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6 6 数字基带传输系统 数字基带传输系统 本章教学要求: 1 、掌握基带信号常用码型、码间干扰的原因和 解决方法、无码间干扰的条件(时域和频域)。 2 、理解数字基带信号的频谱特性和无码间干扰 的基带传输系统的抗噪声性能。 3 、了解均衡原理、部分响应原理和眼图原理。

第 6 章 数字基带传输系统

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第 6 章 数字基带传输系统. 本章教学要求: 1、掌握基带信号常用码型、码间干扰的原因和解决方法、无码间干扰的条件(时域和频域)。 2、理解数字基带信号的频谱特性和无码间干扰 的基带传输系统的抗噪声性能。 3、了解均衡原理、部分响应原理和眼图原理。. 主要外语词汇. 数字基带传输 Digital Baseband Transmission 码间干扰 ISI(Intersymbol Interference) 不归零码 NRZ(Non-Return-to-Zero) 归零码 RZ(Return-to-Zero) - PowerPoint PPT Presentation

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  • 6 12 3

  • Digital Baseband Transmission ISIIntersymbol Interference NRZNon-Return-to-Zero RZReturn-to-Zero AMIAlternate Mark Inversion Code HDB3High Density Bipolar 3 Code PSTPaired Selected Ternary Code Eye Patterns

  • 6.16.26.36.46.56.66.76.8

  • P134283 671015

    1718

  • 6.16.2123

  • 6.1

  • AWGN

  • 0

  • 6.2

  • 1. UNRZ (Unipolar Non-Return-to-Zero) 1102= Tb

  • 2. URZ (Unipolar Return-to-Zero)110 2 Tb

  • 3. BNRZ (Bipolar Non-Return-to-Zero)1102 = Tb

  • 4. BRZ (Bipolar Return-to-Zero)110 2 Tb

  • 5. 1 0 1 1 0 0

  • 6. 01+E00+3E4

  • 1 2 3

  • 1 0 0 1 1 0 0 0 0 0 0 0 1 1 0 0 1 1 AMI Alternate Mark Inversion 10 21AMI +1 0 0 1 +1 0 0 0 0 0 0 0 -1 +1 0 0 -1 +1

  • 2. HDB3 High Density Bipolar 3 code (1) 03AMIHDB3 (2)03401V1V0 (3)V V1 V1 1 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 l 1 AMI -1 0 0 0 0 +1 0 0 0 0 -1 +1 0 0 0 0 -1 +1 HDB3 -1 0 0 0 -V +1 0 0 0 +V - 1 +1 0 0 0 +V -1 +1

  • 1 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 l 1 AMI -1 0 0 0 0 +1 0 0 0 0 -1 +1 0 0 0 0 -1 +1 HDB3 -1 0 0 0 -V +1 0 0 0 +V - 1+1 -B 0 0 -V +1 -14V1400VB B3 VB1VB01

  • 1 AMIHDB3 HDB3 1 0 -1 0 0 0 -V 0 1 -1 B 0 0 V -1 +1AMI 1 0 -1 0 0 0 0 0 1 -1 0 0 0 0 1 -1 1 0 1 0 0 0 0 0 1 1 0 0 0 0 1 1

  • 3. PST ( + 0 )

    00 010 0 10 0 011

  • PST 0 1 0 0 1 1 1 0 1 0 1 1 0 1 PST(+) 0 + - + + - - 0 + 0 + - 0 - PST(- ) 0 - - + + - + 0 - 0 + - 0 +

    00 010 0 10 0 011

  • 4. (Manchester )01 001110 1 1 0 0 1 0 1 10 10 01 01 10 01 10

  • 5. Miller Miller110010000 0011

  • 6. CMI 11100 001 CMI 103 CMI21B2B

  • (b)

    (c)CMI

  • 2 1010000011000011CMIAMIHDB3

  • 34140AMIHDB3

  • 6.3123

  • 1 2 3 4

  • .

  • g ( t - n Ts ) psn(t) = g ( t - n Ts) (1-p) 12

  • 1v(t)S(t)V(t)

  • 2u(t)u(t)

  • UT()u(t)UT(t):

  • m=n

    mn

  • 3s(t)PS(f)Pu(f)Pv(f)

  • RB(2)

    mfs

  • G1(mfs)G2(mfs)

  • 1 g1(t)=0G1( f )=0 g2(t)=G(t)=TS p=1/2,

  • 2 g1(t)= G(t) g2(t)=G(t) G1(f)= G2(f) p=1/2

  • 3g1(t)=0G1()=0 g2(t)=G(t)=TS /2

    p=1/2

  • 4 g1(t)= G(t) g2(t)=G(t)=TS /2 G1(f)= G2(f)

