decreasing point on curve

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Housam
Housam on 30 Sep 2019
Commented: Housam on 8 Oct 2019
Hello,
What could determine the point where the curve start decreasing? how to best specify this point?
Thank you.

Accepted Answer

Aubai
Aubai on 1 Oct 2019
Edited: Aubai on 1 Oct 2019
Hi,
I think the best way to do so has already been described here and here:
basically there are two ways described there:
1- triangle thresholding
2- local maxima of curvature
i prefer the first one as i think it directly answer your question (the second one is for more spesific tasks)
% here is a small code to help you start
% Lets say your data are defined in (x,y) values
% then do the following
time = x;% here you should enter your data name
data = y;% here you should enter your data name
% Basically it draws a line from the peak to the tail and then draws
% perpendicular lines from that hypotenuse line to the curve.
% The longest (smollest) perpendicular line from the hypotenuse to the curve
% indicates the "corner" of the curve. Maybe that will work for you. (this discribtion was taken from the reference before)
% Two endpoints on the curve "data"
x = [time(1) time(end)];
y = [data(1) data(end)];
% The slope of the line connecting the two endpoints
m = ( y(2) - y(1) )/( x(2) - x(1) );
pm= - 1 / m;
figure();hold all
hP = plot(time,data,'b',x,y,'g',x(1),y(1), 'ro');grid on
yl = ( (m * time) + (m^2 * data) - (m * x(1)) + y(1) )/(1+m^2);
xl = T1 - m*(yl - data);
d2 = (xl - time).^2 + (yl - data).^2;
perpDist = d2;
[val_max, idx_max] = max(perpDist);
[val_min, idx_min] = min(perpDist(2:end-1));
[val_maxL, idx_maxL] = max(perpDist(1:idx_min));
plot(time,data,'b',time(idx_max),data(idx_max),'ro',time(idx_min),data(idx_min),'ro',time(idx_maxL),data(idx_maxL),'ro');grid on;hold all; plot(time,perpDist); hold off
I hope this answer your question (how accurate the point will depend on your massy data)
Turnin_points_Curve.emf
  3 Comments
Housam
Housam on 8 Oct 2019
There is an easier, straight forward solution. by using comparision> instead of -subtraction. Now, the result are absoultly perfect and better than the one from the built in function. Thanks a lot.

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More Answers (2)

Steven Lord
Steven Lord on 1 Oct 2019
The ischange function may be of use to you. The different method inputs detect different change points. Experiment with the various options.
>> x = -10:0.125:10;
>> y = -tanh(x);
>> plot(x, y)
>> ic = ischange(y, 'linear');
>> hold on
>> plot(x(ic), y(ic), 'ro')
If you're using release R2019b or later, you can facilitate this experimentation using the Find Change Points Live Editor Task in the Live Editor. This will allow you to interactively change the method and the parameters and see which change points MATLAB detected immediately.
  1 Comment
Housam
Housam on 1 Oct 2019
Much appreciated Steven. Indeed I am using the latest Matlab version, 2019, which admittedly using such functions, saves a lot of work and delivers performance.

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KALYAN ACHARJYA
KALYAN ACHARJYA on 30 Sep 2019
Edited: KALYAN ACHARJYA on 30 Sep 2019
One way:
Find the y, when y(i)<y(i-1) (First Case) iterate i from 2 to end
Second way:
data=y(i+1)-y(i); %Create the data, then apply diff for first positive/negative y value
  1 Comment
Housam
Housam on 30 Sep 2019
Unfortunately, this method does not work- The result to this method as shown below. However, this is not the point i am looking for.untitled2.png

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