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📁 通信原理的经典讲义
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      <P align=right><I><A 
      href="http://www.yobology.info/harbin/part1/index.htm">index</A></I></P>
      <P align=center>Part 1 </P>
      <P align=center>Chapter 1 &nbsp;&nbsp;Basic Preliminary (Shannon's 
      Theory)<BR></P></TD></TR>
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      <P>仒1 &nbsp;&nbsp;BIT<BR>&nbsp;When we see an event that occurs with 
      probability <I>p</I>, we get information</P>
      <P align=center><IMG src="Part1_1.files/part_1_1_htm_eqn3223.gif" border=0 
      NAMO_EQN__><!--NAMO_EQN__ 224 1I=\: -\log _{2}\: p\quad bit--></P>
      <P align=center>&nbsp; <IMG height=347 src="Part1_1.files/img5.gif" 
      width=563 border=0></P>
      <P><SPAN style="FONT-SIZE: 12pt"><FONT color=#660000>Note: In coin 
      tossing, let probability of head or tail be 1/2. Then we get 1 bit from 
      single trial. BIT is acronym of binary information unit.</FONT></SPAN></P>
      <P>仒2 &nbsp;Entropy<BR>&nbsp;Consider a memoryless source outputting the 
      symbol "0" or&nbsp;"1" and let <I>p</I> denote the probability of symbol 
      "0". The average information coming out from this source is given by</P>
      <P align=center><IMG src="Part1_1.files/part_1_1_htm_eqn27974.gif" 
      border=0 NAMO_EQN__><!--NAMO_EQN__ 208 1H(X)=-p\log _{2}p-(1-p)\log _{2}(1-p)\quad bit\slash symbol--></P>
      <P>This increases monotonously from <IMG 
      src="Part1_1.files/part_1_1_htm_eqn8136.gif" border=0 NAMO_EQN__><!--NAMO_EQN__ 192 1p=0--> to <IMG 
      src="Part1_1.files/part_1_1_htm_eqn8199.gif" border=0 NAMO_EQN__><!--NAMO_EQN__ 192 1p=1\slash 2-->, hence implies randomness 
      of the symbol sequence.</P>
      <P align=center><IMG height=317 alt=img15.gif 
      src="Part1_1.files/img15.gif" width=514 border=0></P>
      <P align=center>THEOREM (source coding)<BR><I>H(X)</I> gives lower bound 
      of the reversible information compression.</P>
      <P align=center>&lt;Context of proof&gt;</P>
      <UL>
        <OL>
          <LI align="left">
          <DIV align=left>There is a compact reversible coding scheme 
          (&nbsp;Huffman) which satisfies<BR>&nbsp;<IMG 
          src="Part1_1.files/part_1_1_htm_eqn9591.gif" border=0 NAMO_EQN__><!--NAMO_EQN__ 192 1H(S)\leq L\leq H(S)+1,\quad where\; L\; is\; average\; of\; the\; code\; length--></DIV>
          <LI align="left">
          <DIV align=left>Apply following formula with the memoryless source to 
          above inequality.<BR><IMG 
          src="Part1_1.files/part_1_1_htm_eqn25655.gif" border=0 NAMO_EQN__><!--NAMO_EQN__ 208 1H(S^{n}\: )=nH(S),\quad where\: S^{n}\: is\: n\, th\: expansion\: of\: the\: source--></DIV>
          <LI align="left">
          <DIV align=left>If <I>L</I>&nbsp;is shorter than <IMG 
          src="Part1_1.files/part_1_1_htm_eqn13709.gif" border=0 NAMO_EQN__><!--NAMO_EQN__ 176 1H(S)-->, such code is not 
          reversible.</DIV></LI></OL></UL>
      <P align=center>&lt;Design rule of Huffman code&gt;</P>
      <P align=center><IMG height=429 src="Part1_1.files/img60.gif" width=783 
      border=0></P>
      <P align=center>&nbsp;</P>
      <P align=center>&lt;Numerical examples for binary source&gt;</P>
      <P align=left>The<I> 2</I>nd expansion of source outputs "00", "01", "10" 
      and "11" respectively by probabilities<I> pp</I>, <I>p(1-p)</I>, 
      <I>(1-p)p&nbsp;</I>and <I>(1-p)(1-p)</I>. For this source, design Huffman 
      code and estimate its average code length <IMG 
      src="Part1_1.files/part_1_1_htm_eqn18510.gif" border=0 NAMO_EQN__><!--NAMO_EQN__ 192 1L_{2}-->. Similarly, by expanding the 
      source we have following graphs. Black lines shows <IMG 
      src="Part1_1.files/part_1_1_htm_eqn20559.gif" border=0 NAMO_EQN__><!--NAMO_EQN__ 192 1H(S^{n}\: )\slash n=H(S)-->. Orange 
      colored curve shows <IMG src="Part1_1.files/part_1_1_htm_eqn21070.gif" 
      border=0 NAMO_EQN__><!--NAMO_EQN__ 192 1(H(S^{n}\: )+1)\slash n=(nH(S)+1)\slash n-->. 
