Nâng cao chất lượng điều khiển cho robot Scara, chương 16

Chia sẻ: Van Teo | Ngày: | Loại File: PDF | Số trang:13

Thêm vào BST

Báo xấu

96
lượt xem 62
download

Download Vui lòng tải xuống để xem tài liệu đầy đủ

Tính toán Quỹ đạo chuyển động chuẩn % Tinh toan Quy dao chuyen dong chuan % Xay dung quy dao chuyen dong chuan cho 3 khop quay cua Robot Serpent % qi0, qic: goc quay dau va cuoi cua khop quay thu i function qdc=qdcdc(u) % Toa do diem dau: DDau=[X0;Y0] X0=u(1); Y0=u(2); % Toa do diem cuoi: DCuoi=[Xc;Yc] Xc=u(3); Yc=u(4); % td: Dat thoi gian chuyen dong td=u(5); % Thoi gian gia toc (hoac giam toc) ta=u(6); % Thong so cua Robot: a1,a2 a1=u(7); %(m) a2=u(8); %(m) t=u(9); % thoi gian thuc nx=u(10); % CHUONG TRINH CHINH % Dong hoc...

Chủ đề:

Bình luận(0) Đăng nhập để gửi bình luận!

Lưu

Nội dung Text: Nâng cao chất lượng điều khiển cho robot Scara, chương 16

Chương 16: CÁC CHƯƠNG TRÌNH M FILES (chạy trên nền Matlab/Simulink) 1. Tính toán Quỹ đạo chuyển động chuẩn % Tinh toan Quy dao chuyen dong chuan % Xay dung quy dao chuyen dong chuan cho 3 khop quay cua Robot Serpent % qi0, qic: goc quay dau va cuoi cua khop quay thu i function qdc=qdcdc(u) % Toa do diem dau: DDau=[X0;Y0] X0=u(1); Y0=u(2); % Toa do diem cuoi: DCuoi=[Xc;Yc] Xc=u(3); Yc=u(4); % td: Dat thoi gian chuyen dong td=u(5); % Thoi gian gia toc (hoac giam toc) ta=u(6); % Thong so cua Robot: a1,a2 a1=u(7); %(m) a2=u(8); %(m) t=u(9); % thoi gian thuc 1
nx=u(10); % CHUONG TRINH CHINH % Dong hoc nguoc va tinh cac goc quay cua tung khop tai diem dau va diem cuoi q20=Gocquay2(X0,Y0,a1,a2); q2c=Gocquay2(Xc,Yc,a1,a2); q10=Gocquay1(X0,Y0,q20,a1,a2); q1c=Gocquay1(Xc,Yc,q2c,a1,a2); q40=Gocquay4(q10,q20,nx); q4c=Gocquay4(q1c,q2c,nx); %Khoi tao cac he so cua quy dao chuan cua 3 khop h1=Init(q10,q1c,ta,td); h2=Init(q20,q2c,ta,td); h4=Init(q40,q4c,ta,td); %Quy dao chuyen dong chuan 3 khop quay cua Robot Serpent qdc1= Quydaochuan(t,h1,ta,td); qdc2= Quydaochuan(t,h2,ta,td); qdc4= Quydaochuan(t,h4,ta,td); qdc=[qdc1;qdc2;qdc4]; %KET THUC CHUONG TRINH CHINH %CHUONG TRINH TINH GIA TRI %Tinh cac gia tri tuong ung goc quay 2: q20,q2c 2
function gq2=Gocquay2(x,y,a1,a2) C2=(x^2+y^2-a1^2-a2^2)/(2*a1*a2); S2=sqrt(1-C2^2); gq2=atan2(S2,C2); %Tinh cac gia tri tuong ung goc quay 1: q10,q1c function gq1=Gocquay1(x,y,q2,a1,a2) C1=((a1+a2*cos(q2))*x+a2*sin(q2)*y)/(x^2+y^2); S1=((a1+a2*cos(q2))*y-a2*sin(q2)*x)/(x^2+y^2); gq1=atan2(S1,C1); %Tinh cac gia tri tuong ung goc quay 4: q40,q4c function gq4=Gocquay4(q1,q2,nx) KK=sqrt(1-nx^2); tetaphu=atan2(KK,nx); gq4=q1+q2-tetaphu; %Xay dung quy dao chuyen dong chuan %qac=a0ac+a3ac*t^3+a4ac*t^4 %qcd=a0cd+a1cd*(t-2ta) %qdf=a0df+a1df*t+a3df*t^3+a4df*t^4 function quydao=Quydaochuan(t,h,ta,td) a0ac=h(1);a3ac=h(2);a4ac=h(3); 3
