Basic 'for' loop iteration

I am struggling with an iteration loop.
I have a vector containing all the velocity values.
I need to create a vector for the displacement values to be plotted against time.
Now mathematically this is rather simple however I can't get my 'for' loop to work at all! It keeps just returning the values of velocity on the graph. The biggest indicator that the script is wrong is the fact that displacement should never decrease (there are no negative velocity values) yet mine clearly does
Here is my current loop:
for k = 2:length(v);
distance(k) = (0.1*(v(k-1))) + (0.1*(v(k)));
end
Note that v = velocity, k is the loop counter, and the time step between v values is 0.1

3 comentarios

Stephen23
Stephen23 el 23 de Mayo de 2016
Editada: Stephen23 el 24 de Mayo de 2016
Your code is equivalent to this (using vectorized code):
[0, 0.1 * (v(1:end-1) + v(2:end))]
basically you take 0.1 times the sum of adjacent pairs of values, with a leading zero stuck on the end. If this is not what you intended to calculate, then you should tell us what you are trying to do, and provide correct input and output examples for us to try our code on.
sgc321
sgc321 el 23 de Mayo de 2016
So would that sit inside the 'for' loop or is it a stand alone statement? as I cant see a k variable
Stephen23
Stephen23 el 23 de Mayo de 2016
Editada: Stephen23 el 23 de Mayo de 2016
No loop is required. I just showed you a simpler way of doing what your code actually does, without any loop. It does not solve your task of implementing the trapezoidal integration!

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Respuestas (1)

Stephen23
Stephen23 el 23 de Mayo de 2016
Editada: Stephen23 el 24 de Mayo de 2016

0 votos

Why bother using a loop? MATLAB code can be so much neater!
>> v = [1,2,3,2,3,1];
>> s = 0.1; % step size
>> x = s*(1:numel(v));
>> sum(s*(v(2:end)+v(1:end-1))/2)
ans = 1.1000
But really there is no point in reinventing the wheel: just use trapz:
>> trapz(x,v)
ans = 1.1000

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el 23 de Mayo de 2016

Editada:

el 24 de Mayo de 2016

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