1
$\begingroup$

So I am wondering why I keep seeing that the rank of the LCMV Broadband Constraint Matrix is equal to the number of constraints, when in practice, I find that the rank is only 1. This is precluding me from performing any sort of 'low rank' approximation since it is already of the lowest rank possible!

The constraint matrix is defined as:

$C = [ \mathbf{d}(\theta, \omega_{0}) | \mathbf{d}(\theta, \omega_{1})| \ldots |\mathbf{d}(\theta, \omega_{r-1})] $

Where each $\mathbf{d}(\theta, \omega_{i})$ is a column vector of the form:

$ \mathbf{d}(\theta, \omega_{i}) = [ e^{-j \omega_{i} \tau_{0}}, \ldots, e^{-j \omega_{i} \tau_{M-1}} | e^{-j \omega_{i} (\tau_{0} + T_{s})}, \ldots, e^{-j \omega_{i} (\tau_{M-1} + T_{s})} | \ldots | e^{-j \omega_{i} (\tau_{0} + (J-1)T_{s})}, \ldots, e^{-j \omega_{i} (\tau_{M-1} + (J-1)T_{s})}]^{T}$

Where $M$ is the number of sensors, $J$ is the length of the attached FIR filter, $T_{s}$ is the sampling interval, and $\tau_{i}$ is the delay of the signal coming from angle $\theta$ to the $i$th sensor.

When I create this matrix and compute the SVD via zgesvd_ , I get only one significant singular value. The literature says the rank should be $r$, ie the same as the number of columns of $C$, but this is not the case -- ie the columns have linear dependence.

Why do they say this? Is there something I am missing? Has anyone else encountered this?

References:

Wei Liu and Stephan Weiss. Wideband Beamforming: Concepts and Techniques. Wiley. 2010. pp. 33-35

Kevin M. Buckley. Spatial/Spectral Filtering with Linearly Constrained Minimum Variance Beamformers. IEEE Trans. Acoustics, Speech, and Signal Processing. March 1987 pp.249-266

$\endgroup$
  • $\begingroup$ I want to do a low rank approximation because I want to add extra constraints to C, making it of a larger rank, and then bring the rank back down to r. $\endgroup$ – The Dude Jan 26 '16 at 17:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.