# Complex Gaussian Noise

2.a) Using the result for P(I) in Question 1, prove that the n-th moment of complex Gaussian noise is

{I^n} = n!(2sigma^2)^n

You may use the following result for the Gamma integral

Gamma(n-1) = n! = integral x^n exp(-x)dx.

b) Hence, obtain an expression for the normalized moments {I^n}/{I}^n

c) Deduce that the variance of the intensity is sigma^2 = {I^2} - {I}^2 = {I}^2

p(I) = 1/{I}exp(-1/{I})

d) Calculate the probability p(Ic) that the intensity exceeds a threshold intensity Ic. Note tthat in radar systems a threshold level is set so that one can decide upon the presence or absence of a target.

e) Hence deduce that the probability that the intensity I exceeds the mean intensity {I} by a factor n is

p(I) = exp(-n)

Note: Question number 2 is predicated upon Question number 1 and as such Question number 1 is ONLY shown for 'background' purposes.

Â© BrainMass Inc. brainmass.com March 4, 2021, 6:01 pm ad1c9bdddfhttps://brainmass.com/engineering/electrical-engineering/complex-gaussian-noise-24418

#### Solution Preview

Probability Density Function - Complex Gaussian Noise

1. a) I think you have done some error in integrating; I am getting a better solution.

I will use "t" for phase angle theta, and "~" for infinity.

p(t) = (1/4*pi*sigma^2)*

Integral(from 0 to ~)[exp-(I/2Sigma^2)] dI

put k = -(I/2Sigma^2)

take derivatives on both sides

dk = -dI/(2Sigma^2)

dI = -(2Sigma^2)dk

and use the fact that integral (e^k dk) = e^k

Hence, p(t) = (-2Sigma^2/4*pi*sigma^2)*

(from 0 to ~)[exp-(I/2Sigma^2)]

p(t) = -1/(2*pi)*[0-1] = 1/(2*pi)

------------------------------------------

b) p(I) = ...

#### Solution Summary

This solution solves for (a) the nth moment [In] of the complex Gaussian noise, (b) normalized moments, (c) variance of the intensity, (d) probability that the intensity exceeds a threshold intensity Ic and (e) the probability that the intensity I exceeds the mean intensity. This solution also provides revised figures for question 1.