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- %DFILDEMO Nonlinear filter design problem using discrete optimization.
- clf
- echo on
- % Example script-file for the design of finite precision filters
- nbits = 8; % How many bits have we to realize filter
- maxbin = 2^nbits-1; % Maximum number
- n=4; % Number of coefficients (Order of filter plus 1)
- Wn = 1.5; % Cut-off frequency for filter
- Rp = 0.2; % Decibels of ripple in the passband
- w = 128; % Number of frequency points to take
-
- % Continuous filter design
- % (could use any of cheby, ellip, yulewalk or remez here.)
- [b1,a1]=cheby1(n-1,Wn, Rp);
-
-
- pause % Strike any key to continue
-
- [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
-
- % Set bounds on the maximum and minimum values.
- if (any(x < 0))
- % If there are negative coefficients - must use sign bit
- maxbin = floor(maxbin/2);
- vlb = -maxbin * ones(1, 2*n);
- vub = maxbin * ones(1, 2*n);
- else
- % otherwise, all positive
- vlb = zeros(1,2*n);
- vub = maxbin * ones(1, 2*n);
- end
-
- % Set the biggest value equal to maxbin.
- [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;
-
- pause % Strike any key to continue
-
- % Set termination criteria to reasonably high values
- % to promote fast convergence.
-
- options = 1;
- options(2) = 0.1;
- options(3) = 1e-4;
- options(4) = 1e-6;
-
- % Need to minimize absolute maximum values so set options(15) to the
- % number of frequency points.
-
- if length(w) == 1
- options(15) = w;
- else
- options(15) = length(w);
- end
-
- pause % Strike any key to continue
-
- % Discretize and eliminate first value
- [x, xmask] = elimone(x, xmask, h, w, n, maxbin)
-
- pause % Strike any key to continue
-
- niters = length(xmask);
-
- pause % Strike any key to continue
-
- disp(sprintf('Performing %g stages of optimization.\n\n', niters));
-
- % This may take a long time ! (Ctrl-C to abort)
-
- for m = 1:niters
- disp(sprintf('Stage: %g \n', m));
- % Least squares solution:
- % x(xmask) = constr('filtfun2', x(xmask), options, vlb(xmask), vub(xmask), [], x, xmask, n, h, maxbin);
-
-
- x(xmask) = minimax('filtfun', x(xmask), options, vlb(xmask), vub(xmask), [], x, xmask, n, h, maxbin);
- [x, xmask] = elimone(x, xmask, h, w, n, maxbin);
- end
-
- % Check nearest integer values for better filter
- pause % Strike any key to continue
-
- 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 of filter
- pause % Strike any key to continue
- 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')
-
- % Compare it to a filter where the coefficients are just rounded
- % up or down.
- xround =round(xorig)
- 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')
- set(gcf,'NextPlot','replace')
-