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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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?屬性????????????大小?????日期????時間???名稱
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????.CA....??????4988??2009-08-26?10:19??MSE_compare.m

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?????????????????4988????????????????????1