a0cd=h(4);a1cd=h(5); a0df=h(6);a1df=h(7);a3df=h(8);a4df=h(9);qcuoi=h(10); if td0)&&(td=0)&&(ttd/2)&&(ttd quydao=qcuoi; else quydao=-1; %Error end else %if td>(4*ta) %Gom 3 giai doan: gia toc, chay deu, giam toc if (t>=0)&&(t=2*ta)&&(t
quydao=a0cd+a1cd*(t-2*ta);%Chay deu elseif (t>(td-2*ta))&&(ttd quydao=qcuoi; else quydao=-1;%Error end end %Tinh cac he so cua quy dao chuan ung voi tung khop function heso=Init(qdau,qcuoi,ta,td) a0ac=0;a3ac=0;a4ac=0;a0cd=0;a1cd=0;a0df=0;a1df=0;a3df=0;a4df=0 ; if (td>0)&&(td
a1cd=0; %Quy dao doan df: a0df=qcuoi-(qcuoi-qdau)/2; a1df=(qcuoi-qdau)/(td/2); a4df=-(3*qcuoi-3*a0df-a1df*td)/(td^4/16); a3df=-(a1df+a4df*(td^3/2))/(3*td^2/4); end if td>(4*ta) %Quy dao doan ac: a0ac=qdau; a3ac=(qcuoi-qdau)/(4*ta^2*(td-2*ta)); a4ac=-(qcuoi-qdau)/(16*ta^3*(td-2*ta)); %Quy dao doan cd: a0cd=qdau+(qcuoi-qdau)*ta/(td-2*ta); a1cd=(qcuoi-qdau)/(td-2*ta); %Quy dao doan df: a0df=qcuoi-(qcuoi-qdau)*ta/(td-2*ta); a1df=(qcuoi-qdau)/(td-2*ta); a4df=-(3*qcuoi-3*a0df-4*a1df*ta)/(16*ta^4); a3df=-(a1df+32*a4df*ta^3)/(12*ta^2); end 6
heso=[a0ac,a3ac,a4ac,a0cd,a1cd,a0df,a1df,a3df,a4df,qcuoi]; 2. Tính momem chuẩn % Tinh toan momem chuan function momem=momemchuan(u) q2d=u(1); dq1d=u(2);ddq1d=u(3); dq2d=u(4);ddq2d=u(5); dq4d=u(6);ddq4d=u(7); mt=u(8);Jt=u(9); % Thong so cua robot J1=0.021; J2=0.0034; J3=0.006; J40=0.001; J4=J40+Jt; J124=J1+J2+J4; J24=J2+J4; m1=4; m2=1.5; m3=2; m40=0.6; m4=m40+mt; m1234=m1+m2+m3+m4; 7
m234=m2+m3+m4; m34=m3+m4; l1=0.25; l2=0.15; % Tinh T T=-(m234*l1*l2*sin(q2d)); % Tinh toan cac gia tri Hij H11=m1234*l1^2+m234*l2^2+J124+2*m234*l1*l2*cos(q2d); H12=m234*l2^2+J24+m234*l1*l2*cos(q2d); H13=J4; H21=H12; H22=m234*l2^2+J24; H23=J4; H31=J4; H32=J4; H33=J4; DH=J4^2*(2*H12-H11-H22)+J4*(H11*H22-H12^2); % Dau ra % Tinh momem chuan quy dao dat u1m=H11*ddq1d+H12*ddq2d+H13*ddq4d+T*dq2d^2+2*T*dq1d*d q2d; u2m=H21*ddq1d+H22*ddq2d+H23*ddq4d-T*dq1d^2; u4m=H31*ddq1d+H32*ddq2d+H33*ddq4d; momem =[u1m;u2m;u4m]; 8
