For millimeter wave massive MIMO communication in 5G, we observe that the number of available multipath that avails communication is much smaller than the maximum connections possible between the transmitter(TX) and receiver(RX). Only a few MPCs reach at receiver with good received signal strength. For example, the number of strong MPCs that reaches the receiver is L and there is N transmitter antenna on the transmitter side and N number of antennas on the receiver side.
So, from the channel matrix of the massive MIMO system, we can say the total number of available paths or connections between TX and RX is equal to, N X N or, N^(2)
Now, L << N^(2)
For simplicity, if the number of possible strong beams from the transmitter and receiver sides are NtBeams and NrBeams, then, L = NtBeams * NrBeams
If we look up the massive MIMO channel matrix, then, H=
Primarily, if the number of available MPCs to avail communication between TX and RX are much lesser than the maximum connections possible, N^(2), then the channel matrix will be sparse. Because MPCs other than stronger L number of paths are so weak that they cannot communicate between TX and RX.
Sparse channel matrix means where the elements are zeros. We have nothing to do with those elements at all. So, we need such an algorithm that can sense the zeros of the channel matrix and we can transmit our signal thru that selected first L number of stronger paths. The orthogonal matching Pursuit (OMP) algorithm helps us to do that. It can sense a sparse matrix. OMP is sometimes termed as the "Compressive Sensing Method".
Orthogonal Matching Pursuit Algorithm:
Let assume, NtBeams = NrBeams = 28
Total stronger connections or paths, L = 28 * 28 = 784 (it is observational)
Also assume, the number of antennas at both the transmitter and receiver side is equal to 32
So, the channel matrix contains 32 * 32 elements or 1024 elements
Now, the Equivalent sensing matrix, Q (say), will have an array dimension of 784 X 1024
Mathematically,
y = Qhb + n
Here, y = received vector
Q = Equivalent sensing matrix
hb = beamspace channel matrix where the number of elements is N X N = N^(2). It is a sparse matrix as there are only L-acceptable elements out of N^(2)
n = noise
Also assume, Matrix Q contains n number of columns, like that,
Q = [q1 q2 q3 ..... qn]
Now, we focus on finding out the maximum in the context of which column with a projectile with y generates maximum value, such that,
i(1) = argmax |(qjH / ||qj||)*y| where, j=1,2,...,n
Let's assume, qi1 column in Q matrix contributes maximum value. Then find out using the least square solution
min ||y - qi1*hb1|| (We try to make it zero)
Where, hb1 is any column of hb
Where, hb1= (qi1H * qi1)^(-1) * qi1H *y
Now, take a residue matrix as,
r1 = y - qi1*hb1
Now, we'll find the maximum correlation of residue matrix r1 by finding out which column of Q with a projectile with r1 generates maximum value, such that,
i(2) = argmax|(qjH / ||qj||)*r1| where, j=1,2,...,n
Let assume, qi2 column in Q generates the maximum value
Now form a matrix, Qn, like that,
Qn = [qi1 qi2]
Now find out hb2
hb2 = (QnH * Qn)^(-1) * QnH *y
Now update the residue matrix as,
r2 = y - Qn*hb2
If norm of ||r(n-1) - rn|| falls below a threshold value, then close the loop
In our case, if we consider r2-r1 falls below the threshold, then we close the loop.
If we find hb2=[3;2] then we do such operations in hb matrix, such that,
and if i(1) = 5 and i(2) = 2
Then hb will be [0;2;0;0;3;0; ... up to 28th row all are 0]
Here, we place the element value of hb2 in hb matrix in such row with row number matches with the value of i(1) and i(2)
Mathematical Example of Orthogonal Matching Pursuit (OMP):
Let's assume, for a MIMO communication system,
The size of the equivalent sensing matrix, Q is 4 X 6
And received signal matrix, y=
Now, y = Qhb
Or,
Where, hb =beamspace matrix =
Let assume, Q = [ q1 q2 q3 q4 q5 q6]
Here, q1 is the first column of Q, q2 is the second column of Q, and so on.
First Iteration of Orthogonal Matching Pursuit:
Now we find the maximum correlation of y by finding out which column in Q generates the maximum value with the projection of y,
Or,
QT*y =
Here, we can see the element in the 5th row is the maximum among all elements. So, we’ll select the 5th column of Q with which y has the maximum correlation value.
Now, hb1 = (q5'*q5)^(-1) * q5'*y
Where q5’ denotes the transpose of q5
Or,
hb1 = 4
Residue Matrix, r1 = y – q5* hb1
Or, r1 =
Here, we observe residual matrix r1 is not a zero matrix. So, we go for 2nd iteration.
Second Iteration of Orthogonal Matching Pursuit
Where we find the maximum correlation of r1 with respect to Q matrix.
Alternatively,
QT*r1=
Now, we see the element in the 2nd row of the above matrix generates the maximum value so r1 has a maximum correlation with the 2nd column of Q Matrix.
Now, we’ll form a new matrix, Qn = [q5 q2]
We find hb2 = (Qn'*Qn)^(-1)*Qn'*y;
Or, hb2 =
Now updated residue matrix, r2 = y – Qn* hb2
Or, r2=
Now we get the desired value in residue matrix r2 where all elements are zeros. So, beamspace matrix, hb will be
Here, we replace the elemental value of hb in that rows which are equal with the number of columns which generates maximum values with projection with y, then r1, and so on.
Now, from the mathematics we can say,
y = Q* hb
Or,
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