Matlab - 乘以矩阵列表

时间:2013-10-11 16:30:30

标签: matlab matrix

我正在尝试创建一个Abeles matrix formalism模型来分析一些实验数据 - 我附上了一个wiki链接以供参考,以便您可以看到我试图实现的目标。

我的问题的关键在于我无法将四组矩阵相互叠加,如:A[1]*B[1]*C[1]*D[1], A[2]*B[2]*C[2]*D[2], ..., A[n]*B[n]*C[n]*D[n]。然后我需要将结果存储为它们自己的单个矩阵 - 每个矩阵代表Qmin:Qstep:Qmax的相应动量转移值。

另外,当我尝试执行最后一步时; R = abs((ABCD(2,1)./ABCD(1,1)).^2)我为每个R值设置了一个值,而不是Q的值。

由于代码的大小,简单的for循环不是一个现实的选择。

my 'test' code is: 
%import data fid = fopen('run_22208_09.dat');
%A = textscan(fid,'%f%f%f',270,'headerlines',0,'delimiter',',');

NQ = size(A{1,1}); 
NQ = NQ(1); 
Qmin = A{1,1}(1); 
Qmax = A{1,1}(NQ); 
Qstep = A{1,1}(2) - A{1,1}(1); 
fclose('all');

s0 = 2e-6; 
s1 = 10e-6; 
s2 = 6e-6; 
s3 = 4e-6; 
s4 = 8e-6; 
sn = 12e-6;

r1 = 2; 
r2 = 10; 
r3 = 3; 
r4 = 7; 
t1 = 10; 
t2 = 45; 
t3 = 5; 
t4 = 20;

Q=Qmin:Qstep:Qmax;

k = 2.*Q; 
k1 = (((k).^2) - 4.*pi.*(s1 - s0)).^0.5; 
k2 = (((k).^2) - 4.*pi.*(s2 - s0)).^0.5; 
k3 = (((k).^2) - 4.*pi.*(s3 - s0)).^0.5; 
k4 = (((k).^2) - 4.*pi.*(s4 - s0)).^0.5; 
kn = (((k).^2) - 4.*pi.*(sn - s0)).^0.5;

layer1 = ((k1 - k2)./(k1 + k2)).*(exp(-2.*k1.*k2.*(r1.^2)));
beta1 = (sqrt(-1)).*k1.*t1;

layer2 = ((k2 - k3)./(k2 + k3)).*(exp(-2.*k2.*k3.*(r2.^2)));
beta2 = (sqrt(-1)).*k2.*t2;

layer3 = ((k3 - k4)./(k3 + k4)).*(exp(-2.*k3.*k4.*(r3.^2)));
beta3 = (sqrt(-1)).*k3.*t3;

layer4 = ((k4 - kn)./(k4 + kn)).*(exp(-2.*k4.*kn.*(r4.^2)));
beta4 = (sqrt(-1)).*k4.*t4;

%general matrix 
C1 = [exp(beta1),layer1.*(exp(beta1));layer1.*exp(-beta1),exp(-beta1)] 
C2 = [exp(beta2),layer2.*(exp(beta2));layer2.*exp(-beta2),exp(-beta2)]; 
C3 = [exp(beta3),layer3.*(exp(beta3));layer3.*exp(-beta3),exp(-beta3)]; 
C4 = [exp(beta4),layer4.*(exp(beta4));layer4.*exp(-beta4),exp(-beta4)];

% CA = bsxfun(@times,C1,C2) 
% CB = bsxfun(@times,CA,C3); 
% C = bsxfun(@times,CB,C4)

% R = abs((C(2,1)./C(1,1)).^2) 

1 个答案:

答案 0 :(得分:0)

对于数组的元素乘法,你可以写

 M = C1 .* C2 .* C3 .* C4;
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