I have started working on a Direction-of-Arrival estimation problem and I am testing various algorithms. I came across Multiple Signal Classification (MuSiC) method. In all textbooks I have looked at, the final step of the method is to search for maximums in the pseudo-spectrum given by

$$ P \left(\theta \right) = \frac{1}{a^{H}\left(\theta\right) V_{n} V_{n}^{H} a\left(\theta\right)}$$

where $P\left(\theta\right)$ is the spatial pseudo-spectrum, $a\left(\theta\right)$ is the array manifold of the array under consideration, $[ \cdot]^{H}$ denotes Hermitian transposition and $V_{n}$ is the matrix whose columns form the noise subspace (the noise eigenvectors).

Now, my question is, why should we look for maxima in $P\left(\theta\right)$ an not for minima in $\frac{1}{P\left(\theta\right)}$? Is there something I am missing here? I believe that the implementation of the algorithm could potentially benefit from avoiding this division.

Any insights and/or hints are most welcome. Thanks in advance.


1 Answer 1


Let's define the null-spectrum $Q(\theta)$ and pseudo-spectrum $P(\theta)$ as

$$Q(\theta) = a(\theta)^H{v_n}{v_n^H}a(\theta)$$

$$P(\theta) = \frac{1}{Q(\theta)}$$

The choice to find the maxima in $P(\theta)$ comes down to personal preference. It is more intuitive to associate a peak in a signal to some phenomenon being observed. This is especially so when reading a text, where the author is trying to teach you how the algorithm works.

You can very well search for the minima in $Q(\theta)$ instead of the maxima in $P(\theta)$ and get the same answer as can be seen in the example below

enter image description here

enter image description here

Nothing is stopping you from using the minima. In general there is no significant advantage to using either except for avoiding division operations and potential divide-by-zero, which usually fine in real systems where some kind of noise is present. Still, certain systems and their implementation can make using one over the other more appropriate.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.