How to locate X value for a given Y value
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David Harra
le 1 Fév 2023
Réponse apportée : Star Strider
le 2 Fév 2023
I have Data which is Time along the x-axis and amplitude along the y -axis.
I want to know the exact time at a given value of amplitude. Its probably just one line of code but I can't seem to figure it out.
Any help would be appreciated.
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Star Strider
le 2 Fév 2023
In the absence of the data, getting an independent variable valuex ‘x’ from an dependent variable value ‘y’ is relatively straightforward.
Try soemthing like this —
x = linspace(0, 20, 201);
y = besselj(0, x);
yval = linspace(-0.2, 0.2, 6);
for k1 = 1:numel(yval)
zxi = find(diff(sign(y-yval(k1))));
for k2 = 1:numel(zxi)
idxrng = max(1,zxi(k2)-1) : min(numel(x), zxi(k2)+1);
xv(k1,k2) = interp1(y(idxrng), x(idxrng), yval(k1));
end
end
for k1 = 1:size(xv,1)
xv(k1,xv(k1,:)==0) = NaN;
end
figure
plot(x, y)
hold on
for k1 = 1:size(xv,1)
plot(xv(k1,:), ones(size(xv(k1,:)))*yval(k1), 's')
end
yline(yval, ':')
hold off
grid
xlabel('X')
ylabel('Y')
.
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the cyclist
le 1 Fév 2023
Modifié(e) : the cyclist
le 1 Fév 2023
It would be helpful if you uploaded the data. You can use the paper clip icon in the INSERT section of the toolbar.
It seems like you need either
% The exact value appears in the amplitude vector, and want to find what Time it occurs
Time(amplitude == value) % be careful here if you are comparing floating point values
or
% The exact value does not appear in the amplitude vector, so you want the
% first Time where you surpass that value
Time(find(amplitude>value,1)) % this assume the values of amplitude and time are sorted appropriately
You might also want to interpolate between times, but be more exact. Then you'll need the interp1 function.
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