Finding the first intersection point between 2 lines

Hello all,

I need to find the time constant for which I plotted a line at y = 63% of the maximum value (y_max)

The other line comes from a Simulink model in which I can switch between different parameters, the output (y) differs every time (the line isn't plotted from a function). I'm using the following code which uses the 'interp1' function:

y_tau = 0.63*y_max;

[y,idx] = unique(y);

t = t(idx);

t_constant = interp1(y,x,y_max)

However, this always gives the second intersection point in the graph (see picture below). In this particular case:

t_constant = 190.2194

.

How can I get the first intersection point using this code or any other method? Thank you in advance.

Answer 1

[Mathieu NOE]

hello

try this

code is fairly simple to use as you only have to specify the y threshold value

threshold = 0.63*max(y); % 63% of peak amplitude

% dummy data
n = 1000;
x = 10*(0:n-1)/n;
y = 1+square(x);
[B,A] = butter(1,0.004);
y = filter(B,A,y);

% main code
threshold = 0.63*max(y); % 63% of peak amplitude

t0_pos1 = find_zc(x,y,threshold);
t0_pos1 = t0_pos1(1); % keep only first crossing point

figure(1)
plot(x,y,'b',x,threshold*ones(size(x)),'k--',t0_pos1,threshold*ones(size(t0_pos1)),'*r','linewidth',2,'markersize',12);grid on
legend('signal','threshold','signal positive slope crossing points');

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [Zx] = find_zc(x,y,threshold)

% positive slope "zero" crossing detection, using linear interpolation
y = y - threshold;
zci = @(data) find(diff(sign(data))>0);  %define function: returns indices of +ZCs
ix=zci(y);                      %find indices of + zero crossings of x
ZeroX = @(x0,y0,x1,y1) x0 - (y0.*(x0 - x1))./(y0 - y1); % Interpolated x value for Zero-Crossing 
Zx = ZeroX(x(ix),y(ix),x(ix+1),y(ix+1));

end

Answer 2

[KSSV]

Try this: https://in.mathworks.com/matlabcentral/fileexchange/22441-curve-intersections

Answer 3

[Torsten]

If you have the t-y values as vectors available, you can use this code to determine the t-values where y-y_tau=0:

y_tau = 0.63*y_max;
y = y - y_tau;
i = find(y(1:end-1).*y(2:end) <= 0)
t_root = t(i)-y(i).*(t(i+1)-t(i))./(y(i+1)-y(i))