Chow bit power allocation algorithm and its MATLAB implementation

Principle introduction

Chow algorithm allocates bits according to the channel capacity of each subchannel. Its optimization criterion is to maximize the system margin on the premise of maintaining the target bit error rate. The algorithm allocates bits step by step through the iterative process, and gradually increases the margin of the system until all bits are allocated.

Bit allocation

(1) Calculate the signal-to-noise ratio of each subchannel, S N R ( i ) , ∀ i SNR\left( i \right),\forall i SNR(i), ∀ i, assuming that the signal energy on all subchannels is normalized, ε ( i ) = 1 , ∀ i \varepsilon \left( i \right)=1,\forall i ε(i)=1,∀i
(2) Order γ m arg ⁡ i n = 0 dB {{\gamma }_{m\arg in}}=0\text{dB} γ margin = 0dB, the number of iterations is I t e r c o u n t = 0 Itercount=0 Itercount=0, the number of subchannels used is U s e d c a r r i e r s = N Usedcarriers=N Usedcarriers=N, where N N N is the maximum number of available subchannels
(3) Sequential calculation b ( i ) b(i) b(i) , b ^ ( i ) \hat{b}(i) b^(i) , d i f f ( i ) diff(i) diff(i) and U s e d c a r r i e r s Usedcarriers Usedcarriers
b ( i ) = log ⁡ 2 ( 1 + S N R ( i ) Γ + γ m arg ⁡ i n ( d B ) ) b(i)={{\log }_{2}}\left( 1+\frac{SNR(i)}{\Gamma +{{\gamma }_{m\arg in}}(dB)} \right) b(i)=log2​(1+Γ+γmargin​(dB)SNR(i)​)
b ^ ( i ) = r o u n d ( b ( i ) ) \hat{b}(i)=round(b(i)) b^(i)=round(b(i))
d i f f ( i ) = b ( i ) − b ^ ( i ) diff(i)=b(i)-\hat{b}(i) diff(i)=b(i)−b^(i)
if b ^ ( i ) = 0 \hat{b}(i)\text{=}0 b^(i)=0, U s e d c a r r i e r s = N − 1 Usedcarriers\text{=}N-1 Usedcarriers=N−1, Γ \Gamma Γ by S N R SNR SNR spacing is defined as the ideal distance when the system reaches the transmission capacity C S N R SNR SNR and actual transmission capacity R R R S N R SNR SNR ratio. Γ \Gamma Γ yes B E R BER As a function of BER and channel coding mode, assuming QAM modulation, then
Γ = − ln ⁡ ( 5 B E R t arg ⁡ e t ) 1.5 \Gamma \text{=}\frac{-\ln \left( 5BE{{R}_{t\arg et}} \right)}{1.5} Γ=1.5−ln(5BERtarget​)​
(4) Calculate R = s u m ( { b ^ ( i ) } ) R\text{=}sum\left( \left\{ \hat{b}\left( i \right) \right\} \right) R=sum({b^(i)}), if R = 0 R=0 If R=0, the channel is too poor to be used
(5) Calculate γ m a r g i n {{\gamma }_{margin}} γmargin​
γ m a r g i n   = γ m a r g i n + 10 log ⁡ 10 ( 2 R − R T U S e d C a r r i e r s ) {{\gamma }_{margin\text{ }}}={{\gamma }_{margin}}+10{{\log }_{10}}\left( {{2}^{\frac{R-{{R}_{T}}}{USedCarriers}}} \right) γmargin ​=γmargin​+10log10​(2USedCarriersR−RT​​)
(6) Number of iterations I t e r c o u n t = I t e r c o u n t + 1 Itercount=Itercount+1 Itercount=Itercount+1
(7) If R ≠ R T R\ne {{R}_{T}} R  = RT , and I t e r c o u n t < M a x c o u n t Itercount<Maxcount Itercount < maxcount, make U s e d c a r r i e r s = N Usedcarriers=N Usedcarriers=N, go to step (3), otherwise go to step (8)
(8) If R > R T R>{{R}_{T}} R> RT, find the smallest d i f f ( i ) diff\left( i \right) diff(i), correspondingly b ^ ( i ) = b ^ ( i ) - 1 \hat{b}(i)\text{=}\hat{b}(i)\text{-}1 b^(i)=b^(i)-1, d i f f ( i ) = d i f f ( i ) + 1 diff\left( i \right)=diff\left( i \right)+1 diff(i)=diff(i)+1, repeat this step until R = R T R\text{=}{{R}_{T}} R=RT​
(9) If R < R T R<{{R}_{T}} R < RT, find the largest d i f f ( i ) diff\left( i \right) diff(i), correspondingly b ^ ( i ) = b ^ ( i ) + 1 \hat{b}(i)\text{=}\hat{b}(i)\text{+}1 b^(i)=b^(i)+1 , d i f f ( i ) = d i f f ( i ) - 1 diff\left( i \right)=diff\left( i \right)\text{-}1 diff(i)=diff(i)-1, repeat this step until R = R T R\text{=}{{R}_{T}} R=RT​
(10) The transmission power on each subchannel is adjusted so that it corresponds to b ( i ) b(i) b(i)
P e ( i ) = P e , t arg ⁡ e t ,   ∀ i {{P}_{e}}\left( i \right)={{P}_{e,t\arg et}},\ \forall i Pe​(i)=Pe,target​, ∀i
(11) Adjust the total transmit power. Multiply all used subchannels by the same scale factor,
Total signal power P = P T P={{P}_{T}} P=PT​

