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

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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!
  3 件のコメント
Ashfaq Ahmed
Ashfaq Ahmed 2022 年 3 月 25 日
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
Ashfaq Ahmed 2022 年 3 月 25 日
And the innitial x0 is 0 for all the rows.

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採用された回答

Torsten
Torsten 2022 年 3 月 25 日
編集済み: Torsten 2022 年 3 月 25 日
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
  1 件のコメント
Ashfaq Ahmed
Ashfaq Ahmed 2022 年 3 月 25 日
編集済み: Ashfaq Ahmed 2022 年 3 月 25 日
Yes, that's exactly what I wanted! Thank you so much Torsten :) You are amazing!

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