信息论编码2new

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1、BlockCodes.Detailedsolutionstoproblems2.1)10[]T101TP=11,G=[111],H=[IP]=,H=01n-k01111Codetablemc00001111SyndrometablerS0000000111010010111010010101011101111100Thereare8errorpatterns,andthemostlikelyonesare:eeoHT00000001110100110010Syndromesassociatedtothemostlikelyerrorpatte

2、rns,thatareerrorpatternsofoneerror,arealldistinct,sothatthecodeisabletocorrecterrorpatternsofonebit(t=1).Itcanalsodetectanyerrorpatternoftwobits(l=2).rTroHerÅemˆ0000000000000011100100000100101000000111010011111001010000001010101011111101100111111110011111112.2)d=11;d³l+1⇒l=10minmind³t2+1⇒t

3、=5min2.3)a)Sincel+t+1³d,then6+4+1³d=11minmind=n-k+1=11,withk=1,n=11minminBlockCodes.Detailedsolutionstoproblemsb)Itcandetectnerasuresandcorrect10erasuresc)d=n-k+1=11,withk=1,n=11minmin2.4)00000000111003101010311011041100013101101401101140001113a)R=3/6=1/2=5.0cb)Byselectingthreelinearlyinde

4、pendentcodevectorsofthecode,wecanformthegeneratormatrix.Thismatrixisformedsothattheidentitysubmatrixisatitsrightside100010011100T001G=[PIk]=101010,H=011110001101110c)Sincethecodeisalinearblockcode,theminimumHammingdistanceofthecodeistheminimumweightseenintheabov

5、etable,thatis,d=3mind)d=3,l=2,t=1minTe)Syndromevectorforthereceivedvectorr=(101011)isS=roH=(110),theerrorpattern-syndrometableisthefollowing:eeoHT100000100010000010001000001000100011000010101000001110Errorisinthesixthbit,andthene=(000001),c=rÅe=(101010)2.5)2a)InalinearblockcodeC2,5(),k=2,n

6、=5,thereare2=4codewordsofblength5.Thereforethecorrespondinggeneratormatrixisoftheform:BlockCodes.Detailedsolutionstoproblems10G=[PI]=Pk01Theallzerocodewordshouldbelongtothiscode.Fromthepossible32codewordsoflength5,wecanselectc=(11110)andc=(10101)toformthegeneratormatrixas:1211110G=

7、10101Thiscodehasthefollowingcodetable:mcw00000000011010131011110411010113Theminimumdistanceofthecodeisd=w=3.minminThecodehasatransposeparitycheckmatrixofsizenx(n-k)=5x3,whichmeansthatthesyndromevectorsareofsize1x3.Thenthenumberofnon-zerosyndromevectorsis7.