Channel Coding/Algebraic and Polynomial Description - Revision history

Guenter at 17:50, 2 December 2022

2022-12-02T17:50:29Z

Hwang at 14:28, 30 November 2022

2022-11-30T14:28:29Z

Hwang at 14:24, 30 November 2022

2022-11-30T14:24:42Z

Hwang at 14:22, 30 November 2022

2022-11-30T14:22:19Z

Hwang at 14:21, 30 November 2022

2022-11-30T14:21:27Z

Hwang at 14:19, 30 November 2022

2022-11-30T14:19:54Z

Hwang at 14:19, 30 November 2022

2022-11-30T14:19:30Z

Hwang at 14:12, 30 November 2022

2022-11-30T14:12:40Z

Hwang at 14:04, 30 November 2022

2022-11-30T14:04:45Z

Hwang at 23:43, 12 November 2022

2022-11-12T23:43:05Z

@@ Line 531: / Line 531: @@
 {{GraueBox|TEXT=
 $\text{Example 8:}$&nbsp;
-The coder '''KORREKTUR: encoder''' of rate &nbsp; $2/3$ &nbsp; considered often in the last sections is not systematic because&nbsp; e.g.&nbsp; $\underline{x}^{(1)} &ne; \underline{u}^{(1)}, \ \underline{x}^{(2)} &ne; \underline{u}^{(2)}$ &nbsp; holds&nbsp; $($see adjacent graphic$)$.<br>
+The encoder  of rate &nbsp; $2/3$ &nbsp; considered often in the last sections is not systematic because&nbsp; e.g.&nbsp; $\underline{x}^{(1)} &ne; \underline{u}^{(1)}, \ \underline{x}^{(2)} &ne; \underline{u}^{(2)}$ &nbsp; holds&nbsp; $($see adjacent graphic$)$.<br>
 [[File:P ID2613 KC T 3 2 S1 neu.png|right|frame|Convolutional encoder of rate&nbsp; $2/3$]]

@@ Line 531: / Line 531: @@
 {{GraueBox|TEXT=
 $\text{Example 8:}$&nbsp;
-The coder of rate &nbsp; $2/3$ &nbsp; considered often in the last sections is not systematic because&nbsp; e.g.&nbsp; $\underline{x}^{(1)} &ne; \underline{u}^{(1)}, \ \underline{x}^{(2)} &ne; \underline{u}^{(2)}$ &nbsp; holds&nbsp; $($see adjacent graphic$)$.<br>
+The coder '''KORREKTUR: encoder''' of rate &nbsp; $2/3$ &nbsp; considered often in the last sections is not systematic because&nbsp; e.g.&nbsp; $\underline{x}^{(1)} &ne; \underline{u}^{(1)}, \ \underline{x}^{(2)} &ne; \underline{u}^{(2)}$ &nbsp; holds&nbsp; $($see adjacent graphic$)$.<br>
 [[File:P ID2613 KC T 3 2 S1 neu.png|right|frame|Convolutional encoder of rate&nbsp; $2/3$]]

@@ Line 394: / Line 394: @@
 Now we extend the result to convolutional encoders with more than one input &nbsp; &#8658; &nbsp; $k &#8805; 2$ &nbsp;$($see graph$)$.<br>
-In order to map a convolutional code of rate&nbsp; $k/n$&nbsp; in the D&ndashdomain,&nbsp; the dimension of the above vector equation must be increased with respect to input and transfer function:
+In order to map a convolutional code of rate&nbsp; $k/n$&nbsp; in the D&ndash;domain,&nbsp; the dimension of the above vector equation must be increased with respect to input and transfer function:
 ::<math>\underline{X}(D) = \underline{U}(D) \cdot { \boldsymbol{\rm G}}(D)\hspace{0.05cm}.</math>

← Older revision		Revision as of 14:22, 30 November 2022
Line 289:		Line 289:
	*The  $($rather simple$)$  $\rm proof$  of this important result can be found in the specification for  [[Aufgaben:Exercise_3.3Z:_Convolution_and_D-Transformation\|"Exercise 3.3Z"]].		*The  $($rather simple$)$  $\rm proof$  of this important result can be found in the specification for  [[Aufgaben:Exercise_3.3Z:_Convolution_and_D-Transformation\|"Exercise 3.3Z"]].

