Given a function and x limits of a function, how do I divide y-coordinates of a function in set increments?

for example,

x = -20:0.4:20; f1 = gaussmf(x, [2 2]);

I am working on fuzzy operations so I need to get the lower and upper bounds of x for incremental values of y, for example, if my y is 0 to 1 i need each x-values of y from 0, 0.01, 0.02.. ..0.99, 1

what's the best way to get the x-values? find() doesn't work because matlab already assigns f1 101 values that's not in increments of 0.01

Answer 1

[Star Strider]

It depends on the nature of your function. You could use a generic regression function such as polyfit or an interpolation function such as interp1, with the x and y arguments reversed from the usual order, since you want the x values for certain values of y.

[soloby]

what if my function is just a gaussian function? f1 = gaussmf(x, [2 2]);

im new to coding can you be a little more specific

[Star Strider]

For the Gaussian function, there would be two values of x for every value of y, one on either side of the centre. Probably the easiest way to do that for the Gaussian function is to use the norminv function (Statistics Toolbox), or using erfcinv:

CV = @(alpha) -sqrt(2) * erfcinv(2*alpha);    % Equivalent to ‘norminv’
alpha = 0.95;
zs = CV([(1-alpha)/2   1-(1-alpha)/2])

[soloby]

this looks very promising, I just have a couple of questions, is there a way i can apply this to the gaussmf(x, [2 2]) function directly? and why did you choose alpha to be .95? I assume it's arbitrary.

thanks very much for your help

[Star Strider]

My pleasure.

To apply it to gaussmf(x,[2 2]) directly, you would have to calculate its inverse. The easiest way to do that is with the fzero function. You would have to do that in a loop, but you would only have to do it for values of x<=2, and create a symmetric vector with values for x>2.

To calculate the x for a particular y, try this:

sig = 2;
c   = 2;
invgaussmf = @(x,y) gaussmf(x, [sig c])-y;              % Inverse Function

N = 10;
y = linspace(1E-4, 1-1E-4, N);                          % Vector Of ‘y’

for k1 = 1:N
    x = fzero(@(x) invgaussmf(x,y(k1)), 0.5);
    mf(k1,:) = [y(k1)  x];                              % [y x] Matrix
end

You will probably have to experiment with it to get the result you want. You might want to plot y as a function of x to be sure it’s doing what you want.