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As shown in following code. inp contains 10*4 matrix. I am processing each row but i am getting processed output for the first row and for remaining row I am getting NaN value. why I am getting like that? where is the error?

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Sachin Patil
Sachin Patil 2016-10-7
评论： Sachin Patil 2016-10-7
% I am using this code in simulink model, where Input matrix(10*4) will change for every 1/60 seconds.
%code
function op = fnc(inp)
% define the filter
op = zeros(10,4);
F = [ 1 0 1 0 ; % A
0 1 0 1 ;
0 0 1 0 ;
0 0 0 1 ];
H = [ 1 0 0 0;0 1 0 0;0 0 0 0;0 0 0 0 ] ;
R = [1 0 0 0 ; 0 1 0 0; 0 0 1 0 ; 0 0 1 0];
X = zeros(1,4);
Q = eye(1)*1e-5;
w = 0.2;
P = eye(4)*10;
for i = 1 : 10
x = inp(i,:);
for m = 1
for n = 2
if (x(m,n)==0)
op(i,:) = x;
elseif (0<x(m,n))&&(x(m,n)<70)
X = X*F ; %predicted state (1*4)
S = H'*H*P' + R; %measurement error covariance (4*4)
K = (H*P')/S; %optimal Kalman gain (4*4)
P = P - K*H*P; %(4*4)
P = F*F'*P + Q; % estimate error covariance (4*4)
y = x - X*H; %measurement error/residual (1*4)
X = X + y*K; %updated state estimate (1*4)
X_1 = X*H;
op(i,:) = X_1;
else
%Predictor equations
X = X*F + w; %predicted state (4*1)
P = F*P*F' + Q; % estimate error covariance (4*4)
op(i,:) = X;
end
end
end
end
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dbmn
dbmn 2016-10-7
When I evaluate your code I always get a warning on the following line: (it is between 0 and 70).
K = (H*P')/S; %optimal Kalman gain (4*4)
Warning: Matrix is singular to working precision.
It seems that the divison at that point is no good idea because the rank of those Matrices is not full. If you change the diagonal Elements of the H Matrix to anything not Zero it runs withouth producing NANs.
  2 个评论
dbmn
dbmn 2016-10-7
Or use the back division Operator with your values of H
K = (H*P')\S;
please cross check if this produces the desired results, as it is a different operation as /
Sachin Patil
Sachin Patil 2016-10-7
I changed H matrix elements but I am getting results with difference of 6. i.e. if my input is 37.3 then my output is 31.2, actually I should get near to 37. Is there problem in h matrix assignment?

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更多回答（1 个）

Massimo Zanetti
Massimo Zanetti 2016-10-7
In these two lines anything is looping,
x = inp(i,:);
for m = 1
for n = 2
put something like this:
x = inp(i,:);
for m = 1:SOMETHING
for n = 2:SOMETHING ELSE

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