Hello, I would like to speed up this code, any sugestions
Rm, Zm and Cm are 1700x1700
f = linspace(1e3,5e6,750); % Hier gibt man den Frequenzbereich an
w = 2*pi*f';
parfor k=1:length(f)
Zm = Rm+w(k)*1i*Lm;
Gama = Kk*(Zm\Kkt);
YYn = w(k)*1i*Cm + Gama;
Zn = inv(YYn);
Amp(k) = abs(1-Zn(jj,1)/Zn(1,1));
end
I do not have a GPU

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Joss Knight
Joss Knight 2019년 8월 16일
편집: Joss Knight 2019년 8월 16일
What is Kkt? What is jj?
Matthias Schneider
Matthias Schneider 2019년 8월 19일
편집: Matthias Schneider 2019년 8월 19일
Sorry I forgot to mention what Kk, KKt and jj are. Kk ist an transformations Matrix, and Kkt = Kk', jj is the index of an outer for loop.
a = 1700
Kk=eye(a);
for i=1:(a-1)
Kk(i+1,i)=-1;
end
this little pice of code describes how Kk is built, but the Kk is built only once and is the used over and over again.
madhan ravi
madhan ravi 2019년 8월 19일
a?
Joss Knight
Joss Knight 2019년 8월 21일
Sorry, nothing occurs to me. You take the trouble to invert YYn but then only use a single row of it, so maybe you could focus on that piece of algebra.

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