# Minimize order of a linear phase lowpass FIR filter

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    % "Filter design" lecture notes (EE364) by S. Boyd
    % (figures are generated)
    %
    % Designs a linear phase FIR lowpass filter such that it:
    % - minimizes the filter order
    % - has a constraint on the maximum passband ripple
    % - has a constraint on the maximum stopband attenuation
    %
    % This is a quasiconvex problem and can be solved using a bisection.
    %
    %   minimize   filter order n
    %       s.t.   1/delta <= H(w) <= delta     for w in the passband
    %              |H(w)| <= atten_level        for w in the stopband
    %
    % where H is the frequency response function and variable is
    % the filter impulse response h (and its order/length).
    % Data is delta (max passband ripple) and atten_level (max stopband
    % attenuation level).
    %
    % Written for CVX by Almir Mutapcic 02/02/06

    %********************************************************************
    % user's filter specifications
    %********************************************************************
    % filter order that is used to start the bisection (has to be feasible)
    max_order = 20;

    wpass = 0.12*pi;        % passband cutoff freq (in radians)
    wstop = 0.24*pi;        % stopband start freq (in radians)
    delta = 0.5;            % max (+/-) passband ripple in dB
    atten = -35;      % stopband attenuation level in dB

    %********************************************************************
    % create optimization parameters
    %********************************************************************
    m = 30*max_order; % freq samples (rule-of-thumb)
    w = linspace(0,pi,m);

    %*********************************************************************
    % use bisection algorithm to solve the problem
    %*********************************************************************

    n_bot = 1;
    n_top = max_order;
    n = Inf;

    disp('Rememeber that we are only considering filters with linear phase, i.e.,')
    disp('filters that are symmetric around their midpoint and have order 2*n+1.')
    disp(' ')

    while( n_top - n_bot > 1)
        % try to find a feasible design for given specs
        n_cur = ceil( (n_top + n_bot)/2 );

        % create optimization matrices (this is cosine matrix)
        A = [ones(m,1) 2*cos(kron(w',[1:n_cur]))];

        % passband 0 <= w <= w_pass
        ind = find((0 <= w) & (w <= wpass));    % passband
        Ap  = A(ind,:);

        % transition band is not constrained (w_pass <= w <= w_stop)

        % stopband (w_stop <= w)
        ind = find((wstop <= w) & (w <= pi));   % stopband
        As  = A(ind,:);

        ptop = 10^(delta/20);

        % This is the feasibility problem:
        % cvx_begin
        %      variable h_cur(n_cur+1,1);
        %      10^(-delta/20) <= Ap * h_cur <=  10^(+delta/20);
        %      abs( As * h_cur ) <= +10^(+atten/20);
        % cvx_end
        % Unfortunately, the solvers often struggle with this formulation. For
        % this model, there is a logical optimization problem: minimize the
        % stopband attenuation. If the minimum attenuation is below the target,
        % then we know the original problem is feasible.
        cvx_begin quiet
             variable h_cur(n_cur+1,1);
             minimize( max( abs( As * h_cur ) ) );
             10^(-delta/20) <= Ap * h_cur <=  10^(+delta/20);
        cvx_end

        % bisection
        if isnan( cvx_optval ),
            fprintf( 1, 'Solver failed for n = %d taps, assuming infeasible\n', n_cur );
            n_bot = n_cur;
        elseif cvx_optval <= 10^(atten/20),
            fprintf(1,'Problem is feasible for n = %d taps\n',n_cur);
            n_top = n_cur;
            if n > n_cur,
                n = n_cur;
                h = h_cur;
            end
        else
            fprintf(1,'Problem not feasible for n = %d taps\n',n_cur);
            n_bot = n_cur;
        end
    end

    h = [ h(end:-1:2); h ];
    fprintf(1,'\nOptimum number of filter taps for given specs is 2n+1 = %d.\n', length(h));

    %********************************************************************
    % plots
    %********************************************************************
    figure(1)
    % FIR impulse response
    plot([-n:n],h','o',[-n:n],h','b:')
    xlabel('t'), ylabel('h(t)')

    figure(2)
    % frequency response
    H = exp(-j*kron(w',[0:2*n]))*h;
    % magnitude
    subplot(2,1,1)
    plot(w,20*log10(abs(H)),...
        [wstop pi],[atten atten],'r--',...
        [0 wpass],[delta delta],'r--',...
        [0 wpass],[-delta -delta],'r--');
    axis([0,pi,-50,10])
    xlabel('w'), ylabel('mag H(w) in dB')
    % phase
    subplot(2,1,2)
    plot(w,angle(H))
    axis([0,pi,-pi,pi])
    xlabel('w'), ylabel('phase H(w)')

    Rememeber that we are only considering filters with linear phase, i.e.,
    filters that are symmetric around their midpoint and have order 2*n+1.

    Problem not feasible for n = 11 taps
    Problem is feasible for n = 16 taps
    Problem is feasible for n = 14 taps
    Problem is feasible for n = 13 taps
    Problem is feasible for n = 12 taps

    Optimum number of filter taps for given specs is 2n+1 = 25.