Unable to perform assignment because the left and right sides have a different number of elements.

I used this code for perfoming kalman filtering for signal data. the measurment are random created.

this massege was appearing (Unable to perform assignment because the left and right sides have a

different number of elements.)

% %Initializations
Pk_prev = [0,0; 0,0];
x_kprev_hat = [0 ; 0];

xk_hat = [];
% Time update equations
A= [0.5 , 0 ;  0 , 0.3];
B = 0;
% Measurement update equations
H = [ 1 , 0];
z = rand(2,200);
E_x=[1,0 ;  0,1]; %Simulating noisy measurements
E_s =[1,0 ;  0,1];
E_y=1
%Measurement model
z_k = H*(z + 1) + E_y;
P = zeros(1,length(z));
% Measurement noise
R =1;%R  = 0.000001 for a very low value
% Process noise covariance
Q =[1, 0; 0, 1];
for k = 1: length(z)
    x_k_hat_minus = A*x_kprev_hat+ E_x   ; %a priori estimate
    Pk_minus = A*Pk_prev * A'+ Q; %A priori estimate error covariance
    Kk = Pk_minus/(Pk_minus + R);%Kalman gain
    
    x_kprev_hat = x_k_hat_minus + Kk*(z_k(k) - x_k_hat_minus); %A priori estimate
    xk_hat = cat(2,xk_hat,x_kprev_hat);%A posteriori estimate
    Pk_prev =  (1 - Kk)*Pk_minus;%A posterirori estimate error covariance
    P(k)  = Pk_prev;
end
plot(z_k,'bx-');
hold on;
plot(xk_hat,'gx-');
plot(z*ones(1,200),'rx-');
figure;
plot(P,'r');

E_y =1

[Alex Mcaulley]

Please copy-paste the full error message

[Lina Alhelo]

Unable to perform assignment because the left and right sides have a

different number of elements.

This was the complete message

Answer 1

[Star Strider]

The ‘Pk_prev’ variable is a (2x2) matrix. You cannot assign that to a scalar array element in ‘P’.

If you preallocate ‘P’ as:

P = zeros(2,2,length(z));

and then assign it as:

P(:,:,k)  = Pk_prev;

your code works. You then need to plot ‘z’ as:

plot(z,'rx-');

Plotting ‘P’ however is not possible with your current code.

You can plot it with this:

figure
hold all
for k = 1:size(P,3)
    surf(P(:,:,k))
end
hold off
grid on
view(-30,30)

Experiment to get the result you want.

[Lina Alhelo]

thanks a lot for your support, I have did what you have suggested as follwoing but another error appeared ,

Unable to perform assignment because the size of the left side is 2-by-2-by-200 and the size
of the right side is 2-by-2.

here is the code

%Process model: x_k = Ax_k-1 + Bu_k + E_x
%Mearuement model:z_k = H(x_k +E_s)+ E_y
%E_x and E_y are process and measurement noise respectively
%Initializations
z = rand(2,200);
P = zeros(2,length(z));
Pk_prev = [1,0; 0,1];
x_kprev_hat = [0,0 ; 0,0];
xk_hat = [];
% Time update equations
A= [0.5 , 0 ;  0 , 0.3];
B = 0;
% Measurement update equations
H = [ 1 , 1];
E_x=[1,0 ;  0,1]; %Simulating noisy measurements
E_s =[1,0 ;  0,1];
E_y=1
%Measurement model
z_k = H*(z + 1) + E_y;
P = zeros(2,2,length(z));
P(:,:,k)  = Pk_prev;
% Measurement noise
R =[1, 0; 0, 1];%R  = 0.000001 for a very low value
% Process noise covariance
Q =[1, 0; 0, 1];
for k = 1: length(z)
    x_k_hat_minus = A*x_kprev_hat+ E_x   ; %a priori estimate
    Pk_minus = A*Pk_prev * A'+ Q; %A priori estimate error covariance
    Kk = (Pk_minus*H')/(H*Pk_minus*H' + R);%Kalman gain
    
