Integral of a matrix (x = integration(v)dt) to find x in each time step - how to get it?

Question

Ashfaq Ahmed 2022-3-25

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1680574-integral-of-a-matrix-x-integration-v-dt-to-find-x-in-each-time-step-how-to-get-it

编辑： Ashfaq Ahmed 2022-3-25

采纳的回答： Torsten

在 MATLAB Online 中打开

Dear altruists,

I have these (5x1 size) time steps -

t1 = 5.000;
t2 = 5.005;
t3 = 5.015;
t4 = 5.020;
t5 = 5.025;

For each of these time steps, I have this (8x5 size) velocity matrix -

0.223886255942431	0.766594808762019	0.777477860918720	1.46402363417739	NaN
NaN	                0.666581311838176	0.440897758695197	0.786687514042304	NaN
NaN	                0.760744913959832	0.726179926996350	1.04378898162688	0.222809516150796
NaN	                0.275556537143772	0.805109395242995	0.500274785771561	1.14165961699158
NaN	                0.388113337549528	0.895797227421048	0.868957410079128	0.906645475357619
0.514726870663309	1.37194840648291	1.17901149918237	0.267171163532900	0.679251749599892
0.700833065925187	0.843516259310205	0.692695783390200	0.206484971298492	0.773597719024071
0.748304643154520	0.433574718009251	0.551547292426642	0.346941328685530	0.738716957530404

How can I intigrate it (x = integration(v)dt) in such way that I get the position of x in every time step? I mean, I want x1,x2,x3,x4, and x5 :)

Thank you so much!

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Ashfaq Ahmed 2022-3-25

Yes, you interpreted it right. Those are 7 different person's data. There are NaN values in the problem, but they were some other values (say, 0) how would I do the calculation?

Ashfaq Ahmed 2022-3-25

And the innitial x0 is 0 for all the rows.

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Answer 1

Torsten 2022-3-25

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1680574-integral-of-a-matrix-x-integration-v-dt-to-find-x-in-each-time-step-how-to-get-it#answer_926899

编辑：Torsten 2022-3-25

在 MATLAB Online 中打开

After replacing the NaN values in the matrix, just do

T = [5 5.005 5.015 5.020 5.025];
A = [0.223886255942431	0.766594808762019	0.777477860918720	1.46402363417739	0 
    0	                0.666581311838176	0.440897758695197	0.786687514042304	0 
    0	                0.760744913959832	0.726179926996350	1.04378898162688	0.222809516150796
    0	                0.275556537143772	0.805109395242995	0.500274785771561	1.14165961699158
    0	                0.388113337549528	0.895797227421048	0.868957410079128	0.906645475357619
    0.514726870663309	1.37194840648291	1.17901149918237	0.267171163532900	0.679251749599892
    0.700833065925187	0.843516259310205	0.692695783390200	0.206484971298492	0.773597719024071
    0.748304643154520	0.433574718009251	0.551547292426642	0.346941328685530	0.738716957530404];
X0 = zeros(size(A,1),1);
for i = 1:size(A,1)
    X(i,:) = X0(i) + cumtrapz(T,A(i,:));
end
X