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minimax optimization -凯发k8网页登录

this example uses:

this example shows how to solve a nonlinear filter design problem using a minimax optimization algorithm, fminimax, in optimization toolbox™. note that to run this example you must have the signal processing toolbox™ installed.

set finite precision parameters

consider an example for the design of finite precision filters. for this, you need to specify not only the filter design parameters such as the cut-off frequency and number of coefficients, but also how many bits are available since the design is in finite precision.

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

continuous design first

this is a continuous filter design; we use cheby1, but we could also use ellip, yulewalk or remez here: 

[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')

figure contains an axes object. the axes object with title frequency response using non-integer variables contains an object of type line.

x = [b1,a1];            % the design variables

set bounds for filter coefficients

we now set bounds on the maximum and minimum values: 

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

scale coefficients

set the biggest value equal to maxbin and scale other filter coefficients appropriately.

[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;

set optimization criteria

using optimoptions, adjust the termination criteria to reasonably high values to promote short running times. also turn on the display of results at each iteration:

options = optimoptions('fminimax', ...
    'steptolerance', 0.1, ...
    'optimalitytolerance', 1e-4,...
    'constrainttolerance', 1e-6, ...
    'display', 'iter');

minimize the absolute maximum values

we need to minimize absolute maximum values, so we set options.minabsmax to the number of frequency points:

if length(w) == 1
   options = optimoptions(options,'absolutemaxobjectivecount',w);
else
   options = optimoptions(options,'absolutemaxobjectivecount',length(w));
end

eliminate first value for optimization

discretize and eliminate first value and perform optimization by calling 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     6.123e-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     2.274e-10            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    -4.575e-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     2.097e-12            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.

check nearest integer values

see if nearby values produce a better filter.

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

frequency response comparisons

we first plot the frequency response of the filter and we compare it to a filter where the coefficients are just rounded up or down:

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')

figure contains 2 axes objects. axes object 1 with title optimized filter versus original contains 2 objects of type line. one or more of the lines displays its values using only markers axes object 2 with title rounded filter versus original contains 2 objects of type line. one or more of the lines displays its values using only markers

fig = gcf;
fig.nextplot = 'replace';

see also

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