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<p>例题:</p>
<p>1、电视频道的带宽是6MHz，如果使用4级数字信号，在无噪声的信道上每秒能发送多少比特？</p>
<p>使用奈奎斯特定理解决：</p>
<p>2、公用电话网（BSTN）的带宽是3KHz，其信噪比为20dB，可以取得的最大数据速率是多少？能取得的最大调制速率是多少？</p>
<p>使用香农定理解决：</p>
<p><span style="mso-spacerun: yes; mso-fareast-font-family: 宋体; mso-hansi-font-family: Times New Roman; layout-flow: vertical; mso-fareast-language: ZH-CN" lang="EN-US">3、下面的表中给出了字母D、E、F的7位ASCII码表示。采用偶校验，求出传送的信息为DEF时检验比特及块检验字符BCC。</span><span style="mso-spacerun: yes; mso-fareast-font-family: 宋体; mso-hansi-font-family: Times New Roman; layout-flow: vertical; mso-fareast-language: ZH-CN" lang="EN-US">
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<table border="0" width="56%">
<tr>
<td width="11%"></td>
<td width="10%">b<sub>6</sub></td>
<td width="10%">b<sub>5</sub></td>
<td width="10%">b<sub>4</sub></td>
<td width="11%">b<sub>3</sub></td>
<td width="11%">b<sub>2</sub></td>
<td width="10%">b<sub>1</sub></td>
<td width="27%">b<sub>0</sub></td>
</tr>
<tr>
<td width="11%">D</td>
<td width="10%">1</td>
<td width="10%">0</td>
<td width="10%">0</td>
<td width="11%">0</td>
<td width="11%">1</td>
<td width="10%">0</td>
<td width="27%">0</td>
</tr>
<tr>
<td width="11%">E</td>
<td width="10%">1</td>
<td width="10%">0</td>
<td width="10%">0</td>
<td width="11%">0</td>
<td width="11%">1</td>
<td width="10%">0</td>
<td width="27%">1</td>
</tr>
<tr>
<td width="11%">F</td>
<td width="10%">1</td>
<td width="10%">0</td>
<td width="10%">0</td>
<td width="11%">0</td>
<td width="11%">1</td>
<td width="10%">1</td>
<td width="27%">0</td>
</tr>
</table>
<p> 解答：</p>
<table border="0" width="52%">
<tr>
<td width="11%"></td>
<td width="11%">b<sub>6</sub></td>
<td width="11%">b<sub>5</sub></td>
<td width="11%">b<sub>4</sub></td>
<td width="11%">b<sub>3</sub></td>
<td width="11%">b<sub>2</sub></td>
<td width="11%">b<sub>1</sub></td>
<td width="11%">b<sub>0</sub></td>
<td width="12%"><b><font size="3">b<sub>7</sub></font></b></td>
</tr>
<tr>
<td width="11%">D</td>
<td width="11%">1</td>
<td width="11%">0</td>
<td width="11%">0</td>
<td width="11%">0</td>
<td width="11%">1</td>
<td width="11%">0</td>
<td width="11%">0</td>
<td width="12%"><b><font size="3">0</font></b></td>
</tr>
<tr>
<td width="11%">E</td>
<td width="11%">1</td>
<td width="11%">0</td>
<td width="11%">0</td>
<td width="11%">0</td>
<td width="11%">1</td>
<td width="11%">0</td>
<td width="11%">1</td>
<td width="12%"><b><font size="3">1</font></b></td>
</tr>
<tr>
<td width="11%">F</td>
<td width="11%">1</td>
<td width="11%">0</td>
<td width="11%">0</td>
<td width="11%">0</td>
<td width="11%">1</td>
<td width="11%">1</td>
<td width="11%">0</td>
<td width="12%"><b><font size="3">1</font></b></td>
</tr>
<tr>
<td width="11%"><b><font size="3">BCC</font></b></td>
<td width="11%"><b><font size="3">1</font></b></td>
<td width="11%"><b><font size="3">0</font></b></td>
<td width="11%"><b><font size="3">0</font></b></td>
<td width="11%"><b><font size="3">0</font></b></td>
<td width="11%"><b><font size="3">1</font></b></td>
<td width="11%"><b><font size="3">1</font></b></td>
<td width="11%"><b><font size="3">1</font></b></td>
<td width="12%"><font size="3"><b>0</b></font></td>
</tr>
</table>
<p>4、信息有效数据M是每个字符用7bit字节编码的ASCII码串“well",即M长28位，其中，w=1110111，e=1100101，l=1101100,取多项式CRC-12=X<sup>12</sup>+X<sup>11</sup>+X<sup>3</sup>+X<sup>2</sup>+X+1做循环冗余检验码，求该码串的冗余部分r.</p>
