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

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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;

采纳的回答

dpb
dpb 2016-10-31
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 个评论
dpb
dpb 2016-10-31
编辑:dpb 2016-10-31
...[Osita Onyejekwe Answer moved to comment--dpb]...
thanks, where do i get the locup function?
dpb
dpb 2016-10-31
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).

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更多回答(2 个)

Osita Onyejekwe
Osita Onyejekwe 2016-10-31
one more question, for the second derivative why did you do
==2
and
==-2
  1 个评论
dpb
dpb 2016-10-31
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.

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Osita Onyejekwe
Osita Onyejekwe 2016-10-31
and for the plot
ylim([-1.05 1.05])
why -1.05 and 1.05 thank you that is all my questions!! :)

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