% "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 = 1;              % max (+/-) passband ripple in dB
atten_level = -30;      % 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;

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,:);

  % formulate and solve the feasibility linear-phase lp filter design
  cvx_begin quiet
    variable h_cur(n_cur+1,1);
    % feasibility problem
    % passband bounds
    Ap*h_cur <= 10^(delta/20);
    Ap*h_cur >= 10^(-delta/20);
    % stopband bounds
    abs( As*h_cur ) <= 10^(atten_level/20);
  cvx_end

  % bisection
  if strfind(cvx_status,'Solved') % feasible
    fprintf(1,'Problem is feasible for n = %d taps\n',n_cur);
    n_top = n_cur;
    % construct the full impulse response
    h = [flipud(h_cur(2:end)); h_cur];
  else % not feasible
    fprintf(1,'Problem not feasible for n = %d taps\n',n_cur);
    n_bot = n_cur;
  end
end

n = n_top;
fprintf(1,'\nOptimum number of filter taps for given specs is 2n+1 = %d.\n',...
           2*n+1);

%********************************************************************
% plots
%********************************************************************
figure(1)
% FIR impulse response
plot([0:2*n],h','o',[0:2*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_level atten_level],'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 is feasible for n = 11 taps
Problem not feasible for n = 6 taps
Problem not feasible for n = 9 taps
Problem is feasible for n = 10 taps

Optimum number of filter taps for given specs is 2n+1 = 21.