# Just Stuff

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## Blocking Matrices

Proof that the Blocking Matrices do Indeed Block Signals from Specified Directions.

For some/certain signal processing tasks we wish to generate beam patterns with nulls in specified directions (or equivalently; filters with narrow notches at specified frequencies). Typically these can be used to null out stationary interference sources, directions to multipath images of a desired source, …

We assume that we have a number of sensors ($M$ )which may be array elements, subarrays or formed beams and we wish to null signals from directions $\theta_1, ..., \theta_k$, and $k < M$, and responses ${\bf{b}}(\theta_i)$ for a source in direction $\theta_i$ (being a column vector of complex signals one element for each sensor). In a narrowband system we may assume there are simply complex numbers representing the sensor amplitude and phase response. We let ${\bf{\widehat{b}}}(\theta_i)$ denote the normalised versions of these signals.

It is obvious (when pointed out anyway) that the matrix:

${\bf{A}}={\bf{I}}_{M\times M} - {\bf{B}} ({\bf{B}}^H{\bf{B}})^{^{-1}} {\bf{B}}^H$

is a blocking matrix for directions $\theta_1, ..., \theta_k$ (where ${{{\bf{B}}}}$ is the matrix with $\widehat{\bf{b}}(\theta_i)$  for its columns). The space of possible sensor outputs constitutes a (complex) vector space of dimension $M$ and the subspace  $\text{span}[{\bf{b}}(\theta_i), i=1,\dots ,k]$  is the null space of ${\bf{A}}$ and ${\bf{A}}$ behaves like the identity transformation on the orthogonal complement of the null space.

To see that this is a blocking matrix for $\widehat{\bf{b}}(\theta_i)$ we need just consider:

${\bf{AB}} = \left[{\bf{I}}_{M\times M} - {\bf{B}} ({\bf{B}}^H{\bf{B}})^{^{-1}} {\bf{B}}^H\right] {\bf{B}}$

so:

${\bf{AB}} = {\bf{B}} - {\bf{B}} ({\bf{B}}^H{\bf{B}})^{^{-1}} {\bf{B}}^H {\bf{B}} ={\bf{0}}$

Now the columns of ${\bf{AB}}$ are the vectors ${\bf{A\widehat{b}}}(\theta_i)$ and so these are zeros vectors as expected.

To see that ${\bf{A}}$ behaves like the identity on the orthogonal complement consider ${\bf{x}}$ in the orthogonal complement of the space spanned by the columns of ${\bf{B}}$. Then ${\bf{B}}^H{\bf{x}}={\bf{0}}$ since each component of this is the dot product of a column of ${\bf{B}}$ and ${\bf{x}}$ and so zero. Hence:

${\bf{Ax}} = \left[{\bf{I}}_{M\times M} - {\bf{B}} ({\bf{B}}^H{\bf{B}})^{^{-1}} {\bf{B}}^H\right] {\bf{x}}$

$={\bf{x}}+{\bf{B}} ({\bf{B}}^H{\bf{B}})^{^{-1}} {\bf{B}}^H {\bf{x}}$

$={\bf{x}}$

(I hope that is all clear, WordPress LaTeX seems to have started deleting all the \ characters from the LaTeX strings again so there may still be some omissions from when I have tried to restore them)