getting Complex double in matlab while excel solves it

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Asliddin Komilov 2019-6-15

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评论： Asliddin Komilov 2019-6-15

I had this formula in excel few years ago with less variation of WL, but now want to have a bigger range and put it into matlab and it gives me Complex double and the curve is totally different from what I expect or that came out in excel, can anyone help me here

WL=0:1E-6:4e-3;

SS=0.06977+7.0625*(1-exp(-(WL-0.26053)./0.15994)).^2.28411.*exp(-(WL-0.26053)./0.2285);

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Answer 1

Stephen23 2019-6-15

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编辑：Stephen23 2019-6-15

在 MATLAB Online 中打开

(1-exp(-(WL-0.26053)./0.15994)).^2.28411
%
%  ^^^^                      ^             positive values (around 5)
%^^^                          ^            negative values (around -4)
%                              ^^^^^^^^^   fractional power of negative values.

What do you expect a fractional power of negative values to give you?

>> (-4).^2.28411
ans =  14.884 + 18.473i

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Stephen23 2019-6-15

编辑：Stephen23 2019-6-15

在 MATLAB Online 中打开

As I explained in my answer, you are calculating the exponent of a negative number using a fractional power, which produces a complex result. Taking a look at your example:

>> WL=1E-3:1E-3:2;
>> numel(WL)
ans = 2000
>> X = 1-exp(-(WL-0.26053)./0.15994);
>> Y = Y.^2.28411;
>> find(X(:)<0) % indices of the negative values!
1
2
3
4
...  lots of sequential indices here
259
260
>> find(imag(Y(:))>0) % indices of where negative values -> complex result.
1
2
3
4
...  lots of sequential indices here
259
260

You can easily avoid taking the exponent of negative values using max:

>> Z = max(0,1-exp(-(WL-0.26053)./0.15994)).^2.28411
>> find(imag(Z)>0)
ans = []

Or set those values to NaN, or whatever suits your algorithm.

"getting Complex double in matlab while excel solves it"

In fact your screenshots make it clear that MATALB solves it (correctly, giving complex results for those negative values), whereas Excel just returns a an error message.

Asliddin Komilov 2019-6-15