I know a certain y1 and I want to find the respective x1.

2 views (last 30 days)
I have a graph where x and y are two vectors.
plot(x,y)
I know a certain y1 and I want to find the respective x1.
thanks

Answers (4)

Torsten
Torsten on 14 Jun 2019
x1 = fzero(@(z)interp1(x,y,z)-y1,1)
  2 Comments
darova
darova on 14 Jun 2019
It's difficult to choose intial condition:
x = linspace(0,3,10);
y = x.^2;
y1 = 4;
x1 = fzero(@(z)interp1(x,y,z)-y1,1)
plot(x,y,x1,y1,'or')

Sign in to comment.


darova
darova on 14 Jun 2019
I'd use polyxpoly or intersections
x = linspace(0,10,20);
y = sin(x);
y1 = 0.6; % find x at y=0.6
xc = [min(x) max(x)]; % boundaries of horizontal line
yc = [y1 y1];
[x1,y1] = polyxpoly(x,y,xc,yc);
plot(x,y,'.-b')
hold on
plot(xc,yc,'.-g')
plot(x1,y1,'or')
hold off

roberta perna
roberta perna on 14 Jun 2019
Sorry not work.

James Browne
James Browne on 15 Jun 2019
Edited: James Browne on 15 Jun 2019
Greetings,
I believe I have a solution to your problem, I wrote a scrip to demonstrate the concept, let me know if this is not what you are looking for and I will try to find a better solution. Here you go:
close all
clear
clc
%set arbitrary values of x, x can be any set of inputs to y(x)
x = [ 1 2 3 4 5 6 7 ];
%generate an arbitrary set of values y(x), y(x) can be any function
y = 2*x;
%Select a known, arbitrary value fromthe y-vector, y1
y1 = y(randi(length(y)))
%Now we first find the index of the value y1
idx = find( y == y1)
%Since every entry in the y-vector is dependant on a corrosponding entry in
%the x-vector, we can use the index value of y1 to find the corrosponding
%value in the x-vector ( x1 ) as follows:
x1 = x(idx)
plot(x,y,x1,y1,'kd')
title('y = f(x) and (x1, y1)')
ylabel('y(x)')
xlabel('x')
legend('y = f(x)','(x1, y1)','location','southeast')
  2 Comments
James Browne
James Browne on 15 Jun 2019
I somehow messed up on the upload, I corrected the previous answer, it should work now.

Sign in to comment.

Tags

Products

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by