−	*As in  »[[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#Frequency_response_.E2.80. 93_Transfer_function\|$\text{system theory}$]]«  commonly,  the  D–transform  $G(D)$  of the impulse response  $\underline{g}$  is also called  "transfer function".}}	+	*As in  »[[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#Frequency_response_.E2.80.93_Transfer_function\|$\text{system theory}$]]«  commonly,  the  D–transform  $G(D)$  of the impulse response  $\underline{g}$  is also called  "transfer function".}}

@@ Line 281: / Line 281: @@
 {{BlaueBox|TEXT=
-$\text{Theorem:}$&nbsp; As with all spectral transformations,&nbsp; the&nbsp;  &raquo;<b>multiplication</b>&laquo;&nbsp; applies to the&nbsp; D&ndash;transform in the image domain,&nbsp; since the&nbsp; $($discrete$)$&nbsp; time functionss&nbsp; $\underline{u}$&nbsp; and&nbsp; $\underline{g}$&nbsp; are interconnected by the&nbsp; &raquo;<b>convolution</b>&laquo;:
+$\text{Theorem:}$&nbsp; As with all spectral transformations,&nbsp; the&nbsp;  &raquo;<b>multiplication</b>&laquo;&nbsp; applies to the&nbsp; D&ndash;transform in the image domain,&nbsp; since the&nbsp; $($discrete$)$&nbsp; time functions&nbsp; $\underline{u}$&nbsp; and&nbsp; $\underline{g}$&nbsp; are interconnected by the&nbsp; &raquo;<b>convolution</b>&laquo;:
 ::<math>\underline{x} = \underline{u} * \underline{g} \quad

@@ Line 254: / Line 254: @@
 X(D) =  \sum_{i = 0}^{\infty} x_i \cdot D\hspace{0.05cm}^i \hspace{0.05cm}.</math>}}<br>
-<u>Note</u>: &nbsp; In the literature,&nbsp; sometimes&nbsp; $x(D)$&nbsp; is used instead of&nbsp; $X(D)$.&nbsp; However,&nbsp; we write  in our learning tutorial all image domain functions &nbsp; &rArr; &nbsp; "spectral domain functions"&nbsp; with capital letters, &nbsp: for example the Fourier transform, the Laplace transform  and the D&ndash;transform:
+<u>Note</u>: &nbsp; In the literature,&nbsp; sometimes&nbsp; $x(D)$&nbsp; is used instead of&nbsp; $X(D)$.&nbsp; However,&nbsp; we write  in our learning tutorial all image domain functions &nbsp; &rArr; &nbsp; "spectral domain functions"&nbsp; with capital letters, &nbsp; for example the Fourier transform, the Laplace transform  and the D&ndash;transform:
 ::<math>x(t) \hspace{0.15cm}