  • 01

  • 6.4H()(t)h(t) y(t)=x(t)*h(t) Y()=X()H()

  • g(t) /2 /2 Sa() G()Sa(t)

  • 6.5 H()=GT()C()GR()

  • d(t)t=mTb

  • t = mTb r(mTb)= am (k=m)

  • Q(mTb) = 0 m-k = n

  • 2 2/Tb

  • =-2m/Tb

  • Heq ()=(2m-1)/ Tb mH()2/ Tb(-/ Tb / Tb) Tb

  • B = 1/2Tb(Hz) h(t)=Sa(t / Tb ) t = nTbTb RB = 1/Tb RB = 2B2RB = 2BN BN = 1/2Tb = RB/2 ,

  • H()(-3/ Tb -/Tb) (-/ Tb / Tb) (/ Tb 3/ Tb)2/ Tb (-/ Tb / Tb) Tb

  • 1h(t)nTb(n0)2h(t) t 3 3B = 1/Tb RB = 2BN BN = 1/2Tb = RB/2

  • 0<
  • 2 Baud/Hzh(t) 2 2Baud/Hz

  • 4.6 TbSa(x)

  • TbSa(x) h (t )t 2Sa(x)th (t ) Sa(x)

  • B= 1 / 2Tb = BN = RB/2 = RB / B = 2 Baud/Hz

  • t =Tb/2h(t)=1h(t)

  • h(t)Tb0A 2A

  • anan+1an+2 0101

  • bncn

  • 1anbn an = bn bn 1 bn = an bn 122 Cn=bn+bn1 Cn2[Cn]mod2=[bn+bn1]mod2=bnbn1=anan=[Cn]mod2 Cn2anan-1

  • 32

    {an}10110001011{bn}11011110010{bn-1} 01101111001{Cn}0+200+2+2+20-200{Cn}0+200+2+2+20000{an}10110001111

  • NTbSa(x)

    R1R2RNR1=1R2=1Ri=0I

    BN = 1 / 2TbRB = 2BN

  • 6.7H()

  • 2N+1Cii = - N -101 N

  • t = t0+kTbi = - N -101 NCn

  • Ci

    2N+1 Cig0 N

  • x(t )x1=1/4x0=1x+1=1/22N+1=3N=1

    y-1= C-1x0+C0x-1+C1x-2 = 0 y0 = C-1x1+C0x0+C1x-1 = 1 y1 = C-1x2+C0x1+C1x0 = 0x C1= 1 / 3C0= 4 / 3C+1= 2 / 3

  • y-1=0y0=1y1=0 y-2=-1/12y2=-1/3

  • =T b(nTb)

  • ab

  • 6.8

  • s(t)n(t)nR(t)x(t) () x(t)=s(t)+nR(t)

  • x(t)=s(t)+nR(t)= A1+nR(t)1 A0+nR(t)0 A1 1A0 0A1 A A0 =0 A1 A/2 A0 = -A/2 Vb x(kTb)Vb 1 x(kTb)Vb 0

  • 1001 (a)(b)(a)

  • P(S1)P(S0)10,VbP[ x< Vb | S1 ]=P(0 | 1)10 P[ x> Vb | S0 ] =P(1 | 0)01 Pe= P(S1) P( 0 | 1 ) + P(S0) P( 1 | 0 )Pe

  • n(t)0n0/2nR(t)0nR(t) VnR(kTb)

  • x(t)=s(t)+nR(t)x(t)

    01 : :

  • P(S1)P(S0)1/2 x(t)

  • A1 A0n2 P(1)P(0)1/2 Vb*= ( A1 +A0 ) / 2

  • Vb( A1 +A0 ) / 2

  • A1 A/2 A0 = -A/2

    P(1) = P(0) = 1/2 A1 A A0 =0 P(1) = P(0) = 1/2

  • Vb*= ( A1 +A0 ) / 2PeA1 A0A

  • P(1)P(0)1/2

  • x0 1A1 = A A2 0A0 = 0 0 S = P(1) S1 + P(0) S0 A2 / 2 ()N = n2 = S / N = A2 / 2n2

  • 1A1 A / 2 A2 / 4 0A0 = - A / 2 A2 / 4 S = P(1) S1 + P(0) S0 A2 / 4 ()N = n2 = S / N = A2 / 4n2

  • 1A1 A A2 0A0 = - A A2 S = P(1) S1 + P(0) S0 A2 ()N = n2 = S / N = A2 / n2

  • Pe(1) (2) 3dB (3) Pe~ Pe Pe

  • 1 2Q

  • AMIHDB3

  • 1 2 3Nyquist

    Heq()

  • RB=2BN