How to detect the fluctuating point (cut-off point) of a signal?

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Meshooo 2014-12-9

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https://ww2.mathworks.cn/matlabcentral/answers/165975-how-to-detect-the-fluctuating-point-cut-off-point-of-a-signal

评论： Meshooo 2015-1-6

采纳的回答： dpb

Dear all,

I have a simple signal that I show below with blue color:

I want to find the point Xc. Any idea how find this point? Is it the same point which I will get from the first derivative of the signal?

Thank you very much.

Meshoo

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Answer 1

dpb 2014-12-9

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https://ww2.mathworks.cn/matlabcentral/answers/165975-how-to-detect-the-fluctuating-point-cut-off-point-of-a-signal#answer_161810

http://www.mathworks.com/matlabcentral/answers/164913-difficulty-in-fitting-two-lines-with-polynomials

The first derivative of your trace is a smooth positive curve that will have 2nd derivative that will approach zero at both ends. But the intersection point isn't at an inflection point, no...

Above shows how to fit the piecewise linear segments; you'll probably want to limit the fitting points to the end areas to get the better approximation to long-term slopes as you've drawn above; otherwise the slopes will tend to "sag" to the middle point by the OLS solution over the full data set.

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Meshooo 2014-12-25

编辑：Meshooo 2014-12-25

在 MATLAB Online 中打开

At first, thank you very much for your kind help.

I will try to explain here the struggle point.

I have a noisy data that generates the following plot:

z = [147  148.3333333  149.6666667  150.3333333  151.3333333  152.3333333  155  155.6666667  156  157.3333333  158  160  162.3333333  163  164.6666667  165  167  167.6666667  170.3333333  171.6666667  173.6666667  174.6666667  175.6666667  177.3333333  179  180.6666667  181.6666667  182.6666667  185.6666667  186.3333333  187.3333333  189  190.6666667  191.6666667  193.3333333  195  196  197  199.6666667  200.3333333  202  203  204.3333333  206.3333333  208  209.6666667  212  213.3333333  215  216.6666667  217.6666667  219.6666667  221.6666667  223.3333333  225  226.6666667  228  230  231.3333333  232.6666667  235.3333333  235.6666667  237.6666667  238.3333333  240  241.6666667  243  243  244  245.3333333  246.6666667  248  249.3333333  251  253.6666667  254.3333333  254.6666667  254.6666667  256.3333333  258  258.3333333  258.6666667  260.6666667  260.6666667  261.6666667  262.3333333  265  265  265  266  266  266.3333333  266.6666667  266.6666667  266.6666667  266.6666667  267  267.6666667  267.6666667  267.6666667  267.6666667  267.6666667  267.6666667  267.6666667  268  268  268  268  268  268  268  268  268  268  268  268  268];
x = 1:117;
plot(x, z, 'r-', 'LineWidth', 2);

From the plot, I want to find the the point Xc in the X axis where the plot change its slop, maybe somewhere in the place circle I show below

First idea:

In the beginning I thought we can find Xc point using the first or second derivatives. But maybe this will not work because I need to smooth the data before applying the derivatives. Is that correct? if so then which smoothing technique to use?

Second idea:

To find the crossing point like the way you showed earlier using piecewise regression or polyfil as suggested by Imageanalyst.

So which way you think is good to find Xc?

Thank you very much.

Meshooo

dpb 2014-12-25

编辑：dpb 2014-12-25

在 MATLAB Online 中打开

Why don't you try it and see what you think suits your needs best? The piecewise takes only a few more line of code (presuming you've copied my referenced function piecewise that I linked to earlier)

>> m1=(z(end)-z(1))/(x(end)-x(1));  % first section slope estimate
>> m2=0;                            % second slope 
>> b1=z(1);                         % first intercept estimate
>> c=0.8*x(end);                    % and breakpoint guess
>> coeff=nlinfit(x.',z.',@piecewise,[b1 m1 c  m2])
coeff =
143.1515    1.4592   82.5534    0.1697

NB: The x,y input vectors to nlinfit functional must be column vectors, hence the ".'" transpose.

Now add the fitted curve to the existing plot--

>> zhat=piecewise(coeff,x);  % the fitted curve result at the input points
>> hold on
>> plot(x,zhat,'b','linewidth',2)

Looks pretty good here. Only question is whether you want the best overall fit (in RMS terms) as this does or the line segments to more nearly match the end tangents to the curve. If the latter use the subset of the data points at the ends to fit instead of the whole thing. I've already showed that for the piecewise method following IA's example. As expected, the two provide the same answer given the same input as the piecewise regression is OLS linear regression for the two sections just finds the intersection as part of the model instead of separately.

It appears there's a little positive curvature at the leading portion of the curve from x=~[0:50] and then a small negative curvature until the breakpoint. Again, it's your choice as to what feature(s) you're most interested in; we have no way to know that.

