Hi everybody,
I would like to plot the CDF of the channel hardening in figure 6 at the kth user in cell-free massive MIMO Which is described in equation 11 in [1].
To plot it I followed these steps :
1- To deploy the access points within a square(L=1000m) and I fixed a guard distance d_guard=9m
2- To develop beta(eq3) using three-slope path-loss described in eq 2+shadowing
3- To develop the correlation matrix via equation 7.33 [2]
4- To develop the channel vector gmk where h ~ CN(0,1)
h=(1/sqrt(2))*(randn(M,numberOfRealizations)+1i*randn(M,numberOfRealizations));
5-the channel gain between access point m and kth user g_mk=sqrt(R)*h
6-Inspired from equation (2.17) from [2] : I developed the channel hardening using 2 alternatives :
-1alternative :
var(sum(abs(g).^2,1)/trace(R),0,2)where var is the matlab function
to calculate the variance
I got positive and negative values of channel hardening coefs
-2nd alternative :
variance(s,it1)= real(trace(R*R))/(trace(R)).^2;
I got only positive values of channel hardening coefs
You can find the plots in the attached file.
Could you please help me to deal with these details :
1/Shall we omit the shadowing when dmk
The variance can by definition not be negative. Then you have done something wrong in your implementation.
Figure 7.26 in [2] considers something similar what you want to plot. Maybe the simulation code that generates that figure can be useful for you.
Dear Emil Björnson,
Thank you for your response and for your recommandation.
In fact, channel hardening coef ChDk is defined in two ways:
-way 1: according to your book[2] it is defined as: ChDk1=var({norm(h)^2}/{E{norm(h)^2}})
-way 2: according to giovanni's paper [1].ChDk2=1-var({norm(h)^2}/{E{norm(h)^2}})
Theorically I found in Eq 17 and 22 in this paper https://arxiv.org/pdf/1606.02348.pdf
that the variance can't be above 1 because it is inferior to 1/M.
But on simulation I got negative values of ChDk2( because I got some values of var({norm(h)^2}/{E{norm(h)^2}}) which are above 1 ).
**Method followed:
In fact, I got inspired from your figure 2.7 in [2] in alternative 2 mentionned above, I got only positive values of ChDk.
And also I got inspired from figure 7.26 in [2] in alternative 1 mentionned above and I got negative and positive values of ChDk.
and what I changed comparing to your figure 7.26 is that :
-I used three-slope model for path-loss.
-Concerning the distance between the access point and the user is defined in function d = distance_cell_free( AP_pos, L, Dmin ):
I distributed the access points within a square (L=1000m) and I fixed a guard distance Dmin=10m:
******
function d = distance_cell_free( AP_pos, L, Dmin )
% AP_pos is an array of access point position
% it is a matrix of size 2xM, each column is the x and y of the access
% point, 0
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