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 Commenti

Stephen23
Stephen23 il 23 Mag 2016
Modificato: Stephen23 il 24 Mag 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 il 23 Mag 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 il 23 Mag 2016
Modificato: Stephen23 il 23 Mag 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!

Accedi per commentare.

Risposte (1)

Stephen23
Stephen23 il 23 Mag 2016
Modificato: Stephen23 il 24 Mag 2016
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

Categorie

Scopri di più su Loops and Conditional Statements in Centro assistenza e File Exchange

Richiesto:

il 23 Mag 2016

Modificato:

il 24 Mag 2016

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by