Minimize stopband ripple of a linear phase lowpass FIR filter
n = 10;
wpass = 0.12*pi;
wstop = 0.24*pi;
atten_level = -30;
N = 30*n+1;
w = linspace(0,pi,N);
A = [ones(N,1) 2*cos(kron(w',[1:n]))];
ind = find((0 <= w) & (w <= wpass));
Ap = A(ind,:);
ind = find((wstop <= w) & (w <= pi));
Us = 10^(atten_level/20)*ones(length(ind),1);
As = A(ind,:);
cvx_begin
variable delta
variable h(n+1,1);
minimize( delta )
subject to
Ap*h <= delta;
inv_pos(Ap*h) <= delta;
abs( As*h ) <= Us;
cvx_end
disp(['Problem is ' cvx_status])
if ~strfind(cvx_status,'Solved')
return
else
h = [flipud(h(2:end)); h];
fprintf(1,'The optimal minimum passband ripple is %4.3f dB.\n\n',...
20*log10(delta));
end
figure(1)
plot([0:2*n],h','o',[0:2*n],h','b:')
xlabel('t'), ylabel('h(t)')
figure(2)
H = exp(-j*kron(w',[0:2*n]))*h;
subplot(2,1,1)
plot(w,20*log10(abs(H)),[wstop pi],[atten_level atten_level],'r--');
axis([0,pi,-40,10])
xlabel('w'), ylabel('mag H(w) in dB')
subplot(2,1,2)
plot(w,angle(H))
axis([0,pi,-pi,pi])
xlabel('w'), ylabel('phase H(w)')
Calling sedumi: 606 variables, 12 equality constraints
For improved efficiency, sedumi is solving the dual problem.
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SeDuMi 1.21 by AdvOL, 2005-2008 and Jos F. Sturm, 1998-2003.
Alg = 2: xz-corrector, Adaptive Step-Differentiation, theta = 0.250, beta = 0.500
eqs m = 12, order n = 570, dim = 644, blocks = 267
nnz(A) = 3407 + 0, nnz(ADA) = 144, nnz(L) = 78
it : b*y gap delta rate t/tP* t/tD* feas cg cg prec
0 : 9.09E+00 0.000
1 : -3.18E+00 2.60E+00 0.000 0.2864 0.9000 0.9000 0.10 1 1 4.4E+01
2 : -1.57E+00 1.21E+00 0.000 0.4644 0.9000 0.9000 4.50 1 1 6.0E+00
3 : -1.09E+00 6.92E-01 0.000 0.5728 0.9000 0.9000 6.37 1 1 1.0E+00
4 : -1.05E+00 2.54E-01 0.000 0.3670 0.9000 0.9000 1.90 1 1 2.8E-01
5 : -1.05E+00 1.08E-01 0.000 0.4268 0.9000 0.9000 1.37 1 1 1.1E-01
6 : -1.05E+00 2.66E-02 0.000 0.2455 0.9000 0.9000 1.16 1 1 2.5E-02
7 : -1.05E+00 9.96E-03 0.000 0.3742 0.9000 0.9000 1.04 1 1 9.2E-03
8 : -1.05E+00 3.73E-03 0.000 0.3741 0.9030 0.9000 1.01 1 1 3.4E-03
9 : -1.05E+00 1.38E-03 0.000 0.3691 0.9023 0.9000 1.00 1 1 1.3E-03
10 : -1.05E+00 4.82E-04 0.000 0.3502 0.9008 0.9000 1.00 1 1 4.4E-04
11 : -1.05E+00 1.38E-04 0.000 0.2873 0.9047 0.9000 1.00 1 1 1.3E-04
12 : -1.05E+00 2.75E-05 0.000 0.1988 0.9125 0.9000 1.00 1 1 2.5E-05
13 : -1.05E+00 5.26E-06 0.000 0.1913 0.9058 0.9000 1.00 1 1 4.9E-06
14 : -1.05E+00 5.72E-07 0.000 0.1087 0.9086 0.9000 1.00 1 1 5.8E-07
15 : -1.05E+00 3.82E-09 0.000 0.0067 0.9990 0.9990 1.00 1 1 4.0E-09
iter seconds digits c*x b*y
15 0.1 9.3 -1.0515780059e+00 -1.0515780064e+00
|Ax-b| = 3.7e-09, [Ay-c]_+ = 4.9E-10, |x|= 1.1e+00, |y|= 1.1e+00
Detailed timing (sec)
Pre IPM Post
1.000E-02 1.100E-01 0.000E+00
Max-norms: ||b||=1, ||c|| = 2,
Cholesky |add|=0, |skip| = 0, ||L.L|| = 6.84906.
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Status: Solved
Optimal value (cvx_optval): +1.05158
Problem is Solved
The optimal minimum passband ripple is 0.437 dB.