Why does the SQRT function or ^0.5 function return negative values in MATLAB?

I am taking square roots and MATLAB is returning, negative values, why is this?

Let me show you the maths and the script that is doing this.

I was implementing the following compositional function (which in the end takes a sqrt of two POSITIVE or zero numbers):

in matlab and generating a lot of values using that function. It happens that when I generate a lot of values of that function, some are (strangely) negative. Why is that? It should never happen since its suppose to be the sqrt of two functions that are always positive because of the sum of two squares is always positive or zero.

The script is:

restoredefaultpath;clear;clc;clear;clc;
%%target function
f_target = struct('h', cell(2,2), 'f', cell(2,2));

h11 = @(A) (1/20)*(1*A(1) + 2*A(2))^4; % ( x1 + x2)
h12 = @(A) (1/10)*(3*A(1) + 4*A(2))^3;
h21 = @(A) (1/100)*(5*A(1) + 6*A(2))^2;
f_target(1,1).h = h11;
f_target(1,2).h = h12;
f_target(2,1).h = h21;

h13 = @(A) (1/20)*(1*A(1) + 2*A(2))^4; % ( x1 + x2)
h14 = @(A) (1/10)*(3*A(1) + 4*A(2))^3;
h22 = @(A) (1/100)*(5*A(1) + 6*A(2))^2;
f_target(1,3).h = h13;
f_target(1,4).h = h14;
f_target(2,2).h = h22;

%h31 = @(A) (1/1)*(A(1) + (1/100)*A(2) + 1)^0.5;
h31 = @(A) (1/500)*sqrt(A(1) + (1/100)*A(2) + 1);
f_target(3,1).h = h31;

% f_target(1,1).f_4D = @f_4D;
% f_target(1,1).f_8D = @f_8D;
% f_target(1,1).f_8D_hard_code = @f_8D_hard_code;
f_target(1,1).f = @f_8D_hard_code;
%%make data set
sigpower = 'measured';
powertype = 'linear';
snr = 8;
low_x = -2;
high_x = 2;
nb_samples = 100000; %100,000
D = 8;
[X,Y] = generate_data_from_function( f_target, snr, low_x,high_x, nb_samples, sigpower, powertype, D);
D = size(X,2);
D_out = size(Y,2);
sum(Y < 0)
save('f8D_all_data_set')
beep;

and the coded function is:

function [ f_val ] = f_8D_hard_code( x, f_target )
%compute left
h(1,1).val = f_target(1,1).h(x(1:2));
h(1,2).val = f_target(1,2).h(x(3:4));
h(2,1).val = f_target(2,1).h( [h(1,1).val, h(1,2).val] );
%compute right
h(1,3).val = f_target(1,3).h(x(5:6));
h(1,4).val = f_target(1,4).h(x(7:8));
h(2,2).val = f_target(2,2).h( [h(1,3).val, h(1,4).val] );
%compute all
h(3,1).val = f_target(3,1).h( [h(2,1).val, h(2,2).val] );
f_val = h(3,1).val;
end

and the generation data function is:

function [ X, Y ] = generate_data_from_function( f_target, snr, low_x,high_x, nb_samples, sigpower, powertype, D)
% sigpower = usually 'measured', powertype = usually 'linear'
X = low_x + (high_x - low_x) * rand(nb_samples,D);
Y = zeros(nb_samples,1); % (N x 1)
for n = 1:nb_samples
    xn = X(n,:);
    fx = f_target(1,1).f( xn, f_target );
    yn = awgn(fx,snr,sigpower, powertype);
    Y(n,:) = yn;
end
end

[Image Analyst]

Note: code above requires the Communications System Toolbox. (So I can't run it.)

Answer 1

[Walter Roberson]

?? What does this have to do with sqrt() ??

You are calling awgn . awgn is Gaussian, which is normally distributed, so it has indefinite tails in both directions. So for any given mean and sigma, there is a finite probability that it will generate a negative value.

[Brando Miranda]

yes, your right. Im an idiot. I forgot the awgn.... Thanks.