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En este ejemplo se muestra cómo resolver un problema de diseño de filtro no lineal mediante un algoritmo de optimización MINIMAX, en Optimization Toolbox™.fminimax
Tenga en cuenta que para ejecutar este ejemplo debe tener la™ de caja de herramientas de procesamiento de señal instalada.
Considere un ejemplo para el diseño de filtros de precisión finita. Para esto, debe especificar no sólo los parámetros de diseño de filtro, como la frecuencia de corte y el número de coeficientes, sino también cuántos bits están disponibles ya que el diseño está en precisión finita.
nbits = 8; % How many bits have we to realize filter maxbin = 2^nbits-1; % Maximum number expressable in nbits bits n = 4; % Number of coefficients (order of filter plus 1) Wn = 0.2; % Cutoff frequency for filter Rp = 1.5; % Decibels of ripple in the passband w = 128; % Number of frequency points to take
Este es un diseño de filtro continuo; que usamos, pero también podríamos usar, o aquí:cheby1
ellip
yulewalk
remez
[b1,a1] = cheby1(n-1,Rp,Wn); [h,w] = freqz(b1,a1,w); % Frequency response h = abs(h); % Magnitude response plot(w, h) title('Frequency response using non-integer variables')
x = [b1,a1]; % The design variables
Ahora establecemos límites en los valores máximos y mínimos:
if (any(x < 0)) % If there are negative coefficients - must save room to use a sign bit % and therefore reduce maxbin maxbin = floor(maxbin/2); vlb = -maxbin * ones(1, 2*n)-1; vub = maxbin * ones(1, 2*n); else % otherwise, all positive vlb = zeros(1,2*n); vub = maxbin * ones(1, 2*n); end
Establezca el valor más grande igual a maxbin y escale otros coeficientes de filtro apropiadamente.
[m, mix] = max(abs(x)); factor = maxbin/m; x = factor * x; % Rescale other filter coefficients xorig = x; xmask = 1:2*n; % Remove the biggest value and the element that controls D.C. Gain % from the list of values that can be changed. xmask(mix) = []; nx = 2*n;
Utilizando, ajuste los criterios de terminación a valores razonablemente altos para promover tiempos de ejecución cortos.optimoptions
También Active la visualización de los resultados en cada iteración:
options = optimoptions('fminimax', ... 'StepTolerance', 0.1, ... 'OptimalityTolerance', 1e-4,... 'ConstraintTolerance', 1e-6, ... 'Display', 'iter');
Necesitamos minimizar los valores máximos absolutos, así que configuramos las opciones. MinAbsMax al número de puntos de frecuencia:
if length(w) == 1 options = optimoptions(options,'AbsoluteMaxObjectiveCount',w); else options = optimoptions(options,'AbsoluteMaxObjectiveCount',length(w)); end
Discretizar y eliminar el primer valor y realizar la optimización mediante una llamada a FMINIMAX:
[x, xmask] = elimone(x, xmask, h, w, n, maxbin)
x = 1×8
0.5441 1.6323 1.6323 0.5441 57.1653 -127.0000 108.0000 -33.8267
xmask = 1×6
1 2 3 4 5 8
niters = length(xmask); disp(sprintf('Performing %g stages of optimization.\n\n', niters));
Performing 6 stages of optimization.
for m = 1:niters fun = @(xfree)filtobj(xfree,x,xmask,n,h,maxbin); % objective confun = @(xfree)filtcon(xfree,x,xmask,n,h,maxbin); % nonlinear constraint disp(sprintf('Stage: %g \n', m)); x(xmask) = fminimax(fun,x(xmask),[],[],[],[],vlb(xmask),vub(xmask),... confun,options); [x, xmask] = elimone(x, xmask, h, w, n, maxbin); end
Stage: 1 Objective Max Line search Directional Iter F-count value constraint steplength derivative Procedure 0 8 0 0.00329174 1 17 0.0001845 3.34e-07 1 0.0143 Local minimum possible. Constraints satisfied. fminimax stopped because the size of the current search direction is less than twice the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance. Stage: 2 Objective Max Line search Directional Iter F-count value constraint steplength derivative Procedure 0 7 0 0.0414182 1 15 0.01649 0.0002558 1 0.261 2 23 0.01544 6.126e-07 1 -0.0282 Hessian modified Local minimum possible. Constraints satisfied. fminimax stopped because the size of the current search direction is less than twice the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance. Stage: 3 Objective Max Line search Directional Iter F-count value constraint steplength derivative Procedure 0 6 0 0.0716961 1 13 0.05943 -1.765e-11 1 0.776 Local minimum possible. Constraints satisfied. fminimax stopped because the size of the current search direction is less than twice the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance. Stage: 4 Objective Max Line search Directional Iter F-count value constraint steplength derivative Procedure 0 5 0 0.129938 1 11 0.04278 1.768e-10 1 0.183 Local minimum possible. Constraints satisfied. fminimax stopped because the size of the current search direction is less than twice the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance. Stage: 5 Objective Max Line search Directional Iter F-count value constraint steplength derivative Procedure 0 4 0 0.0901749 1 9 0.03867 -1.765e-10 1 0.256 Local minimum possible. Constraints satisfied. fminimax stopped because the size of the current search direction is less than twice the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance. Stage: 6 Objective Max Line search Directional Iter F-count value constraint steplength derivative Procedure 0 3 0 0.11283 1 7 0.05033 3.192e-16 1 0.197 Local minimum possible. Constraints satisfied. fminimax stopped because the size of the current search direction is less than twice the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance.
Vea si los valores cercanos producen un mejor filtro.
xold = x; xmask = 1:2*n; xmask([n+1, mix]) = []; x = x + 0.5; for i = xmask [x, xmask] = elimone(x, xmask, h, w, n, maxbin); end xmask = 1:2*n; xmask([n+1, mix]) = []; x = x - 0.5; for i = xmask [x, xmask] = elimone(x, xmask, h, w, n, maxbin); end if any(abs(x) > maxbin) x = xold; end
Primero trazamos la respuesta de frecuencia del filtro y lo comparamos con un filtro donde los coeficientes se redondean hacia arriba o hacia abajo:
subplot(211) bo = x(1:n); ao = x(n+1:2*n); h2 = abs(freqz(bo,ao,128)); plot(w,h,w,h2,'o') title('Optimized filter versus original') xround = round(xorig)
xround = 1×8
1 2 2 1 57 -127 108 -34
b = xround(1:n); a = xround(n+1:2*n); h3 = abs(freqz(b,a,128)); subplot(212) plot(w,h,w,h3,'+') title('Rounded filter versus original')
fig = gcf; fig.NextPlot = 'replace';