Filter design by convex iteration

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<math>\begin{array}{lll}
<math>\begin{array}{lll}
-
min |h|_\infty \\
+
\textrm{min}& |h|_\infty & \\
-
subject to & \frac{1}{\delta_1}\leq|H(\omega)|\leq\delta_1, & \omega\in[0,\omega_p]\\
+
\textrm{subject to} & \frac{1}{\delta_1}\leq|H(\omega)|\leq\delta_1, & \omega\in[0,\omega_p]\\
& |H(\omega)|\leq\delta_2, & \omega\in[\omega_s,\pi]
& |H(\omega)|\leq\delta_2, & \omega\in[\omega_s,\pi]
\end{array}</math>
\end{array}</math>

Revision as of 16:39, 23 August 2010

LaTeX: H(\omega) = h(0) + h(1)e^{-j\omega} + \cdots + h(n-1)e^{-j(N-1)\omega}

where LaTeX: h \in \texttt{C}^\texttt{N}


For low pass filter, the frequency domain specifications are: 

LaTeX: 
</pre>
<p>\begin{array}{ll}
\frac{1}{\delta_1}\leq|H(\omega)|\leq\delta_1, & \omega\in[0,\omega_p]\\
|H(\omega)|\leq\delta_2, & \omega\in[\omega_s,\pi]
\end{array}
</p>
<pre>


To minimize the maximum magnitude of LaTeX: h , the problem becomes

LaTeX: \begin{array}{lll} \textrm{min}& |h|_\infty & \\ \textrm{subject to} & \frac{1}{\delta_1}\leq|H(\omega)|\leq\delta_1, & \omega\in[0,\omega_p]\\ & |H(\omega)|\leq\delta_2, & \omega\in[\omega_s,\pi] \end{array}


A new vector LaTeX: g \in \texttt{C}^\texttt{N*N} is defined as concatenation of time-shifted versions of LaTeX: h , \emph{i.e.} 

LaTeX: 
</pre>
<p>g =            \left[
</p>
<pre>              \begin{array}{c}
                h(t) \\
                h(t-1) \\
                \vdots \\
                h(t-N) \\
              \end{array}
            \right]

Then LaTeX: gg^\texttt{H} is a positive semidefinite matrix of size LaTeX: \texttt{N}^2 \times \texttt{N}^2 with rank 1. Summing along each 2N-1 subdiagonals gives entries of the autocorrelation function of LaTeX: h . In particular, the main diagonal holds squared entries of LaTeX: h . Minimizing LaTeX: |h|_\infty is equivalent to minimizing the trace of LaTeX: gg^\texttt{H} .


Using spectral factorization, an equivalent problem is 

LaTeX: 
</pre>
<p>\begin{array}{lll}
\hbox{min} & |r|_\infty & \\
\hbox{subject to} & \frac{1}{\delta_1^2}\leq R(\omega)\leq\delta_1^2, & \omega\in[0,\omega_p]\\
& R(\omega)\leq\delta_2^2, & \omega\in[\omega_s,\pi]\\
& R(\omega)\geq0, & \omega\in[0,\pi]
\end{array}
</p>
<pre>
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