第8章MATLAB系統(tǒng)空間分析法ppt課件

1、第8章 系統(tǒng)狀態(tài)空間分析法8.4節(jié)和8.5節(jié)內(nèi)容n系統(tǒng)特征方程及解n關(guān)于系統(tǒng)相似變換n關(guān)于系統(tǒng)可觀性、可控性判別的n狀態(tài)反饋極點(diǎn)配置n狀態(tài)觀測(cè)器8.1 系統(tǒng)狀態(tài)方程的解n狀態(tài)轉(zhuǎn)移矩陣)(11AsILetAt若狀態(tài)方程是齊次的,即有:)0()(xetxAxxAtd)(BueXe) t (Xt0)t (A0AtEX1a=0 1 0;0 0 1;-6 -11 -6;x0=1;1;1;t=0:0.1:10;for i=1:length(t) x(:,i)=expm(a*t(i)*x0;endplot3(x(1,:),x(2,:),x(3,:);grid onAxx 1110 x6116100010A系

2、統(tǒng)的特征方程、特征值及特征向量n特征方程：|sI-A|=0n特征值及特征向量：V，D=eigA）特征向量矩陣特征值矩陣A*V = V*DEX2 已知控制系統(tǒng)求控制系統(tǒng)的特征方程A=2 1 -1;1 2 -1;-1 -1 2;I=1 0 0;0 1 0;0 0 1;syms s %符號(hào)計(jì)算det(s*I-A)s=solve(det(s*I-A) %求解ans = s3-6*s2+9*s-4s = 4 1 1211121112AEX2 求控制系統(tǒng)的特征值及特征向量V = -0.4082 0.7071 0.5774 -0.4082 -0.7071 0.5774 -0.8165 0 -0.5774D

3、= 1.0000 0 0 0 1.0000 0 0 0 4.0000V,D=eig(A)Veig=inv(V)*A*VVeig = 1.0000 0 -0.0000 -0.0000 1.0000 0.0000 -0.0000 0 4.0000符號(hào)計(jì)算Symbolic Toolboxans = 4 1 1eigensys(A)特征值及特征向量s=determ(s*I-A) s = s3-6*s2+9*s-4矩陣行列式的值EIGENSYS Obsolete Symbolic Toolbox function. V,D = EIGENSYS(A) is the same as V,D = eig(s

4、ym(A)8.2 傳遞矩陣GCsI-A-1B+DA=0 1;0 -2;B=1 0;0 1;C=1 0;0 1;D=0;syms sI=1 0;0 1;G=C*inv(s*I-A)*BG = 1/s, 1/s/(s+2) 0, 1/(s+2)8.3 線性變換n狀態(tài)方程的線性變換n ss2ss(sys,T)DuzCTyuTBzTATz11DuCXyBuAXXDuZCTyBuZATZT111DuZCyuBZAZ11CTCTBBTATATxz zTx1EX3A=0 -2;1 -3;B=2 0;C=0 3;P=6 2;2 0;%變換矩陣變換矩陣x=PzP1=inv(P);A1=P1*A*P %z坐標(biāo)系的

5、模型坐標(biāo)系的模型B1=P1*BC1=C*PA1 = 0 1 -2 -3B1 = 0 1C1 = 6 0The eigenvalues of system are unchanged by the linear transformation: (線性變換不改變系統(tǒng)的特征值）)det()det(AsIAsI約當(dāng)標(biāo)準(zhǔn)形ncanon(sys,model)ncanon(sys,companion)EX4利用特征值及范德蒙特矩陣求約當(dāng)陣A=0 1 0;0 0 1;2 -5 4;V,D=eig(A)P=1 0 1;1 1 2;1 2 4P1=inv(P)；J=P1*A*PV = -0.5774 0.5774

6、 -0.2182 -0.5774 0.5774 -0.4364 -0.5774 0.5774 -0.8729D = 1.0000 0 0 0 1.0000 0 0 0 2.0000P = 1 0 1 1 1 2 1 2 4J = 1 1 0 0 1 0 0 0 2符號(hào)計(jì)算Jo=jordan(A)Jo = 2 0 0 0 1 1 0 0 18. 4 系統(tǒng)的可控性和可觀性MATLAB提供函數(shù)分別計(jì)算能控性矩陣和能觀測(cè)性矩陣可控性矩陣CO=ctrb(A,B)可觀測(cè)性矩陣OB=obsv(A,C)BABAABBCO1n21n2CACACACOB可控性判定A=1 1 0;0 1 0;0 1 1;B=0 1

