find value corresponding X value to a particular Y value

a = 5;
n = 1:4;
theta_teta = -10;
delta_theta = 40;
theta = -10:1:30;

for j = 1: length (n);
    for i = 1:length(theta)
        f(i) = 1-exp(-a*((theta(i)-theta_teta)/delta_theta)^n(j));
    end
    plot (theta,f)
    max_f = sum(f)
    hold on
end

Hi;

I have a plot which is shown in the image above. I want to find the x value corresponding to 0.5 and 0.9 of the maximum y values for the lines i.e. at 0.5 and 0.9 what is the respective x values for the different lines.

Answer 1

[Image Analyst]

Would interp1 be accurate enough for you? If so, try this:

a = 5;
n = 1:4;
theta_teta = -10;
delta_theta = 40;
theta = -10:1:30;

for j = 1: length (n);
  for i = 1:length(theta)
    f(i) = 1-exp(-a*((theta(i)-theta_teta)/delta_theta)^n(j));
  end
  plot (theta, f, 'LineWidth', 2);
  max_f = sum(f);
  fprintf('For n = %d, the sum of f (for some reason called max_f) = %f.\n', j, max_f);
  grid on;
  hold on;
  
  % Find x value for y = 0.5
  x50(j) = interp1(f, theta, 0.5);
  line([x50(j), x50(j)], [0, 0.5], 'LineWidth', 2, 'Color', 'k');
  fprintf('For n = %d, y = 0.5 at x = %f.\n', j, x50(j));
  % Find x value for y = 0.9
  x90(j) = interp1(f, theta, 0.9);
  line([x90(j), x90(j)], [0, 0.9], 'LineWidth', 2, 'Color', 'k');
  fprintf('For n = %d, y = 0.9 at x = %f.\n', j, x90(j));
end
legend('n=1', 'n=2', 'n=3', 'n=4', 'Location', 'east');
% Put lines across at y = 0.9 and 0.5
line(xlim, [0.5, 0.5], 'LineWidth', 2, 'Color', 'k');
line(xlim, [0.9, 0.9], 'LineWidth', 2, 'Color', 'k');
% Label the axes
xlabel('x', 'FontSize', 20);
ylabel('y', 'FontSize', 20);

For n = 1, the sum of f (for some reason called max_f) = 32.540191.
For n = 1, y = 0.5 at x = -4.439359.
For n = 1, y = 0.9 at x = 8.435958.
For n = 2, the sum of f (for some reason called max_f) = 24.668279.
For n = 2, y = 0.5 at x = 4.894287.
For n = 2, y = 0.9 at x = 17.153014.
For n = 3, the sum of f (for some reason called max_f) = 19.624252.
For n = 3, y = 0.5 at x = 10.702175.
For n = 3, y = 0.9 at x = 20.896785.
For n = 4, the sum of f (for some reason called max_f) = 16.262926.
For n = 4, y = 0.5 at x = 14.406551.
For n = 4, y = 0.9 at x = 22.955276.

[sri satya ravi]

Hi, Thanks for your help. That works for me for this problem.

If i have an image something like this I want to know the 0.98th value of max power. That occurs on either side of the curve. How can I get the intercepts for 2 close numbers.

max_power = 523.6947