    x_kprev_hat = x_k_hat_minus + Kk*(z_k(k) - x_k_hat_minus); %A priori estimate
    xk_hat = cat(2,xk_hat,x_kprev_hat);%A posteriori estimate
    Pk_prev =  (1 - Kk)*Pk_minus;%A posterirori estimate error covariance
    P(:,:,k)  = Pk_prev;

end
plot(z,'rx-');
hold on;
plot(xk_hat,'gx-');
plot(z,'rx-');
figure;
hold all;
for k = 1:size(P,3);
    surf(P(:,:,k));
end
hold off;
grid on;
view(-30,30);

E_y =

     1

Unable to perform assignment because the size of the left side is 2-by-2-by-200 and the size
of the right side is 2-by-2.

[Star Strider]

I do not know what line that error refers to. It would help to know.

You changed your code significantly. It worked before, with the changes I noted. I deleted the first assignment of:

P(:,:,k)  = Pk_prev;

that immediately follows your preallocation of ‘P’ because you have not yet defined ‘k’. (The assignment belongs only in the loop.)

That problem corrected, I got different errors, the most significant being ‘matrix dimensions must agree’ with this line:

Kk = (Pk_minus*H')/(H*Pk_minus*H' + R);%Kalman gain

The numerator is a (2x1) vector Pk_minus*H', and in the denominator, you are adding the (2x2) matrix ‘R’ to scalar H*Pk_minus*H'. MATLAB uses ‘automatic singular expansion’ to create the denominator, producing a (2x2) matrix. That still leaves the problem of dividing a (2x1) vector by a (2x2) matrix. That is not possible.

What do you want to do?

[Lina Alhelo]

thank you very much for your contribution. I have solved this issues as your comments but now the drawings are not as wanted.

%z - voltage to be measured
z = rand(2,800);
%Process model: x_k = Ax_k-1 + E_x
%Mearuement model:z_k = H(x_k+E_s) + E_y
%E_x and E_y are process and measurement noise respectively
%Initializations
Pk_prev = [1 0;
    0 1];
x_kprev_hat = [0;
    0];
xk_hat = [];
% Time update equations for z from 1-200
A = [0.5 0;
    0 0.3];
E_y = 1; % Measurement noise (R)
% Measurement update equations
H = [1 1];

%Simulating noisy measurements
%Measurement model
E_s= [25 ;
    25 ];
z_k = H*(z+E_s) + E_y ;
P = zeros(2,2,length(z));
% Process noise covariance
E_x= [1 0;
    0 1]; % Q 
for k = 1: length(z)
    x_k_hat_minus = A*x_kprev_hat+E_x; %a priori estimate
    Pk_minus = A*Pk_prev*A' + E_x; %A priori estimate error covariance
    Kk = Pk_minus*(H')/(H*Pk_minus*(H') + E_y);%Kalman gain
    x_kprev_hat = x_k_hat_minus + Kk*(z_k(k) - H*x_k_hat_minus); %A priori estimate
    xk_hat = cat(2,xk_hat,x_kprev_hat);%A posteriori estimate
    Pk_prev =  (1 - Kk*H)*Pk_minus;%A posterirori estimate error covariance
    P(:,:,k)  = Pk_prev;
end
plot(z_k,'bx-');
hold on;
plot(xk_hat,'gx-');
plot(z,'rx-');

[Lina Alhelo]

this what is expected to ahave

[Star Strider]

I cannot follow what you are doing.

I have no suggestions on how to make your code produce the plot you expect.

Answer 2

[Alex Mcaulley]

Running your code, it throws an error in this line

P(k)  = Pk_prev;

The problem is that PK_prev is of size 2x2 and P is 1x200, then the assignment is incorrect.

[Lina Alhelo]

how I can make Pk_prev to be of the same size?