Filter design by convex iteration
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Our goal here is to find a minimum peak magnitude <math>B_1(t)</math> for a given magnitude profile response. Since the application is for saturation, the phase of the profile response is not important but the passband should be flat and transition region should be sharp. | Our goal here is to find a minimum peak magnitude <math>B_1(t)</math> for a given magnitude profile response. Since the application is for saturation, the phase of the profile response is not important but the passband should be flat and transition region should be sharp. | ||
- | Plan: follow the above 4 steps. Except in step 2, instead of using Parks-McClellan algorithm to find <math>B_N(z)</math>, use optimization | + | Plan: follow the above 4 steps. Except in step 2, instead of using Parks-McClellan algorithm to find <math>B_N(z)</math>, use optimization. |
+ | -------------------------------------------- | ||
Revision as of 15:10, 22 October 2010
RF pulse design with small flip angle
Flip angle is the amount of devriation 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 transversed magnetization pulse profile and lognitudinal 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 choosen 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 opetator.
4. Once and are found, RF waveform can be computed by inverse SLR transform.
Note: upon applying this RF pulse, the resulting transerve magnetization is
where and is the initial magnetization.
If represents a rectangular profile, will also has 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, the phase of the profile response is not important but the passband should be flat and transition region should be sharp.
Plan: follow the above 4 steps. Except in step 2, instead of using Parks-McClellan algorithm to find , use optimization.
where denotes impulse response.
For a low pass filter, frequency domain specifications are:
To minimize peak magnitude of , the problem becomes
But this problem statement is nonconvex.
So instead,
is a positive semidefinite rank 1 matrix.
Summing along each of subdiagonals gives entries of the autocorrelation function of .
In particular, the main diagonal of holds squared entries of .
Minimizing is therefore equivalent to minimizing .
Define and define as a 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.
demonstration in Matlab
% CVX code by Almir Mutapcic in 2006. % Adapted in 2010 for impulse response peak-minimization by convex iteration. % % S.-P. Wu, S. Boyd, and L. Vandenberghe, % "FIR Filter Design via Spectral Factorization and Convex Optimization": % % minimize max |H(w)| for w in stopband % s.t. 1/delta <= |H(w)| <= delta for w in passband % % 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 of 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. % delta2 is specified stopband attenuation. 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 the passband wstop = 0.20*pi; % start of the stopband delta = 1; % maximum passband ripple in dB (+/- around 0 dB) delta2 = -20; % 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 = find((0 <= w) & (w <= wpass)); % passband Lp = 10^(-delta/20)*ones(length(ind),1); Up = 10^(+delta/20)*ones(length(ind),1); Ap = A(ind,:); % stopband (w_stop <= w) ind = find((wstop <= w) & (w <= pi)); % stopband Sp = 10^(delta2/20)*ones(length(ind),1); As = A(ind,:); % 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); % peak impulse response peak = 0.125; %******************************************************************** % optimization %******************************************************************** % formulate and solve the magnitude design problem iteration = 1; while 1 tic fprintf('\nMinimizing impulse response peak: iteration %d\n', iteration); cvx_quiet(true); % cvx_solver('sdpt3'); cvx_begin variable r(N,1); variable G(N,N) symmetric; minimize(trace(W'*G)); % passband constraints Ap*r >= Lp.^2; Ap*r <= Up.^2; %stopband constraint As*r <= Sp.^2; % nonnegative-real constraint A*r >= 0; % relate r to h r == B*vect(G); G == semidefinite(N); %minimize peak of h diag(G) <= peak^2; cvx_end toc %computer 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)'; % W = v(:,2:N)*diag(rand(N-1,1))*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)); if (rankG == 1) break end figure(1) % FIR impulse response h = G(:,1)/sqrt(G(1,1)); plot([0:N-1],h','o',[0:N-1],h','b:') xlabel('t'), ylabel('h(t)') pause(1) % check if problem was successfully solved disp(['Problem is ' cvx_status]) if ~strfind(cvx_status,'Solved') return 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 = G(:,1)/sqrt(G(1,1)); H = [exp(-j*kron(w,[0:N-1]))]*h; figure(2) % 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]) %compare impulse response designed by conventional method figure(3) h = spectral_fact(r); %from CVX distribution, Examples subdirectory plot([0:N-1],h','o',[0:N-1],h','b:') xlabel('t'), ylabel('h(t)') figure(1) % FIR impulse response h = G(:,1)/sqrt(G(1,1)); plot([0:N-1],h','o',[0:N-1],h','b:') xlabel('t'), ylabel('h(t)')