      Purple colored curve shows&nbsp;<IMG 
      src="Part1_1.files/part_1_1_htm_eqn26187.gif" border=0 NAMO_EQN__><!--NAMO_EQN__ 192 1L_{n}\slash n--> &nbsp;of Huffman 
      code.</P>
      <P align=center><IMG height=359 src="Part1_1.files/img61.gif" width=581 
      border=0></P>
      <P align=center><IMG height=253 src="Part1_1.files/img63.gif" width=82 
      border=0> &nbsp;<IMG height=253 src="Part1_1.files/img64.gif" width=209 
      border=0></P>
      <P align=center><IMG height=369 src="Part1_1.files/img62.gif" width=598 
      border=0></P>
      <P align=center><IMG height=253 src="Part1_1.files/img65.gif" width=82 
      border=0> &nbsp;<IMG height=253 src="Part1_1.files/img66.gif" width=173 
      border=0></P>
      <P align=left>&nbsp;</P>
      <P>仒3 &nbsp;Mutual information<BR>I am a messenger to inform you source's 
      output "0" or "1". But I am careless and sometimes tell&nbsp;you inverse 
      of the output. Let the probability of my mistake be 1-<I>r</I>. </P>
      <P align=center><IMG height=301 src="Part1_1.files/img4.gif" width=582 
      border=0></P>
      <P>Mutual information means amount of information about source's output 
      contained in my message&nbsp;and given by</P>
      <P align=center><IMG src="Part1_1.files/part_1_1_htm_eqn7702.gif" border=0 
      NAMO_EQN__><!--NAMO_EQN__ 208 1M(X,Y)=H(Y)-H(Y\mid X)=H(X)-H(X\mid Y)--><BR><IMG 
      src="Part1_1.files/part_1_1_htm_eqn12584.gif" border=0 NAMO_EQN__><!--NAMO_EQN__ 208 1H(Y\mid X)\; is\; entropy\; of\; conditional\; probability\; P(my\: report\mid output)--><BR><IMG 
      src="Part1_1.files/part_1_1_htm_eqn7856.gif" border=0 NAMO_EQN__><!--NAMO_EQN__ 192 1H(X\mid Y)\; is\; entropy\; of\; a\; posteriory\; probability\; P(output\mid my\: report)--></P>
      <P><SPAN style="FONT-SIZE: 12pt"><FONT color=#660000>Note: Small value of 
      <IMG src="Part1_1.files/part_1_1_htm_eqn12960.gif" border=0 NAMO_EQN__><!--NAMO_EQN__ 160 1H(X\mid Y)--> or <IMG 
      src="Part1_1.files/part_1_1_htm_eqn8368.gif" border=0 NAMO_EQN__><!--NAMO_EQN__ 176 1H(X\mid Y)--> implies that my report is 
      reliable. </FONT></SPAN></P>
      <P align=center><IMG height=401 alt=img16.gif 
      src="Part1_1.files/img16.gif" width=535 border=0></P>
      <P align=center><IMG height=511 src="Part1_1.files/img70.gif" width=590 
      border=0></P>
      <P>&nbsp;</P>
      <P align=center>THEOREM (channel coding)<BR>Assume that transmission speed 
      of the output sequence<BR>丒丒丒丒0, 1, 1, 0, 1, 0, 1, 丒丒丒丒<BR>is <IMG 
      src="Part1_1.files/part_1_1_htm_eqn16731.gif" border=0 NAMO_EQN__><!--NAMO_EQN__ 208 1V\; bit\slash sec-->. There exits a 
      perfect error correcting scheme with lower speed than</P>
      <P align=center><IMG src="Part1_1.files/part_1_1_htm_eqn28487.gif" 
      border=0 NAMO_EQN__><!--NAMO_EQN__ 208 1M(X,Y)\cdot V\quad bit\slash sec--> .</P>
      <P align=center>Proof<BR>Refer Shannon's original paper (difficult to 
      prove in general cases)</P>
      <P>For completely randomized data sequence (<IMG 
      src="Part1_1.files/part_1_1_htm_eqn21566.gif" border=0 NAMO_EQN__><!--NAMO_EQN__ 192 1p=0.5-->), the channel capacity is 
      plotted as follows.</P>
      <P align=center><IMG height=317 src="Part1_1.files/img7.gif" width=513 
      border=0></P>
      <P align=left>Refer graphs of other average information, <I>H(Y)</I>, 
      <I>H(Y|X)</I>, <I>H(X|Y)</I> and <I>H(X,Y)</I>.</P>
      <P align=center><IMG height=388 src="Part1_1.files/img71.gif" width=517 
      border=0></P>
      <P align=center><IMG height=388 src="Part1_1.files/img72.gif" width=517 
      border=0></P>
      <P align=center><IMG height=388 src="Part1_1.files/img73.gif" width=517 
      border=0></P>
      <P align=center><IMG height=388 src="Part1_1.files/img74.gif" width=517 
      border=0></P>
      <P align=center>&nbsp;</P></TD></TR></TBODY></TABLE>
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