I am attempting to create a Abeles matrix formalism model to analyse some experimental data - I have attached a wiki link to this for reference so that you can see what I an attempting to achieve.
The crux of my issue is that I am unable to multiply four sets of matrices against each other, as in: A[1]*B[1]*C[1]*D[1], A[2]*B[2]*C[2]*D[2], ..., A[n]*B[n]*C[n]*D[n]. I then need to store the results as individual matrices of their own - each matrix represents the corresponding momentum transfer value from Qmin:Qstep:Qmax.
Also, when I attempt to carry out the final step; R = abs((ABCD(2,1)./ABCD(1,1)).^2) I end up with a single value rather than a value of R for each Q value.
Due to the size of the code a simple for loop isn't a realistic option.
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)
For element-wise multiplications of arrays, you write
M = C1 .* C2 .* C3 .* C4;
Related
This is my Approximate entropy Calculator in MATLAB. https://en.wikipedia.org/wiki/Approximate_entropy
I'm not sure why it isn't working. It's returning a negative value.Can anyone help me with this? R1 being the data.
FindSize = size(R1);
N = FindSize(1);
% N = input ('insert number of data values');
%if you want to put your own N in, take away the % from the line above
and
%insert the % before the N = FindSize(1)
%m = input ('insert m: integer representing length of data, embedding
dimension ');
m = 2;
%r = input ('insert r: positive real number for filtering, threshold
');
r = 0.2*std(R1);
for x1= R1(1:N-m+1,1)
D1 = pdist2(x1,x1);
C11 = (D1 <= r)/(N-m+1);
c1 = C11(1);
end
for i1 = 1:N-m+1
s1 = sum(log(c1));
end
phi1 = (s1/(N-m+1));
for x2= R1(1:N-m+2,1)
D2 = pdist2(x2,x2);
C21 = (D2 <= r)/(N-m+2);
c2 = C21(1);
end
for i2 = 1:N-m+2
s2 = sum(log(c2));
end
phi2 = (s2/(N-m+2));
Ap = phi1 - phi2;
Apen = Ap(1)
Following the documentation provided by the Wikipedia article, I developed this small function that calculates the approximate entropy:
function res = approximate_entropy(U,m,r)
N = numel(U);
res = zeros(1,2);
for i = [1 2]
off = m + i - 1;
off_N = N - off;
off_N1 = off_N + 1;
x = zeros(off_N1,off);
for j = 1:off
x(:,j) = U(j:off_N+j);
end
C = zeros(off_N1,1);
for j = 1:off_N1
dist = abs(x - repmat(x(j,:),off_N1,1));
C(j) = sum(~any((dist > r),2)) / off_N1;
end
res(i) = sum(log(C)) / off_N1;
end
res = res(1) - res(2);
end
I first tried to replicate the computation shown the article, and the result I obtain matches the result shown in the example:
U = repmat([85 80 89],1,17);
approximate_entropy(U,2,3)
ans =
-1.09965411068114e-05
Then I created another example that shows a case in which approximate entropy produces a meaningful result (the entropy of the first sample is always less than the entropy of the second one):
% starting variables...
s1 = repmat([10 20],1,10);
s1_m = mean(s1);
s1_s = std(s1);
s2_m = 0;
s2_s = 0;
% datasample will not always return a perfect M and S match
% so let's repeat this until equality is achieved...
while ((s1_m ~= s2_m) && (s1_s ~= s2_s))
s2 = datasample([10 20],20,'Replace',true,'Weights',[0.5 0.5]);
s2_m = mean(s2);
s2_s = std(s2);
end
m = 2;
r = 3;
ae1 = approximate_entropy(s1,m,r)
ae2 = approximate_entropy(s2,m,r)
ae1 =
0.00138568170752751
ae2 =
0.680090884817465
Finally, I tried with your sample data:
fid = fopen('O1.txt','r');
U = cell2mat(textscan(fid,'%f'));
fclose(fid);
m = 2;
r = 0.2 * std(U);
approximate_entropy(U,m,r)
ans =
1.08567461184858
I'm stuck on K2 as it brought up this error:
In an assignment A(I) = B, the number of elements in B and
I must be the same.
I ran the debugger and I found out that J4 alone was a vector while other variables were all scalar.
How can I resolve this error to have the plot?