<p>解：用被除数：1110111 1100101 11011000 000000000000</p>
<p> 除数：1100000001111</p>
<p>做二进制除法，得商：1011010 1010101 1001000 1010110</p>
<p>
余数：111100010010</p>
<p>因此，该码串的冗余部分： r=111100010010</p>
<p> </p>
<p><span style="font-size: 12.0pt; font-family: 宋体; mso-ascii-font-family: Times New Roman; mso-hansi-font-family: Times New Roman; mso-bidi-font-family: Times New Roman; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">5、</span><span style="font-size:12.0pt;font-family:宋体;
mso-ascii-font-family:"Times New Roman";mso-hansi-font-family:"Times New Roman";
mso-bidi-font-family:"Times New Roman";mso-font-kerning:1.0pt;mso-ansi-language:
EN-US;mso-fareast-language:ZH-CN;mso-bidi-language:AR-SA">在</span><span lang="EN-US" style="font-size:12.0pt;font-family:"Times New Roman";mso-fareast-font-family:
宋体;mso-font-kerning:1.0pt;mso-ansi-language:EN-US;mso-fareast-language:ZH-CN;
mso-bidi-language:AR-SA">50KHz</span><span style="font-size:12.0pt;font-family:
宋体;mso-ascii-font-family:"Times New Roman";mso-hansi-font-family:"Times New Roman";
mso-bidi-font-family:"Times New Roman";mso-font-kerning:1.0pt;mso-ansi-language:
EN-US;mso-fareast-language:ZH-CN;mso-bidi-language:AR-SA">线路上使用</span><span lang="EN-US" style="font-size:12.0pt;font-family:"Times New Roman";mso-fareast-font-family:
宋体;mso-font-kerning:1.0pt;mso-ansi-language:EN-US;mso-fareast-language:ZH-CN;
mso-bidi-language:AR-SA">T1</span><span style="font-size:12.0pt;font-family:
宋体;mso-ascii-font-family:"Times New Roman";mso-hansi-font-family:"Times New Roman";
mso-bidi-font-family:"Times New Roman";mso-font-kerning:1.0pt;mso-ansi-language:
EN-US;mso-fareast-language:ZH-CN;mso-bidi-language:AR-SA">载波需要多大的信噪比？（</span><span lang="EN-US" style="font-size:12.0pt;font-family:"Times New Roman";mso-fareast-font-family:
宋体;mso-font-kerning:1.0pt;mso-ansi-language:EN-US;mso-fareast-language:ZH-CN;
mso-bidi-language:AR-SA"> lg(2<sup>31</sup>-1)</span><span style="font-size:
12.0pt;font-family:宋体;mso-ascii-font-family:"Times New Roman";mso-hansi-font-family:
"Times New Roman";mso-bidi-font-family:"Times New Roman";mso-font-kerning:1.0pt;
mso-ansi-language:EN-US;mso-fareast-language:ZH-CN;mso-bidi-language:AR-SA">≈</span><span lang="EN-US" style="font-size:12.0pt;font-family:"Times New Roman";mso-fareast-font-family:
宋体;mso-font-kerning:1.0pt;mso-ansi-language:EN-US;mso-fareast-language:ZH-CN;
mso-bidi-language:AR-SA">9.3 </span><span style="font-size:12.0pt;font-family:
宋体;mso-ascii-font-family:"Times New Roman";mso-hansi-font-family:"Times New Roman";
mso-bidi-font-family:"Times New Roman";mso-font-kerning:1.0pt;mso-ansi-language:
EN-US;mso-fareast-language:ZH-CN;mso-bidi-language:AR-SA">）</span></p>
<p><span style="font-size: 12.0pt; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">为发送T1信号,我们需要达到的信噪比为S/N。</span></p>
<p><span style="font-size: 12.0pt; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">50000*log<sub>2</sub>(1+S/N)=1.544*10<sup>6</sup></span></p>
<p><span style="font-size: 12.0pt; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">(S/N)=2<sup>31</sup>-1</span></p>
<p><span style="font-size: 12.0pt; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA">10*log<sub>10</sub>(2<sup>31</sup>-1)=93
(dB)</span></p>
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