SVD-Based MIMO Transmission & Optimal Power Allocation
Optimal power allocation is defined as in a MIMO communication system we need to allocate more power to an independent stronger path and allocation of less power to a weaker path. By following this method we can achieve high throughput. Firstly, we talk about SVD-based MIMO. Then we discussed step by step how to find stronger or weaker communication paths between two MIMO antennas.
Channel Matrix,
Let's assume, the first column in the above matrix is c1 . c2 and c3
are the 2nd and 3rd columns, respectively.
Here, columns are orthogonal for instance, i.e., c1Hcj =0
Here, r=3, t=3 (r=number of Rx antenna; t=number of Tx antenna)
Now, c1
Now, c1Hc2
*
=0
Multiplication is 0 since the columns are orthogonal.
Step 1: We normalize each column
We get, H=
Here singular values are not in decreasing order.
Step 2: Now we arrange the singular values in decreasing order
H=
That implies,
Again assume, the first matrix is U (unitary matrix), the middle one is Σ (eigenmatrix), and 3rd matrix is V (unitary matrix).
Alternatively, UUH=I, VHV=VVH=I
Σ =
At receiver side
y ̃= UHy =
At the transmitter side
͞x =V x ̃
Here, notation "x1~, x2~, x3~" represents original message signal vector
Transmit pre-processing or precoding at the receiver side
ỹ= Σx̃ + w̃
Here, "y~" represents the received signal vector and "w~" represents the noise vector
Now, 3 decoupled channel spatial multiplexing are as follows
ỹ1 = √52x̃1 + w̃1
ỹ3 = 2x̃3 + w̃3
Optimal Power allocation
To maximize sum-rate and to achieve the Shannon capacity,
P1=(1/λ- σ2/σ12)= (1/λ- σ2/52)
P2=(1/λ- σ2/σ22)= (1/λ- σ2/13)
P3=(1/λ- σ2/σ32)= (1/λ- σ2/4)
Consider the noise power, σ2= 0dB
So, 10log10 σ2=0
σ2=10^(0/10)=1
let P=total power=3dB
So, 10log10 P=3
P=10^(3/10)=2 (approx.)
So, we must have
P1+P2+P3= 2
(1/λ-1/52)+ (1/λ-1/13)+ (1/λ-1/4)=2
Or, 1/λ=.7821
Now,
P1=10log10(1/λ- σ2/52)= 10log10(0.7821- 1/52)=-1.1755 dB
P2=10log10(1/λ- σ2/13)= 10log10(0.7821- 1/13)=-1.517 dB
P1=10log10(1/λ- σ2/4)= 10log10(0.7821- 1/4)=-2.74 dB
Power allocation decreases as gain σ2 decreases. So, we can say poor power to poor channel , more power to strong channel.