Information Theory/Entropy Coding According to Huffman - Revision history

Guenter at 17:09, 20 February 2023

2023-02-20T17:09:14Z

2023-02-19T15:26:49Z

2023-02-19T15:24:45Z

2023-02-16T13:21:35Z

2023-02-16T13:20:28Z

2023-02-16T13:19:52Z

2023-02-16T13:17:56Z

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2023-02-16T13:08:02Z

2023-02-16T13:06:35Z

@@ Line 210: / Line 210: @@
 \boldsymbol{\rm F} \hspace{0.05cm} \rightarrow \hspace{0.15cm}$'''1100'''$\hspace{0.05cm}.$
-[[File:EN_Inf_T_2_3_S4c.png|right|frame|On the error propagation of the Huffman coding '''KORREKTUR: falsified''']]
+[[File:EN_Inf_T_2_3_S4c_v2.png|right|frame|On the error propagation of the Huffman coding]]
 The source symbol sequence&nbsp; $\rm ADABD$...&nbsp; (upper diagram) is thus represented by the encoded sequence&nbsp; '''0'100'0'111'100' '''...&nbsp;.&nbsp; The inverted commas are only for the reader's orientation.

@@ Line 249: / Line 249: @@
 {{GraueBox|TEXT=
 $\text{Example 6:}$&nbsp; Let a memoryless binary source $(M = 2)$ with the symbols&nbsp; $\{$ $\rm X$,&nbsp;  $\rm Y \}$ be given:
-*Let the symbol probabilities be&nbsp; $p_{\rm X} = 0.8$ &nbsp;and&nbsp; $p_{\rm Y} = 0.2$.&nbsp; Thus the source entropy is&nbsp; $H = 0.722$&nbsp; bit/source\hspace{0.15cm} '''KORREKTUR: ?''' symbol.
+*Let the symbol probabilities be&nbsp; $p_{\rm X} = 0.8$ &nbsp;and&nbsp; $p_{\rm Y} = 0.2$.&nbsp; Thus the source entropy is&nbsp; $H = 0.722$&nbsp; bit/source symbol.
 *We consider the symbol sequence &nbsp; $\{\rm XXXYXXXXXXXXYYXXXXXYYXXYXYXXYX\ \text{...} \}$&nbsp; with only a few&nbsp; $\rm Y$ symbols at positions 4, 13, 14, ...

@@ Line 214: / Line 214: @@
 The source symbol sequence&nbsp; $\rm ADABD$...&nbsp; (upper diagram) is thus represented by the encoded sequence&nbsp; '''0'100'0'111'100' '''...&nbsp;.&nbsp; The inverted commas are only for the reader's orientation.
-During transmission, the first bit is distorted '''KORREKTUR: falsified''':&nbsp; Instead of&nbsp; '''01000111100'''...&nbsp; one receives&nbsp; '''11000111100'''...
+During transmission, the first bit is falsified:&nbsp; Instead of&nbsp; '''01000111100'''...&nbsp; one receives&nbsp; '''11000111100'''...
 *After decoding, the first two source symbols&nbsp; $\rm AD$&nbsp; →&nbsp; '''0100'''&nbsp; become the sink symbol&nbsp; $\rm F$&nbsp; →&nbsp; '''1100'''.
 *The further symbols are then detected correctly again, but now no longer starting at position&nbsp; $ν = 3$, but at&nbsp; $ν = 2$.

@@ Line 249: / Line 249: @@
 {{GraueBox|TEXT=
 $\text{Example 6:}$&nbsp; Let a memoryless binary source $(M = 2)$ with the symbols&nbsp; $\{$ $\rm X$,&nbsp;  $\rm Y \}$ be given:
-*Let the symbol probabilities be&nbsp; $p_{\rm X} = 0.8$ &nbsp;and&nbsp; $p_{\rm Y} = 0.2$.&nbsp; Thus the source entropy is&nbsp; $H = 0.722$&nbsp; bit/source $\hspace{0.15cm}$ symbol.
+*Let the symbol probabilities be&nbsp; $p_{\rm X} = 0.8$ &nbsp;and&nbsp; $p_{\rm Y} = 0.2$.&nbsp; Thus the source entropy is&nbsp; $H = 0.722$&nbsp; bit/source\hspace{0.15cm} '''KORREKTUR: ?''' symbol.
 *We consider the symbol sequence &nbsp; $\{\rm XXXYXXXXXXXXYYXXXXXYYXXYXYXXYX\ \text{...} \}$&nbsp; with only a few&nbsp; $\rm Y$ symbols at positions 4, 13, 14, ...

@@ Line 249: / Line 249: @@
 {{GraueBox|TEXT=
 $\text{Example 6:}$&nbsp; Let a memoryless binary source $(M = 2)$ with the symbols&nbsp; $\{$ $\rm X$,&nbsp;  $\rm Y \}$ be given:
-*Let the symbol probabilities be&nbsp; $p_{\rm X} = 0.8$ &nbsp;and&nbsp; $p_{\rm Y} = 0.2$.&nbsp; Thus the source entropy is&nbsp; $H = 0.722$&nbsp; bit/source \hspace{0.15cm} symbol.
+*Let the symbol probabilities be&nbsp; $p_{\rm X} = 0.8$ &nbsp;and&nbsp; $p_{\rm Y} = 0.2$.&nbsp; Thus the source entropy is&nbsp; $H = 0.722$&nbsp; bit/source $\hspace{0.15cm}$ symbol.
 *We consider the symbol sequence &nbsp; $\{\rm XXXYXXXXXXXXYYXXXXXYYXXYXYXXYX\ \text{...} \}$&nbsp; with only a few&nbsp; $\rm Y$ symbols at positions 4, 13, 14, ...