碼分多址cdma基本原理-在線瀏覽
2025-07-17 22:30本頁面
【正文】 ( ) c o s ( )i i j i j i c ijz t A c t t b t t t?????? ? ? ??????式中包含了所需的信號和其它用戶的信號。 ? Tb/Tc越大，干擾降低越多 1 ( ) ( )2 i i i i iA c t t b t t??練習(xí) ? 在 CDMA系統(tǒng)上行鏈路中，請畫出相關(guān)各點的波形。 (3) b1(t)=“01100”, b2(t)=“00100”, b3(t)=“01001” (4)c1(t)=000011101100101 c2(t)=110110010100001 c3(t)=001010000111011 RAKE接收機 … Path1 Path2 PathN ? ?tc?cosc1(t) c1(t⊿ 1) c1(t⊿ N1) … ∫ Tb ∫ ∫ Hold until Tb+⊿ N Hold until Tb+⊿ N Hold until Tb+⊿ N Tb Tb+⊿ 1 Tb+⊿ N1 … … ∑ decide ?Multipath has been treated as causing delayed versions of the signal to add to the system noise when the differential delay exceeds the chip time. ?Substantial performance improvement can occur by detecting each additional path separately,thereby enabling the signals to be bined z(t) 例：兩路 RAKE接收機 Downlink Multiuser with Two Equal Strength Paths t=0 y(t) y(t2tc) z(t) 例：兩路 RAKE接收機 Downlink Multiuser with Two Equal Strength Paths t=0 z(t) c1(t) Tb 2Tb 3Tb 4Tb 5Tb t 0 1 1 0 0 z(t)c1(t) 0 1 例：兩路 RAKE接收機 Downlink Multiuser with Two Equal Strength Paths t=0 z(t) c1(t2Tc) Tb 2Tb 3Tb 4Tb 5Tb t 0 1 1 0 0 z(t)c1(t2Tc) 2Tc 例：兩路 RAKE接收機 Bit 1 Bit 2 Bit 3 Bit 4 Bit 5 Path1 integrator output +4 +4 8 +8 +12 Path2 integrator output +8 8 12 4 +12 Sum of integrator outputs 12 4 20 +4 +24 Detected bit values 0 1 1 0 0 發(fā)送的比特 0 1 1 0 0 蜂窩系統(tǒng)的通信容量 在蜂窩移動通信系統(tǒng)中，系統(tǒng)的容量有多種衡量方法， 如： 每小區(qū)可用信道數(shù)（ ch/cell） 每小區(qū)每兆赫茲可用信道數(shù)（ ch/cell/MHz） 每小區(qū)愛爾蘭（ Erl/cell） 每平方公里用戶數(shù)（用戶數(shù) /km
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