ofdm信道估计的MSE性能比较阐述了几种经典算法仿真性能的对比

信道估计的MSE性能比较,阐述了几种经典算法仿真性能的对比

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%   Author:      Vinay Uday Prabhu
%   E-mail:      vinay_u_prabhu@yahoo.co.uk
%   Function:    Comparison of the performances of the LS and the MMSE channel estimators
%                for a 64 sub carrier OFDM system based on the parameter of Mean square error
%  Assumptions: The channel is assumed to be g（t）=delta（t-0.5 Ts）+delta（t-3.5 Ts）
%               {Fractionally spaced}
%For more information on the theory and formulae used  please do refer to the paper On
%“Channel Estimation In OFDM systems“ By Jan-Jaap van de Beek Ove Edfors Magnus Sandell
% Sarah Kate wilson and Petr Ola Borjesson In proceedings Of VTC‘95 Vol 2 pg.815-819
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clc;
clear all;
%Generation of a naive training sequence..
%Assuming BPSK modulation ...symbols:+1/-1
X=zeros（6464）;
d=rand（641）;
      for i=1:64
       if（d（i）>=0.5）
           d（i）=+1;
       else
           d（i）=-1;
       end
    end
 for i=1:64
     X（ii）=d（i）;
 end
%Calculation of G[The channel Matrix]
 %The channnel is... 
 %tau=[0 0.2 0.4 0.6 15 17.2]; %HT 
 tau=[0 0.2 0.6 1.6 2.4 5.0]; %TU The fractionally spaced taps..
 %tau=[0.5 3.5];%two-ray
 %for q=1:2
  %   a（q）=3;
 %end
 for q=1:6
   a（q）=sqrt（exp（-tau（q）））;%TU
 end
 %for q=1:4
 %    a（q）=sqrt（exp（-3.5*tau（q）））;%HT
 %end
 %for q=5:6
 %   a（q）=sqrt（0.1*exp（15-tau（q）））;%HT
 %end
  %Generation of the G matrix...
for k=1:64
      s=0;
      for m=1:6%TUHT
      %for m=1:2%2-ray
      s=s+a（m）*（exp（-j*pi*（1/64）*（k+63*tau（m）） * （ sin（pi*tau（m）） / sin（pi*（1/64）*（tau（m）-k）））））;
      %Go through the above cited paper for the theory behind the formula
      end
g（k）=s/sqrt（64）;
end
g（5）=g（4）;%TU
%g（15）=g（14）;%HT
G=g‘;%Thus the channel vector is evaluated..
H=fft（G）;% In the freq domain..
u=rand（6464）;
F=fft（u）*inv（u）;% ‘F‘ is the twiddle factor matrix..

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% Evaluation of the autocovariance matrix of G-Rgg
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gg=zeros（6464）;
for i=1:64
    gg（ii）=G（i）;
end
gg_myu = sum（gg 1）/64;                    
gg_mid = gg - gg_myu（ones（641）:）;        
sum_gg_mid= sum（gg_mid 1）;
Rgg = （gg_mid‘ * gg_mid- （sum_gg_mid‘  * sum_gg_mid） / 64） / （64 - 1）;
hh=zeros（6464）;
for i=1:64
    hh（ii）=H（i）;
end
hh_myu = sum（hh 1）/64;                    
hh_mid = hh - hh_myu（ones（641）:）;        
sum_hh_mid= sum（hh_mid 1）;
Rhh = （hh_mid‘ * hh_mid- （sum_hh_mid‘  * sum_hh_mid） / 64） / （64 - 1）;
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 属性            大小     日期    时间   名称
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    .CA....      4988  2009-08-26 10:19  MSE_compare.m

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                 4988                    1