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grouping the users thereby sharing the subcarriers

2 vues (au cours des 30 derniers jours)
Prabha Kumaresan
Prabha Kumaresan le 14 Déc 2017
Clôturé : MATLAB Answer Bot le 20 Août 2021
Nuser=4;
Nsubcarrier=10;
G is a matrix of (users,subcarriers)%image attached below
C is subcarrier allocation to users %image attached below
could anyone tell me how to group the users as a result sharing of subcarriers takes place.
  3 commentaires
Prabha Kumaresan
Prabha Kumaresan le 15 Déc 2017
Modifié(e) : Walter Roberson le 15 Déc 2017
As a result of grouping the subcarriers can be shared among the users which results in the improvement of throughput performance.
Walter Roberson
Walter Roberson le 15 Déc 2017
That description of grouping tells us nothing about how it needs to be implemented.

Réponses (1)

Walter Roberson
Walter Roberson le 14 Déc 2017
which_user = randi(Nuser, 1, Nsubcarrier);
newC = 0 * C;
idx = ((1:Nsubcarrier)-1) * Nuser + which_user;
newC(idx) = C(idx);
  4 commentaires
Prabha Kumaresan
Prabha Kumaresan le 15 Déc 2017
Modifié(e) : Walter Roberson le 18 Déc 2017
before grouping if C was
[ 1 0 2 0 0 0 0 0;0 0 0 3 4 0 0 0;0 5 0 0 0 0 0 6;0 0 0 0 0 7 8 0]
after grouping C should be
[1 0 2 3 4 0 0 0;1 0 2 3 4 0 0 0;0 5 0 0 0 0 7 8 6;0 5 0 0 0 0 7 8 6]
Which means C is a matrix of [4*8]
4 users sharing 2 different subcarriers each.
Prabha Kumaresan
Prabha Kumaresan le 18 Déc 2017
Modifié(e) : Walter Roberson le 18 Déc 2017
If G =
0.0070 0.0103 0.0066 0.0033 0.0041 0.0027 0.0004 0.0050 0.0040 0.0046;
0.0020 0.0016 0.0001 0.0018 0.0019 0.0008 0.0024 0.0036 0.0047 0.0044;
0.0015 0.0011 0.0009 0.0011 0.0011 0.0022 0.0019 0.0018 0.0013 0.0028;
0.0061 0.0033 0.0022 0.0037 0.0038 0.0011 0.0017 0.0029 0.0029 0.0019;
and C=
0 0 0.0066 0 0 0 0 0 0 0;
0.0020 0.0016 0 0.0018 0 0 0 0 0 0;
0 0 0 0 0 0 0.0019 0 0 0;
0 0 0 0 0.0038 0.0011 0 0.0029 0.0029 0.0019;
Now I want to group the users with respect to subcarriers. Suppose if I group user 1 with user 2 and user 3 with user 4 the corresponding subcarrier need to get shared which results in D=
0.0020 0.0016 0.0066 0.0018 0 0 0 0 0 0;
0.0020 0.0016 0.0066 0.0018 0 0 0 0 0 0;
0 0 0 0 0.0038 0.0011 0.0019 0.0029 0.0029 0.0019;
0 0 0 0 0.0038 0.0011 0.0019 0.0029 0.0029 0.0019;

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