How to back interpolation to find x from z and y?

I would like to do "back" interpolation to find an x value from the given values of z and y. (2Dimensional x,y,z matrix table is available). Is there any prepared MATLAB function for it?/ And how to do it in Simulink(Lookup Table Dynamic block is for 1D. I need 2D)?

As an example: x =[01 2 3 4 5]; y=[ 30 40 50 60]

z=[
1 2  3  4  5
3 4  5  6  7
5 6  7  8 10
7 8 10 10 13]

Table=

       1  2  3  4  5
      ---------------
  30 ! 1  2  3  4  5
  40 ! 3  4  5  6  7
  50 ! 5  6  7  8 10
  60 ! 7  8 10 10 13

According to this given values, I would like to find the x value for a given data of y=43, and z=6.8

Thanks in advance.

[Stephen23]

See also later question:

https://www.mathworks.com/matlabcentral/answers/325704-how-to-get-a-y-z-x-table-from-a-x-y-z-table-data

Answer 1

[Stephen23]

This is easy with interp2 and fzero:

>> X = [1,2,3,4,5]
>> Y = [30,40,50,60]
>> Z = [1,2,3,4,5;3,4,5,6,7;5,6,7,8,10;7,8,10,10,13]
>> y = 43;
>> z = 6.8;
>> fun = @(x)interp2(X,Y,Z,x,y)-z;
>> x = fzero(fun,mean(X))
x =  4.1538

And checking:

>> interp2(X,Y,Z,x,y)
ans =  6.8000

[ercan duzgun]

Thank you. Yes, this works.

Answer 2

[John D'Errico]

Is there something already written in MATLAB to do this? No. Given that your problem is not terribly common, it would be unexpected if it was already written.

Is it possible to do? Yes. Not even that terribly difficult.

You have an array that essentially defines the relation z(x,y).

1. Use interp2 to interpolate the array along the line y = 43, so for each value of x, determine z at the given y value. The result will be a vector of z values, one for each location in x.

2. Find any pairs of locations that cross the desired z level.

3. Interpolate linearly to find x that yields the given z level exactly. (Well, exactly to within the accuracy of linear interpolation.)

[ercan duzgun]

Thank you. Nice programming algorithm.

Answer 3

[Phillip Resor]

Just sorted this out using griddedInterpolant and thought others might find it useful

x = [1,2,3,4,5];
y = [30,40,50,60];
[X, Y] = ndgrid(x,y);
Z = [1,2,3,4,5;3,4,5,6,7;5,6,7,8,10;7,8,10,10,13]'; % transpose to match ndgrid
yp = 43;
zp = 6.8;
F = griddedInterpolant(X, Y, Z);
fun = @(xp)F(xp,yp)-zp;
xp = fzero(fun,mean(x))

And checking:

F(xp,yp)

[ercan duzgun]

This code also works fine. Thank you Phillip Resor.