资源-网络工程师知识点
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资源-网络工程师知识点 <link href="/image.php?url=https://csdnimg.cn/release/download_crawler_static/css/base.min.css" rel="stylesheet"/><link href="/image.php?url=https://csdnimg.cn/release/download_crawler_static/css/fancy.min.css" rel="stylesheet"/><link href="/image.php?url=https://csdnimg.cn/release/download_crawler_static/90403929/3/raw.css" rel="stylesheet"/><div id="sidebar" style="display: none"><div id="outline"></div></div><div class="pf w0 h0" data-page-no="1" id="pf1"><div class="pc pc1 w0 h0"><img alt="" class="bi x0 y0 w1 h1" src="/image.php?url=https://csdnimg.cn/release/download_crawler_static/90403929/bg1.jpg"/><div class="t m0 x1 h2 y1 ff1 fs0 fc0 sc0 ls0 ws0">2018<span class="_ _0"> </span><span class="ff2">网工考点总<span class="_ _1"></span>结</span></div><div class="t m0 x1 h2 y2 ff2 fs0 fc0 sc0 ls0 ws0">海明码</div><div class="t m1 x1 h3 y3 ff3 fs1 fc0 sc0 ls0 ws0">(1)<span class="ff4 fc1 sc1">海明距<span class="_ _1"></span>离</span><span class="ff4">：<span class="_ _2"></span>码字之间的海<span class="_ _1"></span>明距离是一<span class="_ _1"></span>个码字要变成<span class="_ _1"></span>另一个码字<span class="_ _1"></span>时必须改变<span class="_ _1"></span>的</span></div><div class="t m1 x1 h3 y4 ff4 fs1 fc0 sc0 ls0 ws0">最小位数。</div><div class="t m1 x1 h3 y5 ff3 fs1 fc0 sc0 ls0 ws0">(2)<span class="ff4 fc1 sc1">海明不<span class="_ _1"></span>等式</span><span class="ff4">：</span>m<span class="_ _0"> </span><span class="ff4">位数据位增加<span class="_ _0"> </span></span>k<span class="_ _3"> </span><span class="ff4">位冗余位组成<span class="_ _0"> </span></span>n=m+k<span class="_ _0"> </span><span class="ff4">位的纠错码，要实现海</span></div><div class="t m1 x1 h3 y6 ff4 fs1 fc0 sc0 ls0 ws0">明码纠错，则<span class="_ _1"></span>需满足不等<span class="_ _1"></span>式：<span class="ff3"> m+k<span class="_ _1"></span>+1</span>≤<span class="ff3">2</span></div><div class="c x2 y7 w2 h4"><div class="t m2 x3 h5 y8 ff5 fs2 fc0 sc2 ls0 ws0">k</div></div><div class="t m1 x1 h3 y9 ff4 fs1 fc2 sc3 ls0 ws0">注<span class="ff3">:</span>∵<span class="ff3">m</span>≥<span class="ff3">1<span class="_ _1"></span> </span>∴<span class="ff3">k</span>≥<span class="ff3">2</span></div><div class="t m0 x1 h2 ya ff2 fs0 fc0 sc0 ls0 ws0">奈奎斯特定<span class="_ _1"></span>律</div><div class="t m1 x4 h3 yb ff4 fs1 fc0 sc0 ls0 ws0">码元速率<span class="ff3">(</span>信<span class="_ _1"></span>道的极限速<span class="_ _1"></span>率<span class="ff3">)</span>等于信<span class="_ _1"></span>道带宽<span class="ff3">(</span>低通<span class="_ _1"></span>信道<span class="ff3">)</span>的<span class="_ _0"> </span><span class="ff3">2<span class="_ _3"> </span></span>倍。