episode 9
\begin
Discover why over 20 million people worldwide trust Overleaf with their work.
%% LyX 2.3.2-2 created this file. For more info, see http://www.lyx.org/.
%% Do not edit unless you really know what you are doing.
\documentclass[english]{article}
\usepackage[T1]{fontenc}
\usepackage[latin9]{inputenc}
\usepackage{geometry}
\geometry{verbose,tmargin=1cm,bmargin=2cm,lmargin=1cm,rmargin=1cm}
\setlength{\parindent}{0bp}
\usepackage{amsmath}
\usepackage{amssymb}
\PassOptionsToPackage{normalem}{ulem}
\usepackage{ulem}
\makeatletter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% User specified LaTeX commands.
\date{}
\makeatother
\usepackage{babel}
\begin{document}
\title{Fundamentals of Signal Enhancement and Array Signal Processing\\
Solution Manual}
\maketitle
\title{ \textbf{ Lidor Malul 318628005}}
\section*{ 9 Differential Beamforming}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%% Q2 %%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{9.2}
Show that the coefficients of the $N$th-order hypercardioid are given by
\begin{eqnarray*}
\mathbf{a}_{N, \mathrm{max}} = \frac{\mathbf{H}_N^{-1} \mathbf{1}}{\mathbf{1}^T\mathbf{H}_N^{-1} \mathbf{1}} .
\end{eqnarray*}
\textbf{\uline{Solution}}\textbf{:}
First we can notice that the rank of the matrix ${{11}^{T}}$ is 1.
So , the rank of the matrix ${{H}_{N}}^{-1}{{11}^{T}}$ is 1 or 0.
Therefor, for the non zero eigenvalue , the eigenvector is the maximum.
lets show that ${{a}_{N}}$ is eigenvector for non zero eigenvalue:
\[\left( {{H}_{N}}^{-1}{{11}^{T}} \right){{a}_{N}}=\left( {{H}_{N}}^{-1}{{11}^{T}} \right)\frac{{{H}_{N}}^{-1}1}{{{1}^{T}}{{H}_{N}}^{-1}1}=\frac{{{H}_{N}}^{-1}1}{{{1}^{T}}{{H}_{N}}^{-1}1}1{{H}_{N}}^{-1}1={{a}_{N}}\left( 1{{H}_{N}}^{-1}1 \right)={{a}_{N}}\cdot \lambda \]
which $\lambda \triangleq 1{{H}_{N}}^{-1}1$ scalar , non zero eigenvalue of $ {{H}_{N}}^{-1}{{11}^{T}}$
${{a}_{N}}$ is the eigenvector corresponding
to the maximum eigenvalue of the matrix.
\[
\blacksquare
\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%% Q3 %%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{9.3}
Using the definition of the frequency-independent FBR of a theoretical $N$th-order DSA (\ref{C9-FBRd}), show that
\begin{eqnarray*}
{\cal F}\left( \mathbf{a}_N \right) = \frac{\mathbf{a}_N^T \mathbf{H}_N'' \mathbf{a}_N}{ \mathbf{a}_N^T \mathbf{H}_N' \mathbf{a}_N } ,
\end{eqnarray*}
where $\mathbf{H}_N'$ and $\mathbf{H}_N''$ are Hankel matrices.
\textbf{\uline{Solution}}\textbf{:}
We know that the FBR define as:
\[F({{a}_{N}})=\frac{\int_{0}^{\frac{\pi }{2}}{{{B}^{2}}({{a}_{N}},\cos \theta )\sin \theta d\theta }}{\int_{\frac{\pi }{2}}^{\pi }{{{B}^{2}}({{a}_{N}},\cos \theta )\sin \theta d\theta }}\]
and B define as:
\[B({{a}_{N}},\cos \theta )=\sum\limits_{n=0}^{n=N}{{{a}_{N,n}}{{\cos }^{n}}\theta }\]
now we substitute B in FBR definition:
\[F({{a}_{N}})=\frac{\int_{0}^{\frac{\pi }{2}}{\sum\limits_{n=0}^{n=N}{{{a}_{N,n}}{{\cos }^{n}}\theta }\sum\limits_{k=0}^{k=N}{{{a}_{N,k}}{{\cos }^{k}}\theta }\sin \theta d\theta }}{\int_{\frac{\pi }{2}}^{\pi }{\sum\limits_{n=0}^{n=N}{{{a}_{N,n}}{{\cos }^{n}}\theta }\sum\limits_{k=0}^{k=N}{{{a}_{N,k}}{{\cos }^{k}}\theta }\sin \theta d\theta }}=\frac{\sum\limits_{k=0}^{k=N}{\sum\limits_{n=0}^{n=N}{{{a}_{N,n}}}{{a}_{N,k}}}\int_{0}^{\frac{\pi }{2}}{{{\cos }^{n}}\theta {{\cos }^{k}}\theta \sin \theta d\theta }}{\sum\limits_{k=0}^{k=N}{\sum\limits_{n=0}^{n=N}{{{a}_{N,n}}}{{a}_{N,k}}}\int_{\frac{\pi }{2}}^{\pi }{{{\cos }^{n}}\theta {{\cos }^{k}}\theta \sin \theta d\theta }}=\otimes \]
using the following solutions:
\[ \int_{0}^{\frac{\pi }{2}}{{{\cos }^{i}}\theta {{\cos }^{j}}\theta \sin \theta d\theta }=\frac{1}{1+i+j}={{\left[ {{{{H}''}}_{N}} \right]}_{ij}} \]
\[ \int_{\frac{\pi }{2}}^{\pi }{{{\cos }^{i}}\theta {{\cos }^{j}}\theta \sin \theta d\theta }=\frac{{{\left( -1 \right)}^{i+j}}}{1+i+j}={{\left[ {{{{H}'}}_{N}} \right]}_{ij}} \]
finally:
\[\otimes =\frac{\sum\limits_{k=0}^{k=N}{\sum\limits_{n=0}^{n=N}{{{a}_{N,n}}}{{a}_{N,k}}}{{\left[ {{{{H}''}}_{N}} \right]}_{nk}}}{\sum\limits_{k=0}^{k=N}{\sum\limits_{n=0}^{n=N}{{{a}_{N,n}}}{{a}_{N,k}}}{{\left[ {{{{H}'}}_{N}} \right]}_{nk}}}=\frac{{{a}_{N}}^{T}{{{{H}''}}_{N}}{{a}_{N}}}{{{a}_{N}}^{T}{{{{H}'}}_{N}}{{a}_{N}}}\]
where ${{{H}''}_{N}}$ and ${{{H}'}_{N}}$ are two Hankel matrices.
\[
\blacksquare
\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%% Q5 %%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{9.5}
Show that the directivity pattern of the first-order hypercardioid can be expressed as
\begin{eqnarray*}
{\cal B}_{1,\mathrm{Hd}} \left( \cos \theta \right) = \frac{1}{4} + \frac{3}{4} \cos \theta .
\end{eqnarray*}
\textbf{\uline{Solution}}\textbf{:}
We know the definition of B:
\[{{B}_{N,Hd}}(\cos \theta )=\frac{{{1}^{T}}{{H}^{-1}}_{N}p(\cos \theta )}{{{1}^{T}}{{H}^{-1}}_{N}1}\]
and for N=1:
\[ p(\cos \theta )={{\left[ \begin{matrix}
1 & \cos \theta \\
\end{matrix} \right]}^{T}} \]
\[ {{1}^{T}}=\left[ \begin{matrix}
1 & 1 \\
\end{matrix} \right] \]
\[ {1}={{\left[ \begin{matrix}
1 & 1 \\
\end{matrix} \right]}^{T}} \]
\[{{H}_{1}}=\left( \begin{matrix}
1 & 0 \\
0 & \frac{1}{3} \\
\end{matrix} \right)\to {{H}_{1}}^{-1}=\left( \begin{matrix}
1 & 0 \\
0 & 3 \\
\end{matrix} \right)\]
substituting:
\[{{B}_{N,Hd}}(\cos \theta )=\frac{\left[ \begin{matrix}
1 & 1 \\
\end{matrix} \right]\left( \begin{matrix}
1 & 0 \\
0 & 3 \\
\end{matrix} \right){{\left[ \begin{matrix}
1 & \cos \theta \\
\end{matrix} \right]}^{T}}}{\left[ \begin{matrix}
1 & 1 \\
\end{matrix} \right]\left( \begin{matrix}
1 & 0 \\
0 & 3 \\
\end{matrix} \right){{\left[ \begin{matrix}
1 & 1 \\
\end{matrix} \right]}^{T}}}=\frac{1+3\cos \theta }{4}=\frac{1}{4}+\frac{3\cos \theta }{4}\]
\[
\blacksquare
\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%% Q6 %%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{9.6}
Show that the directivity pattern of the first-order supercardioid can be expressed as
\begin{eqnarray*}
{\cal B}_{1,\mathrm{Sd}} \left( \cos \theta \right) = \frac{\sqrt{3}-1}{2} + \frac{3 - \sqrt{3}}{2} \cos \theta .
\end{eqnarray*}
\textbf{\uline{Solution}}\textbf{:}
The beampattern of the Nth-order supercardioid is
\[{{B}_{N,Sd}}(\cos \theta )=\frac{{a}_{N,\max }^{'T}
p(\cos \theta )}{{a}_{N,\max }^{'T}
p(1)}\]
when ${a}_{N,\max }^{'T}$ is the eigenvector corresponding to the maximum eigenvalue of $ {H}_{N}^{'-1}{{H}_{1}}^{\prime \prime }$
\\
for N=1:
\[ {{H}_{1}}^{\prime }=\left( \begin{matrix}
1 & -\frac{1}{2} \\
-\frac{1}{2} & \frac{1}{3} \\
\end{matrix} \right)\to {{H}_{1}}{{^{\prime }}^{-1}}=\left( \begin{matrix}
4 & 6 \\
6 & 12 \\
\end{matrix} \right) \]
\[ {{H}_{1}}^{\prime \prime }=\left( \begin{matrix}
1 & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{3} \\
\end{matrix} \right) \]
\[ {H}_{N}^{'-1}{{H}_{1}}^{\prime \prime }=\left( \begin{matrix}
4 & 6 \\
6 & 12 \\
\end{matrix} \right)\left( \begin{matrix}
1 & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{3} \\
\end{matrix} \right)=\left( \begin{matrix}
7 & 4 \\
12 & 7 \\
\end{matrix} \right) \]
the max eigenvalue of the matrix $\left( \begin{matrix}
7 & 4 \\
12 & 7 \\
\end{matrix} \right)
$ is ${{\lambda }_{\max }}=7+4\sqrt{3}$
the eigenvector for the max eigenvalue is
: $\left( \begin{matrix}
\sqrt{3}-1 \\
3-\sqrt{3} \\
\end{matrix} \right)$
So:
\[{a}_{N,\max }^{'}=\left( \begin{matrix}
\sqrt{3}-1 \\
3-\sqrt{3} \\
\end{matrix} \right)\]
And finally:
\[ {{B}_{1,Sd}}(\cos \theta )=\frac{\left( \begin{matrix}
\sqrt{3}-1 & 3-\sqrt{3} \\
\end{matrix} \right){{\left( \begin{matrix}
1 & \cos \theta \\
\end{matrix} \right)}^{T}}}{\sqrt{3}-1+3-\sqrt{3}}=\frac{\sqrt{3}-1}{2}+\frac{3-\sqrt{3}}{2}\cos \theta \]
\[
\blacksquare
\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%% Q7 %%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{9.7}
Show that the directivity pattern of the second-order hypercardioid can be expressed as
\begin{eqnarray*}
{\cal B}_{2,\mathrm{Hd}} \left( \cos \theta \right) = -\frac{1}{6} + \frac{1}{3} \cos \theta + \frac{5}{6} \cos^2 \theta .
\end{eqnarray*}
\textbf{\uline{Solution}}\textbf{:}
We know the definition of B:
\[{{B}_{N,Hd}}(\cos \theta )=\frac{{{1}^{T}}{{H}^{-1}}_{N}p(\cos \theta )}{{{1}^{T}}{{H}^{-1}}_{N}1}\]
and for N=2:
\[ p(\cos \theta )={{\left[ \begin{matrix}
1 & \cos \theta & {{\cos }^{2}}\theta \\
\end{matrix} \right]\text{ }}^{T}} \]
\[ \text{ }{{1}^{T}}=\left[ \begin{matrix}
1 & 1 & 1 \\
\end{matrix} \right]\text{ } \]
\[ 1={{\left[ \begin{matrix}
1 & 1 & 1 \\
\end{matrix} \right]\text{ }}^{T}} \]
\[ {{H}_{1}}=\left( \begin{matrix}
1 & 0 & \frac{1}{3} \\
0 & \frac{1}{3} & 0 \\
\frac{1}{3} & 0 & \frac{1}{5} \\
\end{matrix} \right)\to {{H}_{1}}^{-1}=\left( \begin{matrix}
\frac{9}{4} & 0 & \frac{-15}{4} \\
0 & \frac{12}{4} & 0 \\
\frac{-15}{4} & 0 & \frac{45}{4} \\
\end{matrix} \right) \]
and now:
\[ {{B}_{2,Hd}}=\frac{\left[ \begin{matrix}
1 & 1 & 1 \\
\end{matrix} \right]\left( \begin{matrix}
\frac{9}{4} & 0 & \frac{-15}{4} \\
0 & \frac{12}{4} & 0 \\
\frac{-15}{4} & 0 & \frac{45}{4} \\
\end{matrix} \right){{\left[ \begin{matrix}
1 & \cos \theta & {{\cos }^{2}}\theta \\
\end{matrix} \right]}^{T}}}{\left[ \begin{matrix}
1 & 1 & 1 \\
\end{matrix} \right]\left( \begin{matrix}
\frac{9}{4} & 0 & \frac{-15}{4} \\
0 & \frac{12}{4} & 0 \\
\frac{-15}{4} & 0 & \frac{45}{4} \\
\end{matrix} \right){{\left[ \begin{matrix}
1 & 1 & 1 \\
\end{matrix} \right]}^{T}}}=\frac{\left[ \begin{matrix}
-\frac{6}{4} & \frac{12}{3} & \frac{30}{4} \\
\end{matrix} \right]{{\left[ \begin{matrix}
1 & \cos \theta & {{\cos }^{2}}\theta \\
\end{matrix} \right]}^{T}}}{\frac{36}{4}}= \]
\[ =-\frac{1}{6}+\frac{1}{3}\cos \theta +\frac{5}{6}{{\cos }^{2}}\theta\]
\[
\blacksquare
\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%% Q8 %%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{9.8}
Show that the directivity pattern of the second-order supercardioid can be expressed as
\begin{eqnarray*}
{\cal B}_{2,\mathrm{Sd}} \left( \cos \theta \right) = \frac{1}{2\left( 3 + \sqrt{7} \right)} +
\frac{\sqrt{7}}{3 + \sqrt{7}} \cos \theta + \frac{5}{2\left( 3 + \sqrt{7} \right)} \cos^2 \theta .
\end{eqnarray*}
\textbf{\uline{Solution}}\textbf{:}
The beampattern of the Nth-order supercardioid is
\[{{B}_{N,Sd}}(\cos \theta )=\frac{{a}_{N,\max }^{'T}
p(\cos \theta )}{{a}_{N,\max }^{'T}
p(1)}\]
when ${a}_{N,\max }^{'T}$ is the eigenvector corresponding to the maximum eigenvalue of $ {H}_{N}^{'-1}{{H}_{1}}^{\prime \prime }$
\\
for N=2:
\[ {{H}_{2}}^{\prime }=\left( \begin{matrix}
1 & -\frac{1}{2} & \frac{1}{3} \\
-\frac{1}{2} & \frac{1}{3} & -\frac{1}{4} \\
\frac{1}{3} & -\frac{1}{4} & \frac{1}{5} \\
\end{matrix} \right)\to {{H}_{2}}{{\prime }^{-1}}=\left( \begin{matrix}
9 & 36 & 30 \\
36 & 192 & 180 \\
30 & 180 & 180 \\
\end{matrix} \right) \]
\[ {{H}_{2}}^{\prime \prime }=\left( \begin{matrix}
1 & \frac{1}{2} & \frac{1}{3} \\
\frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\
\frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\
\end{matrix} \right) \]
\[ H_{2}^{'-1}{{H}_{2}}^{\prime \prime }=\left( \begin{matrix}
9 & 36 & 30 \\
36 & 192 & 180 \\
30 & 180 & 180 \\
\end{matrix} \right)\left( \begin{matrix}
1 & \frac{1}{2} & \frac{1}{3} \\
\frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\
\frac{1}{3} & \frac{1}{4} & \frac{1}{5} \\
\end{matrix} \right)=\left( \begin{matrix}
37 & 24 & 18 \\
192 & 127 & 96 \\
180 & 120 & 91 \\
\end{matrix} \right) \]
the max eigenvalue of the matrix $H_{2}^{'-1}{{H}_{2}}^{\prime \prime }$ is ${{\lambda }_{\max }}=127+48\sqrt{7}$
the eigenvector for the max eigenvalue is
: $\left( \begin{matrix}
1 \\
2\sqrt{7} \\
5 \\
\end{matrix} \right)$
So:
\[a_{2,\max }^{'}=\left( \begin{matrix}
1 \\
2\sqrt{7} \\
5 \\
\end{matrix} \right)\]
And finally:
\[{{B}_{2,Sd}}(\cos \theta )=\frac{{{\left( \begin{matrix}
1 \\
2\sqrt{7} \\
5 \\
\end{matrix} \right)}^{T}}{{\left( \begin{matrix}
1 & \cos \theta & {{\cos }^{2}}\theta \\
\end{matrix} \right)}^{T}}}{1+2\sqrt{7}+5}=\frac{1}{2(3+\sqrt{7})}+\frac{2\sqrt{7}}{2(3+\sqrt{7})}\cos \theta +\frac{5}{2(3+\sqrt{7})}{{\cos }^{2}}\theta \]
\[
\blacksquare
\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%% Q9 %%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{9.9}
Show that the directivity pattern of the third-order hypercardioid can be expressed as
\begin{eqnarray*}
{\cal B}_{3,\mathrm{Hd}} \left( \cos \theta \right) = -\frac{3}{32} - \frac{15}{32} \cos \theta + \frac{15}{32} \cos^2 \theta
+ \frac{35}{32} \cos^3 \theta .
\end{eqnarray*}
\textbf{\uline{Solution}}\textbf{:}
We know the definition of B:
\[{{B}_{N,Hd}}(\cos \theta )=\frac{{{1}^{T}}{{H}^{-1}}_{N}p(\cos \theta )}{{{1}^{T}}{{H}^{-1}}_{N}1}\]
and for N=3:
\[ p(\cos \theta )={{\left[ \begin{matrix}
1 & \cos \theta & {{\cos }^{2}}\theta & {{\cos }^{3}}\theta \\
\end{matrix} \right]}^{T}}\text{ } \]
\[ \text{ }{{1}^{T}}=\left[ \begin{matrix}
1 & 1 & 1 & 1 \\
\end{matrix} \right]\text{ } \]
\[ 1={{\left[ \begin{matrix}
1 & 1 & 1 & 1 \\
\end{matrix} \right]}^{T}} \]
\[ \text{ }{{H}_{1}}=\left( \begin{matrix}
1 & 0 & \frac{1}{3} & 0 \\
0 & \frac{1}{3} & 0 & \frac{1}{5} \\
\frac{1}{3} & 0 & \frac{1}{5} & 0 \\
0 & \frac{1}{5} & 0 & \frac{1}{7} \\
\end{matrix} \right)\to {{H}_{1}}^{-1}=\left( \begin{matrix}
\frac{9}{4} & 0 & \frac{-15}{4} & 0 \\
0 & \frac{75}{4} & 0 & \frac{-105}{4} \\
\frac{-15}{4} & 0 & \frac{45}{4} & 0 \\
0 & \frac{-105}{4} & 0 & \frac{175}{4} \\
\end{matrix} \right)\text{ } \]
substituting:
\[ {{B}_{3,Hd}}=\frac{\left[ \begin{matrix}
1 & 1 & 1 & 1 \\
\end{matrix} \right]\left( \begin{matrix}
\frac{9}{4} & 0 & \frac{-15}{4} & 0 \\
0 & \frac{75}{4} & 0 & \frac{-105}{4} \\
\frac{-15}{4} & 0 & \frac{45}{4} & 0 \\
0 & \frac{-105}{4} & 0 & \frac{175}{4} \\
\end{matrix} \right){{\left[ \begin{matrix}
1 & \cos \theta & {{\cos }^{2}}\theta & {{\cos }^{3}}\theta \\
\end{matrix} \right]}^{T}}\text{ }}{\left[ \begin{matrix}
1 & 1 & 1 & 1 \\
\end{matrix} \right]\left( \begin{matrix}
\frac{9}{4} & 0 & \frac{-15}{4} & 0 \\
0 & \frac{75}{4} & 0 & \frac{-105}{4} \\
\frac{-15}{4} & 0 & \frac{45}{4} & 0 \\
0 & \frac{-105}{4} & 0 & \frac{175}{4} \\
\end{matrix} \right){{\left[ \begin{matrix}
1 & 1 & 1 & 1 \\
\end{matrix} \right]}^{T}}}= \]
\[ =\frac{-\frac{3}{2}-\frac{15}{2}\cos \theta +\frac{15}{2}{{\cos }^{2}}\theta +\frac{35}{2}{{\cos }^{3}}\theta }{16}=-\frac{3}{32}-\frac{15}{32}\cos \theta +\frac{15}{32}{{\cos }^{2}}\theta +\frac{35}{32}{{\cos }^{3}}\theta \]
\[
\blacksquare
\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%% Q11 %%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{9.11}
Show that the beampattern, the DF, and the WNG of the first-order DSA can be approximated as
\begin{align*}
{\cal B} \left[ \mathbf{h}_1(f), \cos \theta \right] &\approx \frac{1}{1-\alpha_{1,1}} \cos \theta - \frac{\alpha_{1,1}}{1-\alpha_{1,1}},\\
{\cal D} \left[ \mathbf{h}_1(f) \right] &\approx \frac{ \left( 1-\alpha_{1,1} \right)^2 }
{ \displaystyle \alpha_{1,1}^2 + \frac{1}{3} },\\
{\cal W} \left[ \mathbf{h}_1(f) \right] &\approx \frac{1}{2} \left[ 2\pi f \tau_0 \left( 1 - \alpha_{1,1} \right) \right]^2 .
\end{align*}
\textbf{\uline{Solution}}\textbf{:}
First we know that the beampattern is:
\[ B[{{h}_{1}}(f),\cos \theta ]=\frac{1-{{e}^{j2\pi f{{\tau }_{0}}(\cos \theta -{{\alpha }_{1,1}})}}}{1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{1,1}})}}} \]
using the approximation:
\[ {{e}^{x}}\approx 1+x \]
we can approximate the beampattern as:
\[ B[{{h}_{1}}(f),\cos \theta ]=\frac{1-{{e}^{j2\pi f{{\tau }_{0}}(\cos \theta -{{\alpha }_{1,1}})}}}{1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{1,1}})}}}\approx \frac{1-1-j2\pi f{{\tau }_{0}}(\cos \theta -{{\alpha }_{1,1}})}{1-1-j2\pi f{{\tau }_{0}}(1-{{\alpha }_{1,1}})}=\frac{1}{1-{{\alpha }_{1,1}}}\cos \theta -\frac{{{\alpha }_{1,1}}}{1-{{\alpha }_{1,1}}} \]
\[
\blacksquare
\]
we know that the DF is:
\[D[{{h}_{1}}(f)]=\frac{1-\cos [2\pi f{{\tau }_{0}}(1-{{\alpha }_{1,1}})]}{1-\sin c(2\pi f{{\tau }_{0}})\cos (2\pi f{{\tau }_{0}}{{\alpha }_{1.1}})}\]
Using the approximations:
\[ \cos x\approx 1-\frac{{{x}^{2}}}{2} \]
\[ \sin cx\approx 1-\frac{{{x}^{2}}}{6}\]
the DF becomes:
\[ D[{{h}_{1}}(f)]\approx \frac{1-1+\frac{{{[2\pi f{{\tau }_{0}}(1-{{\alpha }_{1,1}})]}^{2}}}{2}}{1-\left( 1-\frac{{{(2\pi f{{\tau }_{0}})}^{2}}}{6} \right)\left( 1-\frac{{{(2\pi f{{\tau }_{0}}{{\alpha }_{1.1}})}^{2}}}{2} \right)}=\frac{\frac{{{[2\pi f{{\tau }_{0}}(1-{{\alpha }_{1,1}})]}^{2}}}{2}}{1-\left( 1-\frac{{{(2\pi f{{\tau }_{0}})}^{2}}}{6}-\frac{{{(2\pi f{{\tau }_{0}}{{\alpha }_{1.1}})}^{2}}}{2}+\frac{{{(2\pi f{{\tau }_{0}})}^{4}}{{\alpha }_{1.1}}^{2}}{12} \right)}\]
\[ \approx \frac{\frac{{{[2\pi f{{\tau }_{0}}(1-{{\alpha }_{1,1}})]}^{2}}}{2}}{\frac{{{(2\pi f{{\tau }_{0}})}^{2}}}{6}+\frac{{{(2\pi f{{\tau }_{0}}{{\alpha }_{1.1}})}^{2}}}{2}}=\frac{{{(1-{{\alpha }_{1,1}})}^{2}}}{\frac{1}{3}+{{\alpha }_{1.1}}^{2}} \]
\[
\blacksquare
\]
The WNG is:
\[W[{{h}_{1}}(f)]=\frac{1}{2}{{\left| 1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{1,1}})}} \right|}^{2}}=1-\cos [2\pi f{{\tau }_{0}}(1-{{\alpha }_{1,1}})]\]
using the cos approximation:
\[W[{{h}_{1}}(f)]\approx 1-1+\frac{{{[2\pi f{{\tau }_{0}}(1-{{\alpha }_{1,1}})]}^{2}}}{2}=\frac{1}{2}[2\pi f{{\tau }_{0}}(1-{{\alpha }_{1,1}})]\]
\[
\blacksquare
\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%% Q12 %%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{9.12}
Show that the inverse of the Vandermonde matrix $\mathbf{V}(f)$ that appears in (\ref{C9-SOD-const-mat}) is given by
\begin{align*}
& \mathbf{V}^{-1}(f) = \nonumber \\
& \setlength\arraycolsep{2pt} \left[ \begin{array}{ccc}
\displaystyle \frac{v_2v_3}{\left(v_2-v_1\right)\left(v_3-v_1\right)} &
\displaystyle -\frac{v_1v_3}{\left(v_2-v_1\right)\left(v_3-v_2\right)} &
\displaystyle \frac{v_1v_2}{\left(v_3-v_1\right)\left(v_3-v_2\right)} \\
\displaystyle -\frac{v_2+v_3}{\left(v_2-v_1\right)\left(v_3-v_1\right)} &
\displaystyle \frac{v_1+v_3}{\left(v_2-v_1\right)\left(v_3-v_2\right)} &
\displaystyle -\frac{v_1+v_2}{\left(v_3-v_1\right)\left(v_3-v_2\right)} \\
\displaystyle \frac{1}{\left(v_2-v_1\right)\left(v_3-v_1\right)} &
\displaystyle -\frac{1}{\left(v_2-v_1\right)\left(v_3-v_2\right)} &
\displaystyle \frac{1}{\left(v_3-v_1\right)\left(v_3-v_2\right)}
\end{array} \right].
\end{align*}
\textbf{\uline{Solution}}\textbf{:}
The inverse of the Vandermonde matrix is given by:
\[{{V}^{-1}}(f)=U(f)L(f)\]
where,
\[ U(f)=\left[ \begin{matrix}
1 & -{{v}_{1}} & {{v}_{1}}{{v}_{2}} \\
0 & 1 & -({{v}_{1}}+{{v}_{2}}) \\
0 & 0 & 1 \\
\end{matrix} \right] \]
\[ L(f)=\left[ \begin{matrix}
1 & 0 & 0 \\
\frac{1}{{{v}_{1}}-{{v}_{2}}} & \frac{1}{{{v}_{2}}-{{v}_{1}}} & 0 \\
\frac{1}{\left( {{v}_{1}}-{{v}_{2}} \right)\left( {{v}_{1}}-{{v}_{3}} \right)} & \frac{1}{\left( {{v}_{2}}-{{v}_{1}} \right)\left( {{v}_{2}}-{{v}_{3}} \right)} & \frac{1}{\left( {{v}_{3}}-{{v}_{1}} \right)\left( {{v}_{3}}-{{v}_{2}} \right)} \\
\end{matrix} \right] \]
substituting U L in order to find the inverse of the Vandermonde matrix :
\[ {{V}^{-1}}(f)=\left[ \begin{matrix}
1 & -{{v}_{1}} & {{v}_{1}}{{v}_{2}} \\
0 & 1 & -({{v}_{1}}+{{v}_{2}}) \\
0 & 0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 0 & 0 \\
\frac{1}{{{v}_{1}}-{{v}_{2}}} & \frac{1}{{{v}_{2}}-{{v}_{1}}} & 0 \\
\frac{1}{\left( {{v}_{1}}-{{v}_{2}} \right)\left( {{v}_{1}}-{{v}_{3}} \right)} & \frac{1}{\left( {{v}_{2}}-{{v}_{1}} \right)\left( {{v}_{2}}-{{v}_{3}} \right)} & \frac{1}{\left( {{v}_{3}}-{{v}_{1}} \right)\left( {{v}_{3}}-{{v}_{2}} \right)} \\
\end{matrix} \right]= \]
\[ =\left[ \begin{matrix}
1-\frac{{{v}_{1}}}{{{v}_{1}}-{{v}_{2}}}+\frac{{{v}_{1}}{{v}_{2}}}{\left( {{v}_{1}}-{{v}_{2}} \right)\left( {{v}_{1}}-{{v}_{3}} \right)} & -\frac{{{v}_{1}}}{{{v}_{2}}-{{v}_{1}}}+\frac{{{v}_{1}}{{v}_{2}}}{\left( {{v}_{2}}-{{v}_{1}} \right)\left( {{v}_{2}}-{{v}_{3}} \right)} & \frac{{{v}_{1}}{{v}_{2}}}{\left( {{v}_{3}}-{{v}_{1}} \right)\left( {{v}_{3}}-{{v}_{2}} \right)} \\
\frac{1}{{{v}_{1}}-{{v}_{2}}}-\frac{({{v}_{1}}+{{v}_{2}})}{\left( {{v}_{1}}-{{v}_{2}} \right)\left( {{v}_{1}}-{{v}_{3}} \right)} & \frac{1}{{{v}_{2}}-{{v}_{1}}}-\frac{({{v}_{1}}+{{v}_{2}})}{\left( {{v}_{2}}-{{v}_{1}} \right)\left( {{v}_{2}}-{{v}_{3}} \right)} & \frac{-({{v}_{1}}+{{v}_{2}})}{\left( {{v}_{3}}-{{v}_{1}} \right)\left( {{v}_{3}}-{{v}_{2}} \right)} \\
\frac{1}{\left( {{v}_{1}}-{{v}_{2}} \right)\left( {{v}_{1}}-{{v}_{3}} \right)} & \frac{1}{\left( {{v}_{2}}-{{v}_{1}} \right)\left( {{v}_{2}}-{{v}_{3}} \right)} & \frac{1}{\left( {{v}_{3}}-{{v}_{1}} \right)\left( {{v}_{3}}-{{v}_{2}} \right)} \\
\end{matrix} \right]= \]
\[ =\left[ \begin{matrix}
\frac{\left( {{v}_{1}}-{{v}_{2}} \right)\left( {{v}_{1}}-{{v}_{3}} \right)-{{v}_{1}}\left( {{v}_{1}}-{{v}_{3}} \right)+{{v}_{1}}{{v}_{2}}}{\left( {{v}_{1}}-{{v}_{2}} \right)\left( {{v}_{1}}-{{v}_{3}} \right)} & \frac{-{{v}_{1}}\left( {{v}_{2}}-{{v}_{3}} \right)+{{v}_{1}}{{v}_{2}}}{\left( {{v}_{2}}-{{v}_{1}} \right)\left( {{v}_{2}}-{{v}_{3}} \right)} & \frac{{{v}_{1}}{{v}_{2}}}{\left( {{v}_{3}}-{{v}_{1}} \right)\left( {{v}_{3}}-{{v}_{2}} \right)} \\
\frac{\left( {{v}_{1}}-{{v}_{3}} \right)-({{v}_{1}}+{{v}_{2}})}{\left( {{v}_{1}}-{{v}_{2}} \right)\left( {{v}_{1}}-{{v}_{3}} \right)} & \frac{\left( {{v}_{2}}-{{v}_{3}} \right)-({{v}_{1}}+{{v}_{2}})}{\left( {{v}_{2}}-{{v}_{1}} \right)\left( {{v}_{2}}-{{v}_{3}} \right)} & \frac{-({{v}_{1}}+{{v}_{2}})}{\left( {{v}_{3}}-{{v}_{1}} \right)\left( {{v}_{3}}-{{v}_{2}} \right)} \\
\frac{1}{\left( {{v}_{1}}-{{v}_{2}} \right)\left( {{v}_{1}}-{{v}_{3}} \right)} & \frac{1}{\left( {{v}_{2}}-{{v}_{1}} \right)\left( {{v}_{2}}-{{v}_{3}} \right)} & \frac{1}{\left( {{v}_{3}}-{{v}_{1}} \right)\left( {{v}_{3}}-{{v}_{2}} \right)} \\
\end{matrix} \right]= \]
\[ =\left[ \begin{matrix}
\frac{{{v}_{1}}{{v}_{1}}-{{v}_{1}}{{v}_{3}}-{{v}_{2}}{{v}_{1}}+{{v}_{2}}{{v}_{3}}-{{v}_{1}}{{v}_{1}}+{{v}_{1}}{{v}_{3}}+{{v}_{1}}{{v}_{2}}}{\left( {{v}_{1}}-{{v}_{2}} \right)\left( {{v}_{1}}-{{v}_{3}} \right)} & \frac{-{{v}_{1}}{{v}_{2}}+{{v}_{1}}{{v}_{3}}+{{v}_{1}}{{v}_{2}}}{\left( {{v}_{2}}-{{v}_{1}} \right)\left( {{v}_{2}}-{{v}_{3}} \right)} & \frac{{{v}_{1}}{{v}_{2}}}{\left( {{v}_{3}}-{{v}_{1}} \right)\left( {{v}_{3}}-{{v}_{2}} \right)} \\
\frac{{{v}_{1}}-{{v}_{3}}-{{v}_{1}}-{{v}_{2}}}{\left( {{v}_{1}}-{{v}_{2}} \right)\left( {{v}_{1}}-{{v}_{3}} \right)} & \frac{{{v}_{2}}-{{v}_{3}}-{{v}_{1}}-{{v}_{2}}}{\left( {{v}_{2}}-{{v}_{1}} \right)\left( {{v}_{2}}-{{v}_{3}} \right)} & \frac{-({{v}_{1}}+{{v}_{2}})}{\left( {{v}_{3}}-{{v}_{1}} \right)\left( {{v}_{3}}-{{v}_{2}} \right)} \\
\frac{1}{\left( {{v}_{1}}-{{v}_{2}} \right)\left( {{v}_{1}}-{{v}_{3}} \right)} & \frac{-1}{\left( {{v}_{2}}-{{v}_{1}} \right)\left( {{v}_{3}}-{{v}_{2}} \right)} & \frac{1}{\left( {{v}_{3}}-{{v}_{1}} \right)\left( {{v}_{3}}-{{v}_{2}} \right)} \\
\end{matrix} \right]= \]
\[ =\left[ \begin{matrix}
\frac{{{v}_{2}}{{v}_{3}}}{\left( {{v}_{2}}-{{v}_{1}} \right)\left( {{v}_{3}}-{{v}_{1}} \right)} & \frac{-{{v}_{1}}{{v}_{3}}}{\left( {{v}_{2}}-{{v}_{1}} \right)\left( {{v}_{3}}-{{v}_{2}} \right)} & \frac{{{v}_{1}}{{v}_{2}}}{\left( {{v}_{3}}-{{v}_{1}} \right)\left( {{v}_{3}}-{{v}_{2}} \right)} \\
\frac{-({{v}_{3}}+{{v}_{2}})}{\left( {{v}_{2}}-{{v}_{1}} \right)\left( {{v}_{3}}-{{v}_{1}} \right)} & \frac{{{v}_{1}}+{{v}_{3}}}{\left( {{v}_{2}}-{{v}_{1}} \right)\left( {{v}_{3}}-{{v}_{2}} \right)} & \frac{-({{v}_{1}}+{{v}_{2}})}{\left( {{v}_{3}}-{{v}_{1}} \right)\left( {{v}_{3}}-{{v}_{2}} \right)} \\
\frac{1}{\left( {{v}_{2}}-{{v}_{1}} \right)\left( {{v}_{3}}-{{v}_{1}} \right)} & \frac{-1}{\left( {{v}_{2}}-{{v}_{1}} \right)\left( {{v}_{3}}-{{v}_{2}} \right)} & \frac{1}{\left( {{v}_{3}}-{{v}_{1}} \right)\left( {{v}_{3}}-{{v}_{2}} \right)} \\
\end{matrix} \right]\]
\[
\blacksquare
\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%% Q13 %%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{9.13}
Show that in the case of a second-order DSA with a zero of multiplicity 2 in the beampattern,
the beamformer is given by
\begin{eqnarray*}
\mathbf{h}_{2,0}(f) = \frac{1}{\left[ 1 - e^{ \jmath 2\pi f \tau_0 \left(1-\alpha_{2,1} \right)} \right]^2}
\left[ \begin{array}{c} 1 \\ -2 e^{-\jmath 2\pi f \tau_0\alpha_{2,1} } \\ e^{-\jmath 4\pi f \tau_0\alpha_{2,1} } \end{array} \right] ,
\end{eqnarray*}
we want to solve the following linear problem:
\[\left[ \begin{matrix}
{{d}^{H}}(f,1) \\
{{d}^{H}}(f,{{\alpha }_{2,1}}) \\
{{\sum{d(f,{{\alpha }_{2,1}})}}^{H}} \\
\end{matrix} \right]h(f)=\left[ \begin{matrix}
1 \\
0 \\
0 \\
\end{matrix} \right]\]
\[\to \left[ \begin{matrix}
\begin{matrix}
1 & {{v}_{1}} & {{v}_{1}}^{2} \\
\end{matrix} \\
\begin{matrix}
1 & {{v}_{2}} & {{v}_{2}}^{2} \\
\end{matrix} \\
{{\sum{d(f,{{\alpha }_{2,1}})}}^{H}} \\
\end{matrix} \right]h(f)=\left[ \begin{matrix}
\begin{matrix}
1 & {{e}^{j2\pi f{{\tau }_{0}}}} & {{e}^{j4\pi f{{\tau }_{0}}}} \\
\end{matrix} \\
\begin{matrix}
1 & {{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{2,1}}}} & {{e}^{j4\pi f{{\tau }_{0}}{{\alpha }_{2,1}}}} \\
\end{matrix} \\
{{\sum{d(f,{{\alpha }_{2,1}})}}^{H}} \\
\end{matrix} \right]h(f)=\left[ \begin{matrix}
\left[ \begin{matrix}
1 & {{e}^{j2\pi f{{\tau }_{0}}}} & {{e}^{j4\pi f{{\tau }_{0}}}} \\
\end{matrix} \right]h(f) \\
\left[ \begin{matrix}
1 & {{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{2,1}}}} & {{e}^{j4\pi f{{\tau }_{0}}{{\alpha }_{2,1}}}} \\
\end{matrix} \right]h(f) \\
{{\sum{d(f,{{\alpha }_{2,1}})}}^{H}}h(f) \\
\end{matrix} \right]=\otimes \]
using that ${{\sum{d(f,{{\alpha }_{2,1}})}}^{H}}h(f)=0$:
\[\to \otimes =\left[ \begin{matrix}
\left[ \begin{matrix}
1 & {{e}^{j2\pi f{{\tau }_{0}}}} & {{e}^{j4\pi f{{\tau }_{0}}}} \\
\end{matrix} \right]h(f) \\
\left[ \begin{matrix}
1 & {{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{2,1}}}} & {{e}^{j4\pi f{{\tau }_{0}}{{\alpha }_{2,1}}}} \\
\end{matrix} \right]h(f) \\
0 \\
\end{matrix} \right]\]
and finally its easy to see that the solution h(f) is:
\[{{h}_{2,0}}(f)=\frac{1}{{{\left[ 1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{2,1}})}} \right]}^{2}}}\left[ \begin{matrix}
1 \\
-2{{e}^{-j2\pi f{{\tau }_{0}}{{\alpha }_{2,1}}}} \\
{{e}^{-j4\pi f{{\tau }_{0}}{{\alpha }_{2,1}}}} \\
\end{matrix} \right]\]
\[
\blacksquare
\]
the beampattern can be written as
\begin{eqnarray*}
{\cal B} \left[ \mathbf{h}_{2,0}(f), \cos \theta \right] \approx
\frac{1}{ \left( 1 - \alpha_{2,1} \right)^2 } \left( \cos\theta - \alpha_{2,1} \right)^2,
\end{eqnarray*}
\textbf{\uline{Solution}}\textbf{:}
the beampattern has the form:
\[B[{{h}_{2,0}}(f),\cos \theta ]=\frac{{{[1-{{e}^{j2\pi f{{\tau }_{0}}(\cos \theta -{{\alpha }_{2,1}})}}]}^{2}}}{{{[1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{2,1}})}}]}^{2}}}\]
using the following approximation:
\[{{e}^{x}}\approx 1+x\]
we get:
\[B[{{h}_{2,0}}(f),\cos \theta ]\approx \frac{{{[1-1-j2\pi f{{\tau }_{0}}(\cos \theta -{{\alpha }_{2,1}})]}^{2}}}{{{[1-1-j2\pi f{{\tau }_{0}}(1-{{\alpha }_{2,1}})]}^{2}}}=\frac{{{[j2\pi f{{\tau }_{0}}(\cos \theta -{{\alpha }_{2,1}})]}^{2}}}{{{[j2\pi f{{\tau }_{0}}(1-{{\alpha }_{2,1}})]}^{2}}}=\frac{{{[\cos \theta -{{\alpha }_{2,1}}]}^{2}}}{{{[1-{{\alpha }_{2,1}}]}^{2}}}\]
\[
\blacksquare
\]
The WNG can be approximated as
\begin{eqnarray*}
{\cal W} \left[ \mathbf{h}_{2,0}(f) \right] \approx \frac{1}{6} \left[ 2\pi f \tau_0 \left( 1 - \alpha_{2,1} \right) \right]^4 .
\end{eqnarray*}
\textbf{\uline{Solution}}\textbf{:}
We find that the WNG is:
\[W[h(f)]=\frac{1}{6}{{\left| 1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{2,1}})}} \right|}^{4}}=\frac{2}{3}{{\left[ 1-\cos [2\pi f{{\tau }_{0}}(1-{{\alpha }_{2,1}})] \right]}^{2}}\]
with the following approximation:
\[\cos x\approx 1-\frac{{{x}^{2}}}{2}\]
The WNG can be approximated as
\[W[h(f)]\approx \frac{2}{3}{{\left[ 1-1+\frac{{{[2\pi f{{\tau }_{0}}(1-{{\alpha }_{2,1}})]}^{2}}}{2} \right]}^{2}}=\frac{2}{3\cdot 4}{{\left[ 2\pi f{{\tau }_{0}}(1-{{\alpha }_{2,1}}) \right]}^{4}}=\frac{1}{6}{{\left[ 2\pi f{{\tau }_{0}}(1-{{\alpha }_{2,1}}) \right]}^{4}}\]
\[
\blacksquare
\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%% Q14 %%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{9.14}
Show that the first column of the inverse of the Vandermonde matrix $\mathbf{V}(f)$ that appears in (\ref{C9-TOD-const-mat}) is given by
\begin{eqnarray*}
\mathbf{V}^{-1} \left(f;:,1 \right)= \left[ \begin{array}{c}
\displaystyle \frac{v_2v_3v_4}{\left(v_2-v_1\right) \left(v_3-v_1\right) \left(v_4-v_1\right)} \\
\displaystyle -\frac{v_2v_3+v_3v_4+v_2v_4}{\left(v_2-v_1\right) \left(v_3-v_1\right) \left(v_4-v_1\right)} \\
\displaystyle \frac{v_2+v_3+v_4}{\left(v_2-v_1\right) \left(v_3-v_1\right) \left(v_4-v_1\right)} \\
\displaystyle -\frac{1}{\left(v_2-v_1\right) \left(v_3-v_1\right) \left(v_4-v_1\right)}
\end{array} \right].
\end{eqnarray*}
\textbf{\uline{Solution}}\textbf{:}
The inverse of the Vandermonde matrix is given by:
\[{{V}^{-1}}(f)=U(f)L(f)\]
where,
\[U(f)=\left( \begin{matrix}
1 & -{{v}_{1}} & {{v}_{1}}{{v}_{2}} & -{{v}_{1}}{{v}_{2}}{{v}_{3}} \\
0 & 1 & -({{v}_{1}}+{{v}_{2}}) & {{v}_{1}}{{v}_{2}}+{{v}_{1}}{{v}_{3}}+{{v}_{3}}{{v}_{2}} \\
0 & 0 & 1 & -({{v}_{1}}+{{v}_{2}}+{{v}_{3}}) \\
0 & 0 & 0 & 1 \\
\end{matrix} \right)\]
\[L(f)=\left[ \begin{matrix}
1 & \cdots \\
\frac{1}{({{v}_{1}}-{{v}_{2}})} & \cdots \\
\frac{1}{({{v}_{1}}-{{v}_{2}})({{v}_{1}}-{{v}_{3}})} & \cdots \\
\frac{1}{({{v}_{1}}-{{v}_{2}})({{v}_{1}}-{{v}_{3}})({{v}_{1}}-{{v}_{4}})} & \cdots \\
\end{matrix} \right]\]
substituting U L in order to find the inverse of the Vandermonde matrix :
\[ {{V}^{-1}}(f)=U(f)L(f)=\left( \begin{matrix}
1 & -{{v}_{1}} & {{v}_{1}}{{v}_{2}} & -{{v}_{1}}{{v}_{2}}{{v}_{3}} \\
0 & 1 & -({{v}_{1}}+{{v}_{2}}) & {{v}_{1}}{{v}_{2}}+{{v}_{1}}{{v}_{3}}+{{v}_{3}}{{v}_{2}} \\
0 & 0 & 1 & -({{v}_{1}}+{{v}_{2}}+{{v}_{3}}) \\
0 & 0 & 0 & 1 \\
\end{matrix} \right)\left[ \begin{matrix}
1 & \cdots \\
\frac{1}{({{v}_{1}}-{{v}_{2}})} & \cdots \\
\frac{1}{({{v}_{1}}-{{v}_{2}})({{v}_{1}}-{{v}_{3}})} & \cdots \\
\frac{1}{({{v}_{1}}-{{v}_{2}})({{v}_{1}}-{{v}_{3}})({{v}_{1}}-{{v}_{4}})} & \cdots \\
\end{matrix} \right]= \]
\[ =\left[ \begin{matrix}
1+\frac{-{{v}_{1}}}{({{v}_{1}}-{{v}_{2}})}+\frac{{{v}_{1}}{{v}_{2}}}{({{v}_{1}}-{{v}_{2}})({{v}_{1}}-{{v}_{3}})}+\frac{-{{v}_{1}}{{v}_{2}}{{v}_{3}}}{({{v}_{1}}-{{v}_{2}})({{v}_{1}}-{{v}_{3}})({{v}_{1}}-{{v}_{4}})} & \cdots \\
\frac{1}{({{v}_{1}}-{{v}_{2}})}+\frac{-({{v}_{1}}+{{v}_{2}})}{({{v}_{1}}-{{v}_{2}})({{v}_{1}}-{{v}_{3}})}+\frac{{{v}_{1}}{{v}_{2}}+{{v}_{1}}{{v}_{3}}+{{v}_{3}}{{v}_{2}}}{({{v}_{1}}-{{v}_{2}})({{v}_{1}}-{{v}_{3}})({{v}_{1}}-{{v}_{4}})} & \cdots \\
\frac{1}{({{v}_{1}}-{{v}_{2}})({{v}_{1}}-{{v}_{3}})}+\frac{-({{v}_{1}}+{{v}_{2}}+{{v}_{3}})}{({{v}_{1}}-{{v}_{2}})({{v}_{1}}-{{v}_{3}})({{v}_{1}}-{{v}_{4}})} & \cdots \\
1+\frac{1}{({{v}_{1}}-{{v}_{2}})}+\frac{1}{({{v}_{1}}-{{v}_{2}})({{v}_{1}}-{{v}_{3}})}+\frac{1}{({{v}_{1}}-{{v}_{2}})({{v}_{1}}-{{v}_{3}})({{v}_{1}}-{{v}_{4}})} & \cdots \\
\end{matrix} \right]=\left[ \begin{matrix}
\frac{-{{v}_{2}}{{v}_{3}}{{v}_{4}}}{({{v}_{1}}-{{v}_{2}})({{v}_{1}}-{{v}_{3}})({{v}_{1}}-{{v}_{4}})} & \cdots \\
\frac{{{v}_{2}}{{v}_{3}}+{{v}_{3}}{{v}_{4}}+{{v}_{2}}{{v}_{4}}}{\left( {{v}_{1}}-{{v}_{2}} \right)\left( {{v}_{1}}-{{v}_{3}} \right)\left( {{v}_{1}}-{{v}_{4}} \right)} & \cdots \\
\frac{-{{v}_{2}}-{{v}_{3}}-{{v}_{4}}}{\left( {{v}_{1}}-{{v}_{2}} \right)\left( {{v}_{1}}-{{v}_{3}} \right)\left( {{v}_{1}}-{{v}_{4}} \right)} & \cdots \\
\frac{1}{({{v}_{1}}-{{v}_{2}})({{v}_{1}}-{{v}_{3}})({{v}_{1}}-{{v}_{4}})} & \cdots \\
\end{matrix} \right]= \]
\[ =\left[ \begin{matrix}
-\frac{{{v}_{2}}{{v}_{3}}{{v}_{4}}}{({{v}_{2}}-{{v}_{1}})({{v}_{3}}-{{v}_{1}})({{v}_{4}}-{{v}_{1}})} & \cdots \\
-\frac{{{v}_{2}}{{v}_{3}}+{{v}_{3}}{{v}_{4}}+{{v}_{2}}{{v}_{4}}}{({{v}_{2}}-{{v}_{1}})({{v}_{3}}-{{v}_{1}})({{v}_{4}}-{{v}_{1}})} & \cdots \\
\frac{{{v}_{2}}+{{v}_{3}}+{{v}_{4}}}{({{v}_{2}}-{{v}_{1}})({{v}_{3}}-{{v}_{1}})({{v}_{4}}-{{v}_{1}})} & \cdots \\
-\frac{1}{({{v}_{2}}-{{v}_{1}})({{v}_{3}}-{{v}_{1}})({{v}_{4}}-{{v}_{1}})} & \cdots \\
\end{matrix} \right] \]
so we get:
\[{{V}^{-1}}(f)=\left[ \begin{matrix}
\frac{{{v}_{2}}{{v}_{3}}{{v}_{4}}}{({{v}_{2}}-{{v}_{1}})({{v}_{3}}-{{v}_{1}})({{v}_{4}}-{{v}_{1}})} & \cdots \\
-\frac{{{v}_{2}}{{v}_{3}}+{{v}_{3}}{{v}_{4}}+{{v}_{2}}{{v}_{4}}}{({{v}_{2}}-{{v}_{1}})({{v}_{3}}-{{v}_{1}})({{v}_{4}}-{{v}_{1}})} & \cdots \\
\frac{{{v}_{2}}+{{v}_{3}}+{{v}_{4}}}{({{v}_{2}}-{{v}_{1}})({{v}_{3}}-{{v}_{1}})({{v}_{4}}-{{v}_{1}})} & \cdots \\
-\frac{1}{({{v}_{2}}-{{v}_{1}})({{v}_{3}}-{{v}_{1}})({{v}_{4}}-{{v}_{1}})} & \cdots \\
\end{matrix} \right]\]
\[
\blacksquare
\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%% Q15 %%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{9.15}
Show that the third-order DSA beamformer is given by
\begin{align*}
& \mathbf{h}_3(f) = \frac{1}{\left[1-e^{\jmath 2\pi f \tau_0\left(1-\alpha_{3,1} \right)}\right]
\left[1-e^{\jmath 2\pi f \tau_0\left(1-\alpha_{3,2} \right)}\right]
\left[1-e^{\jmath 2\pi f \tau_0\left(1-\alpha_{3,3} \right)}\right]} \times \\
& \left[ \begin{array}{c} 1 \\ -e^{-\jmath 2\pi f \tau_0 \alpha_{3,1}}-
e^{-\jmath 2\pi f \tau_0 \alpha_{3,2}}- e^{-\jmath 2\pi f \tau_0 \alpha_{3,3}} \\
e^{-\jmath 2\pi f \tau_0 \left( \alpha_{3,1}+\alpha_{3,2}\right)}+
e^{-\jmath 2\pi f \tau_0 \left( \alpha_{3,2}+\alpha_{3,3}\right)}+
e^{-\jmath 2\pi f \tau_0 \left( \alpha_{3,1}+\alpha_{3,3}\right)} \\
-e^{-\jmath 2\pi f \tau_0 \left( \alpha_{3,1}+\alpha_{3,2}+\alpha_{3,3}
\right)} \end{array} \right] .
\end{align*}
\textbf{\uline{Solution}}\textbf{:}
first we know from (9.82):
\[ V(f){{h}_{3}}(f)=\left[ \begin{matrix}
1 \\
0 \\
0 \\
0 \\
\end{matrix} \right] \]
\[ {{h}_{3}}(f)={{V}^{-1}}(f)\left[ \begin{matrix}
1 \\
0 \\
0 \\
0 \\
\end{matrix} \right]={{V}^{-1}}(f;:,1) \]
as we know from problem 9.14 :
\[ \to {{h}_{3}}(f)=\left[ \begin{matrix}
\frac{{{v}_{2}}{{v}_{3}}{{v}_{4}}}{({{v}_{2}}-{{v}_{1}})({{v}_{3}}-{{v}_{1}})({{v}_{4}}-{{v}_{1}})} \\
-\frac{{{v}_{2}}{{v}_{3}}+{{v}_{3}}{{v}_{4}}+{{v}_{2}}{{v}_{4}}}{({{v}_{2}}-{{v}_{1}})({{v}_{3}}-{{v}_{1}})({{v}_{4}}-{{v}_{1}})} \\
\frac{{{v}_{2}}+{{v}_{3}}+{{v}_{4}}}{({{v}_{2}}-{{v}_{1}})({{v}_{3}}-{{v}_{1}})({{v}_{4}}-{{v}_{1}})} \\
-\frac{1}{({{v}_{2}}-{{v}_{1}})({{v}_{3}}-{{v}_{1}})({{v}_{4}}-{{v}_{1}})} \\
\end{matrix} \right] \]
Now let's simplify the phrase:
\[ {{h}_{3}}(f,1)=\frac{{{v}_{2}}{{v}_{3}}{{v}_{4}}}{({{v}_{2}}-{{v}_{1}})({{v}_{3}}-{{v}_{1}})({{v}_{4}}-{{v}_{1}})}=\frac{{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}}{({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})}= \]
\[ =\frac{{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}}{({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})}\frac{\frac{1}{{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}}}{\frac{1}{{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}}} \]
\[ \to {{h}_{3}}(f,1)=\frac{1}{(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,1}})}})(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,2}})}})(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,3}})}})} \]
\[ {{h}_{3}}(f,2)=-\frac{{{v}_{2}}{{v}_{3}}+{{v}_{3}}{{v}_{4}}+{{v}_{2}}{{v}_{4}}}{({{v}_{2}}-{{v}_{1}})({{v}_{3}}-{{v}_{1}})({{v}_{4}}-{{v}_{1}})}=-\frac{{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}+{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}+{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}}{({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})}= \]
\[ =-\frac{{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}+{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}+{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}}{({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})}\frac{\frac{1}{{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}}}{\frac{1}{{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}}} \]
\[ \to {{h}_{3}}(f,2)=-\frac{{{e}^{-j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}+{{e}^{-j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}+{{e}^{-j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}}{(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,1}})}})(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,2}})}})(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,3}})}})} \]
\[ {{h}_{3}}(f,3)=\frac{{{v}_{2}}+{{v}_{3}}+{{v}_{4}}}{({{v}_{2}}-{{v}_{1}})({{v}_{3}}-{{v}_{1}})({{v}_{4}}-{{v}_{1}})}=\frac{{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}+{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}+{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}}{({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})}= \]
\[ =\frac{{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}+{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}+{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}}{({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})}\frac{\frac{1}{{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}}}{\frac{1}{{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}}} \]
\[ \to {{h}_{3}}(f,3)=\frac{{{e}^{-j2\pi f{{\tau }_{0}}({{\alpha }_{3,1}}+{{\alpha }_{3,2}})}}+{{e}^{-j2\pi f{{\tau }_{0}}({{\alpha }_{3,1}}+{{\alpha }_{3,3}})}}+{{e}^{-j2\pi f{{\tau }_{0}}({{\alpha }_{3,2}}+{{\alpha }_{3,3}})}}}{(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,1}})}})(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,2}})}})(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,3}})}})} \]
\[ {{h}_{3}}(f,4)=-\frac{1}{({{v}_{2}}-{{v}_{1}})({{v}_{3}}-{{v}_{1}})({{v}_{4}}-{{v}_{1}})}=-\frac{1}{({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})}= \]
\[ =-\frac{1}{({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})({{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}-{{e}^{j2\pi f{{\tau }_{0}}}})}\frac{\frac{1}{{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}}}{\frac{1}{{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}{{e}^{j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}}} \]
\[ \to {{h}_{3}}(f,4)=-\frac{{{e}^{-j2\pi f{{\tau }_{0}}({{\alpha }_{3,1}}+{{\alpha }_{3,2}}+{{\alpha }_{3,3}})}}}{(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,1}})}})(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,2}})}})(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,3}})}})} \]
and finally:
\[ {{h}_{3}}(f)=\left[ \begin{matrix}
\frac{1}{(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,1}})}})(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,2}})}})(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,3}})}})} \\
-\frac{{{e}^{-j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}+{{e}^{-j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}+{{e}^{-j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}}}{(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,1}})}})(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,2}})}})(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,3}})}})} \\
\frac{{{e}^{-j2\pi f{{\tau }_{0}}({{\alpha }_{3,1}}+{{\alpha }_{3,2}})}}+{{e}^{-j2\pi f{{\tau }_{0}}({{\alpha }_{3,1}}+{{\alpha }_{3,3}})}}+{{e}^{-j2\pi f{{\tau }_{0}}({{\alpha }_{3,2}}+{{\alpha }_{3,3}})}}}{(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,1}})}})(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,2}})}})(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,3}})}})} \\
-\frac{{{e}^{-j2\pi f{{\tau }_{0}}({{\alpha }_{3,1}}+{{\alpha }_{3,2}}+{{\alpha }_{3,3}})}}}{(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,1}})}})(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,2}})}})(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,3}})}})} \\
\end{matrix} \right] \]
\[ {{h}_{3}}(f)=\frac{1}{(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,1}})}})(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,2}})}})(1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,3}})}})}\left[ \begin{matrix}
1 \\
-{{e}^{-j2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}-{{e}^{-j2\pi f{{\tau }_{0}}{{\alpha }_{3,2}}}}-{{e}^{-j2\pi f{{\tau }_{0}}{{\alpha }_{3,3}}}} \\
{{e}^{-j2\pi f{{\tau }_{0}}({{\alpha }_{3,1}}+{{\alpha }_{3,2}})}}+{{e}^{-j2\pi f{{\tau }_{0}}({{\alpha }_{3,1}}+{{\alpha }_{3,3}})}}+{{e}^{-j2\pi f{{\tau }_{0}}({{\alpha }_{3,2}}+{{\alpha }_{3,3}})}} \\
-{{e}^{-j2\pi f{{\tau }_{0}}({{\alpha }_{3,1}}+{{\alpha }_{3,2}}+{{\alpha }_{3,3}})}} \\
\end{matrix} \right] \]
\[
\blacksquare
\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%% Q16 %%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{9.16}
Show that in the case of a third-order DSA with a zero of multiplicity 3 in the beampattern,
the beamformer is given by
\begin{eqnarray*}
\mathbf{h}_{3,0}(f) = \frac{1}{\left[ 1 - e^{ \jmath 2\pi f \tau_0 \left(1-\alpha_{3,1} \right)} \right]^3}
\left[ \begin{array}{c} 1 \\ -3 e^{-\jmath 2\pi f \tau_0\alpha_{3,1} } \\
3 e^{-\jmath 4\pi f \tau_0\alpha_{2,1} } \\ - e^{-\jmath 6\pi f \tau_0\alpha_{2,1} } \end{array} \right] ,
\end{eqnarray*}
\textbf{\uline{Solution}}\textbf{:}
we want to solve the following linear problem:
\[\left[ \begin{matrix}
{{d}^{H}}(f,1) \\
{{d}^{H}}(f,{{\alpha }_{3,1}}) \\
\sum{{{d}^{H}}{{(f,{{\alpha }_{3,2}})}^{H}}} \\
\sum{^{2}{{d}^{H}}{{(f,{{\alpha }_{3,3}})}^{H}}} \\
\end{matrix} \right]h(f)\]
so:
\[\left[ \begin{matrix}
{{d}^{H}}(f,1) \\
{{d}^{H}}(f,{{\alpha }_{3,1}}) \\
\sum{{{d}^{H}}{{(f,{{\alpha }_{3,2}})}^{H}}} \\
\sum{^{2}{{d}^{H}}{{(f,{{\alpha }_{3,3}})}^{H}}} \\
\end{matrix} \right]h(f)=\left[ \begin{matrix}
\begin{matrix}
1 & {{v}_{1}} & {{v}_{1}}^{2} & {{v}_{1}}^{3} \\
\end{matrix} \\
\begin{matrix}
1 & {{v}_{2}} & {{v}_{2}}^{2} & {{v}_{2}}^{3} \\
\end{matrix} \\
\sum{{{d}^{H}}{{(f,{{\alpha }_{3,2}})}^{H}}} \\
\sum{^{2}{{d}^{H}}{{(f,{{\alpha }_{3,3}})}^{H}}} \\
\end{matrix} \right]{{h}_{3,0}}(f)=\left[ \begin{matrix}
\begin{matrix}
[1 & {{v}_{1}} & {{v}_{1}}^{2} & {{v}_{1}}^{3}]{{h}_{3,0}}(f) \\
\end{matrix} \\
\begin{matrix}
[1 & {{v}_{2}} & {{v}_{2}}^{2} & {{v}_{2}}^{3}]{{h}_{3,0}}(f) \\
\end{matrix} \\
\sum{{{d}^{H}}{{(f,{{\alpha }_{3,2}})}^{H}}{{h}_{3,0}}(f)} \\
\sum{^{2}{{d}^{H}}{{(f,{{\alpha }_{3,3}})}^{H}}{{h}_{3,0}}(f)} \\
\end{matrix} \right]= (a) \]
from 9.86 we know:
\[ j2\pi f{{\tau }_{0}}\text{ }\sum{{{d}^{H}}{{(f,{{\alpha }_{3,2}})}^{H}}{{h}_{3,0}}(f)}\text{ }=\text{ }0\to \sum{{{d}^{H}}{{(f,{{\alpha }_{3,2}})}^{H}}{{h}_{3,0}}(f)}=0 \]
and from 9.87 we know:
\[ {{\left( j2\pi f{{\tau }_{0}} \right)}^{2}}\text{ }\sum{^{2}{{d}^{H}}{{(f,{{\alpha }_{3,3}})}^{H}}{{h}_{3,0}}(f)}\text{ }=0\to \sum{^{2}{{d}^{H}}{{(f,{{\alpha }_{3,3}})}^{H}}{{h}_{3,0}}(f)}=0 \]
substituting:
\[(a) =\frac{1}{{{\left[ 1-{{e}^{\jmath 2\pi f{{\tau }_{0}}\left( 1-{{\alpha }_{3,1}} \right)}} \right]}^{3}}}\left[ \begin{matrix}
\begin{matrix}
[1 & {{e}^{\jmath 2\pi f{{\tau }_{0}}}} & {{e}^{\jmath 4\pi f{{\tau }_{0}}}} & {{e}^{\jmath 6\pi f{{\tau }_{0}}}}]\left[ \begin{matrix}
1 \\
-3{{e}^{-\jmath 2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}} \\
3{{e}^{-\jmath 4\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}} \\
-{{e}^{-\jmath 6\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}} \\
\end{matrix} \right] \\
\end{matrix} \\
\begin{matrix}
[1 & {{e}^{\jmath 2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}} & {{e}^{\jmath 4\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}} & {{e}^{\jmath 6\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}]\left[ \begin{matrix}
1 \\
-3{{e}^{-\jmath 2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}} \\
3{{e}^{-\jmath 4\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}} \\
-{{e}^{-\jmath 6\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}} \\
\end{matrix} \right] \\
\end{matrix} \\
0 \\
0 \\
\end{matrix} \right]= \]
\[ =\frac{1}{{{\left[ 1-{{e}^{\jmath 2\pi f{{\tau }_{0}}\left( 1-{{\alpha }_{3,1}} \right)}} \right]}^{3}}}\left[ \begin{matrix}
1-3{{e}^{-\jmath 2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}{{e}^{\jmath 2\pi f{{\tau }_{0}}}}+3{{e}^{-\jmath 4\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}{{e}^{\jmath 4\pi f{{\tau }_{0}}}}-{{e}^{-\jmath 6\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}{{e}^{\jmath 6\pi f{{\tau }_{0}}}} \\
1-3{{e}^{-\jmath 2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}{{e}^{\jmath 2\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}+3{{e}^{-\jmath 4\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}{{e}^{\jmath 4\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}-{{e}^{-\jmath 6\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}}{{e}^{\jmath 6\pi f{{\tau }_{0}}{{\alpha }_{3,1}}}} \\
0 \\
0 \\
\end{matrix} \right]= \]
\[ =\frac{1}{{{\left[ 1-{{e}^{\jmath 2\pi f{{\tau }_{0}}\left( 1-{{\alpha }_{3,1}} \right)}} \right]}^{3}}}\left[ \begin{matrix}
1-3{{e}^{\jmath 2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,1}})}}+3{{e}^{\jmath 4\pi f{{\tau }_{0}}(1-{{\alpha }_{3,1}})}}-{{e}^{\jmath 6\pi f{{\tau }_{0}}(1-{{\alpha }_{3,1}})}} \\
1-3+3-1 \\
0 \\
0 \\
\end{matrix} \right]= \]
\[ =\frac{1}{{{\left[ 1-{{e}^{\jmath 2\pi f{{\tau }_{0}}\left( 1-{{\alpha }_{3,1}} \right)}} \right]}^{3}}}\left[ \begin{matrix}
{{\left[ 1-{{e}^{\jmath 2\pi f{{\tau }_{0}}\left( 1-{{\alpha }_{3,1}} \right)}} \right]}^{3}} \\
0 \\
0 \\
0 \\
\end{matrix} \right]=\left[ \begin{matrix}
1 \\
0 \\
0 \\
0 \\
\end{matrix} \right] \]
\[
\blacksquare
\]
the beampattern can be approximated as
\begin{eqnarray*}
{\cal B} \left[ \mathbf{h}_{3,0}(f), \cos \theta \right] \approx
\frac{1}{ \left( 1 - \alpha_{3,1} \right)^3 } \left( \cos\theta - \alpha_{3,1} \right)^3 ,
\end{eqnarray*}
\textbf{\uline{Solution}}\textbf{:}
the beampattern has the form:
\[B[{{h}_{3,0}}(f),\cos \theta ]=\frac{{{[1-{{e}^{j2\pi f{{\tau }_{0}}(\cos \theta -{{\alpha }_{3,1}})}}]}^{3}}}{{{[1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,1}})}}]}^{3}}}\]
using the following approximation:
\[{{e}^{x}}\approx 1+x\]
we get:
\[B[{{h}_{3,0}}(f),\cos \theta ]\approx \frac{{{[1-1-j2\pi f{{\tau }_{0}}(\cos \theta -{{\alpha }_{3,1}})]}^{3}}}{{{[1-1-j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,1}})]}^{3}}}=\frac{{{[j2\pi f{{\tau }_{0}}(\cos \theta -{{\alpha }_{3,1}})]}^{3}}}{{{[j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,1}})]}^{3}}}=\frac{{{[\cos \theta -{{\alpha }_{3,1}}]}^{3}}}{{{[1-{{\alpha }_{3,1}}]}^{3}}}\]
\[
\blacksquare
\]
and the WNG can be approximated as
\begin{eqnarray*}
{\cal W} \left[ \mathbf{h}_{3,0}(f) \right] \approx \frac{1}{20} \left[ 2\pi f \tau_0 \left( 1 - \alpha_{3,1} \right) \right]^6 .
\end{eqnarray*}
\textbf{\uline{Solution}}\textbf{:}
We find that the WNG is:
\[W[h(f)]=\frac{1}{20}{{\left| 1-{{e}^{j2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,1}})}} \right|}^{6}}=\frac{2}{5}{{\left[ 1-\cos [2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,1}})] \right]}^{3}}\]
with the following approximation:
\[\cos x\approx 1-\frac{{{x}^{2}}}{2}\]
The WNG can be approximated as
\[W[h(f)]\approx \frac{2}{5}{{\left[ 1-1+\frac{{{[2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,1}})]}^{2}}}{2} \right]}^{3}}=\frac{2}{5\cdot 8}{{\left[ 2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,1}}) \right]}^{6}}=\frac{1}{20}{{\left[ 2\pi f{{\tau }_{0}}(1-{{\alpha }_{3,1}}) \right]}^{6}}\]
\[
\blacksquare
\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%% Q18 %%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{9.18}
Show that the beampattern, the WNG, and the DF of the minimum-norm beamformer are given by
\begin{align*}
{\cal B} \left[ \mathbf{h}_{\mathrm{MN}}\left( f, \mbox{\boldmath $\alpha$}, \mbox{\boldmath $\beta$} \right), \cos\theta \right] &= \mathbf{d}^H\left(f,\cos\theta \right) \mathbf{D}^H\left( f, \mbox{\boldmath $\alpha$} \right) \times \\
&~~~~~ \left[ \mathbf{D}\left( f, \mbox{\boldmath $\alpha$} \right) \mathbf{D}^H\left( f, \mbox{\boldmath $\alpha$} \right) \right]^{-1}
\mbox{\boldmath $\beta$}, \\
{\cal W}\left[ \mathbf{h}_{\mathrm{MN}}\left( f, \mbox{\boldmath $\alpha$}, \mbox{\boldmath $\beta$} \right) \right] &= \frac{1}
{ \mbox{\boldmath $\beta$}^T \left[ \mathbf{D}\left( f, \mbox{\boldmath $\alpha$} \right)
\mathbf{D}^H\left( f, \mbox{\boldmath $\alpha$} \right) \right]^{-1} \mbox{\boldmath $\beta$} },\\
{\cal D}\left[ \mathbf{h}_{\mathrm{MN}}\left( f, \mbox{\boldmath $\alpha$}, \mbox{\boldmath $\beta$} \right) \right] &= \frac{1}
{ \mathbf{h}_{\mathrm{MN}}^H\left( f, \mbox{\boldmath $\alpha$}, \mbox{\boldmath $\beta$} \right) \mathbf{\Gamma}_{0,\pi}(f)
\mathbf{h}_{\mathrm{MN}}\left( f, \mbox{\boldmath $\alpha$}, \mbox{\boldmath $\beta$} \right) } .
\end{align*}
\textbf{\uline{Solution}}\textbf{:}
First we know:
\[(1) {{h}_{MN}}(f,\alpha ,\beta )={{D}^{H}}(f,\alpha ){{[D(f,\alpha ){{D}^{H}}(f,\alpha )]}^{-1}}\beta \]
\[ (2){{h}_{MN}}{{(f,\alpha ,\beta )}^{H}}d(f,1)=1 \]
It is easy to see that the beampattern is:
\[ B[{{h}_{MN}}(f,\alpha ,\beta ),\cos \theta ]={{d}^{H}}(f,\cos \theta ){{h}_{MN}}(f,\alpha ,\beta )= \]
\[ ={{d}^{H}}(f,\cos \theta ){{D}^{H}}(f,\alpha ){{[D(f,\alpha ){{D}^{H}}(f,\alpha )]}^{-1}}\beta \]
\[
\blacksquare
\]
The WNG defined as:
\[W[h(f)]=\frac{{{\left| {{h}^{H}}(f)d(f,1) \right|}^{2}}}{{{h}^{H}}(f)h(f)}\]
substitute (1) and (2) to the definition of WNG:
\[W[{{h}_{MN}}(f)]=\frac{{{\left| {{h}_{MN}}^{H}(f,\alpha ,\beta )d(f,1) \right|}^{2}}}{{{h}_{MN}}^{H}(f,\alpha ,\beta ){{h}_{MN}}(f,\alpha ,\beta )}=\]
\[ =\frac{{{1}^{2}}}{{{\beta }^{H}}(f,\alpha ){{[D(f,\alpha ){{D}^{H}}(f,\alpha )]}^{-1}}[D(f,\alpha ){{D}^{H}}(f,\alpha )]{{[D(f,\alpha ){{D}^{H}}(f,\alpha )]}^{-1}}\beta }= \]
\[ =\frac{1}{{{\beta }^{H}}(f,\alpha ){{[D(f,\alpha ){{D}^{H}}(f,\alpha )]}^{-1}}\beta } \]
\[
\blacksquare
\]
The DF of the minimum-norm beamformer defined as:
\[D[h(f)]=\frac{{{\left| {{h}^{H}}(f)d(f,1) \right|}^{2}}}{{{h}^{H}}(f){{\Gamma }_{0,2\pi }}h(f)}\]
substitute (1) and (2) to the definition of the DF;
\[ D[{{h}_{Mn}}(f,\alpha ,\beta )]=\frac{{{\left| {{h}_{Mn}}{{(f,\alpha ,\beta )}^{H}}d(f,1) \right|}^{2}}}{{{h}_{Mn}}{{(f,\alpha ,\beta )}^{H}}{{\Gamma }_{0,2\pi }}{{h}_{Mn}}(f,\alpha ,\beta )}=\frac{1}{{{h}_{Mn}}{{(f,\alpha ,\beta )}^{H}}{{\Gamma }_{0,2\pi }}{{h}_{Mn}}(f,\alpha ,\beta )} \]
\end{document}