How to find the (x,y) combinations that make z=0?

13 views (last 30 days)
Hi Everyone
I have a matrix Z(i,j) that is a function of X and Y. Surface and contour plot look fine, but are difficult to interpret.
Is there a way to plot a function of X and Y , such that Z is equal to zero (or as close to zero as possible, since non of the elements in the Z matrix is actually zero)? What I'd like to find out is for which combinations of X and Y the Z-value is above and below zero.
My code is very long, but I did something like this to get matrix Z:
for i = 1:numel(X)
Z(:,i) = functionZ(X(i))
%Functionhandle functionZ:
Z(j) = %functionZ(X)
for j = 1:numel(Y)
Z(j)= %function with parameter X
thanks in advance for any tips!
Matt J
Matt J on 16 Sep 2019
Is there a way to plot a function of X and Y , such that Z is equal to zero
But you say you've looked at contour plots. Shouldn't this already be given to you by the zero isocontour?
Franziska on 17 Sep 2019
Edited: Franziska on 17 Sep 2019
It does, but it does not give me the 'exact' zero-isocontour, and the contour plot is difficult to interpret, since there is one extreme peak. what I am looking for is a good way to report the result, that is the XY values for which Z is above / below zero.
Edit: Found a way:
to find the iso-countourline for Z=0
contour(X, Y, Z, v)
Thanks so much for giving me the crucial hint!

Sign in to comment.

Accepted Answer

Kevin Phung
Kevin Phung on 16 Sep 2019
Edited: Kevin Phung on 18 Sep 2019
Correct me if I am misinterpreting your goal:
You have a function z(x,y).
You are trying to find the values of x and y that give you some particular values of Z, in this case very close to 0. You can try something like this:
% define some condition for Z values, in this case I want to look for all Z values in my
% Z matrix that are between 0 and 0.001.
% find() those values, and use ind2sub in order to grab the [row,col] pairs.
[x_ind,y_ind] = ind2sub(size(Z),find(and(Z=<0.001,Z>0 )));
xy_pairs = [X(x_ind) Y(y_ind)] % this will give you the x and y pairs
edit: you have already accepted my answer, but I forgot to include an '=' sign

More Answers (1)

Matt J
Matt J on 16 Sep 2019

Community Treasure Hunt

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

Start Hunting!