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I was studying Rayleigh channels from Wireless Communications, Second Edition by Andreas F. Molisch. The book states that by adding different in-phase and quadrature components multipath components the distribution of amplitudes of in phase and quadrature should be Gaussian due to central limit theorem, but in-phase component I(t) and quad components q(t) are random processes but central theorem is for random variables so it can't be applied. What is wrong here in my thinking?

adding different multipath components, page 79.

I also wrote a MATLAB program to simulate addition of many random inphase components. The resultant amplitude distribution didn't fellow Gaussian distribution and the resultant inphase component wasn't even a cosine function.

clc
clear
av_function=0
for k=1:1000
n=5001;
x=randn(1,n);
y=randn(1,n);
a=randn(1,n);
t=0:0.001:10;
sum_out=0;
for i=1:n
   y=a(i)*cos(x(i)*t+y(i));
   sum_out=sum_out+y;
max_A(k)=max(sum_out);
end
av_function=av_function+sum_out;
end 
av_fucntion=av_function/k;
mean_max_A=mean(max_A);
figure
plot(max_A)
figure
plot(t,av_function)   

distribution of amplitude of inphase component

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  • $\begingroup$ If you are a computational-thinking type, the Box-Muller transform Wiki article would clear up your confusion. Read the Wiki article, and compare the phrase R2 is the square of the norm of the standard bivariate normal variable (X, Y), it has the chi-squared distribution with two degrees of freedom with that in the Rayleigh distribution Wiki article: the Rayleigh distribution ... [u]p to rescaling ... coincides with the chi distribution with two degrees of freedom. $\endgroup$
    – V.V.T
    Apr 15, 2023 at 11:52

2 Answers 2

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1.- Definition of Random Variable

from BRITANNICA [1]

[quote]

RANDOM VARIABLE: A random variable is a numerical description of the outcome of a statistical experiment. ... The probability distribution for a random variable describes how the probabilities are distributed over the values of the random variable.

[end of quote]

2.- Definition of Random Process

again from BRITANNICA [2]

[quote]

A stochastic process refers to a family of random variables indexed against some other variable or set of variables.

[end_of_quote]

So, as obvious as it may seem:

  • Stochastic and Random are the same : (totally or partially) unknown.

  • 1 random variable is a random or stochastic process. It's a 1-variable random process.

3.- Fitting probability Density Functions

You have plotted attached to question a time graph of the signal, or sum of signals, but there's little to be seen or inferred from the signal time plot alone.

It's the PDF, the shape of it, that tells you how to classify a given x(t), whether it's Gaussian (aka Normal), or Rayleigh, or else.

4.- A Single Rayleigh Process has Rayleigh PDF

t = [0:.001:5];

N=30
D=.1+.1*randi([1 10],1,N);

x=zeros(N,numel(t));
for k=1:1:N
    x(k,:)=raylrnd(D(k),1,numel(t));
end

% sample
figure
histfit(x(1,:),N,'rayleigh')

enter image description here

When attempting to fit a Normal/Gaussian PDF to Rayleigh PDF data, notice that the PDF is offset away from the actual data : the PDF is not really fitting this dataset.

figure
histfit(x(1,:),N,'normal')

enter image description here

5.- Averaging all Rayleigh traces produces a Normal PDF

x2=mean(x,1);

figure
histfit(x2,N,'normal')

enter image description here

The more traces one puts, the more Normal it's going to look.

6.- Alternative View

In Sales/Marketing the Central Limit Theorem is extensively used:

Think of it as if you are selling train tickets in till, or hot dogs in a shop, and customers come to you, right on a queue.

They are all different with their own statistics widely scattered, but they all want some tickets and you have them.

So you can infer that using a normal distribution on all of them, and there's no need to even know any customer at all, if you center your price/ticket and time span for keeping the till open, you are going to maximize sales.

Additional Reading:

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It's not that Random Rayleigh processes add to get a Gaussian process, but that if we consider the Magnitude of a complex Gaussian process with zero mean: THAT will be Rayleigh distributed. If we add ANY independent random processes together, they converge to a Gaussian given the Central Limit Theorem. So when we consider waveforms in complex form (meaning we have a waveform with both complex magnitude and phase), and if we add enough of them (as few as 6 would be reasonably sufficient in most cases), the result will be a complex Gaussian process, meaning the real and imaginary components are independent and Gaussian distributed. If we have that condition without an offset (mean), then the Magnitude is Rayleigh distributed and the phase is uniformly distributed. If there is an offset, then the magnitude will be Ricean distributed. That's the bottom line.

