How to do this more efficiently

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S. David
S. David el 24 de Mayo de 2014
Comentada: S. David el 7 de Jun. de 2014
Hello,
I have this piece of code in MATLAB:
for kk=0:N-1
for mm=0:N-1
for pp=1:Np
for qq=1:Np
if ((kk*Ts+tau(pp)))<=(mm*Ts+tau(qq))) && ((kk+1)*Ts+tau(pp))>(mm*Ts+tau(qq)))
thetaSR=(((kk+1)*Ts+tau(pp)))-((mm*Ts+tau(qq))));
F_SR_MR(kk+1,mm+1)=F_SR_MR(kk+1,mm+1)+conj(H(pp))*H(qq)*(thetaSR*exp(1i*pi*fc**thetaSR)*sinc(fc*thetaSR));
end
which obviously is not very efficient. How can I re-write it more efficiently?
Thanks
  6 comentarios
the cyclist
the cyclist el 25 de Mayo de 2014
Do you have a reference (e.g. an image you can post) that shows the mathematical formula you are trying to replicate? It might be easier to build the code directly from that rather than optimizing yours.
S. David
S. David el 25 de Mayo de 2014
I attached the formula. g(t) in it is a rectangular pulse of magnitude one over the period [0,Ts).

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Respuesta aceptada

Roger Stafford
Roger Stafford el 26 de Mayo de 2014
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
kk-mm == d
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
  12 comentarios
S. David
S. David el 29 de Mayo de 2014
Editada: S. David el 29 de Mayo de 2014
I didn't see your last response when I replied the last time.
I am glad you asked about this, because I was going to mention it next.
Actually, your are right, according to the equation attached earlier, the matrix should be circular. However, my code is a bit more complicated, and I though writing it this way would make things easier to understand. So, yes, the thetas here are meaningless.
The (k,m)th element of F is given by the equation attached below. You can notice the presence of the exponential inside the integral changes the situation, and the matrix is no longer circular. I think this answers all your questions.
In this case the code would be:
clear all;
clc
N=512;
fc=12*10^3;
B=8000;
df=B/N;
T=1/df;
Ts=T/N;
tau=[0 5 7 20].*10^-3;
h=[1 0.8 0.7 0.5];
Np=length(tau);
H=h.*exp(-1i*2*pi*fc*tau);
F=zeros(N);
a=[0.001 0.0012 0.0024 0.003];
for kk=0:N-1
for mm=0:N-1
for pp=1:Np
for qq=1:Np
if ((kk*Ts+tau(pp))/(1+a(pp)))<=(mm*Ts+tau(qq))/(1+a(qq)) && ((kk+1)*Ts+tau(pp))/(1+a(pp))>(mm*Ts+tau(qq))/(1+a(qq))
theta1=(((kk+1)*Ts+tau(pp))/(1+a(pp)))-((mm*Ts+tau(qq))/(1+a(qq)));
theta2=(((kk+1)*Ts+tau(pp))/(1+a(pp)))+((mm*Ts+tau(qq))/(1+a(qq)));
F(kk+1,mm+1)=F(kk+1,mm+1)+conj(H(pp))*H(qq)*(theta1*exp(1i*pi*fc*(a(qq)-a(pp))*theta2)*sinc(fc*(a(qq)-a(pp))*theta1));
elseif (mm*Ts+tau(qq))/(1+a(qq))<=(kk*Ts+tau(pp))/(1+a(pp)) && (((mm+1)*Ts+tau(qq))/(1+a(qq)))>((kk*Ts+tau(pp))/(1+aSR))
theta1=(((mm+1)*Ts+tau(qq))/(1+a(qq)))-((kk*Ts+tau(pp))/(1+a(pp)));
theta2=(((mm+1)*Ts+tau(qq))/(1+a(qq)))+((kk*Ts+tau(pp))/(1+a(pp)));
F(kk+1,mm+1)=F(kk+1,mm+1)+conj(H(pp))*H(qq)*(theta1*exp(1i*pi*fc*(a(qq)-a(pp))*theta2)*sinc(fc*(a(qq)-a(pp))*theta1));
end
end
end
end
end
Hope it makes more sense now.
Can we still follow the same approach you did previously in this case?
Thanks
S. David
S. David el 7 de Jun. de 2014
Any hint?

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Más respuestas (1)

the cyclist
the cyclist el 24 de Mayo de 2014
Especially if N is large, you might get a huge speedup if you preallocate the memory for F_SR_MR. Put the line
F_SR_MR = zeros(N,N);
ahead of the loops.

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