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1、word倒立摆小车的建模系统参数：(M) 小车质量 0.5 kg(m) 倒立摆质量 0.2 kg(b) 小车的阻尼系数 0.1 N/m/sec(l) 倒立摆长度 0.3 m(I) 倒立摆转动惯量 0.006 kg.m2(F) 施加在小车上的力(x) 小车的位置坐标(theta) 倒立摆与垂直方向的夹角从垂直向下开始计动力学建模： 考虑水平方向的力矩平衡：以上两式合并得到：考虑垂直施加在倒立摆上的力与质心力矩平衡方程：消去 P N ，得到如下公式简化： 简化后模型：传递函数模型：设：PID 控制：或者如下表示：M = 0.5; m = 0.2; b = 0.1; I = 0.006; g = 9
2、.8; l = 0.3; q = (M+m)*(I+m*l2)-(m*l)2; s = tf(s); P_pend = (m*l*s/q)/(s3 + (b*(I + m*l2)*s2/q - (M + m)*m*g*l)*s/q - b*m*g*l/q);Kp = 1; Ki = 1; Kd = 1; C = pid(Kp,Ki,Kd); T = feedback(P_pend,C);t=0:0.01:10; impulse(T,t) title(Response of Pendulum Position to an Impulse Disturbance under PID Control
3、: Kp = 1, Ki = 1, Kd = 1);Kp = 100; Ki = 1; Kd = 1; C = pid(Kp,Ki,Kd); T = feedback(P_pend,C); t=0:0.01:10; impulse(T,t) axis(0, 2.5, -0.2, 0.2); title(Response of Pendulum Position to an Impulse Disturbance under PID Control: Kp = 100, Ki = 1, Kd = 1);Kp = 100; Ki = 1; Kd = 20; C = pid(Kp,Ki,Kd); T
4、 = feedback(P_pend,C); t=0:0.01:10; impulse(T,t) axis(0, 2.5, -0.2, 0.2); title(Response of Pendulum Position to an Impulse Disturbance under PID Control: Kp = 100, Ki = 1, Kd = 20); 这时，小车的位置?传递函数： P_cart = (I+m*l2)/q)*s2 - (m*g*l/q)/(s4 + (b*(I + m*l2)*s3/q - (M + m)*m*g*l)*s2/q - b*m*g*l*s/q); T2
5、= feedback(1,P_pend*C)*P_cart; t = 0:0.01:5; impulse(T2, t); title(Response of Cart Position to an Impulse Disturbance under PID Control: Kp = 100, Ki = 1, Kd = 20);基于状态空间方程的倒立摆小车的控制方法设计 开环根 (LQR) 增加补偿 基于观测器的控制1 状态空间方程设计目标：1 倒立摆倒立状态2首先假设系统状态全部测量获得：求开环的根，即A的特征值M = 0.5;m = 0.2;b = 0.1;I = 0.006;g = 9.
6、8;l = 0.3;p = I*(M+m)+M*m*l2; %denominator for the A and B matricesA = 0 1 0 0; 0 -(I+m*l2)*b/p (m2*g*l2)/p 0; 0 0 0 1; 0 -(m*l*b)/p m*g*l*(M+m)/p 0;B = 0; (I+m*l2)/p; 0; m*l/p;C = 1 0 0 0; 0 0 1 0;D = 0; 0;states = xx_dotphiphi_dot;inputs = u;outputs = x; phi;sys_ss = ss(A,B,C,D,statename,states,in
7、putname,inputs,outputname,outputs);poles = eig(A)poles = 0线性二次型整定 (LQR)1 检查可控性co = ctrb(sys_ss);controllability = rank(co)controllability = 4设 R=1Q = C*CQ = 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0Q = C*C;R = I;K = lqr(A,B,Q,R)Ac = (A-B*K);Bc = B;Cc = C;Dc = D;states = xx_dotphiphi_dot;inputs = r;outputs = x
8、; phi;sys_cl = ss(Ac,Bc,Cc,Dc,statename,states,inputname,inputs,outputname,outputs);t = 0:0.01:5;r =0.2*ones(size(t);y,t,x=lsim(sys_cl,r,t);AX,H1,H2 = plotyy(t,y(:,1),t,y(:,2),plot);set(get(AX(1),Ylabel),String,cart position (m)set(get(AX(2),Ylabel),String,pendulum angle (radians)title(Step Response
