrolledcar .Considering that the input r(s)=0，theblockdiagramcanbetransformedas :The output of the system is .num the numerator of the objectden-denominatoroftheobjectnum PID-thenumeratorofthepidcontrollertransferfunctionden PID-thedenominatorofthepidcontrollertransferfunctionThe object transfer function is，in whichPID控制器transferfunctionis现在weaddthecarspositionasanotheroutput，we get a1pos (126，268 ) 运输功能为theoutputofthecarposition isin which，num1，den1，num2，den2areseparatelymeanthecontrolledobject1and object2andpidcontrollernumeratoranddenominator。我得到了那个In which。wecaneastilysimplifiedtheequationas3. Matlab Simulation设计，thecartspositionwillbeignored.underthedesigncriteriatrare :1 ) settlingtimeislessthan5seconds2 ) pendulumshouldnotmovemorethan0. 05 radiansawayfromtheverticalWhen kd=1，k=1，ki=1:numc1=4.5455 0 0 0，denc2=1 4.7273 -26.6364 0.0909 0，num2=-1.8182 0 44.5455 0，denc2=1 4.7273 -26.6364 0.0909 0222222222222222222222222222652 a1pos (126，268 ) k，ki which is the second situation .2. When kd=20，k=300，ki=1:Numc1=4.5455 0 0 0、denc1=0.001 0.09111.3325 0.0001 0 0、numc2=-1.8182 0 44.5455 0、denc2=0.001 0.09111.3325 0.0001 0 0results :MATLAB代码M=0.5；m=0.2；b=0.1；I=0.006；g=9.8；l=0.3；q=(M m)*(I m*l2)/q-(m*l)2；num1=m*l/q 0 0；de n1= 1b * (im * l 2)/q-(m ) * m * g * l/q-b * m * g * l/Q0 ；num2=-(I m*l2)/q 0 m*g*l/q；den2=den1；kd=1；k=1；ki=1；numPID=kd k ki；denPID=1 0；numci=conv (num1 denPID )denc1=polyadd(conv(denPID，den1 )，conv(numPID，num1) )t=0:0.1:20；figure(1)impulse(numc1，denc1，t )title (angle)figure(2)impulse(numc2，denc2，t )title (position)M=0.5；m=0.2；b=0.1；I=0.006；g=9.8；l=0.3；q=(M m)*(I m*l2)/q-(m*l)2；num1=m*l/q 0 0；de n1= 1b * (im * l 2)/q-(m ) * m * g * l/q-b * m * g * l/Q0 ；num2=-(I m*l2)/q 0 m*g*l/q；den2=den1；kd=20；k=300；ki=1；numPID=kd k ki；denPID=1 0；numci=conv (num1 denPID )denc1=polyadd(conv(denPID，den1 )，conv(numPID，num1) )t=0:0.1:20；figure(1)impulse(numc1，denc1，t )title (angle)figure(2)impulse(numc2，denc2，t )title (position)Simulation:输入封闭循环模型和引用以及addistrubbaneceforcepropecepationaddddistrubbaneforceapplied