doing derivative using diff(Y)/dT makes the vector shorter
47 views (last 30 days)
Show older comments
Yuji Zhang
on 14 Jun 2013
Commented: Felipe Padua
on 12 Oct 2021
Hi everybody,
I'm doing derivative of a curve Y-T I think it's:
T = linspace(-t, t, n); Y = somefuction; dT = T(2)-T(1); DY1 = diff(Y)/dT;
But then DY1 is one element shorter than Y. How do people usually deal with this?
I'm currently dealing with this by shorten the T axis:
plot( T(1:end-1), DY1 );
I don't know whether this is the best way... Is there are relative standard way? Let me know. Thanks everyone~
0 Comments
Accepted Answer
Walter Roberson
on 14 Jun 2013
>> dT = 1;
>> x = 1:dT:3;
>> y = x.^2;
>> diff(y) ./ dT
3 5
>> diff('x^2', 'x')
2 * x
>> subs( diff('x^2', 'x'), 'x', x(1:length(y)) )
2 4
Thus we can see that using diff(y)/dT does not give us the same results as if we worked symbolically.
Question then: at what x values are [3 5] the correct derivative according to symbolic methods ? 2 * xp = [3 5]... by examination, xp must be [3/2 5/2]. Which, by no coincidence at all is the evaluation at the midpoints between x(n) and x(n+1) -- (x(n) + x(n+1))/2, or compactly (x(1:end-1) + x(2:end))/2
If we step back for a few seconds, we can see that using the numeric formula diff(y) ./ dT assigns the entire difference y(n) to y(n+1) as if it were at x(n), but that is not how derivatives work: derivatives are the tangent around x(n) and so y(n-1) must be taken into account, not just y(n) and y(n+1). Easiest resolution is to use x(n) and x(n+1) and y(n) and y(n+1) to construct the slope associated with the midpoint (x(n) + x(n+1))/2
plot( (T(1:end-1)+T(2:end))/2, DY1 )
If, however, you need to a slope at each x(n), then you have problems with the definition of slope at the endpoints. You might, in that case, wish to use the definition predefined:
plot( T, gradient(T) )
2 Comments
Walter Roberson
on 15 Jun 2013
I have no recommendation. Both approaches are valid in different situations. When a task requires a derivative at every point, I study why it requires that in the circumstances, and use whatever endpoint calculations are most suitable for the circumstances. More often, perhaps, I would use the interior points only, in order to avoid the problem.
More Answers (2)
Azzi Abdelmalek
on 14 Jun 2013
Edited: Azzi Abdelmalek
on 14 Jun 2013
Edit
(x(2)-x(1))/(t(2)-t(1)) correspond to the approximative right derivative at the point(t(1),x(1)). The last point is (t(n-1),x(n-1)), which means that you are doing right
Iain
on 14 Jun 2013
How I normally do it:
average_slope_between_y1_and_y2 = diff(Y)./diff(t);
middle_of_the_time_between_y1_and_y2 = (t(2:end)+t(1:end-1))./2;
Alternatively, fit a curve to your data, and differentiate that.
3 Comments
Iain
on 15 Jun 2013
Two consecutive points on your curve "y" :P
Maybe I should have them put my name in all caps...
Felipe Padua
on 12 Oct 2021
You could also use
middle_of_the_time_between_y1_and_y2 = movemean(t, 2, 'endpoints', 'discard')
See Also
Categories
Find more on Calculus in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!