第14講 基帶傳輸：均衡算法、部分響應(yīng)系統(tǒng)

1、2, sisssiHTTTT12 Baud/Hz2sfssfRWWTTWcos122ssWTTcos21( )( ) ( )( )1 ( )( )( )( )0 TRTRHGCGWHGGCW升余弦升余弦( ) ( )( )HTH2, sisssiHTTTTsT2),(ssTT2( ), 2ssssisTiTTTTTiHT222siisssiiiHHTTTTTsT2( )sjnTnnTe1式( )222ssjnTjnTsssnisTTTTedediHT2式1( )()2sssTjnTTnnsTnnhedtnT ( )()NisiNe ttiT ( )sNjiTiiNEe( )( )( )()Ni

2、siNy tx te tx tiT000()()()NNsissisiNiNy kTtx kTtiTx ki Tt,Nkik iik ik iiN k iyxx101101112101211,1,4211,1,42311,0,0,1644kik iixxxyxyyyyy 其余為0001kkkDyy222001kkkyy0001kkkDxx00kkkDx01NiiiNyx0001NiiiNixx001NiiiNix Nkik iiNyx00000001()NNNkkik iiikik iiNiNiNiiiNNNkiikik ikik iikiNiNiNiiiyxxxxxxx xxxxx x000

3、001()()NNkik iikkik iikkiNkiNkikiDxxx xxxx xy0 zeros010 zeros0NYN 101NNNNNAN Nkik iiNyx0121210222121220122110NNNNNNNNxxxxxxxxXxxxxxxxxYXA10111,1,44xxx 010Y 101A0121012101104111441014xxxXxxxxxxYXA()sa mTmyma()sy mTNmkm kkNayNmmmmkm kkNeaaay2mJE e22Nmmkm kkNJE eEaykkJ 0k 2222, 1, 2,mmkmmmmm kkkkeaJE e

4、E eE eE e ykN mm kE e y( )eymm kRkE e y2( ), 1, 2,keykJRkkN 2( )0( )0, 1, 2,keyeykJRkRkkN ( )()()0, 1, 2,eymm kmmm kNNmnm nm km kmm knm nnNnNNm kmnm km nnNRkE e yEaayEayyE yayyE yaE yykN , 1, 2,Nm kmnm km nnNE yaE yykN ( )(), 1, 2,NNaynm nm knynNnNRkE yyR nkkN ( )(), 1, 2,NaynynNRkR nkkN 1ayyyayRR A

5、ARR()( 1)(0)(1)()ayayayayayayRNRRRRNR101NNA(0)( 1)( 21)( 2)(1)(0)( 22)( 21)(21)(22)(0)( 1)(2)(21)(1)(0)yyyyyyyyyyyyyyyyyRRRNRNRRRNRNRNRNRRRNRNRRR(0)(1)(21)(2)(1)(0)(22)(21)(21)(22)(0)(1)(2)(21)(1)(0)yyyyyyyyyyyyyyyyyRRRNRNRRRNRNRNRNRRRNRNRRR ( )0 sssTTHTsin()( )Sa()ssst Th tt Tt T1sin()2( )()2ssssTt

6、Tg tTtT2sin()2( )()2ssssTtTg tTtT22sin()sin()22( )()()22cos()441ssssssssssTTttTTg tTTttTTtTtT2cos, 2( )0, ssssTTTGT21 t2t1 t1sT1 2sT1sin()2( )()2ssssTtTg tTtT22cos()4( )41sstTg ttTkkkaac11kkkaca11kkkkkkbabbba1kkkbbc2mod kkca 1 0 1 1 0 1 0 0 1 1 0 0 0 1 1 0 1 1 0 1 2 2 1 1 2 2 2 1 1 0 1 1 0 1 2 2 1 1 2 2 2 1 kbkakc1 0 0 1 0 0 1 1 1 0 1 1 1 1 0 1 0 1 1 0 1 0 0 1 1 0 0 0 1 kc ka 1 0 1 1 0 1 2 2 0 1 2 2 2 1 1 0 1 1 0 1 0 0 0 1 0 0 0 1 kc ka 212( )2c