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    發布日期: 2021-05-09
  • 語言: Matlab
  • 標簽: BCJR??軟判決??

資源簡介

BCJR的譯碼,軟判決.卷積碼的編碼和譯碼.

資源截圖

代碼片段和文件信息

%?BCJR?algorithm?for?a?half-rate?systematic?recursive?convolutional?code
%?having?1?memory?element?a?generator?polynomial?of?[10]?and?a?feedback
%?polynomial?of?[11].?For?more?information?see?Section?1.3.2.2?of?Rob‘s
%?thesis?(http://eprints.ecs.soton.ac.uk/14980)?or?the?BCJR?paper?
%?(http://ieeexplore.ieee.org/xpls/abs_all.jsp?arnumber=1055186).
%?Copyright?(C)?2008??Robert?G.?Maunder

%?This?program?is?free?software:?you?can?redistribute?it?and/or?modify?it?
%?under?the?terms?of?the?GNU?General?Public?License?as?published?by?the
%?Free?Software?Foundation?either?version?3?of?the?License?or?(at?your?
%?option)?any?later?version.

%?This?program?is?distributed?in?the?hope?that?it?will?be?useful?but?
%?WITHOUT?ANY?WARRANTY;?without?even?the?implied?warranty?of?
%?MERCHANTABILITY?or?FITNESS?FOR?A?PARTICULAR?PURPOSE.??See?the?GNU?General?
%?Public?License?for?more?details.

%?The?GNU?General?Public?License?can?be?seen?at?http://www.gnu.org/licenses/.



function?[aposteriori_uncoded_llrs?aposteriori_encoded1_llrs?aposteriori_encoded2_llrs]?=?bcjr_decoder(apriori_uncoded_llrs?apriori_encoded1_llrs?apriori_encoded2_llrs)

????if(length(apriori_uncoded_llrs)?~=?length(apriori_encoded1_llrs)?||?length(apriori_encoded1_llrs)?~=?length(apriori_encoded2_llrs))
????????error(‘LLR?sequences?must?have?the?same?length‘);
????end


????%?All?calculations?are?performed?in?the?logarithmic?domain?in?order?to
????%?avoid?numerical?issues.?These?occur?in?the?normal?domain?because?some?of?
????%?the?confidences?can?get?smaller?than?the?smallest?number?the?computer?can
????%?store.?See?Section?1.3.2.4?of?Rob‘s?thesis?for?more?information?on?this.
????%
????%?A?multiplication?of?two?confidences?is?achieved?using?the?addition?of?the
????%?corresponding?log-confidences.?If?A?=?log(a)?and?B?=?log(b)?then
????%?log(a*b)?=?A+B?(Equation?1.17?in?Rob‘s?thesis).
????%
????%?An?addition?of?two?confidences?is?achieved?using?the?Jacobian?logarithm
????%?of?the?corresponding?log-confidences.?The?Jacobian?logarithm?is?defined
????%?in?the?jac.m?file.?If?A?=?log(a)?and?B?=?log(b)?then?
????%?log(a+b)?=?max(AB)?+?log(1+exp(-abs(A-B)))?(Equation?1.19?in?Rob‘s
????%?thesis).

????%?Matrix?to?describe?the?trellis
????%?Each?row?describes?one?transition?in?the?trellis
????%?Each?state?is?allocated?an?index?123...?Note?that?this?list?starts?
????%?from?1?rather?than?0.
????%??????????????FromState???ToState?????UncodedBit??Encoded1Bit?Encoded2Bit
????transitions?=?[1???????????1???????????0???????????0???????????0;
???????????????????1???????????2???????????1???????????1???????????1;
???????????????????2???????????1???????????1???????????1???????????0;
???????????????????2???????????2???????????0???????????0???????????1];

????%?Find?the?largest?state?index?in?the?transitions?matrix???????????
????%?In?this?example?we?have?two?states?since?the?code?has?one?memory?element
????state_count?=?max(max(transitions(:1))max(transitions(:2)));

????%?Calc

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