9、 with LQR Control)K =Q = C*C;Q(1,1) = 5000;Q(3,3) = 100R = 1;K = lqr(A,B,Q,R)Ac = (A-B*K);Bc = B;Cc = C;Dc = D;states = xx_dotphiphi_dot;inputs = r;outputs = x; phi;sys_cl = ss(Ac,Bc,Cc,Dc,statename,states,inputname,inputs,outputname,outputs);t = 0:0.01:5;r =0.2*ones(size(t);y,t,x=lsim(sys_cl,r,t);A
10、X,H1,H2 = plotyy(t,y(:,1),t,y(:,2),plot);set(get(AX(1),Ylabel),String,cart position (m)set(get(AX(2),Ylabel),String,pendulum angle (radians)title(Step Response with LQR Control)Q = 5000 0 0 0 0 0 0 0 0 0 100 0 0 0 0 0K =增加补偿系数： = 1 0 0 0;sys_ss = ss(A,B,0);Nbar = rscale(sys_ss,K)Nbar =sys_cl = ss(Ac
11、,Bc*Nbar,Cc,Dc,statename,states,inputname,inputs,outputname,outputs);t = 0:0.01:5;r =0.2*ones(size(t);y,t,x=lsim(sys_cl,r,t);AX,H1,H2 = plotyy(t,y(:,1),t,y(:,2),plot);set(get(AX(1),Ylabel),String,cart position (m)set(get(AX(2),Ylabel),String,pendulum angle (radians)title(Step Response with Prepensat
12、ion and LQR Control)基于能观器的控制检查能观性：ob = obsv(sys_ss);observability = rank(ob)observability = 4能观器的根要比控制器的根快5到10倍的收敛控制器的根：poles = eig(Ac)poles =置根法设置 能观器的根：P = -40 -41 -42 -43;L = place(A,C,P)L = 1.0e+03 *Ace = (A-B*K) (B*K); zeros(size(A) (A-L*C);Bce = B*Nbar; zeros(size(B);Cce = Cc zeros(size(Cc);Dc
13、e = 0;0;states = xx_dotphiphi_dote1e2e3e4;inputs = r;outputs = x; phi;sys_est_cl = ss(Ace,Bce,Cce,Dce,statename,states,inputname,inputs,outputname,outputs);t = 0:0.01:5;r = 0.2*ones(size(t);y,t,x=lsim(sys_est_cl,r,t);AX,H1,H2 = plotyy(t,y(:,1),t,y(:,2),plot);set(get(AX(1),Ylabel),String,cart positio
14、n (m)set(get(AX(2),Ylabel),String,pendulum angle (radians)title(Step Response with Observer-Based State-Feedback Control)FunctionNbar=rscale(a,b,c,d,k) % Given the single-input linear system: % . % x = Ax + Bu % y = Cx + Du % and the feedback matrix K, % % the function rscale(sys,K) or rscale(A,B,C,
15、D,K) % finds the scale factor N which will % eliminate the steady-state error to a step reference % for a continuous-time, single-input system % with full-state feedback using the schematic below: % % /- % R + u | . | % - N -() -| X=Ax+Bu |- y=Cx - y % -| -/ % | | % |- K -| % %8/21/96 Yanjie Sun of
16、the University of Michigan % under the supervision of Prof. D. Tilbury %6/12/98 John Yook, Dawn Tilbury revised error(nargchk(2,5,nargin); % - Determine which syntax is being used - nargin1 = nargin; if (nargin1=2),% System form A,B,C,D = ssdata(a); K=b; elseif (nargin1=5), % A,B,C,D matrices A=a; B=b; C=c; D=d; K=k; else error(Input must be of the form (sys,K) or (A,B,C,D,K) end; % pute Nbar s = size(A,1); Z = zeros(1,s) 1; N = inv(A,B;C,D)*Z; Nx = N(1:s); Nu = N(1+s); Nbar=Nu + K*Nx;21 / 21
