Filter design by convex iteration

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RF pulse design with small flip angle

Flip angle is the amount of deviation one applies to net magnetization vector from its transverse axis. A LaTeX: 90^o pulse will tip the net magnetization vector to the longitudinal axis. If the desired flip angle LaTeX: \theta is small LaTeX: (\sin(\theta)\simeq\theta), then the time domain RF waveform and the pulse response can be approximated by Fourier Transform.

RF pulse design with large flip angle

For design RF pulses with large flip angles, a technique called Shinnar-Le Roux (SLR) transform is often used. SLR transform relates the desired pulse response to the design of FIR filters.

The goal of RF pulse design is: find the time domain waveform LaTeX:  B_1(t) that will produce the desired transverse magnetization pulse profile and longitudinal magnetization pulse profile. SLR transform relates the RF waveform to two complex polynomials LaTeX: A_N(z) and LaTeX: B_N(z).

i.e. LaTeX:  B_1(t)\begin{array}{c}\mbox{SLR}\\\Leftrightarrow \end{array}\, A_N(z),\, B_N(z)

Thus, the task of filter design becomes finding these two polynomials, which can be done via FIR filter design methods.

A typical design procedure involves the following steps:

1. Establish a set of design parameters: e.g. waveform duration, pulse bandwidth, passband and stopband ripples. These parameters are converted to their FIR filter design counterparts.

2. Use Parks-McClellan algorithm to come up with a waveform, LaTeX: B_N(z), where LaTeX: z=e^{\jmath\,\gamma\,G_x\,\Delta\,t}

3. LaTeX: A_N(z) and LaTeX: B_N(z) are related by LaTeX: |A_N(z)| = \sqrt{ 1-B_N(z)\,B_N^\ast\,(z)}. If LaTeX: A_N(z) is chosen to be a minimum-phase polynomial, then it is also an analytic signal. Analytic signals have the property that their log-magnitude and phase are a Hilbert transform pair. For the given LaTeX: B_N(z), the unique minimum-phase LaTeX: A_N(z) is LaTeX: A_N(z)=|A_N(z)|\,e^{\jmath\,\mathcal{H}(\mathrm{log}|A_N(z)|)}, where LaTeX: \mathcal{H} is the Hilbert transform operator.

4. Once LaTeX: A_N\,(z) and LaTeX: B_N\,(z) are found, RF waveform LaTeX:  B_1(t) can be computed by inverse SLR transform.

Note: upon applying this RF pulse, the resulting transverse magnetization is LaTeX: |M_{xy}(\omega)| = 2\sqrt{1-|B_N\,(\omega)|^2}\,|B_N\,(\omega)|\,M_o where LaTeX:  z=e^{\jmath\,\omega} and LaTeX: M_o is the initial magnetization. If LaTeX: |B_N\,(\omega)| represents a rectangular profile, LaTeX: |M_{xy}(\omega)| will also have a rectangular profile.

Minimum peak LaTeX: B_1(t) RF waveform for saturation pulse

Our goal here is to find a minimum peak magnitude LaTeX: B_1(t) for a given magnitude profile response.  Since the application is for saturation, phase of the profile response is not important. But the passband should be flat and transition region sharp.

Plan: Follow the four steps above except instead of using Parks-McClellan to find LaTeX: B_N(z), in step 2, use Optimization:

Minimum peak time-domain response by Convex Optimization

LaTeX: H(\omega) = h(0) + h(1)e^{-\jmath\,\omega} + ... + h(N-1)e^{-\jmath(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}{1+\delta_1}\leq|H(\omega)|\leq1+\delta_1\,, &\omega\in[0,\omega_p]\\
|H(\omega)|\leq\delta_2\,, &\omega\in[\omega_s\,,\pi]

To minimize peak amplitude of impulse response LaTeX: h\, , the problem becomes

LaTeX: \begin{array}{cll}
\textrm{minimize} &\|h\|_\infty\\
\textrm{subject~to} &\frac{1}{1+\delta_1}\leq|H(\omega)|\leq1+\delta_1\,, &\omega\in[0,\omega_p]\\
&|H(\omega)|\leq\delta_2\,, &\omega\in[\omega_s\,,\pi]

But this problem statement is nonconvex because LaTeX: \frac{1}{1+\delta_1}\leq|H(\omega)| specifies a nonconvex set of feasible LaTeX: h\,.

So instead,

LaTeX: G=h(n)h(n)^{\rm H} 
\in\mathbb{C}^{N\times N}

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\, where LaTeX: r\,= r_{\rm re}+\jmath\,r_{\rm im}\,\in\mathbb{C}^N  and LaTeX: r(n)=r(-n)^*\,.

