How to find the indices of a point on a curve

Hi I have a curve with one maximum,I need to find the indices of the the two 0.707 points. Thanks

Answer 1

[dpb]

If you have the curve as computed points, see

doc interp1

Switch the normal X,Y meanings to interpolate with Y as the independent variable instead of X in this case. Also, NB: You'll need to do this in two separate calls, one with the values to the left and another with those to the right of the maximum as interp1 must have unique values for the interpolating function.

This will solve for the precise location; not necessarily integer. If you want the nearest index, then either a) round the above results or b) use

[~,ix]=min(abs(y-sqrt(2)/2*ymax));

Again you'll have to do the above piecewise accounting for the length of the subvectors in the returned indices as there's no guarantee the locations will be symmetric or the same on both sides except under very particular circumstances.

[yousef Yousef]

Sorry but I don't understand your answer Thanks

[dpb]

Well, I don't know what form your data are in...

[yousef Yousef]

Hi My data is in the form of vector

Answer 2

[Thorsten]

If y is your curve, in general you will not have values that are exactly 0.707 of your maximum. So the idea is to use the 2 values that differ the least from the desired value 0.707*max(y):

 [~, idx] = sort(abs(y-0.707*max(y)));
 idx = idx(1:2);

[yousef Yousef]

Thanks.Yes my curve is Y. I tried your suggestion but the results were not as my figure shows,may be because you use absolute value while I have plus &minus.Can we find the closest value to 0.707? Thanks

[Thorsten]

I can see no reason why the code should not work for this curve. The abs is important because else the smallest values would be negative values. What values do you get if you run the my code? Could you provide the data of the curve?

[yousef Yousef]

I have got 64 and 96

[Thorsten]

I see. These are the index values into your x. So if you have

 x = -90:0.5:90;
 x707 = x(idx)

 x([64 96])
 ans =
        -58.50        -42.50

Note that I figured out x=-90:0.5:90 by eye and trial and error from your plot. You have to use the actual values of x of course.

Answer 3

[yousef Yousef]

clear all clc Angle=[5 10 20 60 180 190 195 300 305]; collision=[]; for i=1:length(Angle)-1 d=[]; targetValue = Angle(i); tolerance = 5; diff = abs(Angle-targetValue); tt=find(diff>0 & diff <= tolerance);

if ( tt>0) d=[d 2]; else d=[d 1]; end collision=[collision;d]; end

[dpb]

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