2
$\begingroup$

I encountered this question which references this question which has since been "deleted by the author". The fact that it was referenced by this other question, and the fact that it appears like it might have important information I find useful tells me that maybe this question shouldn't have even been allowed to be deleted by the author. But to my understanding those kinds of questions (ones that are useful) typically get answers and can't be deleted. I don't have 10k rep, so can someone show me what was in this question? Can this question be revived if shown to be useful?

$\endgroup$

1 Answer 1

2
$\begingroup$

OK, I've undeleted that question and its answer. The general SE consensus seems to be that SE owns the rights to the information and can do with it what it wants. In this case, I think it's better to have the information available than not.


I'll do some investigation to see whether undeleting it is OK. Here's a copy of the question and its answer. I'll delete this answer if I believe I can validly undelete the original.


Compared to standard (conventional) beamformers, adaptive beamformers such as minimum variance distortionless response (MVDR) have much better resolution and interference nulling capabilities. Briefly, the MVDR beamformer is as follows:

If $\mathbf{a}_\theta$ is the look-vector in the direction $\theta$ and if $\widehat{\mathbf{R}}$ is the covariance matrix of observations, then the adaptive weight vectors are calculated as

$$\mathbf{w}_\theta=\frac{\widehat{\mathbf{R}}^{-1} \mathbf{a}_\theta}{\mathbf{a}^H_\theta \widehat{\mathbf{R}}^{-1} \mathbf{a}_\theta}$$

and the output power of the MVDR beamformer is

$$B_{mvdr}(\theta)=\mathbf{w}_\theta^H \widehat{\mathbf{R}}\mathbf{w}_\theta$$

For the sake of completeness, the conventional beamformer output power is $B_{conv}(\theta)=\mathbf{a}_\theta^H \widehat{\mathbf{R}}\mathbf{a}_\theta$.

However, these adaptive beamformers are also prone to severe performance hits when there is a signal mismatch i.e., when the steering vector model for the signal of interest doesn't match what's in the data or when the actual array positions are perturbed from the array positions used in the steering/look vectors (which is what happens in practice). The performance degradation can be so large that it becomes inferior to the conventional beamformer.

How can one increase the robustness of adaptive beamformers (in this case, the MVDR, but applicable more generally) to signal mismatches?


Example:

Here's the result from a simulation with two plane wave signals, one at $+30^\circ$ and the other at $-22.5^\circ$ and 10 dB down from the first, impinging on a 10 element uniform line array (spaced at $\lambda/2$, where $\lambda$ is the wavelength) in a noisy environment (the exact value of the noise power is immaterial here and you can assume it to be much lower than the signal powers). The covariance matrix is formed from 50 observations (snapshots). The top two panels show the outputs of the MVDR and the conventional beamformers and you can clearly see that the MVDR has high resolution compared to the conventional. Both do a good job of estimating the signal power.

However, when the element positions are slightly perturbed (perturbation is unknown to the beamformer), causing a mismatch in the actual signal steering vector and the model, the signal power estimates of the MVDR beamformer goes kaput. It can hardly measure them correctly, although it still does a good job of localising the signals. Comparatively, the conventional beamformer gets the relative powers right, although its harder to localise the signal due to increased sidelobes. (Note: The plots have all been normalized by the maximum power so although you can't see it, the peak at $30^\circ$ in the top left plot is at 0 dB. So this doesn't truly show how the actual power estimate degrades due to mismatches, as I just wanted to paint a qualitative picture.)

Plots with MVDR Perturbed Array sloppily corrected in Gimp http://dsp.stackexchange.com/questions/146/how-can-one-improve-the-robustness-of-adaptive-beamformers-to-signal-mismatches#comment-308 :

The title for the lower left plot should read MVDR: Perturbed array. I'll correct the mistake sometime soon. – yoda Sep 3 at 5:28 -->

Is it possible for me to improve the estimates of the signal powers using the MVDR beamformer in the perturbed case?


Answer:

One approach that I'm aware of is to "load" the diagonal of $\widehat{\mathbf{R}}$ by some value $\sigma_d^2$ in order to stabilize the covariance matrix and make the adaptive vectors more robust to signal mismatches. This is often done in matched field processing and is referred to as diagonal loading or white noise constraint beamforming.

In other words, the robustified adaptive weight vectors are calculated as

$$\widehat{\mathbf{w}}_\theta=\frac{\left(\widehat{\mathbf{R}}+\sigma_d^2\mathbf{I}\right)^{-1} \mathbf{a}_\theta}{\mathbf{a}^H_\theta \left(\widehat{\mathbf{R}}+\sigma_d^2\mathbf{I}\right)^{-1} \mathbf{a}_\theta}$$

and the output beam power is now computed as

$$\widehat{B}_{mvdr}(\theta)=\widehat{\mathbf{w}}_\theta^H \left(\widehat{\mathbf{R}}+\sigma_d^2\mathbf{I}\right)\widehat{\mathbf{w}}_\theta$$

Here's the result comparing the no mismatch, mismatch due to perturbed array and diagonally loaded cases for the MVDR beamformer for the situation in the question. There is an improvement in the signal power estimation, at the expense of increased background noise power.

enter image description here

$\endgroup$
2
  • $\begingroup$ This is definitely the kind of answer and question that I find useful, I'm surprised the author was able to delete it at all, I wonder if it was so old that current SE measures stopping this from happening didn't apply? $\endgroup$
    – Krupip
    Commented Dec 4, 2019 at 19:59
  • $\begingroup$ @whn Possibly! It was deleted on Feb 11 '12 at 13:15. And Lorem Ipsum was last seen on Mar 9 '14 at 16:53, so I can't really ask them. $\endgroup$
    – Peter K. Mod
    Commented Dec 4, 2019 at 20:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .