section 10
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\begin{document}
\title{Fundamentals of Signal Enhancement and Array Signal Processing\\
Solution Manual}
\maketitle
\title{ \textbf{ Lidor Malul 318628005}}
\section*{ 10 Beampattern Design}
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%%%%%%%%%% Q1 %%%%%%%%%%%%%%%%%%%%%%%%
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\subsection*{10.1}
Show that the minimization of the LSE criterion yields
\begin{eqnarray*}
\mathbf{c}_N = \mathbf{M}_{\mathrm{C}}^{-1} \mathbf{v}_{\mathrm{C}}\left( \jmath \overline{f}_m \right) .
\end{eqnarray*}
\textbf{\uline{Solution}}\textbf{:}
First ,from (10.12) we know:
\[LSE({{c}_{N}})=1-v{{c}^{H}}(j{{\bar{f}}_{m}}){{c}_{N}}-c_{N}^{H}vc+c_{N}^{H}{{M}_{C}}{{c}_{N}}\]
we want to find the optimal solution for LSE :
\[ \frac{\partial LSE({{c}_{N}})}{\partial {{c}_{N}}}=0 \]
\[ \to \frac{\partial LSE({{c}_{N}})}{\partial {{c}_{N}}}=-vc(j{{{\bar{f}}}_{m}})-vc(j{{{\bar{f}}}_{m}})+2{{c}_{N}}{{M}_{C}}=0 \]
\[ \to {{c}_{N}}={{M}_{C}}^{-1}vc(j{{{\bar{f}}}_{m}}) \]
\[
\blacksquare
\]
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%%%%%%%%%% Q2 %%%%%%%%%%%%%%%%%%%%%%%%
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\subsection*{10.2}
Show that the elements of the vector $\mathbf{v}_{\mathrm{C}}\left( \jmath \overline{f}_m \right)$ are
\begin{eqnarray*}
\left[ \mathbf{v}_{\mathrm{C}}\left( \jmath \overline{f}_m \right) \right]_{n+1} = \jmath^n J_n \left( \overline{f}_m \right),
\end{eqnarray*}
where $J_n \left( z \right)$ is the Bessel function of the first kind.
\textbf{\uline{Solution}}\textbf{:}
we can write the vector vc as:
\[ vc(j{{{\bar{f}}}_{m}})=\frac{1}{\pi }\int_{0}^{\pi }{{{e}^{j{{{\bar{f}}}_{m}}\cos \theta }}{{P}_{c}}(\cos \theta )d\theta } \]
\[ \to vc{{(j{{{\bar{f}}}_{m}})}_{n+1}}=\frac{1}{\pi }\int_{0}^{\pi }{{{e}^{j{{{\bar{f}}}_{m}}\cos \theta }}\cos (n\theta )d\theta } \]
let's define:
\[{{J}_{n}}(z)\triangleq \frac{-{{j}^{-n}}}{\pi }\int_{0}^{\pi }{{{e}^{jz\cos \theta }}\cos (n\theta )d\theta }\]
so we can get:
\[vc{{(j{{\bar{f}}_{m}})}_{n+1}}={{j}^{n}}\cdot {{J}_{n}}({{\bar{f}}_{m}})\]
\[
\blacksquare
\]
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%%%%%%%%%% Q3 %%%%%%%%%%%%%%%%%%%%%%%%
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\subsection*{10.3}
Show that the elements of the matrix $\mathbf{M}_{\mathrm{C}}$ are
\begin{eqnarray*}
\left[ \mathbf{M}_{\mathrm{C}} \right]_{i+1,j+1} = \frac{1}{\pi} \int_{0}^{\pi}
\cos \left( i \theta \right) \cos \left( j \theta \right) d\theta .
\end{eqnarray*}
\textbf{\uline{Solution}}\textbf{:}
The matrix $M_c$ defined as following:
\[{{M}_{C}}=\frac{1}{\pi }\int_{0}^{\pi }{{{P}_{c}}(\cos \theta )P_{c}^{T}(\cos \theta )}d\theta \]
using $P_c$ deifinition:
\[ {{\left[ {{P}_{c}}(\cos \theta )P_{c}^{T}(\cos \theta ) \right]}_{i+1,j+1}}=\cos (i\theta )\cos (j\theta ) \]
\[ \to {{\left[ {{M}_{C}} \right]}_{i+1,j+1}}=\frac{1}{\pi }\int_{0}^{\pi }{\cos (i\theta )\cos (j\theta )}d\theta \]
\[
\blacksquare
\]
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%%%%%%%%%% Q4 %%%%%%%%%%%%%%%%%%%%%%%%
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\subsection*{10.4}
Prove the Jacobi-Anger expansion, i.e.,
\begin{eqnarray*}
e^{ \jmath \overline{f}_m \cos \theta } = \sum_{n=0}^{\infty} \jmath_n J_n \left( \overline{f}_m \right) \cos \left( n \theta \right),
\end{eqnarray*}
where
\begin{eqnarray*}
\jmath_n = \left\{ \begin{array}{ll} 1, & n=0 \\
2 \jmath^n, & n=1,2,\ldots,N \end{array} \right. .
\end{eqnarray*}
\textbf{\uline{Solution}}\textbf{:}
from 10.11 we know:
\[{{e}^{j{{f}_{m}}\cos \theta }}=\underset{N\to \infty }{\mathop{\lim }}\,\sum\limits_{n=0}^{N}{{{c}_{N}}\cos (n\theta )}\]
where,
\[ {{c}_{N}}={{[\begin{matrix}
{{c}_{0}} & {{c}_{1}} & \cdots & {{c}_{N}} \\
\end{matrix}]}^{T}} \]
\[ {{c}_{N}}={{M}_{C}}^{-1}vc(j{{{\bar{f}}}_{m}}) \]
from problem 10.2:
\[vc{{(j{{\bar{f}}_{m}})}_{n+1}}={{j}^{n}}{{J}_{n}}({{\bar{f}}_{m}})\]
from problem 10.3:
\[ {{M}_{C}}=\left( \begin{matrix}
1 & 0 & \cdots & 0 \\
0 & \frac{1}{2} & 0 & \vdots \\
\vdots & 0 & \ddots & 0 \\
0 & \cdots & 0 & \frac{1}{2} \\
\end{matrix} \right)\to {{M}^{-1}}_{C}=\left( \begin{matrix}
1 & 0 & \cdots & 0 \\
0 & 2 & 0 & \vdots \\
\vdots & 0 & \ddots & 0 \\
0 & \cdots & 0 & 2 \\
\end{matrix} \right) \]
so we get:
\[ {{c}_{N}}=\left\{ \begin{matrix}
\begin{matrix}
{{J}_{0}}({{f}_{m}}) & n=0 \\
\end{matrix} \\
\begin{matrix}
2{{j}^{n}}{{J}_{n}}({{f}_{m}}) & n\ge 1 \\
\end{matrix} \\
\end{matrix} \right. \]
subtituting all above:
\[{{e}^{j{{f}_{m}}\cos \theta }}={{J}_{0}}({{f}_{m}})+\sum\limits_{n=1}^{\infty }{2{{j}^{n}}{{J}_{n}}({{f}_{m}})\cos (n\theta )}=\sum\limits_{n=0}^{\infty }{{{j}_{n}}\cdot {{J}_{n}}({{f}_{m}})\cos (n\theta )}\]
where,
\[{{j}_{n}}=\left\{ \begin{matrix}
\begin{matrix}
1 & n=0 \\
\end{matrix} \\
\begin{matrix}
2{{j}^{n}} & n\ge 1 \\
\end{matrix} \\
\end{matrix} \right.\]
\[
\blacksquare
\]
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\subsection*{10.6}
Show that with the nonrobust filter, $ \mathbf{h}_{\mathrm{NR}}(f)$, the first-order beampattern is given by
\begin{eqnarray*}
{\cal B}_{1} \left[ \mathbf{h}(f), \cos\theta \right] = H_1(f) + J_0 \left( \overline{f}_2 \right) H_2(f)
+ 2 \jmath J_1 \left( \overline{f}_2 \right) H_2(f) \cos \theta .
\end{eqnarray*}
\textbf{\uline{Solution}}\textbf{:}
let's use 10.20 with M=2:
\[ B[h(f),\cos \theta ]=\sum\limits_{n=0}^{\infty }{\cos (n\theta )[\sum\limits_{m=1}^{\infty }{{{j}_{n}}{{J}_{n}}({{f}_{m}}){{H}_{m}}}]}= \]
\[ =\sum\limits_{n=0}^{\infty }{\cos (n\theta )[{{j}_{n}}{{J}_{n}}({{f}_{1}}){{H}_{1}}+{{j}_{n}}{{J}_{n}}({{f}_{2}}){{H}_{2}}]}= \]
\[ ={{J}_{0}}({{f}_{1}}){{H}_{1}}(f)+{{J}_{0}}({{f}_{2}}){{H}_{2}}+\sum\limits_{n=1}^{\infty }{\cos (n\theta )}2{{j}^{n}}[{{J}_{n}}({{f}_{1}}){{H}_{1}}+{{J}_{n}}({{f}_{2}}){{H}_{2}}] \]
We know that :
\[ {{J}_{0}}({{f}_{1}})=1 \]
\[ {{J}_{n}}({{f}_{n}})=0 \]
substitute:
\[B[h(f),\cos \theta ]={{H}_{1}}(f)+{{J}_{0}}({{f}_{2}}){{H}_{2}}+\cos (\theta )2j{{J}_{1}}({{f}_{2}}){{H}_{2}}\]
\[
\blacksquare
\]
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\subsection*{10.8}
Show that by minimizing $J_{\mathrm{FI}} \left[ \mathbf{h}(f) \right]$ subject to $\overline{\mathbf{B}}_N(f) \mathbf{h}(f) = \mathbf{b}_N$ and $\mathbf{h}^H(f) \mathbf{h}(f) = \delta_{\epsilon}$, we obtain the filter:
\begin{eqnarray*}
\mathbf{h}_{\mathrm{FI},\epsilon}(f) = \mathbf{\Gamma}_{\mathrm{C},\epsilon}^{-1}(f) \overline{\mathbf{B}}_N^H(f) \left[ \overline{\mathbf{B}}_N(f) \mathbf{\Gamma}_{\mathrm{C},\epsilon}^{-1}(f) \overline{\mathbf{B}}_N^H(f) \right]^{-1} \mathbf{b}_N ,
\end{eqnarray*}
where $\mathbf{\Gamma}_{\mathrm{C},\epsilon}(f) = \mathbf{\Gamma}_{\mathrm{C}}(f) + \epsilon \mathbf{I}_M$.
\textbf{\uline{Solution}}\textbf{:}
in order to find corresponding filter we will solve the following minimization:
\[\begin{matrix}
\min & {{h}^{H}}(f){{\Gamma }_{c}}(f)h(f) & subject & to & {{B}_{N}}(f)h(f)={{b}_{n}} & and & {{h}^{H}}(f)h(f)={{\delta }_{c}} \\
\end{matrix}\]
using Lagrange multiplier we defined the next function:
\[L(h,\lambda ,\varepsilon )=f(h)+\lambda g(h)+\varepsilon k(h)\]
where $\lambda$ and $\epsilon$ is a 1xM vector and :
\[ f(h)={{h}^{H}}(f){{\Gamma }_{c}}(f)h(f) \]
\[ g(h)={{B}_{N}}(f)h(f)-{{b}_{n}} \]
\[ k(h)={{h}^{H}}(f)h(f)-{{\delta }_{c}} \]
now, finding the min of L:
\[ \frac{\partial L(h,\lambda ,\varepsilon )}{\partial h}=0=2h(f){{\Gamma }_{c}}(f)+B_{N}^{H}(f)\lambda +2h(f)\varepsilon \]
\[ \to 2({{\Gamma }_{c}}(f)+\varepsilon I)h(f)=-B_{N}^{H}(f)\lambda \]
\[ \to h(f)=-\frac{1}{2}{{({{\Gamma }_{c}}(f)+\varepsilon I)}^{-1}}B_{N}^{H}\lambda \]
\[ \frac{\partial L(h,\lambda ,\varepsilon )}{\partial \lambda }=0\to {{B}_{N}}(f)h(f)={{b}_{n}}\]
\[ {{B}_{N}}(f)h(f)=-\frac{1}{2}{{B}_{N}}(f){{({{\Gamma }_{c}}(f)+\varepsilon I)}^{-1}}B_{N}^{H}\lambda ={{b}_{n}} \]
\[ \to \lambda =-2{{\left( {{B}_{N}}(f){{({{\Gamma }_{c}}(f)+\varepsilon I)}^{-1}}B_{N}^{H} \right)}^{-1}}{{b}_{n}} \]
\[ \to h(f)={{({{\Gamma }_{c}}(f)+\varepsilon I)}^{-1}}B_{N}^{H}{{\left( {{B}_{N}}(f){{({{\Gamma }_{c}}(f)+\varepsilon I)}^{-1}}B_{N}^{H} \right)}^{-1}}{{b}_{n}}= \]
\[ ={{\Gamma }_{c,\varepsilon }}^{-1}(f)B_{N}^{H}{{\left( {{B}_{N}}(f){{\Gamma }_{c,\varepsilon }}^{-1}(f)B_{N}^{H} \right)}^{-1}}{{b}_{n}} \]
where,
\[{{\Gamma }_{c,\varepsilon }}^{-1}(f)\triangleq {{\Gamma }_{c}}(f)+\varepsilon I\]
\[
\blacksquare
\]
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\subsection*{10.9}
Show that the LSE between the array beampattern and the desired directivity pattern can be written as
\begin{align*}
\mathrm{LSE}\left[ \mathbf{h}(f) \right] &= \mathbf{h}^H(f) \mathbf{\Gamma}_{\mathrm{C}}(f) \mathbf{h}(f) -
\mathbf{h}^H(f) \mathbf{\Gamma}_{\mathbf{d}\mathbf{p}_{\mathrm{C}}}(f) \mathbf{b}_N - \nonumber \\
& ~~~~~ \mathbf{b}_N^T \mathbf{\Gamma}_{\mathbf{d}\mathbf{p}_{\mathrm{C}}}^H(f) \mathbf{h}(f)
+ \mathbf{b}_N^T \mathbf{M}_{\mathrm{C}} \mathbf{b}_N.
\end{align*}
\textbf{\uline{Solution}}\textbf{:}
let's remember the definition of LSE:
\[LSE[h(f)]=\frac{1}{\pi }\int_{0}^{\pi }{{{\left| \varepsilon [h(f),\cos \theta ] \right|}^{2}}d\theta }\]
where,
\[ {{\left| \varepsilon [h(f),\cos \theta ] \right|}^{2}}={{\left| {{d}^{H}}(f,\cos \theta )h(f)-{{P}_{c}}^{T}(\cos \theta ){{b}_{N}} \right|}^{2}}= \]
\[ =\left( {{h}^{H}}(f)d(f,\cos \theta )-b_{N}^{H}{{P}_{c}}(\cos \theta ) \right)\left( {{d}^{H}}(f,\cos \theta )h(f)-{{P}_{c}}^{T}(\cos \theta ){{b}_{N}} \right)= \]
\[ ={{h}^{H}}(f)d(f,\cos \theta ){{d}^{H}}(f,\cos \theta )h(f)-{{h}^{H}}(f)d(f,\cos \theta ){{P}_{c}}^{T}(\cos \theta ){{b}_{N}}-b_{N}^{H}{{P}_{c}}(\cos \theta ){{d}^{H}}(f,\cos \theta )h(f)+ \]
\[ +b_{N}^{H}{{P}_{c}}(\cos \theta ){{P}_{c}}^{T}(\cos \theta ){{b}_{N}} \]
substituting:
\[ LSE[h(f)]=\frac{1}{\pi }\int_{0}^{\pi }{{{\left| \varepsilon [h(f),\cos \theta ] \right|}^{2}}d\theta }= \]
\[ =\frac{1}{\pi }\int_{0}^{\pi }{{{h}^{H}}(f)d(f,\cos \theta ){{d}^{H}}(f,\cos \theta )h(f)d\theta }-\frac{1}{\pi }\int_{0}^{\pi }{{{h}^{H}}(f)d(f,\cos \theta ){{P}_{c}}^{T}(\cos \theta ){{b}_{N}}d\theta }- \]
\[ -\frac{1}{\pi }\int_{0}^{\pi }{b_{N}^{H}{{P}_{c}}(\cos \theta ){{d}^{H}}(f,\cos \theta )h(f)d\theta }+\frac{1}{\pi }\int_{0}^{\pi }{b_{N}^{H}{{P}_{c}}(\cos \theta ){{P}_{c}}^{T}(\cos \theta ){{b}_{N}}d\theta }= \]
\[ ={{h}^{H}}(f)\left[ \frac{1}{\pi }\int_{0}^{\pi }{d(f,\cos \theta ){{d}^{H}}(f,\cos \theta )d\theta } \right]h(f)-{{h}^{H}}(f)\left[ \frac{1}{\pi }\int_{0}^{\pi }{d(f,\cos \theta ){{P}_{c}}^{T}(\cos \theta )d\theta } \right]{{b}_{N}}- \]
\[ -b_{N}^{H}\left[ \frac{1}{\pi }\int_{0}^{\pi }{{{P}_{c}}(\cos \theta ){{d}^{H}}(f,\cos \theta )d\theta } \right]h(f)+b_{N}^{H}\left[ \frac{1}{\pi }\int_{0}^{\pi }{{{P}_{c}}(\cos \theta ){{P}_{c}}^{T}(\cos \theta )d\theta } \right]{{b}_{N}} \]
using the following definitions:
\[ {{\Gamma }_{c}}(f)\triangleq \frac{1}{\pi }\int_{0}^{\pi }{d(f,\cos \theta ){{d}^{H}}(f,\cos \theta )d\theta } \]
\[ {{\Gamma }_{dpc}}(f)\triangleq \frac{1}{\pi }\int_{0}^{\pi }{d(f,\cos \theta ){{P}_{c}}^{T}(\cos \theta )d\theta } \]
\[ {{M}_{C}}\overset{\wedge}{=}\frac{1}{\pi }\int_{0}^{\pi }{{{P}_{c}}(\cos \theta ){{P}_{c}}^{T}(\cos \theta )d\theta } \]
so,
\[LSE[h(f)]={{h}^{H}}(f){{\Gamma }_{c}}(f)h(f)-{{h}^{H}}(f){{\Gamma }_{dpc}}(f){{b}_{N}}-b_{N}^{H}\Gamma _{dpc}^{H}(f)h(f)+b_{N}^{H}{{M}_{C}}{{b}_{N}}\]
\[
\blacksquare
\]
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\subsection*{10.10}
Show that by minimizing the LSE with a constraint on the coefficients, we obtain the regularized LS filter:
\begin{eqnarray*}
\mathbf{h}_{\mathrm{LS},\epsilon}(f) = \mathbf{\Gamma}_{\mathrm{C},\epsilon}^{-1}(f) \mathbf{\Gamma}_{\mathbf{d}\mathbf{p}_{\mathrm{C}}}(f) \mathbf{b}_N .
\end{eqnarray*}
\textbf{\uline{Solution}}\textbf{:}
in order to find the LS filter we will solve the following minimization:
\[ \begin{matrix}
\min & {{h}^{H}}(f){{\Gamma }_{c}}(f)h(f)-{{h}^{H}}(f){{\Gamma }_{dpc}}(f){{b}_{N}}-{{b}_{N}}^{H}(f){{\Gamma }^{H}}_{dpc}(f)h(f)+{{b}_{N}}^{T}{{M}_{c}}{{b}_{N}} \\
\end{matrix} \]
\[ \begin{matrix}
subject & to & {{h}^{H}}(f)h(f)={{\delta }_{c}} \\
\end{matrix} \]
using Lagrange multiplier we defined the next function:
\[L(h,\lambda ,\varepsilon )=f(h)+\varepsilon g(h)\]
where $\epsilon$ is a 1xM vector and :
\[\ f(h)={{h}^{H}}(f){{\Gamma }_{c}}(f)h(f) \]
\[ g(h)={{h}^{H}}(f)h(f)-{{\delta }_{c}} \]
now, finding the min of L:
\[ \frac{\partial L(h,\varepsilon )}{\partial h}=0\to 2{{\Gamma }_{c}}(f)h(f)-{{\Gamma }_{dpc}}(f){{b}_{N}}-{{\Gamma }_{dpc}}(f){{b}_{N}}+2\varepsilon h(f) \]
\[ \to 2({{\Gamma }_{c}}(f)+\varepsilon I)h(f)=2{{\Gamma }_{dpc}}(f){{b}_{N}} \]
\[ \to {{h}_{LS,e}}={{({{\Gamma }_{c}}(f)+\varepsilon I)}^{-1}}{{\Gamma }_{dpc}}(f){{b}_{N}}={{\Gamma }_{c,\varepsilon }}{{(f)}^{-1}}{{\Gamma }_{dpc}}(f){{b}_{N}} \]
\[
\blacksquare
\]
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\subsection*{10.11}
Show that by minimizing the LSE subject to the distortionless constraint and a constraint on the coefficients, we obtain the regularized CLS filter:
\begin{align*}
\mathbf{h}_{\mathrm{CLS},\epsilon}(f) &= \mathbf{h}_{\mathrm{LS},\epsilon}(f) -
\frac{1 - \mathbf{d}^H\left(f,1 \right) \mathbf{h}_{\mathrm{LS},\epsilon}(f) }
{ \mathbf{d}^H\left(f,1 \right) \mathbf{\Gamma}_{\mathrm{C},\epsilon}^{-1}(f) \mathbf{d}\left(f,1 \right) }
\mathbf{\Gamma}_{\mathrm{C},\epsilon}^{-1}(f) \mathbf{d}\left(f,1 \right) .
\end{align*}
\textbf{\uline{Solution}}\textbf{:}
in order to find the CLS filter we will solve the following minimization:
\[\begin{matrix}
\min & LSE{{\left| \varepsilon \right|}^{2}} & subject & to & {{h}^{H}}(f)d(f,\cos \theta )=1 & and & {{h}^{H}}(f)h(f)={{\delta }_{c}} \\
\end{matrix}\]
using Lagrange multiplier we defined the next function:
\[L(h,\lambda ,\varepsilon )=f(h)+\lambda g(h)+\varepsilon k(h)\]
where $\lambda$ and $\epsilon$ is a 1xM vector and
\[ f(h)=LSE[h(f)] \]
\[ g(h)={{h}^{H}}(f)d(f,\cos \theta )-1 \]
\[ k(h)={{h}^{H}}(f)h(f)-{{\delta }_{c}} \]
now, finding the min of L:
\[ \frac{\partial L(h,\lambda ,\varepsilon )}{\partial h}=0\to 2{{\Gamma }_{c}}(f)h(f)-{{\Gamma }_{dpc}}(f){{b}_{N}}-{{\Gamma }_{dpc}}(f){{b}_{N}}+2\varepsilon h(f)+\lambda d(f,\cos \theta ) \]
\[ \to 2({{\Gamma }_{c}}(f)+\varepsilon I)h(f)=2{{\Gamma }_{dpc}}(f){{b}_{N}}-\lambda d(f,\cos \theta ) \]
\[ \to {{h}_{CLS}}={{({{\Gamma }_{c}}(f)+\varepsilon I)}^{-1}}[{{\Gamma }_{dpc}}(f){{b}_{N}}-\frac{1}{2}\lambda d(f,\cos \theta )]={{h}_{LS,e}}(f)-\frac{1}{2}\lambda {{\Gamma }^{-1}}_{c,e}d(f,\cos \theta ) \]
\[ \frac{\partial L(h,\lambda ,\varepsilon )}{\partial \lambda }=0\to {{h}^{H}}(f)d(f,\cos \theta )=1\to {{d}^{H}}(f,\cos \theta )h(f)=1 \]
\[ {{d}^{H}}(f,\cos \theta )h(f)={{d}^{H}}(f,\cos \theta ){{h}_{LS,e}}(f)-\frac{1}{2}\lambda {{d}^{H}}(f,\cos \theta ){{\Gamma }^{-1}}_{c,e}d(f,\cos \theta )=1 \]
\[ \to \lambda =-2\frac{1-{{d}^{H}}(f,\cos \theta ){{h}_{LS,e}}(f)}{{{d}^{H}}(f,\cos \theta ){{\Gamma }^{-1}}_{c,e}d(f,\cos \theta )} \]
\[ \to {{h}_{CLS}}={{h}_{LS,e}}(f)+\frac{1-{{d}^{H}}(f,\cos \theta ){{h}_{LS,e}}(f)}{{{d}^{H}}(f,\cos \theta ){{\Gamma }^{-1}}_{c,e}d(f,\cos \theta )}{{\Gamma }^{-1}}_{c,e}d(f,\cos \theta ) \]
\[
\blacksquare
\]
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\subsection*{10.12}
Show that with the constraint $\overline{\mathbf{B}}_N(f) \mathbf{h}(f) = \mathbf{b}_N$, the error signal between the array beampattern and the desired directivity pattern can be expressed as
\begin{align*}
{\cal E}\left[ \mathbf{h}(f), \cos \theta \right] &= \sum_{i=N+1}^{\infty} \cos \left( i \theta \right) \overline{\mathbf{b}}_i^T(f) \mathbf{h}(f).
\end{align*}
\textbf{\uline{Solution}}\textbf{:}
from 10.55 we know:
\[ \varepsilon [h(f),\cos \theta ]=\sum\limits_{i=0}^{\infty }{\cos (i\theta ){{{\bar{b}}}_{i}}^{T}h(f)}-\sum\limits_{i=0}^{N}{\cos (i\theta ){{{\bar{b}}}_{N,i}}}= \]
\[ =\sum\limits_{i=N+1}^{\infty }{\cos (i\theta ){{{\bar{b}}}_{i}}^{T}h(f)}+\sum\limits_{i=0}^{N}{\cos (i\theta ){{{\bar{b}}}_{i}}^{T}h(f)}-\sum\limits_{i=0}^{N}{\cos (i\theta ){{{\bar{b}}}_{N,i}}} \]
using the following constraint:
\[ {{B}_{N}}(f)h(f)={{b}_{N}} \]
\[ \to {{b}_{i}}^{T}h(f)={{{\bar{b}}}_{N,i}} \]
substituting:
\[ \varepsilon [h(f),\cos \theta ]=\sum\limits_{i=N+1}^{\infty }{\cos (i\theta ){{{\bar{b}}}_{i}}^{T}h(f)}+\sum\limits_{i=0}^{N}{\cos (i\theta ){{{\bar{b}}}_{i}}^{T}h(f)}-\sum\limits_{i=0}^{N}{\cos (i\theta ){{b}_{i}}^{T}h(f)}= \]
\[ =\sum\limits_{i=N+1}^{\infty }{\cos (i\theta ){{{\bar{b}}}_{i}}^{T}h(f)} \]
\[
\blacksquare
\]
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\subsection*{10.13}
Using the orthogonality property of the Chebyshev polynomials, show that the criterion $J_{\mathrm{FI}} \left[ \mathbf{h}(f) \right]$ can be expressed as
\begin{align*}
J_{\mathrm{FI}} \left[ \mathbf{h}(f) \right] &= \mathrm{LSE}\left[ \mathbf{h}(f) \right] +
\frac{1}{\pi} \int_{0}^{\pi} \left| {\cal B}\left( \mathbf{b}_N, \cos \theta \right) \right|^2 d\theta ,
\end{align*}
where
\begin{align*}
\mathrm{LSE}\left[ \mathbf{h}(f) \right] &= \frac{1}{\pi} \int_{0}^{\pi} \left|
\sum_{i=N+1}^{\infty} \cos \left( i \theta \right) \overline{\mathbf{b}}_i^T(f) \mathbf{h}(f) \right|^2 d\theta .
\end{align*}
\textbf{\uline{Solution}}\textbf{:}
first we know that:
\[LSE[h(f)]=\frac{1}{\pi }\int_{0}^{\pi }{{{\left| \varepsilon [h(f),\cos \theta ] \right|}^{2}}d\theta }=\frac{1}{\pi }\int_{0}^{\pi }{{{\left| \sum\limits_{i=N+1}^{\infty }{\cos (i\theta ){{b}_{i}}^{T}h(f)} \right|}^{2}}d\theta }\]
now, using 10.56 the criterion defined in 10.40 can be expressed as:
\[{{J}_{FI}}[h(f)]=\frac{1}{\pi }\int_{0}^{\pi }{{{\left| \varepsilon [h(f),\cos \theta ]+B[{{b}_{N}},\cos \theta ] \right|}^{2}}d\theta =}\frac{1}{\pi }\int_{0}^{\pi }{{{\left| \sum\limits_{i=N+1}^{\infty }{\cos (i\theta ){{b}_{i}}^{T}h(f)}+\sum\limits_{i=0}^{N}{\cos (i\theta ){{b}_{N,i}}} \right|}^{2}}d\theta }\]
using the orthogonality property:
\[\begin{matrix}
\int_{0}^{\pi }{\cos (i\theta )\cos (j\theta )d\theta =0} & i\ne j \\
\end{matrix}\]
and now:
\[ {{J}_{FI}}[h(f)]=\frac{1}{\pi }{{\int_{0}^{\pi }{\left| \sum\limits_{i=N+1}^{\infty }{\cos (i\theta ){{b}_{i}}^{T}h(f)} \right|}}^{2}}d\theta +\frac{1}{\pi }\int_{0}^{\pi }{{{\left| \sum\limits_{i=0}^{N}{\cos (i\theta ){{b}_{N,i}}} \right|}^{2}}d\theta }= \]
\[ =\frac{1}{\pi }{{\int_{0}^{\pi }{\left| \varepsilon [h(f),\cos \theta ] \right|}}^{2}}d\theta +\frac{1}{\pi }\int_{0}^{\pi }{{{\left| B[{{b}_{N}},\cos \theta ] \right|}^{2}}d\theta } \]
\[ \to {{J}_{FI}}[h(f)]=LSE[h(f)]+\frac{1}{\pi }\int_{0}^{\pi }{{{\left| B[{{b}_{N}},\cos \theta ] \right|}^{2}}d\theta } \]
\[
\blacksquare
\]
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%%%%%%%%%% Q14 %%%%%%%%%%%%%%%%%%%%%%%%
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\subsection*{10.14}
Show that the filters defined in (\ref{C10-filt-hM}) preserve the nulls of $\mathbf{h}'(f) = \mathbf{h}_{\mathrm{NR}}(f)$, i.e., if $\theta_0$ is a null of $\mathbf{h}'(f)$, then
\begin{eqnarray*}
\mathbf{h}^H(f) \mathbf{d}\left(f, \cos \theta_0 \right) =
\mathbf{g}^H(f) \widetilde{\mathbf{d}}\left(f, \cos \theta_0 \right) \times 0 = 0 ,
\end{eqnarray*}
where
\begin{align*}
\widetilde{\mathbf{d}}\left(f, \cos \theta_0 \right) &= \left[ \begin{array}{cccc}
1 & e^{-\jmath 2\pi f \tau_0 \cos \theta_0 } & \cdots & e^{-\jmath (M-N-1) 2\pi f \tau_0 \cos \theta_0 } \end{array} \right]^T .
\end{align*}
\textbf{\uline{Solution}}\textbf{:}
we know the form of h is:
\[h(f)={{H}^{'}}(f)g(f)\to {{h}^{H}}(f)={{g}^{H}}(f){{H}^{'H}}(f)\]
let's define:
\[\tilde{d}(f,\cos \theta )\triangleq d(f,\cos \theta ){{H}^{'H}}={{[\begin{matrix}
1 & {{e}^{-j2\pi f{{\tau }_{0}}\cos \theta }} & \cdots & {{e}^{-j(M-N-1)2\pi f{{\tau }_{0}}\cos \theta }} \\
\end{matrix}]}^{T}}\]
if ${{\theta }_{0}}$ is a null of $h'(f)={{h}_{NR}}(f)$ so:
\[{{h}^{H}}(f)d(f,\cos \theta )={{g}^{H}}(f)\tilde{d}(f,\cos {{\theta }_{0}})\times 0=0\]
\[
\blacksquare
\]
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%%%%%%%%%% Q15 %%%%%%%%%%%%%%%%%%%%%%%%
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\subsection*{10.15}
Show that by maximizing the WNG subject to the distortionless constraint, we obtain the MWNG filter:
\begin{eqnarray*}
\mathbf{h}_{\mathrm{MWNG}}(f) =
\frac{\mathbf{H}'(f) \left[ \mathbf{H}'^H(f) \mathbf{H}'(f) \right]^{-1} \widetilde{\mathbf{d}}\left(f,1 \right) }
{ \widetilde{\mathbf{d}}^H\left(f,1 \right) \left[ \mathbf{H}'^H(f) \mathbf{H}'(f) \right]^{-1}
\widetilde{\mathbf{d}}\left(f,1 \right) } .