Here is the code that I ran.
h1 = 1*10^-6;
h2 = (10:10:1500)*10^-6;
a = 62.5*10^-6;
b = a+h1;
c = b+h2;
alpha_1 = 0.55*10^-6;
alpha_2 = 17.2*10^-6;
alpha_3 = 14.2*10^-6;
zeta = 6.3*10^-6;
P11 = 0.121;
P12 = 0.27;
neff = 1.456;
U1 = 0.17;
U2 = 0.32;
U3 = 0.31;
E1 = 0.74*10^11;
E2 = 1.08*10^11;
E3 = 1.96*10^11;
n = 1;
while(n<=150)
J1(n) = E2*(b^2-a^2)*(1-U1)-E1*a^2*(1-U2)-E1*b^2*(1+U2);
J2(n) = 2*E1*b^2;
J3(n) = E1*E2*(b^2-a^2)*(alpha_2 - alpha_1);
J4(n) = 2*E3*(c(n)^2-b^2)*a^2;
J5(n) = E2*(b^2-a^2)*(1-U3)*b^2+E2*(b^2-a^2)*(1+U3)*c(n)^2-E3*(c(n)^2-b^2)*(1+U2)*a^2-E3*(c(n)^2-b^2)*(1-U2)*b^2;
J6(n) = E2*E3*(c(n)^2 - b^2)*(b^2-a^2)*(alpha_2-alpha_3);
K1(n) = ((alpha_3-alpha_1)*E3*(c(n)^2-b^2)+(alpha_2-alpha_1)*E2*(b^2-a^2))/(E1*a^2+E2*(b^2-a^2)+E3*(c(n)^2-b^2));
K2(n) = (J2*J6-J3*J5)/(J2*J4-J1*J5);
Sr(n) = (neff^2/2)*(P11+P12)*(((1-U1)*K2/E1)-U1*K1);
Sz(n) = (1+P12)*(K1-(2*U2*K2/E1));
St(n) = alpha_1+zeta;
Km(n) = St+Sz+Sr;
n=n+1;
end
plot(h2,Km)
To recap what was already said in one answer, here's how I would modify the code:
h1 = 1e-6;
h2 = (10:10:1500)*1e-6;
a = 62.5*1e-6;
b = a+h1;
c = b+h2;
alpha_1 = 0.55*1e-6;
alpha_2 = 17.2*1e-6;
alpha_3 = 14.2*1e-6;
zeta = 6.3*1e-6;
P11 = 0.121;
P12 = 0.27;
neff = 1.456;
U1 = 0.17;
U2 = 0.32;
U3 = 0.31;
E1 = 0.74*1e11;
E2 = 1.08*1e11;
E3 = 1.96*1e11;
% pre-allocate variables
J1 = zeros(size(h2));
J2 = zeros(size(h2));
J3 = zeros(size(h2));
J4 = zeros(size(h2));
J5 = zeros(size(h2));
J6 = zeros(size(h2));
K1 = zeros(size(h2));
K2 = zeros(size(h2));
Sr = zeros(size(h2));
Sz = zeros(size(h2));
for n=1:length(h2)
J1(n) = E2*(b^2-a^2)*(1-U1)-E1*a^2*(1-U2)-E1*b^2*(1+U2);
J2(n) = 2*E1*b^2;
J3(n) = E1*E2*(b^2-a^2)*(alpha_2 - alpha_1);
J4(n) = 2*E3*(c(n)^2-b^2)*a^2;
J5(n) = E2*(b^2-a^2)*(1-U3)*b^2+E2*(b^2-a^2)*(1+U3)*c(n)^2-E3*(c(n)^2-b^2)*(1+U2)*a^2-E3*(c(n)^2-b^2)*(1-U2)*b^2;
J6(n) = E2*E3*(c(n)^2 - b^2)*(b^2-a^2)*(alpha_2-alpha_3);
K1(n) = ((alpha_3-alpha_1)*E3*(c(n)^2-b^2)+(alpha_2-alpha_1)*E2*(b^2-a^2))/(E1*a^2+E2*(b^2-a^2)+E3*(c(n)^2-b^2));
K2(n) = (J2(n)*J6(n)-J3(n)*J5(n))/(J2(n)*J4(n)-J1(n)*J5(n));
Sr(n) = (neff^2/2)*(P11+P12)*(((1-U1)*K2(n)/E1)-U1*K1(n));
Sz(n) = (1+P12)*(K1(n)-(2*U2*K2(n)/E1));
end
St = alpha_1+zeta;
Km = Sz+Sr+St;
plot(h2,Km)
Notes:
I have used a for loop to ensure the vector lengths are consistent with h2
I have pre-allocated the variables for speed
I have added various (n) to K1, K2, J1, J2, etc... in the equations to have only scalar operations
I have moved stuff out of the for loop that didn't need to be there
This gives the following plot (in Octave)
I ran your code and I found out that the dimensions of the vectors used to compute K2 are not compatible, e.g. each of J2 and J6 is a 1-row by 2-columns vector and you cannot multiply those. This also applies for the other multiplications. Depending on what you want to compute, you should transpose either one of them in each multiplication.