Meshooo 2014-12-26

OK, I understand your comment. Indeed thank you very much.

Meshooo 2015-1-6

I have a further related question posted here, please check it when you have time..

http://www.mathworks.com/matlabcentral/answers/169030-how-to-find-the-second-interestion-between-two-plots

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Answer 2

Image Analyst 2014-12-21

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https://ww2.mathworks.cn/matlabcentral/answers/165975-how-to-detect-the-fluctuating-point-cut-off-point-of-a-signal#answer_162995

在 MATLAB Online 中打开

Try this code. It will produce the image that follows it.

clc;    % Clear the command window.
close all;  % Close all figures (except those of imtool.)
clear;  % Erase all existing variables. Or clearvars if you want.
workspace;  % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 25;
% Make up some demo equation.
x = 1 : 500;
y = sqrt(x);
% Plot it.
plot(x, y, 'b-', 'LineWidth', 2);
grid on;
title('Y = Sqrt(X)', 'FontSize', fontSize);
xlabel('X', 'FontSize', fontSize);
ylabel('Y', 'FontSize', fontSize);
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);
% Give a name to the title bar.
set(gcf, 'Name', 'Demo by ImageAnalyst', 'NumberTitle', 'Off') 
% Extract left and right segments of the curve.
numPoints = 15;  % Whatever.
x1 = x(1:numPoints);
y1 = y(1:numPoints);
x2 = x(end-numPoints : end);
y2 = y(end-numPoints : end);
% Draw a green line going down from there.
line([x1(end), x1(end)], [0, y1(end)], 'Color', [0, .5, 0], 'LineWidth', 2);
line([x2(1), x2(1)], [0, y2(1)], 'Color', [0, .5, 0], 'LineWidth', 2);
% Find the equations of the line for each segment.
leftCoefficients = polyfit(x1,y1,1)
rightCoefficients = polyfit(x2,y2,1)
% Plot the lines in red.
yl = ylim(); % Save original y axis range.
hold on;
yLeft = polyval(leftCoefficients, x);
plot(x, yLeft, 'r-', 'LineWidth', 2);
yRight = polyval(rightCoefficients, x);
plot(x, yRight, 'r-', 'LineWidth', 2);
ylim(yl);
% Then set the two equations to each other and solve for xc:
xc = (rightCoefficients (2) - leftCoefficients (2)) / (leftCoefficients(1) - rightCoefficients(1))
yc = leftCoefficients(1) * xc + leftCoefficients(2);
% Draw a green line going down from there.
line([xc, xc], [0, yc], 'Color', [0, .5, 0], 'LineWidth', 3);
plot(xc, yc, 'o', 'MarkerSize', 25, 'Color', [0, .5, 0], 'LineWidth', 3);
% Put up text
message = sprintf('(xc, yc) = (%.1f, %.1f)', xc, yc);
text(1.1*xc, 0.5*yc, message, 'Color', [0, .5, 0], 'FontSize', fontSize);

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Meshooo 2014-12-22

Seems to be what I want and will test it soon. Thank you very very much..

dpb 2014-12-22

编辑：dpb 2014-12-22

在 MATLAB Online 中打开

I used the piecewise regression on your example dataset, IA, with initial guesses for the two slopes as

>> m1=Y(npts)/X(npts);
>> m2=(Y(end)-Y(end-npts+1))/(X(end)-X(end-npts+1));
>> b1=0;
>> c=X(end)/10;
>> coeff=nlinfit(X,Y,@piecewise,[b1 m1 c  m2])
coeff =
  1.1459    0.1940   58.0548    0.0225

The yc value is then

>> xc=coeff(3); yc=xc*coeff(2)+coeff(1)
yc =
 12.4087
>>

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Answer 3

Image Analyst 2014-12-20

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/165975-how-to-detect-the-fluctuating-point-cut-off-point-of-a-signal#answer_162941

在 MATLAB Online 中打开

Take the first and last 10% or so of the points and fit a line.

numPoints = 10;  % Whatever.
x1 = x(1:numPoints);
y1 = y(1:numPoints);
x2 = x(end-numPoints : end);
y2 = y(end-numPoints : end);
leftCoefficients = polyfit(x1,y1,1);
rightCoefficients = polyfit(x2,y2,1);

Then set the two equations to each other and solve for xc:

xc = (rightCoefficients (2) - leftCoefficients (2)) / (leftCoefficients(1) - rightCoefficients(1));

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Image Analyst 2014-12-21

All right, I made it easy for you. See the code below in my other answer. Vote for my answer if it's useful to you.

Meshooo 2015-1-6

I have a further related question posted here, please check it when you have time..

http://www.mathworks.com/matlabcentral/answers/169030-how-to-find-the-second-interestion-between-two-plots

请先登录，再进行评论。

How to detect the fluctuating point (cut-off point) of a signal?

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