Problem description
When applying robust estimation in spp, when calculating the covariance matrix of the post-test residual v of the observation value, there is a problem, and Qvv has a negative value for the diagonal element.
First of all, I suspect that there is a problem with my own code, and then debugging. Debugging the matrix in VS is really daunting.
Secondly, I printed out the coefficient matrix H, constant matrix l, covariance matrix Qxx, parameter vector dx in spp, and verified the comparison in matlab.
1. Print the above matrix in matlab
>> H=[ 0.497179 -0.767231 0.405178 1.000000 0.000000 0.000000 0.000000;
0.954274 0.112689 -0.276879 1.000000 0.000000 0.000000 0.000000;
0.387339 -0.481553 -0.786178 1.000000 0.000000 0.000000 0.000000;
-0.580737 -0.519400 -0.626872 1.000000 0.000000 0.000000 0.000000;
-0.287620 -0.601084 -0.745636 1.000000 0.000000 0.000000 0.000000;
0.394442 -0.847597 -0.354957 1.000000 0.000000 0.000000 0.000000;
0.479621 -0.224119 -0.848371 1.000000 0.000000 0.000000 0.000000;
-0.266211 -0.887121 0.377026 1.000000 0.000000 0.000000 0.000000;
0.000000 0.000000 0.000000 0.000000 1.000000 0.000000 0.000000;
0.000000 0.000000 0.000000 0.000000 0.000000 1.000000 0.000000;
0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 1.000000];
>> Q=[ 1.444742 -1.205538 -0.559907 -1.256723 0.000000 0.000000 0.000000;
-1.205538 4.239855 1.510516 3.308947 0.000000 0.000000 0.000000;
-0.559907 1.510516 1.847995 1.891790 0.000000 0.000000 0.000000;
-1.256723 3.308947 1.891790 3.215663 0.000000 0.000000 0.000000;
0.000000 0.000000 0.000000 0.000000 0.010000 0.000000 0.000000;
0.000000 0.000000 0.000000 0.000000 0.000000 0.010000 0.000000;
0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.010000];
>> dx=[ 0.000009
-0.000020
-0.000022
-0.000041
0.000000
0.000000
0.000000];
>> l=[0.123075
0.433912
0.828791
0.067140
0.026320
-0.289038
-0.862615
-0.198928
0.000000
0.000000
0.000000];
>> H'
ans =
Columns 1 through 8
0.497179000000000 0.954274000000000 0.387339000000000 -0.580737000000000 -0.287620000000000 0.394442000000000 0.479621000000000 -0.266211000000000
-0.767231000000000 0.112689000000000 -0.481553000000000 -0.519400000000000 -0.601084000000000 -0.847597000000000 -0.224119000000000 -0.887121000000000
0.405178000000000 -0.276879000000000 -0.786178000000000 -0.626872000000000 -0.745636000000000 -0.354957000000000 -0.848371000000000 0.377026000000000
1.000000000000000 1.000000000000000 1.000000000000000 1.000000000000000 1.000000000000000 1.000000000000000 1.000000000000000 1.000000000000000
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
Columns 9 through 11
0 0 0
0 0 0
0 0 0
0 0 0
1.000000000000000 0 0
0 1.000000000000000 0
0 0 1.000000000000000
>> H*Q*H'
ans =
Columns 1 through 8
1.332860500714645 0.645546014871716 -0.098580562413659 -0.064031507548390 -0.165752578512155 0.396295832699334 -0.141008818175153 1.168889713304825
0.645546014871717 1.968804844653495 0.053246247795955 -0.005468449432181 -0.265950607478022 -0.319534503408733 0.574091433817809 0.243277031690377
-0.098580562413659 0.053246247795955 0.357273653443425 -0.021840446651663 0.164659427109121 0.283415207166032 0.318871615198842 -0.284958778729781
-0.064031507548390 -0.005468449432181 -0.021840446651663 1.072132904685670 0.641476084869937 -0.226446903091478 0.079281778351138 0.679077103029250
-0.165752578512155 -0.265950607478023 0.164659427109121 0.641476084869937 0.515321629013082 0.067397424711531 0.125308500906703 0.247806596987045