FYI: J.' is the transposed version of J in MATLAB.
I would like to run the following for 30 to 40 values of the current (I0) with increasing magnitude. Can someone please tell me how to make this happen?
GNa = 400;
GK = 200;
GL = 2;
ENa = 99;
EK = -85;
VL = -65;
C = 2;
dt = .01;
t = 0:dt:200;
I0 = 200;
It= I0 + 0*t;
It(1:40/dt)=0;
m = zeros(1,length(t));
n = zeros(1,length(t));
h = zeros(1,length(t));
V = zeros(1,length(t));
V(1) = -65;
am = zeros(1,length(t));
bm = zeros(1,length(t));
an = zeros(1,length(t));
bn = zeros(1,length(t));
ah = zeros(1,length(t));
bh = zeros(1,length(t));
for i=1:length(t)-1
I=It(i);
am(i) = (0.1*(V(i) + 40))/(1 - exp(-0.1*(V(i)+40)));
bm(i) = 4*exp(-0.0556*(V(i)+65));
ah(i) = 0.07*exp(-0.05*(V(i)+65));
bh(i) = 1./(1+exp(-0.1*(V(i)+35)));
an(i) = (0.01*(V(i)+55))/(1 - exp(-0.1*(V(i)+55)));
bn(i) = 0.125*exp(-0.0125*(V(i)+65));
m(i+1) = m(i) + dt*(am(i)*(1 - m(i)) - bm(i)*m(i));
h(i+1) = h(i) + dt*(ah(i)*(1 - h(i)) - bh(i)*h(i));
n(i+1) = n(i) + dt*(an(i)*(1 - n(i)) - bn(i)*n(i));
V(i+1) = V(i) + dt*(((-GL*(V(i) - VL) - GNa*(m(i).^3)*h(i)*(V(i) - ENa) - GK* (n(i).^4)*(V(i) - EK)+ I))/C);
end
You can always wrap this in another for loop, no?
I0list = 200:2.5:300 % 41 values including endpoints
for j = 1:length(IOlist)
I0 = IOlist(j);
% Rest of your code, minus setting I0, goes here
end
If you need to store all of the values, make m etc. be m = zeros(length(IOlist),length(t)); instead, and then use both indices when assigning, e.g.
am(j,i) = ...
I asked a question a few days before but I guess it was a little too complicated and I don't expect to get any answer.
My problem is that I need to use ANN for classification. I've read that much better cost function (or loss function as some books specify) is the cross-entropy, that is J(w) = -1/m * sum_i( yi*ln(hw(xi)) + (1-yi)*ln(1 - hw(xi)) ); i indicates the no. data from training matrix X. I tried to apply it in MATLAB but I find it really difficult. There are couple things I don't know:
should I sum each outputs given all training data (i = 1, ... N, where N is number of inputs for training)
is the gradient calculated correctly
is the numerical gradient (gradAapprox) calculated correctly.
I have following MATLAB codes. I realise I may ask for trivial thing but anyway I hope someone can give me some clues how to find the problem. I suspect the problem is to calculate gradients.
Many thanks.
