Честотна лента на канала

Analogchannel

ThebandwidthoftheanalogchannelW=f2-f1wheref1isthelowestfrequencythatthechannelcanpass,andf2isthehighestfrequencythatthechannelcanpass.BotharedeterminedbythechannelDeterminedbyphysicalcharacteristics.Whenthecircuitsthatmakeupthechannelaremade,thebandwidthofthechannelisdetermined.Inordertomakethesignaltransmissionlessdistortion,thechannelmusthaveenoughbandwidth.

Digitalchannel

Digitalchannelisadiscretechannel,whichcanonlytransmitdigitalsignalsofdiscretevalues.Thebandwidthofthechanneldeterminesthehighestpossibleundistortedtransmissionpulsesequenceinthechannelrate.

Adigitalpulseiscalledasymbol,andweusethesymbolratetorepresentthenumberofsignalwaveformtransformationsperunittime,thatis,thenumberofsymbolstransmittedthroughthechannelinaunittime.IfthesignalsymbolwidthisTseconds,thesymbolrateB=1/T.Theunitofsymbolrateiscalledbaud(Baud),sothesymbolrateisalsocalledbaudrate.Asearlyas1924,HenryNyquist,aresearcheratBellLaboratories,deducedthelimitbaudrateofanoiselesschannelwithlimitedbandwidth,whichiscalledtheNyquisttheorem.IfthechannelbandwidthisW,theNyquisttheoremstatesthatthemaximumsymbolrateisB=2W(Baud).ThechannelcapacityspecifiedbytheNyquisttheoremisalsocalledtheNyquistlimit,whichisdeterminedbythephysicalcharacteristicsofthechannel..ItisimpossibletotransmitpulsesignalsbeyondtheNyquistlimit,sothechannelbandwidthmustbeimprovedtofurtherincreasethebaudrate.

Theamountofinformationcarriedbythecodeelementisdeterminedbythenumberofdiscretevalues​​takenbythecodeelement.Ifthesymboltakestwodiscretevalues,onesymbolcarries1bitofinformation.Ifthesymbolcantakefourdiscretevalues,onesymbolcarries2bitsofinformation.Thatis,theamountofinformationn(bit)carriedbyasymbolhasthefollowingrelationshipwiththenumberoftypesofsymbolsN:n=log2N

Theamountofinformation(numberofbits)transmittedonthechannelperunittimeiscalledthedatarate.Thewaytoincreasetherateatacertainbaudrateistouseonesymboltorepresentmorebits.Iftwobitsareencodedintoonesymbol,thedataratecanbedoubled.

За това имаме формулата:

R=Blog2N=2Wlog2N(b/s)

където Rпредставлява скоростта на данните и единицата за секундабит, съкратено като bpsorb/s

Datarateandbaudratearetwodifferentconcepts.Thetwoareequalonlywhenthesymboltakestwodiscretevalues.Forordinarytelephonelines,thebandwidthis3000HZandthehighestbaudrateis6000Baud.Themaximumdataratecantakedifferentvalues​​dependingontheencodingmethod.Thesearethelimitvalues​​underidealconditionswithoutnoise.Theactualchannelwillbeinterferedbyvariousnoises,soitisfarfromreachingthedatatransferratecalculatedaccordingtotheNyquisttheorem.Shannon'sresearchshowsthatthelimitdataratewithnoisecanbecalculatedbythefollowingformula:

C=Wlog2(1+s/n)

Тази формула се нарича теорема на Шанън, където Wisthechannelwidthwisthechannelwisthesredmobilityofthesignal,Nisthesredmobilityofthenoise,and/se нарича съотношение сигнал-шум.ОткактоStoNistoolaширока реална употреба,се вземат децибели(db).Връзкатамеждудецибелиисигнал-към -коефициент на шум: db=10lgs/n

Например, когато/nis1000, съотношението сигнал/шум е 30db. Тази формула няма нищо общо с дискретната стойност на сигнала, тоест, без значение какъв метод е използван за модулация, докато съотношението сигнал/шум е дадено, се определя максималното количество предаване на информация за единица време. Например, ако чан nelbandwidтова е 3000HZ и съотношението сигнал/шум е 30db, максималната скорост на предаване на данни е

C=3000log2(1+1000)≈3000×9,97≈30000b/s

ThisisthelimitValuehasonlytheoreticalmeaning.Infact,itisverygoodthatthedataratecanreach9600b/sonatelephonelinewithabandwidthof3000HZ.

Tosumup,wehavetwoconceptsofbandwidth.Intheanalogchannel,thebandwidthiscalculatedaccordingtotheformulaW=f2-f1.Forexample,thebandwidthofaCATVcableis600HZor1000HZ;thebandwidthofadigitalchannelisthechannelThemaximumdataratethatcanbeachieved,forexample,thebandwidthofEthernetis10MB/Sor100MB/S,andthetwocanbeconvertedtoeachotherbyShannon'stheorem.

Свързани разлики

Широчина на честотната лента на канала

Ширината на честотната лента е този стандарт за изпращане на безжични сигнали. В често използваната честотна лента 2,4-2,4835 GHz, широчината на честотната лента на всеки канал е 20 MHz; първата работи с b/g/nprоколи, а втората имаsac/a/n.

Thehigherthefrequency,theeasieritistobedistorted.Inthecaseof11n,20MHzcanreachabandwidthof144Mbps(howtocalculate?).Ithasbetterpenetrationandalongtransmissiondistance(about100meters);40MHzInthecaseof11n,itcanreachabandwidthof300Mbps,withaslightlypoorerpenetrationandashorttransmissiondistance(about50meters).

Ширина на честотната лента на сигнала

Signalbandwidthreferstoelectromagneticwavesintheradiofrequencybandthatpropagateinfreespace(includingairandvacuum).

Electromagneticwavesincludemanytypes,arrangedintheorderoffrequencyfromlowtohigh:radiowaves,infraredrays,visiblelight,ultravioletrays,X-raysandgammarays.Radiowavesaredistributedinthefrequencyrangeof3Hzto3000GHz.Inthisspectrum,itcanbedividedinto12bands.Thelowerthefrequency,thesmallerthepropagationloss,thelongerthecoveragedistance,andthestrongerthediffractionability.However,thefrequencyresourcesofthelowfrequencybandaretight,andthesystemcapacityislimited.Therefore,theradiowavesofthelowfrequencybandaremainlyusedinsystemssuchasbroadcasting,television,andpaging.

Thehighfrequencybandisrichinfrequencyresourcesandthesystemcapacityislarge.Butthehigherthefrequency,thegreaterthepropagationloss,thecloserthecoveragedistance,andtheweakerthediffractionability.Inaddition,thehigherthefrequency,thegreaterthetechnicaldifficultyandthecorrespondingincreaseinthecostofthesystem.