0.396295832699334 -0.319534503408734 0.283415207166032 -0.226446903091478 0.067397424711531 0.647339251042735 0.015423699735895 0.151774558340723
-0.141008818175153 0.574091433817809 0.318871615198843 0.079281778351138 0.125308500906704 0.015423699735896 0.481685375510297 -0.331346855770643
1.168889713304826 0.243277031690377 -0.284958778729780 0.679077103029250 0.247806596987045 0.151774558340723 -0.331346855770643 1.674731658216493
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
Columns 9 through 11
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0 0 0
0.010000000000000 0 0
0 0.010000000000000 0
0 0 0.010000000000000
>>
2. Print the above matrix in vs
2.1 Print Q_ matrix
I have a problem when calculating HQH' in vs. The code is as follows:
printf("**********************\n");
Q_ = mat(nv, nv);
matmul("TN", nv, NX, NX, 1.0, H_, Q, 0.0, Q);
matmul("NN", nv, nv, NX, 1.0, Q, H_, 0.0, Q);
printmat(Q, nv, nv, "Q");
Among them, matmul is the function of matrix multiplication, T means transpose, N means no transpose; H_ is the coefficient matrix, the unweighted coefficient matrix is the copy of H directly returned by the rescode function, and Q is the return value of lsq, Q_ means HQH', then the printing result obtained according to the above code is as shown in the figure below:
the diagonal element has a negative value, and it is not a diagonal matrix. Obviously it is wrong. Hahaha
2.2. The printing coefficient matrix H is
in rescode In the function, H is placed in an array or matrix according to [-e1 -e2 -e3 1.0 0.0 0.0 0.0], which is nv * NX elements. I print it according to NX * nv defined by mat, and then follow nv * NX to print, the result is as follows. The
coefficient matrix H is said to be NX * nv. The printed result shows that the internal data arrangement should be the transposed form of the second printing matrix above, which means that the form of H is actually the form of H'in matlab in the form of
2.3, print H '* H
in order to verify the above conclusion 2.2, on the right H' * H print results, as shown below:
2.4, 2.1 for the matrix multiplication operation revision verification
prior to phase H and the matrix Multiply, calculate H', and then perform H'QH operation
// 3、根据协方差传播定律可知，Qvv=Qll-B*Q*B',Q=(B'PB)^-1
// lsq: n<m,n=NX m=nv
// matmul("NN",n,1,m,1.0,A,y,0.0,Ay); /* Ay=A*y */
// matmul("NT", n, n, m, 1.0, A, A, 0.0, Q); /* Q=A*A' */
printf("**********************\n");
Q_ = mat(nv, nv);
double *eyeI;
double *tH;
eyeI = mat(NX, NX);
tH = mat(nv, NX);
for (j = 0; j < NX;j++)
{
for (k = 0; k < NX;k++)
{
eyeI[j + k*NX] = (j==k)?1.0:0.0;
}
}
matmul("TN", nv, NX, NX, 1.0, H_, eyeI, 0.0, tH); /* 求转置矩阵H' */
printmat(tH, NX, nv, "tH");// 打印转置矩阵
double *Q1,*Q2;
Q1 = mat(nv, NX);
Q2 = mat(nv, nv);
matmul("NN", nv, NX, NX, 1.0, tH, Q, 0.0, Q1);
matmul("NN", nv, nv, NX, 1.0, Q1, H_, 0.0, Q_);
printmat(Q_, nv, nv, "Q_");
The print result is as follows:
In fact, this tH is the transposed form of H in Matlab. Once again, debugging methods have been used to verify that the rtklib matrix is in the first column and the second column. . . Filling the matrix in the form of the nth column is different from our usual habit of writing code by ourselves. The printed result is consistent with Matlab.
Robust Code
robust code spp added also finished, the next test validation, plus anti-poor indeed rms and std without robust than small, so I put the comment plus good code, will test data +code+result graph Let it out, let's talk about it.