intersection of lines and curves

i have data in which line is plotted on a signal having pulses. I want to find out all points of intersections between the line and the curve. Right now i am able to get single point of intersection on only one side of pulse. I also need point of intersection on the other side.

[SHANTANU KSHIRSAGAR]

also in some cases getting wrong intersection, in remaining getting error of not having unique points(intersect1).

Answer 1

[Star Strider]

Try this:

D = load('intersect1.mat');
x_curve = D.x_curve;
y_curve = D.y_curve;
x_line  = D.x_line;
y_line  = D.y_line;

x_curve(x_curve == 0) = [];                                             % Eliminate Spurious Points
y_curve(y_curve == 0) = [];                                             % Eliminate Spurious Points

zci = @(v) find(v(:).*circshift(v(:), -1, 1) <= 0);                     % Returns Approximate Zero-Crossing Indices Of Argument Vector (>= R2016b)
icix = zci(y_curve - y_line(1:end-2));

figure
plot(x_curve, y_curve,    x_line, y_line)
hold on
plot(x_curve(icix), y_curve(icix), 'dr')
hold off
grid

producing:

The x-values of the intersections are about 0.135 and 0.149. (They are: x_curve(icix).)

The points are dense enough that it is not necessary to interpolate them to get reasonably precise values for the intersections.

.

[SHANTANU KSHIRSAGAR]

thanks Star strider it works great.

[Star Strider]

As always, my pleasure!

I am somewhat concerned by Walter Roberson’s question about this being homework. If it is, you cannot use my code (or any code that you did not write yourself). If it is legitmate research, then it is permissible to use it.

[SHANTANU KSHIRSAGAR]

No it is not a homework, i am using it in addition to the code i have formulated and tring to complete an algrithm with possible techniques.

If i am using this technique in some cases i am getting multiple points can i eliminate them to find a single nearest point

[Star Strider]

The code appears to be operating correctly, according to the plot image. I do not know what to suggest because I do not know what points you want to keep and what points (if any) you want to discard, or if you simply want to classify them somehow.

[SHANTANU KSHIRSAGAR]

If i get a single point at each state , here the line crossing the curve at intersections i may be able to find the time required to pulse to reach from first state(bottom line) to the top state(upper line) that will be coordinates of the points obtained. But in case there are multiple point it cannot be calculated.

needed is the a single point at each intersection nearest to the intersection, which we are getting multiple points due to zero crossing calculations i think.

[Star Strider]

Rather than write specific code for this, if you have the Signal Processing Toolbox, use the risetime function. (There are also related functions.) It is likely more robust than anything I can write.

Try this:

D = load('intersect3.mat');
x_curve = D.x_curve;
y_curve = D.y_curve;
% x_line = D.x_line;
% y_line = D.y_line;

stlvls = [0.75 3.1];
tolv = [min(y_curve) max(y_curve)];
pcrl = 100 * stlvls/diff(tolv);
[r,lt,ut,ll,ul] = risetime(y_curve, x_curve, 'PercentReferenceLevels',pcrl);

figure
plot(x_curve, y_curve)
hold on
plot([lt ut], [ll ul], 'dr')
hold off
grid

Here, the first output ‘r’ is the rise time between levels, the next two outputs are the times for both levels, and the last two are the levels themselves.

If you just want to get the intersections, my earlier code will do what you want. If you want transition times, state levels, and other parameters, the Signal Processing Toolbox functions are the best options.

(My spologies for the delay. I was locked out of my account for a while this morning. I still have no idea what that was about.)

[SHANTANU KSHIRSAGAR]

thanks you Star Strider, the answer was helpful . I will try to add some robustness to it as i do not have access Signal Processing toolbox.

[Star Strider]

As always, my pleasure!

O.K. In that event, see if this does what you want.

Try this:

D = load('intersect3.mat');
x_curve = D.x_curve;
y_curve = D.y_curve;
x_line = D.x_line;
y_line = D.y_line;

stlvls = [0.75 3.1];
[ymin,mnix] = min(y_curve);
[ymax,mxix] = max(y_curve);

tlvl = interp1(y_curve(mnix:mxix), x_curve(mnix:mxix), stlvls, 'linear');
fprintf('risetime = %.6f\n', diff(tlvl))

figure
plot(x_curve, y_curve)
hold on
plot(tlvl, stlvls, 'dr')
hold off
grid

producing:

The rise time between the two levels is the difference between the ‘tlvl’ elements, and the times are those elements. This is subject to the constraints of interp1, that are explained in its documentation. The code works here, although it may not be robust for all waveforms.

For instances such as the earlier waveform, it would be necessary to sort the indices of the ‘icix’ vector for each intersecting line (assuming two different levels and four elements, so concatenate the elements in one vector, then sort), and then use interp1 if necessary to get the associated times. If the lines and curve have a large number of points, it may not be necessary to interpolate them to get accurate time estimates. Just use the indices, and do the appropriate arithmetic on the times derived from the indices to get the risetime and falltime.

[SHANTANU KSHIRSAGAR]

Thank you star , I worked on your code for some time,made some changes, additions it is really helpful .

[Star Strider]

As always, my pleasure!

Answer 2

[KALYAN ACHARJYA]

Have you tried this way?

data1
data2
diff=data1-data2;
small_value=??  % Define small value to deal with samll differenece also 
% Make the lower as possible, though ensure that it donot missed the interconnection
idx=find(diff<small_value);

If the data1 and data2 are non floating numbers, you can do the following

idx=find(data1==data2)

[SHANTANU KSHIRSAGAR]

ok can't we find the actual point of intersections seen in the figure. According to data all those points dont match to get exact intersection.

[KALYAN ACHARJYA]

Please refer

https://www.mathworks.com/matlabcentral/answers/30915-finding-intersection-point-of-the-lines

https://www.mathworks.com/help/matlab/ref/double.intersect.html;jsessionid=75d3602afe695818146b692cc138

[SHANTANU KSHIRSAGAR]

I have seen these answers previously. I wanted to know if there is intersection in the figure but in data we do not have a comman point then can we extract that exact position (point of intersection), instead of getting approximate or wrong point.

[SHANTANU KSHIRSAGAR]

If i am using the suggested method in the following link:

https://www.mathworks.com/matlabcentral/fileexchange/22441-curve-intersections

i am getting an error as : Unrecognized function or variable 'InterX'.

how do i try to make use of this function, would it help?

[Walter Roberson]

You need to install InterX for that. You can use the Add-On Explorer to do that. In the top menu ribbon, Add-Ons -> Get Add-Ons

[SHANTANU KSHIRSAGAR]

can we have other solution than using the interx

for solving the error and getting the accurate intersected coordinates in attached figures

[Walter Roberson]

https://www.mathworks.com/matlabcentral/fileexchange/11837-fast-and-robust-curve-intersections

https://www.mathworks.com/matlabcentral/fileexchange/8908-curve-intersect-2

[SHANTANU KSHIRSAGAR]

i am not able to use the fuctions which are mentioned in the links. If we need to do it without these external functions is it possible .

[Walter Roberson]

Do you mean that you are not permitted to use the functions there, or that you have technical problems when you try to use those functions?

If you are not permitted to use the functions there, then I would be concerned that if I were to give you code for this purpose, that you would not be permitted to use it.

[SHANTANU KSHIRSAGAR]

thanks @Walter Roberson and @KALYAN ACHARJYA