i tried this.but the result that i obtained is different from actual X..i think the system has no unique solution..whether the chance of existence of unique solution depends on the dimension of the matrix or its elements? can anyone help me?
solve AX=B
3 views (last 30 days)
Show older comments
hi..i am working on OFDMA.i have represented the received vector in the form of AX=B where A has 1024*2048 elements ,X has 2048*1 elements so that the received vector has 1024*1 elements. But many of the elements in the X are zeros.So in order to solve the equation ,i only considered the non-zero elements in X and corresponding elements in A .so now i have 1024*1023 elements in A,1023*1 elements in X.How can i solve it? i used pinv command.But the result obtained is not matching with my actual X..Can anyone help me?
2 Comments
Torsten
on 8 Feb 2016
You can easily check:
If both A*X and A*X_actual give B, the system does not have a unique solution.
Best wishes
Torsten.
Answers (0)
See Also
Categories
Find more on Digital Filter Analysis in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!