The OP has shown the distribution of magnitude NOT amplitude. A histogram of the amplitude will certainly reveal a Gaussian distribution. The central limit theorem is applicable to the samples returned from a random process, which is applicable the waveform as received in a fading channel with multiple paths (we can consider fast fading vs slow fading to determine the time variation separately). It would be sufficient to add as few as 6 paths to result in a reasonable approximation to a Gaussian. To simulate the Rayleigh distribution, a complex waveform is used to include both magnitude and phase.

The magnitude distribution IS considered for the Rayleigh distribution, which is applicable to complex signals that can be described in terms of magnitude and phase components. It is the magnitude of a complex Gaussian distribution specifically (given with real and imaginary components each being samples from an independent and equally distributed Gaussian random process) as having a Rayleigh distribution, and the phase as having a uniform distribution.

The OP's simulation is using a Gaussian distribution to assign a frequency and phase variable. The point the author made in the text is that any random distribution can be used, as long as we add enough of them the result will be Gaussian. That said, to keep this similar to the OP's result the following demonstrates this with the sum as few as 6 paths (what we would also find is the cosine function has no effect, we could easily sum the individual random sequences, which then need not be Gaussian themselves!). Note how I also made use of MATLAB's vector processing rather than doing this all in a loop (much faster). I first did this using x=randn(paths,n) as a 6 x 5001 matrix where each row is an independent random sequence 5001 long representing the amplitude for each path. So in that case, as the OP has done, the amplitude is already Gaussian distributed even if we just used one path. To make this more meaningful, I changed it to rand()-0.5 as a zero mean uniform distribution to demonstrate the convergence toward a Gaussian:

n=5001;
paths = 6
x=rand(paths,n)-.5;
y=rand(paths,n)-.5;
a=rand(paths,n)-.5;
t=0:0.001:10;
y=a*cos(x*t+y);
total = sum(y);
figure
subplot(2,1,1)
plot(total)
title("Time Domain Waveform")
subplot(2,1,2)
hist(total, 512)
title("Histogram")

result

Note that the absolute magnitude of the process above will NOT result in a Rayleigh distribution, for that we need to simulate the complex baseband waveform such as $I+jQ$ with $I$ and $Q$ as independent Gaussian distributions (the author that the OP referenced simulates a complex passband waveform which is also applicable, but the carrier has no effect on the result and it takes a lot more simulation cycles, needlessly, to simulate a passband signal!).

An intuitive description I found of how the Rayleigh distribution results is from this very useful App-note from Keysight (back when it was HP), https://www.keysight.com/us/en/assets/7018-06765/application-notes/5966-4008.pdf, with the graphic copied below. Here we see the samples plotted on a complex plane which when given in Cartesian form as $I + jQ$ would have $I$ representing the real component and $Q$ representing the imaginary component. Both $I$ and $Q$ are each Gaussian distributed and independent of each other. We can also describe this in polar form at magnitude and phase. The rings shown on the plot depict uniform magnitude spacing, and by counting the number of samples between any two rings we can create the resulting histogram which is the Rayleigh distribution. Even though both $I$ and $Q$ have a peak in their Gaussian distributions at 0, the relative area of that inner most circle is smallest compared to the area of the space between the magnitude rings as we move outward. Thus the Rayleigh distribution approaches 0 as the magnitude approaches 0 and has a peak as the magnitude increases, and then starts to go down as the magnitude decreases further.

Rayleigh

I provide further details about fading channels at this post which includes the reference to the plot given above.

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  • $\begingroup$ I don't understand shouldn't every path gives only one random amplitude and phase ? In your code in a,x,y random variables every row represents a path that many values for path which should give only one value of amplitude and phase $\endgroup$ Apr 14, 2023 at 22:48
  • $\begingroup$ @Mahmoud I thought I was duplicating what you did with your code but adding just 6 paths rather than 5001 paths- did I interpret that incorrectly? Regardless when every path gives one amplitude and phase that isn’t changing the result will be a fixed magnitude out of the distribution and is referred to as “slow fading” — and the statistic will be how that may change with position. When the amplitude and phase changes more quickly (relative to our symbol rate)- the result is fast fading. The distribution is how that varies with time and postion $\endgroup$ Apr 15, 2023 at 0:59
  • $\begingroup$ But regardless of that I thought your question was how adding random events converges to a Gaussian and how that relates to a Rayleigh distribution which is what I tried to answer $\endgroup$ Apr 15, 2023 at 1:01

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