How to make a random circularly-symmetric gaussian distributed additive noise component with mean 0 and variance (sigma^2)/2?

yBob(t) = xAlice(t) hab(t) + nb(t)

where t denotes the time, denotes the convolution operator, hab(t) denotes the circularly-symmetric Gaussian-distributed channel response with mean 0 and variance σ2 h /2 in each dimension, and nb(t) denotes the circularly-symmetric Gaussian-distributed additive noise component with mean 0 and variance σ2 n /2 in each dimension.

hab(t) and nb(t) should be different, but how?

If I write this code then I get same hab(t) and nb(t):

N=16;
t = linspace(0, 2*pi, N);
j = sqrt(-1);
hab = cos(t) + j*sin (t);
plot(x, y, 'bo-', 'LineWidth', 2);
grid on;
title('hab');
xlabel('Real Component');
ylabel('Imaginary Component');


nb = cos(t) + j*sin(t);
plot(x, y, 'bo-', 'LineWidth', 2);
grid on;
title('nab');
xlabel('Real Component');
ylabel('Imaginary Component');

But I want different hab and nb. hab and nb should be circularly symmetric gaussian-distributed with 0 mean.

Using below code I didn't get circular symmetric gaussian but only gaussian with mean 0. But I want both.

hab = (1/sqrt(2))*(randn(N, 1) +1i*randn(N,1));  
plot(hab, 'bo-', 'LineWidth', 2);
grid on;
title('hab');
xlabel('Real Component');
ylabel('Imaginary Component');

nb = (1/sqrt(2))*(randn(N, 1) +1i*randn(N,1));
plot(nb, 'bo-', 'LineWidth', 2);
grid on;
title('nb');
xlabel('Real Component');
ylabel('Imaginary Component');

Through this I got the different value of hab and nb, but it is not circularly-symmetric gaussian.

[Jon]

Why do you say that it is not circularly symmetric gaussian? How are you checking this and deciding it is not circularly symmetric gaussian?

Answer 1

[Jon]

Using your code and plotting results, qualitatively it looks like you are getting two circularly symmetric distriuted sets of data, what is it that you think isn't working?

N = 1000;
hab = (1/sqrt(2))*(randn(N, 1) +1i*randn(N,1));  %circularly-symmetric Gaussian-distributed and channel coefficient
nb = (1/sqrt(2))*(randn(N, 1) +1i*randn(N,1));
plot(hab,'o'),
xlim([-3,3])
ylim([-3,3])
axis equal
title('hab')

plot(nb,'o')
xlim([-3,3])
ylim([-3,3])
axis equal
title('nb')

[Himanshi]

Sir, I am new in this field. So, I think it is not circularly symmetric gaussian. It is just gaussian. Sir, will you please explain me how they are circularly symmetric gaussian. As when I write this code

hab = cos(t) + j*sin (t); than I get a circular points and it make a circle. But using this

hab = (1/sqrt(2))*(randn(N, 1) +1i*randn(N,1)); for small value of N, I get a scattered ponts that are seems like they are not circular, when we increase the value of N, then it seems like circular.

Sir, If I am wrong please tell me. Sir, is using this hab = (1/sqrt(2))*(randn(N, 1) +1i*randn(N,1)); we get circular symmetric gaussian?

[Jon]

According to definition of http://web.eng.ucsd.edu/~massimo/ECE278/Lectures_files/Lec12_Probability_3.pdf

"A complex gaussian random variable z = x + iy with independent, zero-mean components x and y of equal variance is a circularly-symmetric gaussian random variable."

I think that your code (repeated below), satisfies this definition.

N = 1000;
hab = (1/sqrt(2))*(randn(N, 1) +1i*randn(N,1));

It sounds like you are descibing a signal with gaussian random phase, this can be generated using:

N = 1000;
z = exp(2*pi*1i*randn(N,1));
plot(z,'o')
axis equal

[Himanshi]

Thank you, Sir. I get it. I have understood. Thank you so much Sir.

[Jon]

Your welcome. Good luck with your project!

Answer 2

[Image Analyst]

Not exactly sure what you want, but how about this:

g = fspecial("gaussian", 201, 40); % Create Gaussian intensity pattern.
imshow(g, []); % Display it as an image.
axis('on', 'image') % Turn on axis tick marks and labels.