Filter design by convex iteration

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Line 63: Line 63:
\textrm{minimize} &\|\textrm{diag}(G)\|_\infty\\
\textrm{minimize} &\|\textrm{diag}(G)\|_\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(n) \triangleq \sum\limits_{t=1}^{N} h(t)h^\mathrm{H}(t-m) & \\
+
& r(m) \triangleq \sum\limits_{n=1}^{N} h(n)h^\mathrm{H}(t-m) & \\
& 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]\\

Revision as of 03:12, 24 August 2010

LaTeX: H(\omega) = h(0) + h(1)e^{-j\omega} + \cdots + h(N-1)e^{-j(N-1)\omega}  where  LaTeX: h(n)\in\mathbb{C}^N denotes impulse response.

For a low pass filter, 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 peak magnitude of LaTeX: h\, , the problem becomes

LaTeX: \begin{array}{cll} \textrm{minimize} &\|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}

But this problem statement is nonconvex.

So instead, a new vector

LaTeX: g \triangleq \left[ </p> <pre> \begin{array}{c} h(n) \\ h(n-1) \\ \vdots \\ h(n-N) \\ \end{array} \right]\in\mathbb{C}^{N^2} </pre> <p>

is defined by concatenation of time-shifted versions of LaTeX: h\, .

Then

LaTeX: G\triangleq gg^\mathrm{H}=\,\left[\begin{array}{*{20}c} h(n)h(n)^{\rm H} & h(n)h(n-1)^{\rm H} & h(n)h(n-2)^{\rm H} & \cdots & h(n)h(n-N)^{\rm H}\\ h(n-1)h(n)^{\rm H} & h(n-1)h(n-1)^{\rm H} & h(n-1)h(n-2)^{\rm H} & \cdots & h(n-1)h(n-N)^{\rm H}\\ h(n-2)h(n)^{\rm H} & h(n-2)h(n-1)^{\rm H} & h(n-2)h(n-2)^{\rm H} & \ddots & h(n-2)h(n-N)^{\rm H}\\ \vdots & \vdots & \ddots & \ddots & \vdots\\ h(n-N)h(n)^{\rm H} & h(n-N)h(n-1)^{\rm H} & h(n-N)h(n-2)^{\rm H} & \cdots & h(n-N)h(n-N)^{\rm H} \end{array}\right]\in\mathbb{C}^{N^2\times N^2}

is a positive semidefinite rank 1 matrix.

Summing along each of LaTeX: 2N-1\, subdiagonals gives entries of the autocorrelation function LaTeX: r\, of LaTeX: h\, .

In particular, the main diagonal of LaTeX: G\, holds squared entries of LaTeX: h\, .

Minimizing LaTeX: \|h\|_\infty is therefore equivalent to minimizing LaTeX: \|\textrm{diag}(G)\|_\infty .


Define LaTeX: I_n\triangleq\,...

By spectral factorization, LaTeX: R(\omega)=|H(\omega)|^2\, , an equivalent problem is:

LaTeX: \begin{array}{cll} \textrm{minimize} &\|\textrm{diag}(G)\|_\infty\\ \textrm{subject~to} &\frac{1}{\delta_1^2}\leq R(\omega)\leq\delta_1^2 &\omega\in[0,\omega_p]\\ & r(m) \triangleq \sum\limits_{n=1}^{N} h(n)h^\mathrm{H}(t-m) & \\ & R(\omega)\leq\delta_2^2 & \omega\in[\omega_s\,,\pi]\\ & R(\omega)\geq0 & \omega\in[0,\pi]\\ & R(\omega) = r(0)+\!\sum\limits_{t=1}^{N-1}2r(t)\cos(\omega t)\\ & r(n)=\frac{1}{n}\textrm{trace}(I_{n-N}G) &n=1\ldots N\\ & r(n)=\frac{1}{n-N}\textrm{trace}(I_{n-N}G) &n=N+1\ldots 2N-1\\ & G\succeq0\\ & \textrm{rank}(G) = 1 \end{array}

Excepting the rank constraint, this problem is convex.

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