How to store values matrix in cell
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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.
Réponse acceptée
Plus de réponses (1)
KSSV
le 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 commentaires
Tipu Sultan
le 13 Mai 2019
KSSV
le 13 Mai 2019
It will work...how you have used it? show us the full code.
Tipu Sultan
le 13 Mai 2019
KSSV
le 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
le 13 Mai 2019
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