Comparative Study on the NOMA Based Optimum Power Allocation Using DLS Algorithm with DNN

Abstract

High throughput, successive interference cancelation, higher cell edge spectrum efficiency, and low latency are the main requirements for future generation technology. Non-orthogonal multiple access (NOMA) technology offers a scale of the multiple numbers of users (multiplexing), extremely high spectral efficiency, greater improvements in user pairing and more than one user shares single resource block; hence, it is a superior technology than orthogonal multiple access schemes (OMA). NOMA-based optimum power allocation, first by using direct techniques and then using (depth limited search) DLS algorithm with (deep neural network) DNN is studied in this research paper. These techniques are first applied to two users and then extended to multi-user communications. Distributing optimum power to the weaker user (who are not getting proper signal strength) is a challenging task and to add to it successive interference cancelation (SIC) also brings difficulty in the proper distribution of source power from a base station. In this work, we try to solve this problem with the help of DLS algorithm with DNN, where optimum power is allocated to the weaker user and minimum power is allocated to the stronger user. Here, the DLS algorithm-based DNN-NOMA technology assists to decode the user without interference, with more accuracy in real-time. DLS algorithm provides greater potential in DNN-NOMA technology by the successful application of successive interference cancelation (SIC). A DLS predicts the position of user equipment and provides the optimum power allocation. In our proposed work, performance is improved compared to the previous existing conventional multi-user case. In the case of multi-user communication, optimum power allocation capacity to the weaker user is very less compared to the two-user case optimum power allocation NOMA. Through the application of the DLS algorithm and DNN, optimum power allocation capacity for the weaker user is improved.

Appendix

For ‘i’ user’s data rate is

$$ \log_2 (1 + a_i \rho_s \beta_i ) + \cdots + \log_2 \left( {1 + \frac{a_1 \rho_s \beta_1 }{{\sum_{l = 1}^i {a_i \rho_s \beta_1 } + 1}}} \right) $$

(16)

If assume i = 2

$$ \log_2 (1 + a_i \rho_s \beta_i ) + \cdots + \log_2 \left( {1 + \frac{a_1 \rho_s \beta_1 }{{\sum_{l = 1}^i {a_i \rho_s \beta_1 } + 1}}} \right) = {\text{sum}} $$

(17)

To find, function dependent values from the above function

$$ \log_2 \left( {1 + \frac{a_1 \rho_s \beta_1 }{{a_2 \rho_s \beta_1 + 1}}} \right) \cdot (a_2 \rho_s \beta_1 + 1) $$

(18)

The Sum of the power coefficient is one

$$ a_2 = - a_1 + 1 $$

(19)

Substitute a₂ in Eq. (18)

$$ \log_2 \left( {1 + \frac{a_1 \rho_s \beta_1 }{{( - a_1 + 1)\rho_s \beta_1 + 1}}} \right) \cdot (( - a_1 + 1)\rho_s \beta_2 + 1) $$

(20)

$$ \log_2 (1 + \rho_s \beta_1 ).\frac{(( - a_1 + 1)\rho_s \beta_2 + 1)}{{(( - a_1 + 1)\rho_s \beta_1 + 1)}} $$

(21)

$1 + \rho_s \beta_1$ is constant

$$ \frac{{\frac{1}{a_1 } + \frac{\rho_s \beta_2 }{{a_1 }} - \frac{\rho_s \beta_2 }{1}}}{{\frac{1}{a_1 } + \frac{\rho_s \beta_1 }{{a_1 }} - \frac{\rho_s \beta_1 }{1}}} $$

(22)

From Eq. (22), if, a₁ = 0 then function becomes invalid, so apply L—‘Hospital’s rule, to make function valid.

Apply partial derivation above and below.

$$ \frac{{\frac{1}{a_1^2 } - \frac{\rho_s \beta_2 }{{a_1^2 }} - 0}}{{\frac{1}{a_1^2 } - \frac{\rho_s \beta_1 }{{a_1^2 }} - 0}} $$

(23)

$$ \frac{1 - \rho_s \beta_2 }{{1 - \rho_s \beta_1 }} $$

(24)

To make the function maximum, consider β2 minimum and β1maximum.

To find, a₁max from the data rate function. Assume data rate is greater than or equal to R₁

$$ \log_2 (1 + r_{x_1 }^{u_1 } ) \ge R_1 $$

(25)

$$ 1 + r_{{x_{1} }}^{{u_{1} }} \cong 2^{{R_{1} }} = \hat{R}_{1} $$

(26)

From Eq. (17)

$$ a_1 \rho_s \beta_1 + \hat{R}_1 a_1 \rho_s \beta_1 = \hat{R}_1 (1 + a_1 \rho_s \beta_1 ) $$

(27)

$$ a_{1\max } \ge \frac{{\hat{R}_{1} (1 + a_1 \rho_s \beta_1 )}}{{\beta_1 \rho_s (1 +\hat{R}_{1} )}} $$

(28)

Similarly, from the ‘i’ users

$$ \log_2 \left( {1 + \frac{a_1 \rho_s \beta_1 }{{\sum_{l = 1}^{i - 1} {a_i \rho_s \beta_1 } + 1}}} \right) $$

(29)

$$ 1 + r_{x_1 }^{u_1 } \cong 2^{R_1 } = \hat{R}_1 $$

(30)

$$ \frac{a_1 \rho_s \beta_1 }{{\sum_{l = 1}^{i - 1} {a_i \rho_s \beta_1 } + 1}} = R_1 - 1 = \hat{R}_1 $$

(31)

The Sum of the power coefficient is one, so

$$ \sum_{l = 2}^i {a_l \cong 1} $$

(32)

$$ \frac{a_1 \rho_s \beta_1 }{{\rho_s \beta_1 + 1}} = R_1 - 1 = \hat{R}_1 $$

(33)

• 1. From the Eq. (28) a_1max is

  $$ a_{1\max } \ge \frac{{\hat{R}_1 (\rho_s \beta_1 + 1) - 1}}{\beta_1 \rho_s } $$

  (34)