Main script:
close all
clear all
L = #(x) (1 + exp(-x)).^(-1);
NN = #(x,theta) theta{2}*[ones(1,size(x,1));L(theta{1}*[ones(size(x,1),1) x]')];
% theta = [10 -30 -30];
x = [0 0; 0 1; 1 0; 1 1];
y = [0.9 0.1 0.1 0.1]';
theta0 = 2*rand(9,1)-1;
options = optimset('gradObj','on','Display','iter');
thetaVec = fminunc(#costFunction,theta0,options,x,y);
theta = cell(2,1);
theta{1} = reshape(thetaVec(1:6),[2 3]);
theta{2} = reshape(thetaVec(7:9),[1 3]);
NN(x,theta)'
Cost function:
function [jVal,gradVal,gradApprox] = costFunction(thetaVec,x,y)
persistent index;
% 1 x x
% 1 x x
% 1 x x
% x = 1 x x
% 1 x x
% 1 x x
% 1 x x
m = size(x,1);
if isempty(index) || index > size(x,1)
index = 1;
end
L = #(x) (1 + exp(-x)).^(-1);
NN = #(x,theta) theta{2}*[ones(1,size(x,1));L(theta{1}*[ones(size(x,1),1) x]')];
theta = cell(2,1);
theta{1} = reshape(thetaVec(1:6),[2 3]);
theta{2} = reshape(thetaVec(7:9),[1 3]);
Dew = cell(2,1);
DewApprox = cell(2,1);
% Forward propagation
a0 = x(index,:)';
z1 = theta{1}*[1;a0];
a1 = L(z1);
z2 = theta{2}*[1;a1];
a2 = L(z2);
% Back propagation
d2 = 1/m*(a2 - y(index))*L(z2)*(1-L(z2));
Dew{2} = [1;a1]*d2;
d1 = [1;a1].*(1 - [1;a1]).*theta{2}'*d2;
Dew{1} = [1;a0]*d1(2:end)';
% NNRes = NN(x,theta)';
% jVal = -1/m*sum(NNRes-y)*NNRes*(1-NNRes);
jVal = -1/m*(a2 - y(index))*a2*(1-a2);
gradVal = [Dew{1}(:);Dew{2}(:)];
gradApprox = CalcGradApprox(0.0001);
index = index + 1;
function output = CalcGradApprox(epsilon)
output = zeros(size(gradVal));
for n=1:length(thetaVec)
thetaVecMin = thetaVec;
thetaVecMax = thetaVec;
thetaVecMin(n) = thetaVec(n) - epsilon;
thetaVecMax(n) = thetaVec(n) + epsilon;
thetaMin = cell(2,1);
thetaMax = cell(2,1);
thetaMin{1} = reshape(thetaVecMin(1:6),[2 3]);
thetaMin{2} = reshape(thetaVecMin(7:9),[1 3]);
thetaMax{1} = reshape(thetaVecMax(1:6),[2 3]);
thetaMax{2} = reshape(thetaVecMax(7:9),[1 3]);
a2min = NN(x(index,:),thetaMin)';
a2max = NN(x(index,:),thetaMax)';
jValMin = -1/m*(a2min-y(index))*a2min*(1-a2min);
jValMax = -1/m*(a2max-y(index))*a2max*(1-a2max);
output(n) = (jValMax - jValMin)/2/epsilon;
end
end
end
EDIT:
Below I present the correct version of my costFunction for those who may be interested.
