Trajectory positioning method-Matlab simulation method
Known plane coordinates of four points (x1,y1)=(0,0), (x2,y2) =(1,0), (x3,y3) =(0,1), (x4,y4) =(0 ,2), someone walks along a straight line, and the coordinates of the points to be located during walking are (a1,b1),……(a5,b5), and the distances from each coordinate point to be located to the known four points are measured separately For (0.71, 0.78, 0.64, 1.52), (1.4, 0.95, 1.05, 1.5), (2.12, 1.61, 1.55, 1.52), (2.87, 2.24, 2.31, 2.1), (3.61, 3, 2.97, 2.57) Find the linear equation for this person to walk.
Positioning method - multi-point positioning
Problem-solving ideas:
(1) Linear regression equation
(2) Least squares method
Program realization:
Location.m
function kk=Location(xx,yy,ll)
xxx = xx(1:end-1);
yyy = yy(1:end-1);
lll = ll(1:end-1);
A = [2*(xxx-xx(end)) 2*(yyy-yy(end))];
b = (xxx.^2-xx(end)^2) + (yyy.^2-yy(end)^2) - (lll.^2-ll(end)^2);
kk=(A.'*A)^(-1)*A.'*b;
drawline.m (linear regression equation matlab realizes drawing a straight line)
function drawline(x,y)
n=length(x);
xb=mean(x);
yb=mean(y);
x2b=sum(x.^2)/n;
xyb=x*y'/n;
b=(xb*yb-xyb)/(xb^2-x2b);
a=yb-b*xb;
y1=a+b.*x;
plot(x,y,'*',x,y1);
serror=sum((y-y1).^2)The first conventional program: (Disadvantage: you have to call the Location() function every time you find an anchor point, the code is a lot of redundant)
heheda_matlab_changui.m
tic
xx=[0 1 0 0].';
yy=[0 0 1 2].';
l1=[0.71 0.78 0.64 1.52].';
l2=[1.4 0.95 1.05 1.5].';
l3=[2.12 1.61 1.55 1.52].';
l4=[2.87 2.24 2.31 2.1].';
l5=[3.61 3 2.97 2.57].';
k1=Location(xx,yy,l1)
k2=Location(xx,yy,l2)
k3=Location(xx,yy,l3)
k4=Location(xx,yy,l4)
k5=Location(xx,yy,l5)
X=[k1(1) 1
k2(1) 1
k3(1) 1
k4(1) 1
k5(1) 1]
Y=[k1(2)
k2(2)
k3(2)
k4(2)
k5(2)]
X1=[k1(1) k2(1) k3(1) k4(1) k5(1)]
Y1=[k1(2) k2(2) k3(2) k4(2) k5(2)]
drawline(X1,Y1)
toc
The second type: simplified procedures
heheda_chengxu.m
tic
xx=[0 1 0 0].';
yy=[0 0 1 2].';
l1=[0.71 0.78 0.64 1.52].';
l2=[1.4 0.95 1.05 1.5].';
l3=[2.12 1.61 1.55 1.52].';
l4=[2.87 2.24 2.31 2.1].';
l5=[3.61 3 2.97 2.57].';
sum=[l1 l2 l3 l4 l5];
kk=[]
i=1
for i=1:5
kk(:,i)=Location(xx,yy,sum(:,i));
end;
k_x=kk(1,:)
k_y=kk(2,:)
X=[]
Y=[]
for i=1:5
X(:,i)=[k_x(i),1]
Y(:,i)=k_y(i)
end
X=X.'
Y=Y.'
a = -5:0.1:2;
b = 0:0.1:10;
drawline(k_x,k_y)
toc
The third type: realize code optimization
code_fast.m
tic
xx=[0 1 0 0].';
yy=[0 0 1 2].';
l1=[0.71 0.78 0.64 1.52].';
l2=[1.4 0.95 1.05 1.5].';
l3=[2.12 1.61 1.55 1.52].';
l4=[2.87 2.24 2.31 2.1].';
l5=[3.61 3 2.97 2.57].';
sum=[l1 l2 l3 l4 l5];
kk=[]
i=1
for i=1:5
kk(:,i)=Location(xx,yy,sum(:,i));
end;
k_x=kk(1,:)
k_y=kk(2,:)
X=[]
Y=[]
for i=1:5
X(:,i)=[k_x(i),1]
Y(:,i)=k_y(i)
end
X=X.'
Y=Y.'
a = -5:0.1:2;
b = 0:0.1:10;
drawline(k_x,k_y)
tocFourth: the implementation code is more concise
true_line.m
tic
xx=[0 1 0 0].';
yy=[0 0 1 2].';
l1=[0.71 0.78 0.64 1.52].';
l2=[1.4 0.95 1.05 1.5].';
l3=[2.12 1.61 1.55 1.52].';
l4=[2.87 2.24 2.31 2.1].';
l5=[3.61 3 2.97 2.57].';
sum=[l1 l2 l3 l4 l5];
kk=[]
i=1
for i=1:5
kk(:,i)=Location(xx,yy,sum(:,i));
end;
k_x=kk(1,:)
k_y=kk(2,:)
X=[]
Y=[]
for i=1:5
X(:,i)=[k_x(i),1]
Y(:,i)=k_y(i)
end
X=X.'
Y=Y.'
%a = -5:0.1:2;
%b = 0:0.1:10;
EK=(X.'*X)^(-1)*X.'*Y
a=EK(1)
b=EK(2)
y1=a.*k_x+b;
plot(k_x,k_y,'*',k_x,y1);
toc
Find the coordinates
X1 =
0.4483 1.0283 1.4512 2.1122 2.5165
Y1 =
0.5487 0.9273 1.5460 1.9580 2.6070
That is, the coordinates of 5 pending points can be determined as a straight line.
The least square method determines the coefficients a and b:
Same idea as above
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Hope to help you
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