How to find the zero crossing in x and time data sets?

How can I find the zero crossing in the data sets?
figure()
plot(x,t)

3 Kommentare

Nice plot. Just look at it -- you can easily find where the line crosses 0. Is there something else you might be looking for?
Yes we dont have 0.000000000000 so we have range is 10^-3 so all the value fall under 10^-3 are accepted. Now how can we do it as we have two sets x& t?

(Answers Dev) Restored edit

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 Akzeptierte Antwort

Scott MacKenzie
Scott MacKenzie am 9 Jun. 2021
Bearbeitet: Scott MacKenzie am 10 Jun. 2021
Here's what I put together. The variable fCross is what you are looking for.
% data from posted matlab.mat files
f = readmatrix('testdata1.txt');
t = readmatrix('testdata2.txt');
tiledlayout(3,1);
nexttile;
plot(t,f);
hold on;
axis([1 10 -1.2 1.2]);
nexttile;
fAbove = f .* (f >= 0);
fBelow = f .* (f <= 0);
area(t, fAbove, 'FaceColor', 'r');
hold on;
area(t, fBelow, 'FaceColor', 'g');
axis([1 10 -1.2 1.2]);
nexttile;
fCrossRaw = find(diff(fAbove>0));
fCross = fCrossRaw ./ length(t)*10; % as per axes
plot(fCross, zeros(1,length(fCross)), '*r');
hold on;
axis([1 10 -1.2 1.2]);
xline(fCross, 'color', [.7 .7 .7]);
yline(0, 'color', [.7 .7 .7]);

2 Kommentare

thank you sir.
@vimal kumar chawda You're welcome. I just updated my answer to make the 3rd plot look a bit better.

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Weitere Antworten (2)

Stefan Schuberth
Stefan Schuberth am 27 Jul. 2022
Bearbeitet: Stefan Schuberth am 27 Jul. 2022

2 Stimmen

If you have (x,y) data and want to do it without using loops try that:
i=find(y(1:end-1).*y(2:end)<0); % index of zero crossings
m=(y(i+1)-y(i))./(x(i+1)-x(i)); % slope
x0=-y(i)./m+x(i); % x coordinates of zero crossings linear interpolated
Joel Lynch
Joel Lynch am 9 Jun. 2021
Bearbeitet: Joel Lynch am 9 Jun. 2021
idx = find( f(2:end).*f(1:end-1)<0 )
Will return the left-hand indicies of cross-over points.
To get the exact X-values where the cross-over occurs, you would have to do some linear intepolation (inverted)
t_zero = zeros(size(idx));
for i=1:numel(idx)
j = idx(i); % Need both the index in the idx/t_zero vector, and the f/t vector
t_zero(i) = interp1( f(j:j+1), t(j:j+1), 0.0, 'linear' );
end
Note: this will fail if the cross-over happens on the last i value (i+1 would extend outside the range of the dataset)

4 Kommentare

yea, true it fails so how to fix it?
Simplest solution is to put the interpolation line in an if statement
if j< numel(f)
% do j:j+1 interpret
else
t_zero(i) = interp1( f(j-1:j), t(j-1;j), 0.0, 'linear' );
end
or something similar
It worked thanks.
Scott MacKenzie
Scott MacKenzie am 11 Jun. 2021
Bearbeitet: Scott MacKenzie am 29 Jun. 2021
Yes, Joel's code gives the exact cross-over point. Bear in mind, however, that this is exact for the linearly interpolated data. The actual data are empirical, so it is not possible to know the exact cross-over point.
It probably doesn't matter much in this case, since the data appear to be gathered at a high sampling rate.

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