Filter design by convex iteration

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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:

\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]

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]

But this problem statement is nonconvex.

So instead, a new vector

g \triangleq   \left[
<pre>              \begin{array}{c}
                h(n) \\
                h(n-1) \\
                \vdots \\
                h(n-N) \\

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


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}
& R(\omega) = r(0)+\!\sum\limits_{n=1}^{N-1}2r(n)\cos(\omega n)\\
&\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]\\
& 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

Excepting the rank constraint, this problem is convex.

N = 32;
delta1 = 1.01;
delta2 = 0.01;
w = linspace(0,pi,N);

  variable h(N,1);
  variable r(2*N-1,1);
  variable g(N^2,1);
  variable G(N^2,N^2);
  variable R(2*N-1,1);
  for I=1:N
    g((I-1)*N+1:I*N,1) = circshift(h,[0,I-1]);
  G = g*g';
%   G >= 0;
  for I=1:N
      r(I) = 1/I*trace(diag(ones(1,N),I-N)*G);
  for I=N+1:2*N-1
      r(I) = 1/(I-N)*trace(diag(ones(1,N),I-N)*G);
  for I=1:N
      temp = 0;
      for J=1:N-1
          temp = temp + 2*r(J)*cos(w(J)*J);
      R(I) = r(1) + temp;
  for I=1:N/2
      1/delta1^2 < R(I);
      R(I) < delta1^2;
  for I=N/2+1:N
      R(I) < delta2^2;

  R > 0;
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