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

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

Define LaTeX: I_0= I\, and define LaTeX: I_n\, as a square zero-matrix having vector LaTeX: \mathbf{1} along the LaTeX: n^{\rm{th}}\, superdiagonal when LaTeX: n\, is positive or LaTeX: \mathbf{1} along the LaTeX: n^{\rm{th}}\, subdiagonal when LaTeX: n\, is negative.

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

LaTeX: \begin{array}{cll}
\textrm{minimize}\limits_{G\,,\,r}&\|{\text diag}(G)\|_\infty\\
& R(\omega) = r_{\rm re}(0)+\!\sum\limits_{n=1}^{N-1}2(r_{\rm re}(n)\cos(\omega n)+r_{\rm im}(n)\sin(\omega n))\\
&\frac{1}{(1+\delta_1)^2}\leq R(\omega)\leq(1+\delta_1)^2 &\omega\in[-\omega_p,\omega_p]\\
& R(\omega)\leq\delta_2^2 & \pm\omega\in[\omega_s\,,\pi]\\
& R(\omega)\geq0 & \omega\in[-\pi,\pi]\\
& r(n)=\textrm{trace}(I_n^{\rm T}G) &n=0\ldots N-1\\
& G\succeq0\\
& {\text rank}(G) = 1

Excepting the rank constraint, this problem is convex.

The technique of Convex Iteration (to find direction vector LaTeX: W) from Dattorro's book is applied:

LaTeX: \begin{array}{cll}
\textrm{minimize}\limits_{G\,,\,r}&\langle G\,,W\rangle\\
& \max{\text diag}(G)\leq\kappa^2\\
& R(\omega) = r_{\rm re}(0)+\!\sum\limits_{n=1}^{N-1}2(r_{\rm re}(n)\cos(\omega n)+r_{\rm im}(n)\sin(\omega n))\\
&\frac{1}{(1+\delta_1)^2}\leq R(\omega)\leq(1+\delta_1)^2 &\omega\in[-\omega_p,\omega_p]\\
& R(\omega)\leq\delta_2^2 & \pm\omega\in[\omega_s\,,\pi]\\
& R(\omega)\geq0 & \omega\in[-\pi,\pi]\\
& r(n)=\textrm{trace}(I_n^{\rm T}G) &n=0\ldots N-1\\
& G\succeq0

where peak impulse response amplitude LaTeX: \kappa\, is a constant chosen by cut-and-try or bisection.

Matlab demonstration

% CVX code by Almir Mutapcic in 2006.  
% Adapted in 2010 for impulse response peak-minimization by convex iteration by Christine Law.
% "FIR Filter Design via Spectral Factorization and Convex Optimization"
% by S.-P. Wu, S. Boyd, and L. Vandenberghe
% Designs an FIR lowpass filter using spectral factorization method with
% constraint on maximum passband ripple and stopband attenuation:
%   minimize   max |H(w)|                      for w in stopband
%       s.t.   1/delta <= |H(w)| <= delta      for w in passband
% We change variables via spectral factorization method and get:
%   minimize   max R(w)                        for w in stopband
%       s.t.   (1/delta)^2 <= R(w) <= delta^2  for w in passband
%              R(w) >= 0                       for all w
% where R(w) is squared magnitude frequency response
% (and Fourier transform of autocorrelation coefficients r).
% Variables are coeffients r and G = hh' where h is impulse response.  
% delta is allowed passband ripple.
% This is a convex problem (can be formulated as an SDP after sampling).
clear all, clc, close all, fclose('all');
% filter specs (for a low-pass filter)
% number of FIR coefficients (including zeroth)
N = 32;
wpass = 0.12*pi;   % end of passband
wstop = 0.20*pi;   % start of stopband
delta0_wpass = 0.125; 
delta0_wstop = 0.125; 
delta  = 20*log10(1 + delta0_wpass)  % maximum passband ripple in dB (+/- around 0 dB)
delta2 = 20*log10(delta0_wstop)      % stopband attenuation desired in dB

% optimization parameters
% rule-of-thumb discretization (from Cheney's Approximation Theory)
m = 15*N;
w = linspace(0,pi,m)';  % omega

% A is the matrix used to compute the power spectrum
% A(w,:) = [1 2*cos(w) 2*cos(2*w) ... 2*cos(N*w)]
A = [ones(m,1) 2*cos(kron(w,[1:N-1]))];

% passband 0 <= w <= w_pass
ind_p = find((0 <= w) & (w <= wpass));    % passband
Lp  = 10^(-delta/20);
Up  = 10^(+delta/20);
Ap  = A(ind_p,:);

% stopband (w_stop <= w)
ind_s = find((wstop <= w) & (w <= pi));   % stopband
Sp  = 10^(delta2/20);
As  = A(ind_s,:);

