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 pulse will tip the net magnetization vector to the longitudinal axis. If the desired flip angle
is small
, 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 that will produce the desired transverse magnetization pulse profile and longitudinal magnetization pulse profile. SLR transform relates the RF waveform to two complex polynomials
and
.
i.e.
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, , where
3.
and
are related by
.
If
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
, the unique minimum-phase
is
, where
is the Hilbert transform operator.
4.
Once and
are found, RF waveform
can be computed by inverse SLR transform.
Note: upon applying this RF pulse, the resulting transverse magnetization is
where
and
is the initial magnetization.
If
represents a rectangular profile,
will also have a rectangular profile.
Minimum peak 
RF waveform for saturation pulse
Our goal here is to find a minimum peak magnitude 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 , in step 2, use Optimization:
Minimum peak time-domain response by Convex Optimization
where
denotes impulse response.
For a low pass filter, frequency domain specifications are:
To minimize peak amplitude of impulse response , the problem becomes
But this problem statement is nonconvex because specifies a nonconvex set of feasible
.
So instead,
is a positive semidefinite rank 1 matrix.
Summing along each of subdiagonals gives entries of the autocorrelation function
of
where
and
.
In particular, the main diagonal of holds squared entries of
.
Minimizing is therefore equivalent to minimizing
.
Define and define
as a square zero-matrix having vector
along the
superdiagonal when
is positive or
along the
subdiagonal when
is negative.
By spectral factorization, ,
an equivalent problem is:
Excepting the rank constraint, this problem is convex.
The technique of Convex Iteration (to find direction vector ) from Dattorro's book is applied:
where peak impulse response amplitude 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');
rand('twister',sum(100*clock));
randn('state',sum(100*clock));
%*********************************************************************
% 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)';
end
%initial direction vector
W = eye(N);
%********************************************************************
% optimization
%********************************************************************
convergence = [];
iteration = 1;
while 1
tic
fprintf('\niteration %d\n', iteration);
cvx_quiet(true);
cvx_solver('sdpt3');
cvx_precision('best');
cvx_begin
variable r(N,1);
variable G(N,N) symmetric;
minimize(trace(W'*G));
%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);
cvx_end
toc
%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));
figure(1)
% FIR impulse response
h = v(:,1)*sqrt(d(1,1));
plot([0:N-1],h','ob:')
xlabel('t'), ylabel('h(t)')
% monitor convergence to 0
convergence = [convergence trace(G*W)];
if iteration > 1
figure(4)
plot(convergence)
set(gcf,'position',[70 200 256 256])
end
pause(1)
% check if problem was successfully solved
disp(['Problem is ' cvx_status])
if (rankG == 1)
break
end
if ~strfind(cvx_status,'Solved')
fprintf(2,'Excuse me.\n')
end
iteration = iteration + 1;
end
% 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;
figure(2);
subplot(121)
% magnitude
plot(w,20*log10(abs(H)), ...
[0 wpass],[delta delta],'r--', ...
[0 wpass],[-delta -delta],'r--', ...
[wstop pi],[Ustop Ustop],'r--')
xlabel('w')
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
subplot(122)
h_sp = spectral_fact(r); %from CVX distribution, Examples subdirectory
plot([0:N-1],h_sp','+r--');
hold on;
plot([0:N-1],h(end:-1:1)','ob:');
legend('conventional','optimal');
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])
figure(1)
% FIR impulse response
plot([0:N-1],h','ob:');
xlabel('t'), ylabel('h(t)')
spectral_fact()
% 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));