• 2022-06-26
    某数字基带传输系统在抽样时刻的抽样值存在码间干扰, 该系统的冲激响应 [tex=1.714x1.357]hAHQ1b+f+jrnIQc4cpjp4QsqWvvwYG0f4/qcOVOClSU=[/tex]的离散抽样值为[tex=10.714x1.214]BvA/wNaTYFOjma+XlcFd9BBe98wGZVsGtujpMrwIgCAEI9Fd8SKkND7u3x2F1fI3[/tex]。若该系统与三抽头迫零均衡器相级联, 请求出此迫零均衡器的抽头系数 [tex=5.0x1.214]JnYd9YyFcYSW99G2tbLz6Q==[/tex] 值, 并计算出均衡前后的峰值畸变值。
  • 解: (1) 包含均衡器后总的系统响应是[tex=8.357x1.357]0kAvukRfItd4NZptbrBjFhVDTStCDMdpGak0sIBVGa40yuEIO+4NrFhkP/ShwyB2LupAxCne06saRTM6XQ4gtSN3Bry9k0XnIrplu59PvT0=[/tex][tex=0.786x1.071]wnxUJMNvT36byk/uBD/yJQ==[/tex]代表卷积, 即[tex=29.214x3.286]4gjvqbN57Bwo6ivALt0lInPRzEPhCynJFwU4MrD+PBwvU9ZaE7uWNbhV6EPrit9Xp/fs+n0wEbgUgpxwzhlCTyNBQ+luAviJ6UKt3MEa41RvwcicrxzWzhkcwMEUFKsP+OXfoNi/bV8Z4H9b1MXVje1VFwrgguLlEL2kKh2kCb4=[/tex]三抽头迫零均衡将使[tex=8.214x2.786]KFMLrdH38Xbaow8UeyINY4ujqEPP+VXfGn0JptLa5wczsY/t1/cd8vTkoPyOqce48eKZBLQJhY/tze/6uZtnIMcEIXmjDb9lnQsfPQVXVTI=[/tex]由此得线性方程组[tex=12.357x4.071]GE56u9QCDTqcLxZ66HADyv04jvaCfJSCNXQ8V4JYrxte+bIKSQKOtTfencgCQbV1dGeWc1Vx4yfOgVM0WL+EMRhcmnk3iHJqFHG2/F+THZiNoRrldsUT72gOpjinnSJyuFbO9FchLk1BMMVYj/R12n5qT/XCIDbhQ8CTmy2prETWKYrwmZqY/MReWk6HKlDp[/tex]得解为[tex=8.357x5.214]GE56u9QCDTqcLxZ66HADypn/pF6ODEPe4uC9wgfVPPTeaab+s8bpe7in3RsRKJrGA74M1y54lxVd59xHJ9UOJuRX/1fuoTfHIgcMAMtCHnxb2aYVQbV2GrB7CjTbMG3yKeyoKI34dJ2sauytOBQKvQ==[/tex](2) 计算均衡前后的峰值畸变(1)均衡前的峰值畸变[tex=12.786x3.929]36wSHgRfw4UGCc+qSSpUu0isUVKits7MvXRbn8ZwUreSo0GfMhoSfWZkJ9SFLGLSSX46UvM2CrzUcGObhTYawSYlkTIcoY5HWqD2AQsW8cWi2ou0COMTTMnPksqaiZF91n2gEcvkmHCMMyEWwSIAjw==[/tex](2)均衡后的总体响应为[tex=21.643x4.786]C/FZmjzme1BWaR6p461zGGqm5qxF/gjMe9PmlxFQkbHpajtDCnqDhjh3TPRvNkVacTUKKoNYHbv+7K4wQiuf1MgP3HfWzv7RJm1XOodTuzbJ4T1K02g6NpzT0oDrJBZheuMbTSuYkYfzgwC+Xp+ixM78ROwRPx5qCCYMd6fYX5I=[/tex]均衡后的峰值畸变值为[tex=11.786x3.929]uGCm/CkjXFAH8QER+Xt1TKZJ2hoFjuFqvknrZWoaoM2VpGyUbSnU0HgmoMjnbaEz8f+5ShWSByf443m3RC6KHG+eeK1A45yn2YfLF/WqFKNnJ+6S4Ud3s52bG7N0CBCKg1hCr76T1kQv7nYNMIN5/Q==[/tex]

    内容

    • 0

      以4,9,1为为插值节点,求\(\sqrt x \)的lagrange的插值多项式 A: \( {2 \over {15}}(x - 9)(x - 1) + {3 \over {40}}(x - 4)(x - 1) + {1 \over {24}}(x - 4)(x - 9)\) B: \( - {2 \over {15}}(x - 9)(x - 1) + {3 \over {40}}(x - 4)(x - 1) + {1 \over {24}}(x - 4)(x - 9)\) C: \( - {2 \over {15}}(x - 9)(x - 1) + {3 \over {40}}(x - 4)(x +1) + {1 \over {24}}(x - 4)(x - 9)\) D: \( - {2 \over {15}}(x - 9)(x - 1) + {3 \over {40}}(x - 4)(x - 1) - {1 \over {24}}(x - 4)(x - 9)\)

    • 1

      【计算题】设有一个三抽头的时域均衡器, x(t) 在各抽样点的值依次为 ( 在其他抽样点均为零 ) , 试求均衡器输入波形 x(t) 峰值失真及输出波形 y(t) 峰值的失真。 (20.0分)

    • 2

      已知 x = {1:2, 2:3, 3:4},那么表达式 sum(x) 的值为____。 A: 6 B: 9 C: 4 D: 3

    • 3

      若某个时域均衡器能够确保消除当前时刻的前后5个时刻抽样值的码间串扰,则其抽头数目至少为() A: 5 B: 7 C: 9 D: 11

    • 4

      假设“☆”是一种新的运算,若3☆2=3×4,6☆3=6×7×8,x☆4=840(x>0),那么x等于: A: 2 B: 3 C: 4 D: 5 E: 6 F: 7 G: 8 H: 9