dimension error that prevents estimation

Gostaria de realizar estimativas, em que eu necessito fazer um comparativo entre as minhas saídas y e minhas saídas estimadas e somente consigo a da primeira interação. Alguém pode ajudar? Segue o meu código até então. As I do not know how to program very well, forgive me any grotesque mistake in the code and questions I make that seem simple!

ps: I edited because I solved the problem and found another one that...

clc; close all; clear;dbstop if error;


R = 0.3456; %---->K
T = 0.9746;
Z = -1.6557;
% a_0 = [0 0 0]';
Q = 0.876;  %----> cov e(k)
H =0;

y = [ 1 2 3 4 5 6 7 8 9 10 11 34 23 55 6;
    2 3 8 1 1 3 5 6 7 4 12 8 5 6  3;
    6 5 2 12 2 4 5 6 8 9 1 2 1 3 9];

[p,q] = size(y);
%inicicalização dados
A = eye(size(T,1));
P_oo = A*A';
P_finito_inicial = zeros(size(T,1));
P_finito = P_finito_inicial;
a_conhecido = zeros(size(T,1),1);
a_inicial = a_conhecido;
a_o = a_inicial;
k=1;
a_predicao(:,:,k) = zeros(q,size(T,1));
var_erro_finito = zeros(q,size(T,1));
inovacao_predicao=zeros(q,size(y,2));
inovacao_predicao(:,1) = y(1,:)'-Z*a_o(:,:,k);
saida_predicao(:,:,k) = zeros(p,q);
saida_predicao(1,:) = Z*a_predicao(1,:,:);

% Inicializacao Exata

 for k =1:q
     % Testa F_oo
     F_oo = Z*P_oo*Z';
     if all(F_oo(:)== 0) % testa para zero 
     %F e M
     F_finito = Z*P_finito*Z' + H;
     F_ini(:,:,k) = [zeros(size(F_finito )) zeros(size(F_finito ))];
     M_finito = P_finito*Z';
     K_o = (T*M_finito)/F_finito;
     L_o = T-K_o*Z;
     L_ini(:,:,k) = [L_o zeros(size(L_o))];
     v_o = y(k,:)' - Z*a_o;
     a_o = T*a_o +K_o*v_o;                % nova estimativa
     P_oo = T*P_oo*T';
     P_finito(:,:,k+1) = T*P_finito*L_o' + R*Q*R';               % nova covariancia
     elseif rank(F_oo)== min(size(F_oo,1),size(F_oo,2)) % testa nao singularidade
     F_finito = Z*P_finito*Z' + H;
     M_oo = P_oo*Z'; 
     M_finito = P_finito*Z';
     F_2 = -(F_oo\F_finito)/F_oo;                       %F_1 = inv(F_oo);
     F_1 = eye(size(F_2))/F_oo;
     F_ini(:,:,k) = [F_1 F_2];
     K_o = (T*M_oo)*F_1;
     K_1 = (T*M_finito)*F_1 +T*M_oo*F_2;
     L_o = T-K_o*Z;
     L_1 = -K_1*Z;
     L_ini(:,:,k) = [L_o L_1];
     v_o = y(k,:)' - Z*a_o;
     a_o = T*a_o(:,:,k) +K_o*v_o; %erro
     P_oo = T*P_oo*L_o';
     else
          error('Matriz F_oo pode ser singular')
     end
%% Guarda valores

a_predicao(:,k) =a_o;
saida_predicao(k,:) = Z*a_predicao(:,k);
var_erro_finito(:,k) = diag(P_finito);
var_erro_oo(:,k) = diag(P_oo); % guarda var P_oo;
P_predicao(:,:,k) = [P_oo P_finito];
if all(P_oo<=1e-7)==1
d = k;
break;
end
 end

 %% FK preditor normal
a_estimado = a_predicao(:,d);
P = P_finito;
for k = 1:q

F = Z*P*Z' + H; 
F_ref(:,:,k) = F; %guarda F para refinamento
K = T*P*Z'/F;
%L
L = T-K*Z; 
% L_ref(:,:,k) = L;
%Inovacao
v= y(k,:)' - Z*a_estimado;
%% Atualizacao
a_estimado= T*a_estimado +K*v; % nova estimativa
P = T*P*L' + R*Q*R'; % nova covariancia
%% Guarda valores
a_predicao(:,k) =a_estimado;
saida_predicao(k,:) = Z*a_predicao(:,k);
inovacao_predicao(:,k) = v;
var_erro_finito(:,k) = diag(P);
P_predicao(:,:,k) = [P_oo P];
end
 

t = 0:1:q;
 
figure(1)
subplot(3,1,1)
plot(t,y(1,:),'r',t,saida_predicao(1,:)','-.b')
subplot(3,1,2)
plot(t,y(2,:),'r', t,saida_predicao(2,:)','-.b')
subplot(3,1,3)
plot(t,y(3,:),'r', t,saida_predicao(3,:)','-.b')



 







 

Answer 1

[Walter Roberson]

y = [ 1 2 3 4 5 6 7 8 9 10 11 34 23 55 6;
    2 3 8 1 1 3 5 6 7 4 12 8 5 6  3;
    6 5 2 12 2 4 5 6 8 9 1 2 1 3 9];
[p,q] = size(y);

so q is the number of columns in y

t = 0:1:q;

so t has q+1 entries. For example if q was 5 then t = [0 1 2 3 4 5], for a total of 6 entries.

plot(t,y(1,:),'r',t,saida_predicao(1,:)','-.b')

t has one more entry than y has columns; it also has one more entry than saida_predicao has columns.

[GTA]

You are right, but here I made a mistake not to put all the code necessary for its diagnosis, in this case I edited the post, this way you can see the code in full.

[Walter Roberson]

for k = 1:q

so k is a column index.

v= y(k,:)' - Z*a_estimado;

but you are using it as a row index.

[GTA]

Even though I am making the change

v= y(:,k)' - Z*a_estimado;

and

a_predicao(:,k) =a_estimado;

The following error occurs:

Assignment has more non-singleton rhs dimensions than non-singleton subscripts

Error in testemil (line 98)
a_predicao(:,k) =a_estimado;
 

[Walter Roberson]

y(:,k)

is 3 x 1. y(:,k)' is 1 x 3. Z*a_estimado is 15 x 1. y(:,k)' - Z*a_estimado is an error in R2016a and earlier, but in R2016b it is like bsxfun(@minus, y(:,k).', Z*a_estimado) which would give you 15 x 3. That is not going to fit into a_predicao(:,k)

You need to rethink what you are doing. - Z*a_estimado is 15 x 1 so you probably want to the term before that to be 15 x 1 giving you a 15 x 1 result. But that would require indexing a row of y, and there are only 3 rows of y. Perhaps your for loop should be k=1:p instead of k=1:q

[GTA]

There is still a dimension error in the same variable.