Performing integral trapz-error invalid permutation index

Question

Jarett 2015-2-24

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/180098-performing-integral-trapz-error-invalid-permutation-index

回答： Lazaros Christoforidis 2019-11-14

I have the following code

Vg = linspace(0, 5, 100);
for t = 1:length(Vg);
Fun(t) =((Vg(t)^2)/(exp(Vg(t)/(k*Ts))-1));
IntNs(t) = trapz(Vg,Fun(t));
end

Every time I attempt to run this I receive the error message "Order contains an invalid permutation index", Im not quite sure what is going awry as I have run an almost identical code in other scripts. Could anyone guide me to the problem and how to correct it.

Thanks

-Jarett

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Answer 1

Roger Stafford 2015-2-24

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https://ww2.mathworks.cn/matlabcentral/answers/180098-performing-integral-trapz-error-invalid-permutation-index#answer_169125

在 MATLAB Online 中打开

Apparently you wish to find the integral of the function 'Fun' with respect to 'Vg'. If so, the line

IntNs(t) = trapz(Vg,Fun(t));

should be located after you have exited the for-loop:

   Vg = linspace(0, 5, 100);
   for t = 1:length(Vg);
     Fun(t) =((Vg(t)^2)/(exp(Vg(t)/(k*Ts))-1));
   end
   IntNs = trapz(Vg,Fun);

As you have it, you are trying to find the integral of the scalar Fun(t) with respect to 'Vg' which 'trapz' doesn't like. See the documentation at:

http://www.mathworks.com/help/matlab/ref/trapz.html

Note where it says, "Q = trapz(X,Y) integrates Y with spacing increment X. By default, trapz operates on the first dimension of Y whose size does not equal 1. length(X) must be equal to the size of this dimension." That is obviously not true in your use of 'trapz', since Fun(t) is a scalar.