%remove redundant contraints
ind_nr = setdiff(1:m,ind_p);   % fullband less passband
Anr = A(ind_nr,:);

% make I matrices
B = zeros(N,N^2);
for i=0:N-1
   C = zeros(N,N);
   C = spdiags(ones(N,1),i,C);
   B(i+1,:) = vect(C)';

%initial direction vector
W = eye(N);

% optimization
convergence = [];
iteration = 1;
while 1
   fprintf('\niteration %d\n', iteration);
      variable r(N,1);
      variable G(N,N) symmetric;
      %peak impulse response by cut and try
      max(diag(G)) <= 0.1053^2;
      % passband constraints
      Ap*r >= Lp.^2;
      Ap*r <= Up.^2;
      %stopband constraint
      As*r <= Sp.^2;
      % nonnegative-real constraint
      Anr*r >= 0;
      % relate r to h
      r == B*vect(G);
      G == semidefinite(N);
   %compute new direction vector
   [v,d,q] = svd(G);
   f = diag(d); fprintf('first few eigenvalues of G:\n%f\n%f\n%f\n%f\n%f\n%f\n%f\n', f(1:7));
   W = v(:,2:N)*v(:,2:N)';
   rankG = sum(diag(d) > max(diag(d))*1e-5);
   fprintf('rank(G)=%d,  trace(W*G)=%f\n', rankG, trace(G*W)); 
   % FIR impulse response
   h = v(:,1)*sqrt(d(1,1));
   xlabel('t'), ylabel('h(t)')

   % monitor convergence to 0
   convergence = [convergence trace(G*W)];
   if iteration > 1
      set(gcf,'position',[70 200 256 256])
   % check if problem was successfully solved
   disp(['Problem is ' cvx_status])
   if (rankG == 1)
   if ~strfind(cvx_status,'Solved')
       fprintf(2,'Excuse me.\n')

   iteration = iteration + 1;

% compute the min attenuation in the stopband (convert to original vars)
Ustop = delta2;
fprintf(1,'Min attenuation in the stopband is %3.2f dB.\n',Ustop);

% plotting routines
% frequency response of the designed filter, where j = sqrt(-1)
H = [exp(-j*kron(w,[0:N-1]))]*h;

% magnitude
plot(w,20*log10(abs(H)), ...
   [0 wpass],[delta delta],'r--', ...
   [0 wpass],[-delta -delta],'r--', ...
   [wstop pi],[Ustop Ustop],'r--')
ylabel('mag H(w) in dB')
axis([0 pi -50 5])
title(sprintf('N=%d, w_p(pi)=%3.2f, w_s(pi)=%3.2f, delta=%3.2f', N, wpass/pi, wstop/pi, delta));

%compare impulse response designed by conventional method
h_sp = spectral_fact(r);  %from CVX distribution, Examples subdirectory
hold on;
xlabel('t'), ylabel('h(t)'); grid
title(sprintf('h_{max} conventional=%3.4f, h_{max} optimal=%3.4f',max(abs(h_sp)),max(abs(h))));
set(gcf,'Outerposition',[300 300 256*4 256*2])
% FIR impulse response
xlabel('t'), ylabel('h(t)')


% Spectral factorization using Kolmogorov 1939 approach.
% (code follows pp. 232-233, Signal Analysis, by A. Papoulis)
% Computes the minimum-phase impulse response which satisfies
% given auto-correlation.
% Input:
%   r: top-half of the auto-correlation coefficients
%      starts from 0th element to end of the auto-corelation
%      should be passed in as a column vector
% Output
%   h: impulse response that gives the desired auto-correlation

function h = spectral_fact(r)

% length of the impulse response sequence
n = length(r);

% over-sampling factor
mult_factor = 100;        % should have mult_factor*(n) >> n 
m = mult_factor*n;

% computation method:
% H(exp(jTw)) = alpha(w) + j*phi(w)
% where alpha(w) = 1/2*ln(R(w)) and phi(w) = Hilbert_trans(alpha(w))

% compute 1/2*ln(R(w))
w = 2*pi*[0:m-1]/m;
R = [ ones(m,1) 2*cos(kron(w',[1:n-1])) ]*r;
alpha = 1/2*log(R);

% find the Hilbert transform 
alphatmp = fft(alpha);
alphatmp(floor(m/2)+1:m) = -alphatmp(floor(m/2)+1:m);
alphatmp(1) = 0;
alphatmp(floor(m/2)+1) = 0;
phi = real(ifft(j*alphatmp));

% now retrieve the original sampling 
index  = find(rem([0:m-1],mult_factor)==0);
alpha1 = alpha(index);
phi1   = phi(index);

% compute the impulse response (inverse Fourier transform)
h = real(ifft(exp(alpha1+j*phi1),n));
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