find change points, point of inflection and concave up and concave down

22 Ansichten (letzte 30 Tage)
This is my code and I want to find the change points of my sign curve, that is all and I want to put points on the graph where it is concave up and concave down. (2 different shapes for concave up and down would be preferred. I just have a simple sine curve with 3 periods and here is the code below. I have found the first and second derivatives. I'm not sure where to go from here. I know have to set it equal to zero and solve for x, any algorithmic help would be appreciated
x = 1:500;
X = x;
T = 1;
Fs = 499;
N = T*Fs;
t = 0: 1/Fs : T;
Fn = 3; % this control the number of cycles/periods
deltax = 0.0020;
y_sine_25HZ = sin(Fn*2*pi*t);
y = y_sine_25HZ ;
drvY = diff(y)/deltax; % one unit shorter than y
drvY = [drvY , drvY(end)]; % making up for the missing point
secondDrvY = diff(drvY)/deltax;

Akzeptierte Antwort

dpb
dpb am 31 Okt. 2016
If you have the Signal Processing Toolbox, the simplest solution is to use findpeaks --
>> [~,locup]=findpeaks(y) % (+)ive peaks
locup =
43 209 375
>> [~,locdn]=findpeaks(-y) % (-)ive peaks (positive negative y)
locdn =
126 292 458
>>
Lacking it, to find the nearest point is--
>> dy=[0 sign(diff(y))];
>> find((diff(dy))==2)
ans =
126 292 458
>> find((diff(dy))==-2)
ans =
43 209 375
>>
are same locations as findpeaks located.
To show 'em,
>> plot(t,y)
>> ylim([-1.05 1.05])
>> hold on
>> scatter(t(locup)',y(locup)',30,'r','*')
>> scatter(t(locdn)',y(locdn)',30,'g','d','filled')
>>
You can also knowing these locations use them and a range on either side with
>> interp1(y(locup-1:locup+1),t(locup-1:locup+1),1,'cubic')
ans =
0.0828
find an approximation for t to be slightly better resolution than perhaps the actual point given the resolution in t.
  2 Kommentare
dpb
dpb am 31 Okt. 2016
Bearbeitet: dpb am 31 Okt. 2016
...[Osita Onyejekwe Answer moved to comment--dpb]...
thanks, where do i get the locup function?
dpb
dpb am 31 Okt. 2016
It's not a function; it's the optional second return from the findpeaks function giving the location of the found peaks, not the values (the first, default return value).

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (2)

Osita Onyejekwe
Osita Onyejekwe am 31 Okt. 2016
one more question, for the second derivative why did you do
==2
and
==-2
  1 Kommentar
dpb
dpb am 31 Okt. 2016
I'll leave that one as "exercise for the student", I think... :)
Look up the sign function and think about what happens when apply it to differences of a function changing slope.

Melden Sie sich an, um zu kommentieren.


Osita Onyejekwe
Osita Onyejekwe am 31 Okt. 2016
and for the plot
ylim([-1.05 1.05])
why -1.05 and 1.05 thank you that is all my questions!! :)
  1 Kommentar
dpb
dpb am 31 Okt. 2016
Just so the maxima/minima will show up on the plot w/o being obscured by the outer box of the axes...

Melden Sie sich an, um zu kommentieren.

Community Treasure Hunt

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

Start Hunting!

Translated by