Lookup Table of 1*2

If I have a 2D Lookup table having x and y inputs with dimension of 1*15 each and output z of 15*15, how can i reverse it in a way that I have z as input and x and y as outputs? What should be the dimensions now for i/p and o/p?

Answer 1

[Image Analyst]

Try interp2().

[Bilal Mustafa]

Could you pl elaborate it a bit?

So, for example if i have

x= [0:1:15] 
y= [0:1:15] 
Z=reshape(z,size(y,2),size(x,2))

where z is 1*225 and Z=15*15

that implements a 2D LUT in simulink. How can I now reverse it as mentioned? And how should be the dimensions of my ("interp2[arguments]") arguments for interp2 command?

[Image Analyst]

Won't work. Since x and y are 16 elements each you'll need z to be 16*16 = 256 elements. Or else x and y can only be 15 elements instead of 16.

Even if it were 16:

x= [0:1:15] 
y= [0:1:15] 
% Make a 1-D vector (for some reason).
z = randi(100, 1, length(y) * length(x))
% Now immediately turn it into a 2-D matrix
Z = reshape(z, length(y), length(x))

you could have the same z at multiple locations so you'd need to specify which z first.

For example if there might be a z with value 0.4 between rows 10 and 11 and columns 14 and 15, but there might be another z with the value of 0.4 between rows 3 and 4 and columns 7 and 8, so for which of those two z values do you want the interpolated x,y coordinates for?

[Bilal Mustafa]

Sorry I wrote it wrong, x and y are [1:1:15] each and i m already implementing a 2D LUT from above equations. Now, i need to reverse these as mentioned above!!

[Image Analyst]

Did you read my last paragraph above? I don't see an answer.

In general, I now think you should use the contour() function as the requested z may occur along contour lines (multiple places) in your array.

[Bilal Mustafa]

Thanks for your replies. Perhaps I should rephrase the question. I already have 'Z' from some other source in dimension of 15*15. and I have also implemented the 2D LUT with x,y as input and Z as output. Now I want to make it such that Z behaves as input and x and y as output of LUT 'with the same data I already have'. Is is possible somehow? Hope it would b more clear now.

[Image Analyst]

It's still not clear to me why contour did not work. Please post your code for it. Also, post code to create the matrix (like I did but that might not be what you have) and give the desired z value that you want to find, and tell us where you expect to find that z value, and if you expect that that z value should occur in only one place or in multiple places. In other words, post an example with the solution.

[Bilal Mustafa]

Well with contour(), I haven't tried yet. I just thought to clarify my question if its not. I'll try with contour and come again, if any problem. Thanks a lot.

[Image Analyst]

See this code:

x= [0:1:15] 
y= [0:1:15] 
% Make a 1-D vector (for some reason).
z = randi(100, 1, length(y) * length(x))
% Now immediately turn it into a 2-D matrix
Z = reshape(z, length(y), length(x)) 
subplot(1, 2, 1);
contour(Z)
title('Random Data', 'FontSize', 20);

subplot(1, 2, 2);
p = peaks(400);
contour(p);
grid on;
title('Peaks Function', 'FontSize', 20);

Each contour line is a line of some constant Z value. Not sure what Z value you want, but the point is that, that Z may show up in lots of places.