3. Tính góc quay ban đầu %Tinh toan goc quay ban dau function init=gocquaybd(u) x=u(1);%m y=u(2);%m a1=u(3);%(m) a2=u(4);%(m) nx=u(5); %q20=Gocquay2(X0,Y0,a1,a2) %q10=Gocquay1(X0,Y0,q20,a1,a2); %q40=Gocquay4(q10,q20,nx); %Tinh cac gia tri tuong ung goc quay 2: q20 C2=(x^2+y^2-a1^2-a2^2)/(2*a1*a2); S2=sqrt(1-C2^2); q20=atan2(S2,C2); %Tinh cac gia tri tuong ung goc quay 1: q10 C1=((a1+a2*cos(q20))*x+a2*sin(q20)*y)/(x^2+y^2); S1=((a1+a2*cos(q20))*y-a2*sin(q20)*x)/(x^2+y^2); q10=atan2(S1,C1); %Tinh cac gia tri tuong ung goc quay 4: q40 KK=sqrt(1-nx^2); 9
thetaphu=atan2(KK,nx); q40=q10+q20-thetaphu; %Dau ra init=[q10;q20;q40]; 4. Tính góc điều khiển thực % Tinh toan goc dieu khien thuc function ddtheta=tinh_ddtheta(u) % Momem dieu khien u1=u(1); u2=u(2); u4=u(3); % Thong so cua tai mt=u(4); Jt=u(5); % Goc quay theta2 thuc te q1tt=u(6);q2tt=u(7);q4tt=u(8); % Dao ham goc quay thuc te dq1tt=u(9);dq2tt=u(10); % Goc quay ban dau q10=u(11);q20=u(12);q40=u(13); % CHUONG TRINH CHINH % Tinh cac he so heso=tinh_heso(q2tt,dq1tt,dq2tt,mt,Jt); 10
% Tinh dao ham cap hai cua goc quay thuc ddtheta=tinh_ddthetathuc(heso,u1,u2,u4,q10,q20,q40); % Chuong trinh tinh cac he so function HS=tinh_heso(q2tt,dq1tt,dq2tt,mt,Jt) % Thong so cua robot J1=0.021; J2=0.0034; J3=0.006; J40=0.001; J4=J40+Jt; J124=J1+J2+J4; J24=J2+J4; m1=4; m2=1.5; m3=2; m40=0.6; m4=m40+mt; m1234=m1+m2+m3+m4; m234=m2+m3+m4; m34=m3+m4; l1=0.25; l2=0.15; % Tinh T T=-(m234*l1*l2*sin(q2tt)); % Tinh toan cac gia tri Hij 11
H11=m1234*l1^2+m234*l2^2+J124+2*m234*l1*l2*cos(q2tt); H12=m234*l2^2+J24+m234*l1*l2*cos(q2tt); H13=J4; H21=H12; H22=m234*l2^2+J24; H23=J4; H31=J4; H32=J4; H33=J4; DH=J4^2*(2*H12-H11-H22)+J4*(H11*H22-H12^2); % Cac gia tri dau ra bij b11=DH^-1*(H22*H33-H23*H32); b12=DH^-1*(H13*H32-H12*H33); b13=DH^-1*(H12*H23-H13*H22); b21=DH^-1*(H23*H31-H21*H33); b22=DH^-1*(H11*H33-H13*H31); b23=DH^-1*(H13*H21-H11*H23); b31=DH^-1*(H21*H32-H31*H22); b32=DH^-1*(H12*H31-H11*H32); b33=DH^-1*(H11*H22-H12^2); % Cac gia tri h1,h2 h1=T*dq2tt^2+2*T*dq1tt*dq2tt; h2=-(T*dq1tt^2); 12
% Cac gia tri ai a1=-(b11*h1+b12*h2); a2=-(b21*h1+b22*h2); a4=-(b31*h1+b32*h2); HS =[b11,b12,b13,b21,b22,b23,b31,b32,b33,a1,a2,a4]; % Chuong trinh tinh dao ham cap hai cua goc quay thuc function d2theta=tinh_ddthetathuc(heso,u1,u2,u4,q10,q20,q40) b11=heso(1);b12=heso(2);b13=heso(3); b21=heso(4);b22=heso(5);b23=heso(6); b31=heso(7);b32=heso(8);b33=heso(9); a1=heso(10);a2=heso(11);a4=heso(12); % Tinh dao ham cap hai cua goc quay thuc d2theta1=a1+b11*u1+b12*u2+b13*u4; d2theta2=a2+b21*u1+b22*u2+b23*u4; d2theta4=a4+b31*u1+b32*u2+b33*u4; d2theta=[d2theta1;q10;d2theta2;q20;d2theta4;q40]; 13