\end{eqnarray*}
\textbf{\uline{Solution}}\textbf{:}
in order to find the MWNG filter we will solve the following minimization:
\[\begin{matrix}
\min & {{g}^{H}}(f){{H}^{'H}}(f)H'(f)g(f) & subject & to & {{g}^{H}}(f)\tilde{d}(f,1)=1 \\
\end{matrix}\]
using Lagrange multiplier we defined the next function:
\[L(g,\lambda )=f(g)+\lambda k(g)\]
where $\lambda$ is a 1xM vector and
\[ f(g)={{g}^{H}}(f){{H}^{'H}}(f)H'(f)g(f) \]
\[ k(g)={{g}^{H}}(f)\tilde{d}(f,1)-1 \]
now, finding the min of L:
\[ \frac{\partial L(g,\lambda )}{\partial g}=0=2{{H}^{'H}}(f){{H}^{'}}(f)g(f)+\tilde{d}(f,1)\lambda \]
\[ \to g(f)=-\frac{1}{2}{{\left[ {{H}^{'H}}(f){{H}^{'}}(f) \right]}^{-1}}\tilde{d}(f,1)\lambda \]
\[ \frac{\partial L(g,\lambda )}{\partial \lambda }=0\to {{g}^{H}}(f)\tilde{d}(f,1)=1\to {{{\tilde{d}}}^{H}}(f,1)g(f)=1 \]
\[ {{{\tilde{d}}}^{H}}(f,1)g(f)=-\frac{1}{2}{{{\tilde{d}}}^{H}}(f,1){{\left[ {{H}^{'H}}(f){{H}^{'}}(f) \right]}^{-1}}\tilde{d}(f,1)\lambda =1 \]
\[ \to \lambda =-\frac{2}{{{{\tilde{d}}}^{H}}(f,1){{\left[ {{H}^{'H}}(f){{H}^{'}}(f) \right]}^{-1}}\tilde{d}(f,1)} \]
\[ \to {{g}_{MWNG}}(f)=\frac{{{\left[ {{H}^{'H}}(f){{H}^{'}}(f) \right]}^{-1}}\tilde{d}(f,1)}{{{{\tilde{d}}}^{H}}(f,1){{\left[ {{H}^{'H}}(f){{H}^{'}}(f) \right]}^{-1}}\tilde{d}(f,1)} \]
use the form of h(f):
\[ h(f)={{H}^{'H}}(f)g(f) \]
\[ \to {{h}_{MWNG}}(f)=\frac{{{H}^{'H}}(f){{\left[ {{H}^{'H}}(f){{H}^{'}}(f) \right]}^{-1}}\tilde{d}(f,1)}{{{{\tilde{d}}}^{H}}(f,1){{\left[ {{H}^{'H}}(f){{H}^{'}}(f) \right]}^{-1}}\tilde{d}(f,1)} \]
\[
\blacksquare
\]
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%%%%%%%%%% Q16 %%%%%%%%%%%%%%%%%%%%%%%%
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\subsection*{10.16}
Show that by minimizing $J_{\aleph}\left[ \mathbf{g}(f) \right]$ subject to the distortionless constraint, we obtain the tradeoff filter:
\begin{align*}
\mathbf{g}_{\mathrm{T},\aleph}(f) &= \mathbf{g}_{\mathrm{U},\aleph}(f) +
\frac{1 - \widetilde{\mathbf{d}}^H\left(f,1 \right) \mathbf{g}_{\mathrm{U},\aleph}(f)}
{\widetilde{\mathbf{d}}^H\left(f,1 \right) \mathbf{R}_{\aleph}^{-1}(f) \widetilde{\mathbf{d}}\left(f,1 \right)}
\mathbf{R}_{\aleph}^{-1}(f) \widetilde{\mathbf{d}}\left(f,1 \right),
\end{align*}
where
\begin{eqnarray*}
\mathbf{g}_{\mathrm{U},\aleph}(f) = \aleph \mathbf{R}_{\aleph}^{-1}(f) \mathbf{H}'^H(f) \mathbf{\Gamma}_{\mathbf{d}\mathbf{p}_{\mathrm{C}}}(f) \mathbf{b}_N
\end{eqnarray*}
is the unconstrained filter obtained by minimizing $J_{\aleph}\left[ \mathbf{g}(f) \right]$ and
\begin{align*}
\mathbf{R}_{\aleph}(f) &= \aleph \mathbf{R}(f) + (1-\aleph) \mathbf{H}'^H(f) \mathbf{H}'(f) .
