Finding the turning point on a curve

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Muhammad Ghani
Muhammad Ghani le 1 Août 2022
Commenté : Muhammad Ghani le 1 Août 2022
Hello;
I have a question about finding the turning points on a graph. See the curve that is attached. Its basically a second derivate of the data, and I can see the maximum values (points of interests) at x=10, x=37. How can I automatically find those two indices for these two maximum points.
The input data is also attached.

Réponse acceptée

Torsten
Torsten le 1 Août 2022
v = [-0.000499915992890149
-0.00136716055731299
-0.00312202879147695
-0.00415143850349407
-0.00372399579905874
-0.00214247189597304
-4.88031918734373e-05
0.00186090040637454
0.00310143550393687
0.00352174551809843
0.00326203341992525
0.00261014952275744
0.00185224852020417
0.00118261260181092
0.000685142369381347
0.000361685775221777
0.000173644481175180
7.45001399801368e-05
2.65154258571087e-05
4.92923050407924e-06
-4.18585620462963e-06
-7.66464037567534e-06
-8.27292576250456e-06
-6.40533966605236e-06
-5.74990859836169e-07
1.30721733050333e-05
4.20520793933117e-05
9.94354287544991e-05
0.000205809520089438
0.000390149987664096
0.000687660786704286
0.00113216341113945
0.00174119870531302
0.00249441537474893
0.00331021806888304
0.00403084496564659
0.00442921778777076
0.00424844768230841
0.00327454998785455
0.00169124545504720
0.000814499263465100];
[pks,locs] = findpeaks(v)
pks = 2×1
0.0035 0.0044
locs = 2×1
10 37

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