# Filter design by convex iteration

(Difference between revisions)
 Revision as of 17:03, 23 August 2010 (edit)← Previous diff Revision as of 17:05, 23 August 2010 (edit) (undo)Next diff → Line 17: Line 17: $\begin{array}{lll} [itex]\begin{array}{lll} \textrm{min}& |h|_\infty & \\ \textrm{min}& |h|_\infty & \\ - \textrm{subject\,\, to} & \frac{1}{\delta_1}\leq|H(\omega)|\leq\delta_1, & \omega\in[0,\omega_p]\\ + \textrm{subjec t\,\, to} & \frac{1}{\delta_1}\leq|H(\omega)|\leq\delta_1, & \omega\in[0,\omega_p]\\ & |H(\omega)|\leq\delta_2, & \omega\in[\omega_s,\pi] & |H(\omega)|\leq\delta_2, & \omega\in[\omega_s,\pi] \end{array}$ \end{array}[/itex] Line 40: Line 40: Using spectral factorization, an equivalent problem is Using spectral factorization, an equivalent problem is - $+ [itex]\begin{array}{lll} - \begin{array}{lll} + \textrm{min} & |\textrm{diag}(gg^\texttt{H})|_\infty & \\ \textrm{min} & |\textrm{diag}(gg^\texttt{H})|_\infty & \\ - \textrm{subject to} & \frac{1}{\delta_1^2}\leq R(\omega)\leq\delta_1^2, & \omega\in[0,\omega_p]\\ + \textrm{subject \,\, to} & \frac{1}{\delta_1^2}\leq R(\omega)\leq\delta_1^2, & \omega\in[0,\omega_p]\\ & R(\omega)\leq\delta_2^2, & \omega\in[\omega_s,\pi]\\ & R(\omega)\leq\delta_2^2, & \omega\in[\omega_s,\pi]\\ & R(\omega)\geq0, & \omega\in[0,\pi]\\ & R(\omega)\geq0, & \omega\in[0,\pi]\\ Line 49: Line 48: & r(n) = frac{1}{n}\textrm{trace(\texttt{I})_{n-N}\,g & for n=1,\hdots,N\\ & r(n) = frac{1}{n}\textrm{trace(\texttt{I})_{n-N}\,g & for n=1,\hdots,N\\ & r(n) = frac{1}{n-N}\textrm{trace(\texttt{I})_{n-N}\,g & for n=N+1,\hdots,2N-1\\ & r(n) = frac{1}{n-N}\textrm{trace(\texttt{I})_{n-N}\,g & for n=N+1,\hdots,2N-1\\ - \end{array} + \end{array}$ - [/itex] +

## Revision as of 17:05, 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:  \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}  $


To minimize the maximum magnitude of $LaTeX: h$, the problem becomes $LaTeX: \begin{array}{lll} \textrm{min}& |h|_\infty & \\ \textrm{subjec t\,\, to} & \frac{1}{\delta_1}\leq|H(\omega)|\leq\delta_1, & \omega\in[0,\omega_p]\\ & |H(\omega)|\leq\delta_2, & \omega\in[\omega_s,\pi] \end{array}$

A new vector $LaTeX: g \in \texttt{C}^\texttt{N*N}$ is defined as concatenation of time-shifted versions of $LaTeX: h$, i.e. $LaTeX:  g = \left[   \begin{array}{c} h(n) \\ h(n-1) \\ \vdots \\ h(n-N) \\ \end{array} \right]$


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 $LaTeX: |\textrm{diag}(gg^\texttt{H})|_\infty$.

Using spectral factorization, an equivalent problem is $LaTeX: \begin{array}{lll} \textrm{min} & |\textrm{diag}(gg^\texttt{H})|_\infty & \\ \textrm{subject \,\, to} & \frac{1}{\delta_1^2}\leq R(\omega)\leq\delta_1^2, & \omega\in[0,\omega_p]\\ & R(\omega)\leq\delta_2^2, & \omega\in[\omega_s,\pi]\\ & R(\omega)\geq0, & \omega\in[0,\pi]\\ & \textrm{trace}(gg^\texttt{H}) = 1 &\\ & r(n) = frac{1}{n}\textrm{trace(\texttt{I})_{n-N}\,g & for n=1,\hdots,N\\ & r(n) = frac{1}{n-N}\textrm{trace(\texttt{I})_{n-N}\,g & for n=N+1,\hdots,2N-1\\ \end{array}$