即：</div><div class="t m1 x5 h3 yc ff3 fs1 fc0 sc0 ls0 ws0">B=2W</div><div class="t m1 x1 h3 yd ff4 fs1 fc0 sc0 ls0 ws0">式中<span class="_ _4"> </span><span class="ff3">B<span class="_ _3"> </span></span>为<span class="_ _1"></span>码元速率<span class="ff3">(<span class="_ _1"></span></span>单位波特<span class="ff3">)<span class="_ _1"></span></span>，<span class="ff3">W<span class="_ _4"> </span></span>为信道带宽<span class="_ _1"></span><span class="ff3">(</span>单位<span class="_ _4"> </span><span class="ff3">Hz)</span></div><div class="t m0 x1 h2 ye ff2 fs0 fc0 sc0 ls0 ws0">奈奎斯特取<span class="_ _1"></span><span class="ff1">(</span>采<span class="ff1">)</span>样定<span class="_ _1"></span>理</div><div class="t m1 x4 h3 yf ff4 fs1 fc0 sc0 ls0 ws0">若取样频率大<span class="_ _1"></span>于模拟信号<span class="_ _1"></span>最高频率的两<span class="_ _1"></span>倍，则可用<span class="_ _1"></span>得到的样本空<span class="_ _1"></span>间恢复原</div><div class="t m1 x1 h3 y10 ff4 fs1 fc0 sc0 ls0 ws0">来的模拟信号<span class="_ _1"></span>，即：</div><div class="t m1 x1 h3 y11 ff3 fs1 fc0 sc0 ls0 ws0"> <span class="_ _1"></span> <span class="_ _1"></span> <span class="_ _1"></span> <span class="_ _5"></span>f=1/T><span class="_ _1"></span>2f</div><div class="c x6 y12 w3 h6"><div class="t m3 x3 h5 y13 ff5 fs2 fc0 sc2 ls0 ws0">max</div></div><div class="t m0 x1 h2 y14 ff2 fs0 fc0 sc0 ls0 ws0">香农公式</div><div class="t m1 x7 h3 y15 ff3 fs1 fc0 sc0 ls0 ws0">C=Wlog</div><div class="c x8 y16 w2 h7"><div class="t m4 x3 h5 y17 ff5 fs2 fc0 sc2 ls0 ws0">2</div></div><div class="t m1 x9 h3 y15 ff3 fs1 fc0 sc0 ls0 ws0">(1+S/N<span class="_ _1"></span>)</div><div class="t m1 x4 h3 y18 ff4 fs1 fc0 sc0 ls0 ws0">式中，<span class="ff3">C<span class="_ _4"> </span></span>为信道的<span class="_ _1"></span>极限信息传<span class="_ _1"></span>输速率<span class="ff3">(</span>单<span class="_ _1"></span>位<span class="_ _4"> </span><span class="ff3">bit/s)</span>，<span class="_ _1"></span><span class="ff3">W<span class="_ _4"> </span></span>为信道的带宽<span class="_ _1"></span><span class="ff3">(</span>单位</div><div class="t m1 x1 h3 y19 ff3 fs1 fc0 sc0 ls0 ws0">Hz)<span class="ff4">，</span>S<span class="_ _4"> </span><span class="ff4">为信号<span class="_ _1"></span>平均功率，<span class="_ _1"></span></span>N<span class="_ _4"> </span><span class="ff4">为噪声平均功率</span></div><div class="t m1 x7 h3 y1a ff4 fs1 fc0 sc0 ls0 ws0">香农公式表明<span class="_ _1"></span>，信道的带<span class="_ _1"></span>宽或信道中的<span class="_ _1"></span>信噪比越大<span class="_ _1"></span>，信息的极限<span class="_ _1"></span>传输速</div><div class="t m1 x1 h3 y1b ff4 fs1 fc0 sc0 ls0 ws0">率就越高</div><div class="t m0 x1 h2 y1c ff2 fs0 fc0 sc0 ls0 ws0">信噪比</div><div class="t m1 x7 h3 y1d ff4 fs1 fc0 sc0 ls0 ws0">信号的平均功<span class="_ _1"></span>率和噪声的<span class="_ _1"></span>平均功率之比<span class="_ _1"></span>，即：</div></div><div class="pi" data-data='{"ctm":[1.611830,0.000000,0.000000,1.611830,0.000000,0.000000]}'></div></div>