\end{align*}
\textbf{\uline{Solution}}\textbf{:}
in order to find the tradeoff filter we will solve the following minimization:
\[\min \text{ }\aleph LSE[g(f)]+\left( 1-\aleph \right){{\text{g}}^{H}}(f){{H}^{'H}}(f){{H}^{'}}(f)\text{g}(f)\text{ }subject\text{ }to\text{ }{{g}^{H}}(f)\tilde{d}(f,1)=1\text{ }\]
using Lagrange multiplier we defined the next function:
\[L(g,\lambda )=f(g)+\lambda k(g)\]
where $\lambda$ is a 1xM vector and
\[ f(g)=\aleph LSE[g(f)]+\left( 1-\aleph \right){{\text{g}}^{H}}(f){{H}^{'H}}(f){{H}^{'}}(f)\text{g}(f) \]
\[ k(g)={{g}^{H}}(f)\tilde{d}(f,1)-1 \]
we know:
\[LSE\left[ g(f) \right]={{\text{g}}^{H}}(f){{H}^{'H}}(f){{\Gamma }_{C}}(f){{H}^{'}}(f)\text{g}(f)-{{\text{g}}^{H}}(f){{H}^{'H}}(f){{\Gamma }_{dpc}}(f){{b}_{N}}-b_{N}^{T}\Gamma _{dpc}^{H}(f){{H}^{'}}(f)\text{g}(f)+b_{N}^{T}{{M}_{c}}{{b}_{N}}\]
now, finding the min of L:
\[\frac{\partial L(g,\lambda )}{\partial g}=0=\aleph \left[ 2\left( {{H}^{'H}}(f){{\Gamma }_{C}}(f){{H}^{'}}(f) \right)g(f)-{{H}^{'H}}(f){{\Gamma }_{dpc}}(f){{b}_{N}}-{{H}^{'H}}(f){{\Gamma }_{dpc}}(f){{b}_{N}} \right]+ \]
\[ +2\left( 1-\aleph \right){{H}^{'H}}(f){{H}^{'}}(f)\text{g}(f)+\tilde{d}(f,1)\lambda =0 \]
\[ \to \left[ 2\aleph \left( {{H}^{'H}}(f){{\Gamma }_{C}}(f){{H}^{'}}(f) \right)+2\left( 1-\aleph \right){{H}^{'H}}(f){{H}^{'}}(f) \right]g(f)=2\aleph {{H}^{'H}}(f){{\Gamma }_{dpc}}(f){{b}_{N}}-\tilde{d}(f,1)\lambda \]
\[ \to g(f)={{\left[ \aleph \left( {{H}^{'H}}(f){{\Gamma }_{C}}(f){{H}^{'}}(f) \right)+\left( 1-\aleph \right){{H}^{'H}}(f){{H}^{'}}(f) \right]}^{-1}}\left[ \aleph {{H}^{'H}}(f){{\Gamma }_{dpc}}(f){{b}_{N}}-\frac{1}{2}\tilde{d}(f,1)\lambda \right] \]
we know from 10.76:
\[ R(f)={{H}^{'H}}(f){{\Gamma }_{C}}(f){{H}^{'}}(f) \]
\[ \to g(f)={{\left[ \aleph R(f)+\left( 1-\aleph \right){{H}^{'H}}(f){{H}^{'}}(f) \right]}^{-1}}\left[ \aleph {{H}^{'H}}(f){{\Gamma }_{dpc}}(f){{b}_{N}}-\frac{1}{2}\tilde{d}(f,1)\lambda \right] \]
let's dfine:
\[ {{R}_{\aleph }}(f)\triangleq \aleph R(f)+\left( 1-\aleph \right){{H}^{'H}}(f){{H}^{'}}(f) \]
\[ \to g(f)=\aleph {{R}_{\aleph }}{{(f)}^{-1}}{{H}^{'H}}(f){{\Gamma }_{dpc}}(f){{b}_{N}}-\frac{1}{2}{{R}_{\aleph }}{{(f)}^{-1}}\tilde{d}(f,1)\lambda \]
find $\lambda$ :
\[ \frac{\partial L(g,\lambda )}{\partial \lambda }=0\to {{g}^{H}}(f)\tilde{d}(f,1)=1\to {{{\tilde{d}}}^{H}}(f,1)g(f)=1 \]
\[ {{{\tilde{d}}}^{H}}(f,1)g(f)=\aleph {{{\tilde{d}}}^{H}}(f,1){{R}_{\aleph }}{{(f)}^{-1}}{{H}^{'H}}(f){{\Gamma }_{dpc}}(f){{b}_{N}}-\frac{1}{2}{{{\tilde{d}}}^{H}}(f,1){{R}_{\aleph }}{{(f)}^{-1}}\tilde{d}(f,1)\lambda \]
\[ \to \lambda =-2\frac{1-\aleph {{{\tilde{d}}}^{H}}(f,1){{R}_{\aleph }}{{(f)}^{-1}}{{H}^{'H}}(f){{\Gamma }_{dpc}}(f){{b}_{N}}}{{{{\tilde{d}}}^{H}}(f,1){{R}_{\aleph }}{{(f)}^{-1}}\tilde{d}(f,1)} \]
substituting
\[ \to {{g}_{T,\aleph }}(f)=\aleph {{R}_{\aleph }}{{(f)}^{-1}}{{H}^{'H}}(f){{\Gamma }_{dpc}}(f){{b}_{N}}+{{R}_{\aleph }}{{(f)}^{-1}}\tilde{d}(f,1)\frac{1-\aleph {{{\tilde{d}}}^{H}}(f,1){{R}_{\aleph }}{{(f)}^{-1}}{{H}^{'H}}(f){{\Gamma }_{dpc}}(f){{b}_{N}}}{{{{\tilde{d}}}^{H}}(f,1){{R}_{\aleph }}{{(f)}^{-1}}\tilde{d}(f,1)} \]
define:
\[\ {{g}_{U,\aleph }}(f)\triangleq \aleph {{R}_{\aleph }}{{(f)}^{-1}}{{H}^{'H}}(f){{\Gamma }_{dpc}}(f){{b}_{N}} \]
\[ \to {{g}_{T,\aleph }}(f)={{g}_{U,\aleph }}(f)+{{R}_{\aleph }}{{(f)}^{-1}}\tilde{d}(f,1)\frac{1-\aleph {{{\tilde{d}}}^{H}}(f,1){{R}_{\aleph }}{{(f)}^{-1}}{{H}^{'H}}(f){{\Gamma }_{dpc}}(f){{b}_{N}}}{{{{\tilde{d}}}^{H}}(f,1){{R}_{\aleph }}{{(f)}^{-1}}\tilde{d}(f,1)} \]
\[
\blacksquare
\]
\end{document}