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

Alex Mcaulley
Alex Mcaulley le 13 Mai 2019
Modifié(e) : Alex Mcaulley le 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 commentaires

Tipu Sultan
Tipu Sultan le 13 Mai 2019
I dont want finction is there any other way as I want to write this inside a function itself!
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 le 13 Mai 2019
Thanks it is working!

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Plus de réponses (1)

KSSV
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
Tipu Sultan le 13 Mai 2019
not working says finction handle!
KSSV
KSSV le 13 Mai 2019
It will work...how you have used it? show us the full code.
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
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
This is my whole code now when I am checking in workspace it says M as function handle.
error as follows:
Undefined operator '*' for input arguments of type 'function_handle'.
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 le 13 Mai 2019
thanks it is working!

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