基于狀態(tài)空間方程和置根法的汽車的勻速巡航控制

1、 基于狀態(tài)空間方程和置根法的汽車的勻速巡航控制 框圖如下： 傳遞函數(shù)為： 利用置根法的Matlab 程序 求 反饋系數(shù)矩陣 K，并輸出響應(yīng)結(jié)果：m = 1000;b = 50;t = 0:0.1:10;r = ones(size(t);A = -b/m;B = 1/m;C = 1;D = 0;sys = ss(A,B,C,D);x0 = 0;p1 = -1.5;K = place(A,B,p1)sys_cl = ss(A-B*K,B,C,D);lsim(sys_cl,r,t,x0);axis(0 10 0 0.35)加比例系數(shù)補(bǔ)償 穩(wěn)態(tài)誤差：求補(bǔ)償系數(shù)的函數(shù)如下：functionNbar =

2、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,D,K) % finds the scale factor N which will % eliminate the steady-state error to a step reference % for a continuous-time, single-input sy

3、stem % with full-state feedback using the schematic below: % /- % R + u | . | % -> N ->() -> | X=Ax+Bu |-> y = Cx -> y % -| -/ % | | % |<- K <-| % % 8/21/96 Yanjie Sun of the University of Michigan % under the supervision of Prof. D. Tilbury % 6/12/98 John Yook, Dawn Tilbury rev

4、ised error(nargchk(2,5,nargin); % - Determine which syntax is being used - nargin1 = nargin; if (nargin=2), % System form A,B,C,D = ssdata(a); K=b; elseif (nargin=5), % A,B,C,D matrices A=a; 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;

5、% compute Nbar s = size(A,1);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; 加入控制系統(tǒng)得到：Nbar = rscale(sys,K);sys_cl = ss(A-B*K,B*Nbar,C,D);rnew = r *10；lsim(sys_cl,rnew,t,x0);axis(0 10 0 11) 直流電機(jī)的速度控制1. 建模：(J) 轉(zhuǎn)子的慣性常量 0.01 kg.m2(b) 電機(jī)摩擦系數(shù) 0.1 N.m.s(Ke) 電

6、生電動勢常數(shù) 0.01 V/rad/sec(Kt) 力矩常數(shù) 0.01 N.m/Amp(R) 電阻 1 Ohm(L) 電感 0.5 H 傳遞函數(shù)模型:狀態(tài)空間模型： 控制算法：1. PID 控制：J = 0.01;b = 0.1;K = 0.01;R = 1;L = 0.5;s = tf('s');P_motor = K/(J*s+b)*(L*s+R)+K2);PID 控制器 模型：（1）比例控制Kp = 100;C = pid(Kp);sys_cl = feedback(C*P_motor,1); t = 0:0.01:5; step(sys_cl,t) grid title

7、('Step Response with Proportional Control')（2） PID控制Kp = 75;Ki = 1;Kd = 1;C = pid(Kp,Ki,Kd);sys_cl = feedback(C*P_motor,1);step(sys_cl,0:1:200)title('PID Control with Small Ki and Small Kd')調(diào)整參數(shù)： Kp = 100; Ki = 200; Kd = 1; C = pid(Kp,Ki,Kd); sys_cl = feedback(C*P_motor,1); step(sys_

8、cl, 0:0.01:4) grid title('PID Control with Large Ki and Small Kd')2. 狀態(tài)空間方程控制J = 0.01;b = 0.1;K = 0.01;R = 1;L = 0.5;A = -b/J K/J -K/L -R/L;B = 0 1/L;C = 1 0;D = 0;sys = ss(A,B,C,D);1. 求反饋控制系數(shù) k(1) 判斷能控性sys_order = order(sys)sys_rank = rank(ctrb(A,B)（2） 利用置根法求反饋矩陣Kp1 = -5 + 1i;p2 = -5 - 1i;Kc = place(A,B,p1 p2)t = 0:0.01:3;sys_cl = ss(A-B*Kc,B,C,D);step(sys_cl,t)gridtitle('Step Response with State-Feedback Controller')（3） 加上比例常數(shù)補(bǔ)償：Nbar