n = 10;
wpass = 0.12*pi;
wstop = 0.24*pi;
max_pass_ripple = 1;
N = 30*n;
w = linspace(0,pi,N);
A = [ones(N,1) 2*cos(kron(w',[1:n]))];
ind = find((0 <= w) & (w <= wpass));
Lp = 10^(-max_pass_ripple/20)*ones(length(ind),1);
Up = 10^(max_pass_ripple/20)*ones(length(ind),1);
Ap = A(ind,:);
ind = find((wstop <= w) & (w <= pi));
As = A(ind,:);
cvx_begin
variable h(n+1,1);
minimize( max( abs( As*h ) ) )
subject to
Lp <= Ap*h;
Ap*h <= Up;
cvx_end
disp(['Problem is ' cvx_status])
if ~strfind(cvx_status,'Solved')
return
else
fprintf(1,'The minimum attenuation in the stopband is %3.2f dB.\n\n',...
20*log10(cvx_optval));
h = [flipud(h(2:end)); h];
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)),...
[0 wpass],[max_pass_ripple max_pass_ripple],'r--',...
[0 wpass],[-max_pass_ripple -max_pass_ripple],'r--');
axis([0,pi,-50,10])
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: 756 variables, 240 equality constraints
For improved efficiency, sedumi is solving the dual problem.
------------------------------------------------------------
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 = 240, order n = 757, dim = 757, blocks = 1
nnz(A) = 6948 + 0, nnz(ADA) = 5844, nnz(L) = 3042
it : b*y gap delta rate t/tP* t/tD* feas cg cg prec
0 : 2.72E+02 0.000
1 : -2.64E-01 1.54E+02 0.000 0.5670 0.9000 0.9000 8.59 1 1 7.7E+01
2 : -1.41E-01 6.07E+01 0.000 0.3934 0.9000 0.9000 5.26 1 1 9.9E+00
3 : -4.93E-02 3.86E+01 0.000 0.6360 0.9000 0.9000 3.88 1 1 2.9E+00
4 : -3.46E-02 2.87E+01 0.000 0.7427 0.9000 0.9000 2.55 1 1 1.7E+00
5 : -2.32E-02 1.56E+01 0.000 0.5430 0.9000 0.9000 1.97 1 1 7.5E-01
6 : -2.01E-02 7.84E+00 0.000 0.5038 0.9000 0.9000 1.37 1 1 3.5E-01
7 : -1.79E-02 2.57E+00 0.000 0.3285 0.9000 0.9000 1.17 1 1 1.1E-01
8 : -1.75E-02 8.33E-01 0.000 0.3235 0.9000 0.9000 1.04 1 1 3.5E-02
9 : -1.75E-02 1.83E-01 0.000 0.2192 0.9080 0.9000 1.01 1 1 8.1E-03
10 : -1.75E-02 3.45E-02 0.000 0.1892 0.9055 0.9000 1.00 1 1 1.6E-03
11 : -1.75E-02 4.21E-03 0.000 0.1219 0.9088 0.9000 1.00 1 1 2.0E-04
12 : -1.75E-02 2.19E-05 0.000 0.0052 0.9990 0.9990 1.00 1 1
iter seconds digits c*x b*y
12 0.1 15.4 -1.7476196636e-02 -1.7476196636e-02
|Ax-b| = 3.2e-16, [Ay-c]_+ = 5.0E-17, |x|= 6.1e-01, |y|= 3.2e-01
Detailed timing (sec)
Pre IPM Post
1.000E-02 5.000E-02 0.000E+00
Max-norms: ||b||=1, ||c|| = 1.122018e+00,
Cholesky |add|=0, |skip| = 0, ||L.L|| = 1.33849.
------------------------------------------------------------
Status: Solved
Optimal value (cvx_optval): +0.0174762
Problem is Solved
The minimum attenuation in the stopband is -35.15 dB.