基于狀態(tài)空間方程和置根法的汽車的勻速巡航控制剖析

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)Linear SimulatiDn Resultso0.10.050Time

2、(seconds加比例系數(shù)補(bǔ)償穩(wěn)態(tài)誤差:rA'I >B+ iJ/-Icy=v .<1/5求補(bǔ)償系數(shù)的函數(shù)如下:fun cti onNbar = rscale(a,b,c,d,k) % Given the single-input linear system:% .% x = Ax + Bu% y = Cx + Du% and the feedback matrix K,% the fun ction rscale(sys,K) or rscale(A,B,C,D,K)% finds the scale factor N which will% elimi nate the

3、steady-state error to a step refere nee% for a con ti nuous-time, sin gle-i nput system% with full-state feedback using the schematic below:% /% R + u | .|% > N >() > | X=Ax+Bu |-> y = Cx > y-I /I I|<K <| % 8/21/96 Yanjie Sun of the Un iversity of Michiga n% un der the supervisi

4、 on of Prof. D. Tilbury% 6/12/98 Joh n Yook, Daw n Tilbury revised error( nargchk(2,5 ,n argi n);% - Determ ine which syn tax is being used - n arg ini = n arg in;if (nargin=2),% System formA,B,C,D = ssdata(a); K=b;elseif(n arg in=5),% A,B,C,D matrices A=a;A = a;B = b;C = c;D = d;K = k;elseerror( &#

5、39;In put must be of the form (sys,K) or (A,B,C,D,K)')end ;% compute Nbar s = size(A,1);s = size(A,1);Z = zeros(1,s) 1;N = in v(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,xO);axis(0 10 0 11)

6、Linear Simulation ResultsTT1234910Time (seconds)(Ke)電生電動勢常數(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 H1.建模:直流電機(jī)的速度控制(b)電機(jī)摩擦系數(shù)0.1 NmsT /；e = A e 0JO be = KidiLF Ri = V lffdt傳遞函數(shù)模型：+ b)e(s) = KjT(s)Ls + H)I(s) = V (s) Ks (3)控制算法:1. PID控制:plantcontrollerj = 0.01；b = 0.1;K = 0.01;R = 1;L = 0.

7、5;s = tf( 's');P_motor = K/(J*s+b)*(L*s+R)+O2);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('Step Response with Proportional Control')Step Response with Proportion曲 ControlQlpn一社 uvTime (sec(2) PID控制Kp = 75;Ki = 1;Kd = 1;C

8、 = 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')1P1D Control with Small Ki and Small Kd9 B 7 a 00.0.2 -0 1 -q IiIIIiII1II0 2Q 40600100120140160160200Time (seconds調(diào)整參數(shù)：Kp = 100;Ki = 200;Kd = 1;C = pid(Kp,Ki,Kd);sys_cl = feedb

9、ack(C*P_motor,1);step(sys_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;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Kpl = -5 + 1i;p2 = -5 - 1i;Kc = place(A,B,p1 p2) t = 0:0.01:3;sys_cl =

10、 ss(A-B*Kc,B,C,D);step(sys_cl,t)gridtitle( 'Step Response with State-Feedback Controller'0.08Step Response with State-Feedback Controller00.511.5Time (seconds)2.S302 o.01 a7 o o.050403 o.ao.0(3) 加上比例常數(shù)補(bǔ)償：Nbar = rscale(sys,Kc) t = 0:0.01:10;step(sys_cl*Nbar,t)grid title( 'Step Response with S