Filter löschen
Filter löschen

How to store values matrix in cell

2 Ansichten (letzte 30 Tage)
Tipu Sultan
Tipu Sultan am 13 Mai 2019
Kommentiert: Tipu Sultan am 13 Mai 2019
Hello,
I have an array as follows:
t= [ 1 2 3 ];
A matrix as follows:
M = [t^2 t 1 0 0 0;
0 0 0 t^2 t 1];
Now I want to store the process the M for 3 values of t i.e in this case [1 2 3].How do I do it.
like if t=1
then M=[1^2 1 1 0 0 0;0 0 0 1^2 1 1];
if t=2
then M=[2^2 2 1 0 0 0;0 0 0 2^2 2 1]; and so on.

Akzeptierte Antwort

Alex Mcaulley
Alex Mcaulley am 13 Mai 2019
Bearbeitet: Alex Mcaulley am 13 Mai 2019
Try this:
t = [1 2 3];
M = arrayfun(@(t) [t^2 t 1 0 0 0; 0 0 0 t^2 t 1],t,'UniformOutput',false);
  3 Kommentare
Alex Mcaulley
Alex Mcaulley am 13 Mai 2019
You can use it in your scrip as:
S = [100 0 0 0 0 0;
0 100 0 0 0 0;
0 0 100 0 0 0;
0 0 0 100 0 0;
0 0 0 0 100 0;
0 0 0 0 0 100];
prev_S = S;
Big_lambda = eye(2);
a=0;
b=0;
c=0;
p=0;
q=0;
s=0;
est_vec=[ a ; b; c; p; q; s];
theta = [ 45 46 48] ;
t= [ 1 2 3 ];
r= [200 210 220];
%Wanted = num2cell(M,1)
%[r,b,distance,angle]=deal(Wanted{:})
x = [ r; theta ; t];
%dif_x=zeros(2,3);
M = arrayfun(@(t) [t^2 t 1 0 0 0; 0 0 0 t^2 t 1],t,'UniformOutput',false);
time=4;
for i=1:3
dif_x = [(-cos(theta(i))) (r(i).*sin(theta(i))) (2*a.*t(i)+b);(-sin(theta(i))) (-r(i).*cos(theta(i))) (2*p.*t(i)+q)]
W = dif_x(i)*Big_lambda(i)*dif_x(i)'
pred_x = x+randn(3)
y = M{i} * est_vec + dif_x * (x(:,i)-pred_x(:,i))
K = prev_S*M{i}'/((W + M{i}*prev_S*M{i}'))
%S =prev_S;
est_vec_new = est_vec + K*(y-M{i}*est_vec)
cond = abs(est_vec_new - est_vec)
if cond < 0.003
break
end
est_vec = est_vec_new
a=est_vec(1,1)
b=est_vec(2,1)
c=est_vec(3,1)
p=est_vec(4,1)
q=est_vec(5,1)
s=est_vec(6,1)
S_new = (eye(6) - K*M{i})*S
S = S_new
r_cosTheta = a*time.^2+b*time+c % To calculate x co-ordinate for t>=3
r_sineTheta = p*time.^2+q*time+s % To calculate y co-ordinate for t>=3
%figure,plot(t,meas_equa1)
%new_t=[t,time];
%figure,plot(r_cosTheta,r_sineTheta)
end
%end
Tipu Sultan
Tipu Sultan am 13 Mai 2019
Thanks it is working!

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

KSSV
KSSV am 13 Mai 2019
M = @(t) [t^2 t 1 0 0 0;
0 0 0 t^2 t 1];
t = [1 2 3] ;
iwant = zeros(2,6,length(t)) ;
for i = 1:length(t)
iwant(:,:,i) = M(t(i)) ;
end
  5 Kommentare
KSSV
KSSV am 13 Mai 2019
clc; clear all ;
S = [100 0 0 0 0 0;
0 100 0 0 0 0;
0 0 100 0 0 0;
0 0 0 100 0 0;
0 0 0 0 100 0;
0 0 0 0 0 100];
prev_S = S;
Big_lambda = eye(2);
a=0;
b=0;
c=0;
p=0;
q=0;
s=0;
est_vec=[ a ; b; c; p; q; s];
theta = [ 45 46 48] ;
t= [ 1 2 3 ];
r= [200 210 220];
%Wanted = num2cell(M,1)
%[r,b,distance,angle]=deal(Wanted{:})
x = [ r; theta ; t];
%dif_x=zeros(2,3);
M = @(t) [t^2 t 1 0 0 0;
0 0 0 t^2 t 1];
t = [1 2 3] ;
iwant = zeros(2,6,length(t)) ;
for i = 1:length(t)
iwant(:,:,i) = M(t(i)) ;
end
time=4;
for i=1:3
M = iwant(:,:,i) ;
dif_x = [(-cos(theta(i))) (r(i).*sin(theta(i))) (2*a.*t(i)+b);(-sin(theta(i))) (-r(i).*cos(theta(i))) (2*p.*t(i)+q)]
W = dif_x(i)*Big_lambda(i)*dif_x(i)'
pred_x = x+randn(3)
y = M * est_vec + dif_x * (x(:,i)-pred_x(:,i))
K = prev_S*M'/((W + M*prev_S*M'))
%S =prev_S;
est_vec_new = est_vec + K*(y-M*est_vec)
cond = abs(est_vec_new - est_vec)
if cond < 0.003
break
end
est_vec = est_vec_new
a=est_vec(1,1)
b=est_vec(2,1)
c=est_vec(3,1)
p=est_vec(4,1)
q=est_vec(5,1)
s=est_vec(6,1)
S_new = (eye(6) - K*M)*S
S = S_new
r_cosTheta = a*time.^2+b*time+c % To calculate x co-ordinate for t>=3
r_sineTheta = p*time.^2+q*time+s % To calculate y co-ordinate for t>=3
%figure,plot(t,meas_equa1)
%new_t=[t,time];
%figure,plot(r_cosTheta,r_sineTheta)
end
%end
Tipu Sultan
Tipu Sultan am 13 Mai 2019
thanks it is working!

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Triangulation Representation finden Sie in Help Center und File Exchange

Produkte


Version

R2015b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by