Simulation parameter setting

Parameter nameParameter value
Average signal-to-noise ratio20
Bit error rate 1 0 − 4 10^{-4} 10−4
Number of subcarriers32
Average transmit power1
Upper limit of total number of bits128

The experimental results are as follows

As can be seen from the above figure, the final number of bits of the system is maximized by gradually adjusting the allocated bits. At the same time, it can also be seen that the result of Chow algorithm allocation has certain advantages. It discards the subcarriers with poor channel conditions, that is, the subcarriers used by the system for final transmission must have information transmission, which may be confirmed by the simulation results.
The code is as follows:
Main function

clear ;
close all;
legend('Channel gain'); 
hold on; 
title('Chow Bit power allocation'); 
ylabel('Bit allocation result'); 
xlabel('Subcarrier serial number'); 
ylabel('Power allocation results'); 
xlabel('Subcarrier serial number'); 


function [bits_alloc, power_alloc,iter_count] = Chow(SNR,N_subc,gap,Rt) 
% ---------------------initialization------------------------- 
margin=0;         %Initialization threshold        
Max_count=10;     %Maximum number of iterations        
iter_count=0;     %Initialization iteration                  
total_bits=0;     %Always more special than            
power_alloc=zeros(1,N_subc);  %Initialization power allocation result
bits_alloc=zeros(1,N_subc);   %Initialization bit allocation result
temp_bits=zeros(1,N_subc);    %Initialize intermediate variable
bit_round=zeros(1,N_subc);    %Initialize rounding bit
diff=zeros(1,N_subc);         %Initialize differential bit
%-----------------------------bits allocation------------------------- 
while (total_bits~=Rt)&&(iter_count<Max_count) 
    N_use=N_subc;                            %Number of available subcarriers during initialization
    temp_bits=log2(1+SNR./(1+margin/gap));   %Initial result of bit allocation
    bit_round=round(temp_bits);              %Number of bits after rounding
    diff=temp_bits-bit_round;                %Bit difference
    total_bits=sum(bit_round);               %Total number of bits after rounding
        disp('the channel is not be used');  %The channel is too poor to use
    N_notuse=length(find(bit_round==0));       %Number of unused subcarriers
    N_use=N_subc-N_notuse;                      %Number of subcarriers used 
    margin=margin+10*log10(2^((total_bits-Rt)/N_use));    %Update threshold
%------------------------------bits alteration-------------------------- 
while(total_bits>Rt)                %If bit_total>Rt,Find the smallest bit_diff(i)
    use_ind=find(bit_round>0);      %Accordingly bit_round Minus 1, bit_total=bit_total-1
    bit_diff=diff(use_ind);         %bit_diff(i)Add 1 until bit_total=Rt
while(total_bits<Rt)               %If bit_total<Rt,Find the largest bit_diff(i)
    use_ind=find(bit_round~=0);     %Accordingly bit_round Plus 1, bit_total=bit_total+1
    bit_diff=diff(use_ind);          %bit_diff(i)Minus 1 until bit_total=Rt
%--------------------------power allocation----------------------------- 

Added by DWaller10 on Thu, 17 Feb 2022 14:10:59 +0200