It isn't necessary to do the inequality test for every possible pair of kk and mm values. Since kk and mm are integers and Ts must surely be a positive number, your pair of inequalities is logically equivalent to
where d = floor((tau(qq)-tau(pp))/Ts). Therefore you can simply add the appropriate vectors along corresponding diagonals of F. For large N, doing it this way should save quite a bit of computation time.
F=zeros(N);
for pp=1:Np
for qq=1:Np
d = floor((tau(qq)-tau(pp))/Ts);
kk = max(d,0):min(N-1,N-1+d);
mm = kk-d;
theta=(d+1)*Ts-(tau(qq)-tau(pp));
ix = d+1+(N+1)*mm;
F(ix) = F(ix)+conj(H(pp))*H(qq)*theta.*exp(1i*pi*fc*theta).*sinc(fc*theta);
end
end
Best Answer