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

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;

Answer 1

[dpb]

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.

[dpb]

...[Osita Onyejekwe Answer moved to comment--dpb]...

thanks, where do i get the locup function?

[dpb]

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).

Answer 2

[Osita Onyejekwe]

one more question, for the second derivative why did you do

==2

and

==-2

[dpb]

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.

Answer 3

[Osita Onyejekwe]

and for the plot

ylim([-1.05 1.05])

why -1.05 and 1.05 thank you that is all my questions!! :)

[dpb]

Just so the maxima/minima will show up on the plot w/o being obscured by the outer box of the axes...