7、;1 0;0 1;n=length(A)CO=ctrb(A,B);rCO=rank(CAM);if rCO=n disp(System is controllable)elseif rCOn disp(System is uncontrollable)end n = 3 CO = 0 1 1 1 2 1 1 0 1 0 1 0 0 1 1 1 2 1 rCO = 2 System is uncontrollable可觀測(cè)性判定A=-3 1;1 -3;B=1 1;1 1;C=1 1;1 1;D=0;n=length(A);OB=obsv(A,C);rOB=rank(OB)if rOB=n dis

8、p(System is observable)elseif rOBn disp(System is unobservable)endOB= 1 1 1 1 -2 -2 -2 -2rOB = 1System is unobservable可控標(biāo)準(zhǔn)形bAbAAbbS1n2若S為非奇異，逆矩陣存在，設(shè)為TnT2T11ssscS那么，變換矩陣為P1nTnTnTnAsAssPA=-2 2 -1;0 -2 0;1 -4 0;B=0 1 1;n=length(A);CAM=ctrb(A,B);if det(CAM)=0 CAM1=inv(CAM);endP=CAM1(3,:);CAM1(3,:)*A;CAM

9、1(3,:)*A*A;P1=inv(P);A1=P*A*P1B1=P*BA1 = 0 1 0 0 0 1 -2 -5 -4B1 = 0 0 1可觀測(cè)標(biāo)準(zhǔn)形那么，變換矩陣為M=PTT1nTTTTCACACV)(若V為非奇異，逆矩陣存在，設(shè)為1nTTnTT2T11AvAvvV)(cMcAMMA18.5 系統(tǒng)狀態(tài)反饋與狀態(tài)觀測(cè)器利用反饋結(jié)構(gòu)，研究在什么條件下能實(shí)現(xiàn)閉環(huán)系統(tǒng)極點(diǎn)的任意配置，以達(dá)到預(yù)期要求。狀態(tài)反饋與狀態(tài)觀測(cè)器原理參見Module24，2524.1 The Structure of State Space Feedback Control（狀態(tài)反饋控制的結(jié)構(gòu)）（狀態(tài)反饋控制的結(jié)構(gòu)）CXy

10、BuAXXu0 AsI1. State Variable Feedback Control SystemCXyBrxBGAGxrBAXX)()(0)(BGAsI,21ngggGn the number of state variableIf the desired location of the closed-loop poles are , the desired characteristic equation will be n,210)()(21nsssThe eigenvalues of the closed-loop system will be given by),(0)(21ng

11、ggGBGAsI0)()()(21nsssBGAsIletWeWe can obtained , to make the closed-loop poles to be located in desired position.), 2, 1(nigiThe principle of designing a state space controller 2. The sufficient and necessary condition of state feedback for closed-loop placement: (狀態(tài)反饋實(shí)現(xiàn)極點(diǎn)配置的充要條件）The state variables

12、 of system are all controllable.25.1 Observer A model of the system under study（P550 Section 2）The approach taken to solve the problem is as following :To construct a model of the system under study;Assume (subject to certain restrictions ) that the computed state variables are good approximations t

13、o the true state variables;From these computed state variables, a suitable controller for the actual system may be constructed using the techniques described in Module 24. CXyBuAXXuWhere, x are assumed to be unmeasured directly.狀態(tài)觀測(cè)器設(shè)計(jì)uuBAxxCyNow, we construct a model to simulate the origin system ,

14、 and assume the parameter matrix are good approximations toCBA,CBA,CBA,CBA,CXyBuAXXCBAModelXCyuBXAXCBAModeltoEquals,But in model is different from in the origin system ,because is/are unmeasured directly. xxCBA,0 x is called the estimated value of .xxTo decrease the error , ( that is error ), we tak

15、e to correct to make well approach : xyyxxyyxxu)2()() 1 (XCyXCKKCXuBXAyyKuBXAXCXyBuAXX)(: )2() 1 (;XXKCAXKCKCXXAAXXXCCBBAAwheneKCAethenXXename)(,Select the matrix K to make the solution of this equation on error be convergent (收斂的), then, XXore , 0)(:XXKCAXXObserverThe gain matrix K is written as: n

16、kkkK21The closed-loop poles of this model (observer) can be selected by selecting the gain matrix K , so that the state variables will be same as in the end.Hence, we can use as the state variables in the state variable feedback system.XXXq The closed-loop system with observerBC AsI+uxyBC AsI+x y K-G-r+狀態(tài)觀測(cè)器狀態(tài)反饋25.2 The sufficient and necessary condition of constructing a state variable observerThe state variables of system are all observable.Observability criterion:A system A, C is state observable if and only ifnCACACACrankn12參見539頁P(yáng)OLE PLACEMENT VIA AKER