For Loop using two variables

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Myles Wright-Walker
编辑： DGM 2024-11-13，15:56
I am new to MATLAB and I am trying to use a for loop using two variables. The question is: Generate a MATLAB program to compute and plot the Fermi function, f(E), and 1- f(E) versus ΔE = E-Ef for values of ΔE that is over the range of -0.5eV ≤ ΔE ≤ 0.5eV for varying temperature settings where Temperature = 150, 250, 350, 450 and 550K.
The code I have written is:
k = 1.3806488*10^-23; %boltzman constant
format shortEng
for dE = linspace (-0.5, 0.5, 5), T1 = 150:100:550
fE = 1/(1+exp(dE/(k*T1))) %Fermi Function
fE2 = 1-fE
end
However, it its only displaying T1

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回答（1 个）

Andrew Newell
Andrew Newell 2014-4-24
编辑：Andrew Newell 2014-4-24
You need to use a couple of nested loops:
k = 1.3806488*10^-23; %boltzman constant
format shortEng
for dE = linspace (-0.5, 0.5, 5)
for T1 = 150:100:550
fE = 1/(1+exp(dE/(k*T1))) %Fermi Function
fE2 = 1-fE
end
end
For plotting these results, however, you'll need to store the values in a matrix, like this:
dE = linspace(-0.5,0.5,5);
T1 = 150:100:550;
fE = zeros(length(dE),length(T1)); % initializing fE speeds up the calculation
for i=1:length(dE)
for j=1:length(T1)
fE(i,j) = 1/(1+exp(dE/(k*T1))); %Fermi Function
end
end
fE2 = 1-fE; % No need to put this in the loop
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显示 2更早的评论
Torsten
Torsten 2024-11-13，14:05
I think fE is meant as a matrix with
fE(i,j) = 1 / (1+exp(deltaE(i)/(k*Temp(j))))
DGM
DGM 2024-11-13，15:41
编辑：DGM 2024-11-13，15:56
Ran in:
Consider:
% inputs
k = 1.3806488*10^-23; %boltzman constant
dE = linspace(-0.5,0.5,5);
T1 = 150:100:550;
% you could use loops
fE = zeros(length(dE),length(T1)); % initializing fE speeds up the calculation
for i = 1:length(dE)
for j = 1:length(T1)
fE(i,j) = 1/(1 + exp(dE(i)/(k*T1(j)))); % Fermi Function
end
end
fE2_0 = 1-fE % No need to put this in the loop
fE2_0 = 5×5
0 0 0 0 0 0 0 0 0 0 0.5000 0.5000 0.5000 0.5000 0.5000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
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% or just calculate the entire output at once
% assuming both inputs are row vectors to begin with
fE2_1 = 1 - 1./(1 + exp(dE.'./(k*T1)))
fE2_1 = 5×5
0 0 0 0 0 0 0 0 0 0 0.5000 0.5000 0.5000 0.5000 0.5000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
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% that wouldn't have been an option when this was posted.
% prior to R2016b, this is how it would be done
fE2_2 = 1 - 1./(1 + exp(bsxfun(@rdivide,dE.',k*T1)))
fE2_2 = 5×5
0 0 0 0 0 0 0 0 0 0 0.5000 0.5000 0.5000 0.5000 0.5000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
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Note that this looks odd because these are tiny numbers (actually several orders of magnitude smaller than floating point resolution for the numerator).
(k*T1)
ans = 1×5
1.0e-20 * 0.2071 0.3452 0.4832 0.6213 0.7594
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As a consequence, these are huge numbers being fed to an exponential function
dE.'./(k*T1)
ans = 5×5
1.0e+20 * -2.4143 -1.4486 -1.0347 -0.8048 -0.6585 -1.2072 -0.7243 -0.5174 -0.4024 -0.3292 0 0 0 0 0 1.2072 0.7243 0.5174 0.4024 0.3292 2.4143 1.4486 1.0347 0.8048 0.6585
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... and so we're excessively sensitive to dE and we lose all sensitivity to T1:
exp(dE.'./(k*T1))
ans = 5×5
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 Inf Inf Inf Inf Inf Inf Inf Inf Inf Inf
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You should probably want to rethink how you want to solve a problem like this. Naive floating point numeric calculations might not be the way to go.

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