function [jVal,gradVal,gradApprox] = costFunction(thetaVec,x,y)
m = size(x,1);
L = #(x) (1 + exp(-x)).^(-1);
NN = #(x,theta) L(theta{2}*[ones(1,size(x,1));L(theta{1}*[ones(size(x,1),1) x]')]);
theta = cell(2,1);
theta{1} = reshape(thetaVec(1:6),[2 3]);
theta{2} = reshape(thetaVec(7:9),[1 3]);
Delta = cell(2,1);
Delta{1} = zeros(size(theta{1}));
Delta{2} = zeros(size(theta{2}));
D = cell(2,1);
D{1} = zeros(size(theta{1}));
D{2} = zeros(size(theta{2}));
jVal = 0;
for in = 1:size(x,1)
% Forward propagation
a1 = [1;x(in,:)']; % added bias to a0
z2 = theta{1}*a1;
a2 = [1;L(z2)]; % added bias to a1
z3 = theta{2}*a2;
a3 = L(z3);
% Back propagation
d3 = a3 - y(in);
d2 = theta{2}'*d3.*a2.*(1 - a2);
Delta{2} = Delta{2} + d3*a2';
Delta{1} = Delta{1} + d2(2:end)*a1';
jVal = jVal + sum( y(in)*log(a3) + (1-y(in))*log(1-a3) );
end
D{1} = 1/m*Delta{1};
D{2} = 1/m*Delta{2};
jVal = -1/m*jVal;
gradVal = [D{1}(:);D{2}(:)];
gradApprox = CalcGradApprox(x(in,:),0.0001);
% Nested function to calculate gradApprox
function output = CalcGradApprox(x,epsilon)
output = zeros(size(thetaVec));
for n=1:length(thetaVec)
thetaVecMin = thetaVec;
thetaVecMax = thetaVec;
thetaVecMin(n) = thetaVec(n) - epsilon;
thetaVecMax(n) = thetaVec(n) + epsilon;
thetaMin = cell(2,1);
thetaMax = cell(2,1);
thetaMin{1} = reshape(thetaVecMin(1:6),[2 3]);
thetaMin{2} = reshape(thetaVecMin(7:9),[1 3]);
thetaMax{1} = reshape(thetaVecMax(1:6),[2 3]);
thetaMax{2} = reshape(thetaVecMax(7:9),[1 3]);
a3min = NN(x,thetaMin)';
a3max = NN(x,thetaMax)';
jValMin = 0;
jValMax = 0;
for inn=1:size(x,1)
jValMin = jValMin + sum( y(inn)*log(a3min) + (1-y(inn))*log(1-a3min) );
jValMax = jValMax + sum( y(inn)*log(a3max) + (1-y(inn))*log(1-a3max) );
end
jValMin = 1/m*jValMin;
jValMax = 1/m*jValMax;
output(n) = (jValMax - jValMin)/2/epsilon;
end
end
end
I've only had a quick eyeball over your code. Here are some pointers.
Q1
should I sum each outputs given all training data (i = 1, ... N, where
N is number of inputs for training)
If you are talking in relation to the cost function, it is normal to sum and normalise by the number of training examples in order to provide comparison between.
I can't tell from the code whether you have a vectorised implementation which will change the answer. Note that the sum function will only sum up a single dimension at a time - meaning if you have a (M by N) array, sum will result in a 1 by N array.
The cost function should have a scalar output.
Q2
is the gradient calculated correctly
The gradient is not calculated correctly - specifically the deltas look wrong. Try following Andrew Ng's notes [PDF] they are very good.
Q3
is the numerical gradient (gradAapprox) calculated correctly.
This line looks a bit suspect. Does this make more sense?
output(n) = (jValMax - jValMin)/(2*epsilon);
EDIT: I actually can't make heads or tails of your gradient approximation. You should only use forward propagation and small tweaks in the parameters to compute the gradient. Good luck!
I would like to perform what follows:
where
The parameters shown in the latter figure can be obtained as follows:
%% Inizialization
time = 614.4; % Analysis Time
Uhub = 11;
HubHt = 90;
alpha = 0.14;
TI = 'A'; % Turbulent Intensity (A,B,C as in the IEC or Specific value)
N1 = 4096;
N2 = 32;
