Optimal Power Allocation in MIMO Transmission using SVD

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 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.,  c₁^Hc_j =0

Here, r=3, t=3  (r=number of Rx antenna; t=number of Tx antenna)

Now, c₁

c₂

Now, c₁^Hc₂

 *

₌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, UU^H=I,     V^HV=VV^H=I

Σ =

In the above matrix, σ₁=√52, σ₂=√13, σ₃=2, and Singular values are in decreasing order.

At receiver side   

            y ̃= U^Hy =

           

At the transmitter side

  ͞x =V x ̃

Or,

Here, notation "x1~, x2~, x3~" represents original message signal vector

Transmit pre-processing or precoding at the receiver side

ỹ= Σx̃ + w̃

Or,

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

ỹ2 = √13x̃2 + w̃2

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.