How can I find the Y value on an X–Y plot that corresponds to the tangent of the flattest part of a curve?

I have plots like the one attached. At Y >0, the curve plateaus (flattens) before it increases sharply. I need to find the Y value at which the plateau/flat area is flattest.

Does anyone know how to do this? I can't figure out a solution that gets the part of the curve that I want. Thank you.

[Turlough Hughes]

Can we depend on that initial pattern always happening? i.e. initially flat, then accelerates then decelerates again to the "flattest" point on the plateau? Can you provide some representative data please.

[Srh Fwl]

Thank you for your reply, Turlough. Yes, you can depend on that inital flat part. There won't be an increase until at least X=1. There is usualy a steeper part at high X. However, there may not always be a plateau at intermediate X. Or it may not be prominent.

I will attach a file that contains example data.

[Srh Fwl]

Here are some example data.

[dpb]

For the example, the only "flat" portion I see is for X<~3; is that the area of interest or is it the inflection region between ~3 < X < ~5?

If it's the latter, looking for the change in slope to locate the "hump" would seem to be the starting point.

[Adam Danz]

If you plot the slope of the line (see gradient) you'll see a dip toward y=0 at the area around ~3.5 but it doesn't quite reach 0 so it's not technically flat. You may want to set a threashold (slope ~2?) and identify the area I think you're refering to by searching for slopes that fall below the threshold after the initial rise of the slope curve. That means you'll exclude all slope values prior to the first peak of the slope curve which can be found by a change in sign of the second derivative of the raw data.

[Srh Fwl]

@Adam Danz, thank you very much for that suggestion. Would you happen to know how to search for slopes on a plot? Is there a Matlab function that does that? Thank you.

[dpb]

@Srh Fwl -- You've still not precisely answered the Q? of just what it is you're after here...is it the end of the "real" initial floor or an inflection point later that is the subject?

[Adam Danz]

@Srh Fwl, looks like Star Strider demonstrated that idea in a comment below the answer. Notice that the dip in the red curve is at the location I think you're refering to. If you need additional specific help, you may want to clarify the foggy areas pointed out by dpb.

Answer 1

[Star Strider]

T1 = readtable('https://www.mathworks.com/matlabcentral/answers/uploaded_files/718334/exampleData.txt')

T1 = 101×2 table
    Var1    Var2
    ____    ____

     NaN     0  
    0.15     0  
    0.25     0  
    0.35     0  
    0.45     0  
    0.55     0  
    0.65     0  
    0.75     0  
    0.85     0  
    0.95     0  
    1.05     0  
    1.15     0  
    1.25     0  
    1.35     0  
    1.45     0  
    1.55     0  

T1 = rmmissing(T1);                                                 % Remove 'NaN' Values
h = mean(diff(T1.Var1))

h = 0.1000

d2d1 = gradient(T1.Var2, h);                                        % Numerical Derivative

flatidx = find(abs(d2d1)<1E-14);                                    % Zero Slope (With Tolerance)
yflat = T1.Var2(flatidx)

yflat = 19×1

0 0 0 0 0 0 0 0 0 0

figure
plot(T1.Var1, T1.Var2)
hold on
plot(T1.Var1(flatidx), T1.Var2(flatidx), 'vr')
hold off
grid
legend('Data','Flat Section', 'Location','best')

This should also work with other data sets, although obviously I cannot test it with them.

.

[Srh Fwl]

@Star Strider, thank you very much. I apologize because my question wasn't clear enough. I need the y value of the flattest area at y >0. The curves start off with y close to 0 and it's where they again flatten (in my example, at y between 3 and 5) that I need the y value. Apologies.

[Star Strider]

Set the conditions in the find call to match what you want to define.

T1 = readtable('https://www.mathworks.com/matlabcentral/answers/uploaded_files/718334/exampleData.txt')

T1 = 101×2 table
    Var1    Var2
    ____    ____

     NaN     0  
    0.15     0  
    0.25     0  
    0.35     0  
    0.45     0  
    0.55     0  
    0.65     0  
    0.75     0  
    0.85     0  
    0.95     0  
    1.05     0  
    1.15     0  
    1.25     0  
    1.35     0  
    1.45     0  
    1.55     0  

T1 = rmmissing(T1);                                                 % Remove 'NaN' Values
h = mean(diff(T1.Var1))

h = 0.1000

d2d1 = gradient(T1.Var2, h);                                        % Numerical Derivative

flatidx = find((abs(d2d1)<1.0) & (T1.Var2 > 0));                    % Define Slope Criteria (With Tolerance)
yflat = T1.Var2(flatidx)

yflat = 6×1

0.0000 0.0004 0.0030 0.0123 0.0395 0.0982

figure
plot(T1.Var1, T1.Var2)
hold on
plot(T1.Var1, d2d1)
plot(T1.Var1(flatidx), T1.Var2(flatidx), '.r')
hold off
grid
legend('Data', 'Numeircal Derivative', 'Flat Section', 'Location','best')

Make appropriate changes to get the result you want.

.

Answer 2

[Turlough Hughes]

How robust this is depends on the consistency of that initial pattern, i.e. the initial acceleration followed by a period of deceleration (starting to plateau) until the "flattest" point where it then begins to accelerate again. This point between the initial deceleration and acceleration is also known as an inflection point, as mentioned by @dpb. It's also the point where where y is closest to being parallel to the x-axis in the region (where it is initially plateauing).

To find the inflection point we find the location where . I understand you want the second one as follows:

T = readmatrix('https://www.mathworks.com/matlabcentral/answers/uploaded_files/718334/exampleData.txt');
x = T(:,1);
y = T(:,2);

subplot(2,1,1)
plot(x,y,'LineWidth',3)
ylabel('f(x)'), xlabel('x')
set(gca,'fontSize',12)

subplot(2,1,2)
ypp = gradient(gradient(y,x),x); % second derivative of y w.r.t. x
plot(x,ypp,'LineWidth',3);

ylabel('f''''(x)')
xlabel('x')
set(gca,'fontSize',12)

idx = ypp > 0;
hold on, plot(x(idx),ypp(idx),'or','LineWidth',2)

iFlat = find(diff(idx)==1)+1; % y is most linear when it's second derivate, ypp, is equal to 0
iInflect = iFlat(2);          % the second time ypp becomes > 0 approximates the second inflection point.

plot(x(iInflect),ypp(iInflect),'sk','MarkerSize',10,'LineWidth',2)

subplot(2,1,1)
hold on, plot(x(iInflect),y(iInflect),'sk','MarkerSize',10,'LineWidth',2)

x(1:iInflect-1) = [];
y(1:iInflect-1) = [];

plot(x,y,'--r','LineWidth',2)
legend('Original Dataset','Second Inflection Point','New Dataset','Location','NorthWest')

[Srh Fwl]

Thank you very much, Turlough. Your solution does work but unfortunately I can only accept one answer and I put the first answerer through more inconvenience so chose that one. I appreciate your help.