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Why does the SGOLAYFILT function not accept zeros as elements of the weighting vector w, in Signal Processing Toolbox 6.1 (R13+)?

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Why does the SGOLAYFILT function not accept zeros as elements of the weighting vector w, in Signal Processing Toolbox 6.1 (R13+)?

henryjames012 asked: matlab , signal , signal processing , SGOLAYFILT function

Understand why MATLABs sgolayfilt function in Signal Processing Toolbox 6.1 (R13) rejects zero values in the weighting vector w. Explore constraints and work

Expert Answer

Profile picture of Kshitij Singh Kshitij Singh answered . 2025-07-10 21:18:05

This is a weighted least-squares problem. The requirement for the weights to be positive is needed in order to ensure the "least-squares matrix" is positive definite and the system has a solution.
For example, if you change the error checking as suggested above, to allow for zero weights, and try this simple case:
 
[b,g]=sgolay(3,5,[1,0,0,0,1]);

you will run into matrix singularity issues as indicated by the following warning messages:

 Warning: Matrix is singular to working precision.

 (Type "warning off MATLAB:singularMatrix" to suppress this warning.)

 > In I:\1186207\sgolay.m at line 59

 Warning: Matrix is singular to working precision.

 (Type "warning off MATLAB:singularMatrix" to suppress this warning.)

 > In I:\1186207\sgolay.m at line 59
This is the reason we have this requirement.
You can make some of the weights "very small" if you want to achieve a similar effect as is required here. But it is important to keep in mind that the smaller the weight, the more ill-conditioned the problem will become.
This is due to the so-called inertia theorem for matrices; if
 
 A = C'*D*C
then A will have the same number of positive eigenvalues as D, the same number of negative eigenvalues, and the same number of zero eigenvalues.
If D (in this case a diagonal matrix with the weights on the diagonal) has a zero eigenvalue (a weight equal to zero in this case), then A will be positive semidefinite (assuming no negative weights) and a solution to the least-squares problem is not guaranteed.


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