早安大家,提高5D矩阵计算
我已经写了这个代码:
clc
clear all
close all
%Box size
Nx=4096;
Ny=15;
Nz=15;
%Spatial gird resolution
delta=6;
%WT/turbulence condition
UHub=11.4;
HubHt=90;
z0=0.03;
IECturbC='B';
%%INITIALISATION
% definition of constants
twopi=2*pi;
fourpi=4*pi;
sqrt2=sqrt(2);
%constants and derived parameters from IEC
gamma = 3.9; %IEC, (B.12)
alpha = 0.2; %IEC, sect. 6.3.1.2
%set delta1 according to guidelines (chap.6)
if HubHt<=60,
delta1=0.7*HubHt;
else
delta1=42;
end;
%IEC, Table 1, p.22
if IECturbC == 'A',
Iref=0.16;
elseif IECturbC == 'B',
Iref=0.14;
elseif IECturbC == 'C',
Iref=0.12;
else
error('IECturbC can be equal to A,B or C;adjust the input value')
end;
%IEC, sect. 6.3.1.3
b=6.5;
sigma1=Iref*(0.75*UHub+b);
%derived constants
l=0.8*delta1; %IEC, (B.12)
sigmaiso=0.55*sigma1; %IEC, (B.12)
%%MAIN PROGRAM
Cij=zeros(3,3,Nx,Ny,Nz);
k = zeros(3,1); %current vector k
for ikx=1:(Nx),
m = -1.*Nx/2+ikx;
k(1)=m*l/(Nx*delta)*twopi;
for iky=1:(Ny),
m= -1.*Ny/2+iky;
k(2)=m*l/(Ny*delta)*twopi;
for ikz=1:(Nz),
m= -1.*Nz/2+ikz;
k(3)=m*l/(Nz*delta)*twopi;
if k(1)==0,
Cij(:,:,ikx,iky,ikz)=0;
else
kabs=sqrt(k(1)^2+k(2)^2+k(3)^2);
beta= gamma./(kabs.^(2/3));
k0(3)=k(3)+beta.*k(1);
k0abs=sqrt(k(1)^2+k(2)^2+k0(3)^2);
Ek0=1.453*k0abs^4/(1.+k0abs.^2)^(17/6);
C1=beta.*k(1)^2*(k0abs.^2 - 2*k0(3)^2 + beta.*k(1)*k0(3))/(kabs.^2*(k(1)^2 + k(2)^2));
C2=k(2).*k0abs.^2./ (exp((3/2).*log(k(1).^2 + k(2).^2))) .* atan2(beta.*k(1).* sqrt(k(1)^2 + k(2)^2) ,(k0abs.^2 - k0(3).*k(1).*beta));
xhsi1=C1 - k(2).*C2./k(1);
xhsi2=k(2).*C1./k(1) + C2;
Cij(1,1,ikx,iky,ikz)= sigmaiso*sqrt(twopi*pi*l^3.*Ek0/(Nx*Ny*Nz*delta^3.*k0abs.^4))*(k(2).*xhsi1);
Cij(1,2,ikx,iky,ikz)= sigmaiso*sqrt(twopi*pi*l^3.*Ek0/(Nx*Ny*Nz*delta^3.*k0abs.^4))*(k(3) - k(1).*xhsi1 + beta.*k(1));
Cij(1,3,ikx,iky,ikz)= sigmaiso*sqrt(twopi*pi*l^3.*Ek0/(Nx*Ny*Nz*delta^3.*k0abs.^4))*(-k(2));
Cij(2,1,ikx,iky,ikz)= sigmaiso*sqrt(twopi*pi*l^3.*Ek0/(Nx*Ny*Nz*delta^3.*k0abs.^4))*(k(2).*xhsi2 - k(3) - beta.*k(1));
Cij(2,2,ikx,iky,ikz)= sigmaiso*sqrt(twopi*pi*l^3.*Ek0/(Nx*Ny*Nz*delta^3.*k0abs.^4))*(-k(1).*xhsi2);
Cij(2,3,ikx,iky,ikz)= sigmaiso*sqrt(twopi*pi*l^3.*Ek0/(Nx*Ny*Nz*delta^3.*k0abs.^4))*(k(1));
Cij(3,1,ikx,iky,ikz)= sigmaiso*sqrt(twopi*pi*l^3.*Ek0/(Nx*Ny*Nz*delta^3.*k0abs.^4))*(k0abs.^2.*k(2) ./ (kabs.^2));
Cij(3,2,ikx,iky,ikz)= sigmaiso*sqrt(twopi*pi*l^3.*Ek0/(Nx*Ny*Nz*delta^3.*k0abs.^4))*(-k0abs.^2*k(1) ./ (kabs.^2));
Cij(3,3,ikx,iky,ikz)= 0;
end;
end;
end;
end;
我想问你: 1.有没有更快的方法来获得Cij的矩阵?当Nx,Ny,Nz增加时,Cij的计算相当慢; 2.有什么方法可以获得情节(kabs,beta)和情节(kabs,Ek0)吗?
请耐心等待,我仍然是matlab世界中的新手。
提前感谢和问候, 弗朗切斯科
看到这个答案:http://stackoverflow.com/a/7973945/907578。此外,您应该通过提供更少的不相关代码来使问题更加通用。 – cyborg 2011-12-20 10:54:00
问题是,如果没有整个代码,就不太容易理解我需要的东西。对不起,这只是2天,我是一个stackoverflow用户:) – fpe 2011-12-20 10:59:57
@cyborg:btw你有任何线索如何改变我的Cij实现根据该答案?我对matlab非常熟悉,需要很长时间才能获得正确的编码。我提前感谢您。 – fpe 2011-12-20 11:20:59