N3 = 32;
N = N1*N2*N3; % Total Number of Point
t = 0:(time/(N1-1)):time; % Sampled Time Vector
L1 = Uhub*time; % Box length along X
L2 = 150; % Box length along Y
L3 = 220; % Box length along Z
dx = L1/N1; % Grid Resolution along X-axis
dy = L2/N2; % Grid Resolution along Y-axis
dz = L3/N3; % Grid Resolution along Z-axis
V = L1*L2*L3; % Analysis Box Volume
gamma = 3.9; % Turbulent Eddies Distorsion Factor
c = 1.476;
b = 5.6;
if HubHt < 60
lambda1 = 0.7*HubHt;
else
lambda1 = 42;
end
L = 0.8*lambda1;
if isequal(TI,'A')
Iref = 0.16;
sigma1 = Iref*(0.75*Uhub + b);
elseif isequal(TI,'B')
Iref = 0.14;
sigma1 = Iref*(0.75*Uhub + b);
elseif isequal(TI,'C')
Iref = 0.12;
sigma1 = Iref*(0.75*Uhub + b);
else
sigma1 = str2num(TI)*Uhub/100;
end
sigma_iso = 0.55*sigma1;
sigma2 = 0.7*sigma1;
sigma3 = 0.5*sigma1;
%% Wave number vectors
ik1 = cat(2,(-N1/2:-1/2),(1/2:N1/2));
ik2 = -N2/2:N2/2-1;
ik3 = -N3/2:N3/2-1;
[x y z] = ndgrid(ik1,ik2,ik3);
k1 = reshape((2*pi*L/L1)*x,N1*N2*N3,1);
k2 = reshape((2*pi*L/L2)*y,N1*N2*N3,1);
k3 = reshape((2*pi*L/L3)*z,N1*N2*N3,1);
k = sqrt(k1.^2 + k2.^2 + k3.^2);
%% Calculation of beta by means of the Energy Spectrum Integration
E = #(k) (1.453*k.^4)./((1 + k.^2).^(17/6));
%//Independent integration on segments
Local_int = arrayfun(#(i)quad(E,i-1,i), 2:(N1*N2*N3));
%//integral additivity + cumulative removal of queues
E_int = 1.5 - [0 fliplr(cumsum(fliplr(Local_int)))]; %//To remove queues
E_int = reshape(E_int,N,1);
S = k.*sqrt(E_int);
beta = (c*gamma)./S;
%% Help Parameters
k30 = k3 + beta.*k1;
k0 = sqrt(k1.^2 + k2.^2 + k30.^2);
C1 = (beta.*k1.^2.*(k1.^2 + k2.^2 - k3.*k30))./(k.^2.*(k1.^2 + k2.^2));
C2 = (k2.*k0.^2./((k1.^2 + k2.^2).^(3/2))).*atan2((beta.*k1.*sqrt(k1.^2 + k2.^2)),(k0.^2 - k30.*k1.*beta));
xhsi1 = C1 - (k2./k1).*C2;
xhsi2 = (k2./k1).*C1 + C2;
E_k0 = (1.453*k0.^4)./((1 + k0.^2).^(17/6));
For instance, typing in
phi_33 = #(k2,k3) (E_k0./(4*pi.*k.^4)).*((k1.^2 + k2.^2));
F_33 = arrayfun(#(i) dblquad(phi_33,k3(i),k3(i+1),k2(i),k2(i+1)), 1:((N1*N2*N3)-1));
Matlab retrieves the following error msg:
Error using +
Matrix dimensions must agree.
Error in #(k2,k3)(E_k0./(4*pi.*k.^4)).*((k1.^2+k2.^2))
Do you have a clue how to overcome this issue?
I really look forward to hearing from you.
Best regards,
FPE
The error is easily explained:
First you define E_k0 then you try to call Ek0.
phi_11 = #(k1,k2,k3) (E_k0./4*pi.*kabs.^4).*(k0abs.^2 - k1.^2 - 2*k1.*k03.*xhsi1 + (k1.^2 + k2.^2).*xhsi1.^2);
I solved it this way:
Code a function for each of the PHI elements, such as (for PHI11)
function phi_11 = phi_11_new(k1,k2,k3,beta,i)
k = sqrt(k1(i).^2 + k2.^2 + k3.^2);
k30 = k3 + beta(i).*k1(i);
k0 = sqrt(k1(i).^2 + k2.^2 + k30.^2);
E_k0 = 1.453.*k0.^4./((1 + k0.^2).^(17/6));
C1 = (beta(i).k1(i).^2).(k1(i).^2 + k2.^2 - k3.k30)./(k.^2.(k1(i).^2 + k2.^2));
C2 = k2.*k0.^2./((k1(i).^2 + k2.^2).^(3/2)).*atan2((beta(i).*k1(i).*sqrt(k1(i).^2 + k2.^2)),(k0.^2 - k30.*k1(i).*beta(i)));
xhsi1 = C1 - k2./k1(i).*C2;
xhsi1_q = xhsi1.^2;
phi_11 = E_k0./(4.*pi.k0.^4).(k0.^2 - k1(i).^2 - 2.*k1(i).*k30.*xhsi1 + (k1(i).^2 + k2.^2).*xhsi1_q);
end
I recall this function within the main code as follows
for l = 1:numel(k1)
phi11 = #(k2,k3) phi_11(k1,k2,k3,l)
F11(l) = integral2(phi,-1000,1000,-1000,1000);
end
at it seems it works.