@@ Line 99: / Line 99: @@
 *The generator matrix&nbsp; $\mathbf{G}$&nbsp; actually extends downwards and to the right to infinity.&nbsp; Explicitly shown,&nbsp; however,&nbsp; are only eight rows and twelve columns.
-*For the temporal information sequence &nbsp; $\underline{u} = (0, 1, 1, 0, 0, 0, 1, 1)$ &nbsp; the drawn matrix part is sufficient.&nbsp; The code sequence is then:
+*For the temporal information sequence &nbsp; $\underline{u} = (0, 1, 1, 0, 0, 0, 1, 1)$ &nbsp; the drawn matrix part is sufficient.&nbsp; The encoded sequence is then:
 :$$\underline{x} = (0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0).$$
@@ Line 136: / Line 136: @@
 	\end{pmatrix}\hspace{0.05cm}.$$
-*For the input sequence &nbsp; $\underline{u} = (1, 0, 1, 1)$,&nbsp; the code sequence starts with &nbsp; $\underline{x} = (1, 1, 0, 1, 1, 1, 1, 0, \ \text{...})$.
+*For the input sequence &nbsp; $\underline{u} = (1, 0, 1, 1)$,&nbsp; the encoded sequence starts with &nbsp; $\underline{x} = (1, 1, 0, 1, 1, 1, 1, 0, \ \text{...})$.
 *This result is equal to the sum of rows&nbsp; '''1''',&nbsp; '''3'''&nbsp; and&nbsp; '''4'''&nbsp; of the generator matrix.<br><br>
@@ Line 165: / Line 165: @@
 	\end{pmatrix}\hspace{0.05cm}.</math>
-*Here the input sequence&nbsp; $\underline{u} = (1, 0, 1, 1)$&nbsp; leads to the code sequence&nbsp; $\underline{x} = (1, 1, 1, 0, 0, 0, 0, 1, \ \text{...})$.<br><br>
+*Here the input sequence&nbsp; $\underline{u} = (1, 0, 1, 1)$&nbsp; leads to the encoded sequence&nbsp; $\underline{x} = (1, 1, 1, 0, 0, 0, 0, 1, \ \text{...})$.<br><br>
 [[File:P ID2604 KC T 3 2 S3c.png|right|frame|Convolutional encoder&nbsp; $(k = 1, \ n = 3, \ m = 3)$]]
@@ Line 195: / Line 195: @@
 	\end{pmatrix}\hspace{0.05cm}.</math>
-*One obtains for &nbsp; $\underline{u} = (1, 0, 1, 1)$ &nbsp; the code sequence&nbsp; $\underline{x} = (1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, \ \text{...})$.<br>
+*One obtains for &nbsp; $\underline{u} = (1, 0, 1, 1)$ &nbsp; the encoded sequence&nbsp; $\underline{x} = (1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, \ \text{...})$.<br>
 == GF(2) description forms of a digital filter ==
@@ Line 347: / Line 347: @@
 ::<math>\underline{G}(D) = \left ( G^{(1)}(D), G^{(2)}(D), \hspace{0.05cm}\text{...}\hspace{0.1cm}, G^{(n)} (D) \right )\hspace{0.05cm}.</math>
-*The vector&nbsp; $\underline{X}(D)$&nbsp; contains the&nbsp; D&ndash;transform&nbsp; of&nbsp; $n$&nbsp; code sequences&nbsp; $\underline{x}^{(1)}, \underline{x}^{(2)}, \ \text{...} \ , \underline{x}^{(n)}$:
+*The vector&nbsp; $\underline{X}(D)$&nbsp; contains the&nbsp; D&ndash;transform&nbsp; of&nbsp; $n$&nbsp; encoded sequences&nbsp; $\underline{x}^{(1)}, \underline{x}^{(2)}, \ \text{...} \ , \underline{x}^{(n)}$:
 ::<math>\underline{X}(D) = \left ( X^{(1)}(D), X^{(2)}(D), \hspace{0.05cm}\text{...}\hspace{0.1cm}, X^{(n)} (D) \right )\hspace{0.05cm}.</math>
@@ Line 460: / Line 460: @@
 \end{pmatrix}\hspace{0.05cm}.</math>
-*This results in the following code sequences in the three strands:
+*This results in the following encoded sequences in the three strands:
 ::<math>{X}^{(1)}(D) = (D + D^3) \cdot (1+D) + (1 + D^3) \cdot  D =D + D^2 +  D^3 + D^4 + D  + D^4 = D^2 + D^3</math>
@@ Line 486: / Line 486: @@
 *This&nbsp; $k &times; n$&nbsp; matrix is used to recognize whether it is a&nbsp; [[Channel_Coding/General_Description_of_Linear_Block_Codes#Systematic_Codes| &raquo;$\text{systematic code}$&laquo;]].
-*This refers to a code where the code sequences&nbsp; $\underline{x}^{(1)}, \ \text{...} \ , \ \underline{x}^{(k)}$&nbsp; are identical with the information sequences&nbsp; $\underline{u}^{(1)}, \ \text{...} \ , \ \underline{u}^{(k)}$.
+*This refers to a code where the encoded sequences&nbsp; $\underline{x}^{(1)}, \ \text{...} \ , \ \underline{x}^{(k)}$&nbsp; are identical with the information sequences&nbsp; $\underline{u}^{(1)}, \ \text{...} \ , \ \underline{u}^{(k)}$.
 *The graph shows an example of a systematic&nbsp; $(n = 4, \ k = 3)$&nbsp; convolutional code.<br>

@@ Line 221: / Line 221: @@
 ::<math>\underline{x} = \underline{u} * \underline{g}\hspace{0.05cm}.</math>
-*Major difference compared to the chapter&nbsp; &raquo;[[Theory_of_Stochastic_Signals/Digital_Filters|$\text{Digital Filters}$]]&laquo;&nbsp; in the book&nbsp; "Theory of Stochastic Signals"&nbsp; is the modulo-2 addition&nbsp; $(1 + 1 = 0)$&nbsp; instead of the conventional addition&nbsp; $(1 + 1 = 2)$.<br><br>
+*Major difference compared to the chapter&nbsp; &raquo;[[Theory_of_Stochastic_Signals/Digital_Filters|"Digital Filters"]]&laquo;&nbsp; in the book&nbsp; "Theory of Stochastic Signals"&nbsp; is the modulo-2 addition&nbsp; $(1 + 1 = 0)$&nbsp; instead of the conventional addition&nbsp; $(1 + 1 = 2)$.<br><br>
 {{GraueBox|TEXT=
@@ Line 281: / Line 281: @@
 {{BlaueBox|TEXT=
-$\text{Theorem:}$&nbsp; As with all spectral transformations,&nbsp; the&nbsp;  <b>multiplication</b>&nbsp; applies to the&nbsp; D&ndash;transform in the image domain,&nbsp; since the&nbsp; $($discrete$)$&nbsp; time functionss&nbsp; $\underline{u}$&nbsp; and&nbsp; $\underline{g}$&nbsp; are interconnected by the&nbsp; &raquo;<b>convolution</b>&laquo;:
+$\text{Theorem:}$&nbsp; As with all spectral transformations,&nbsp; the&nbsp;  &raquo;<b>multiplication</b>&laquo;&nbsp; applies to the&nbsp; D&ndash;transform in the image domain,&nbsp; since the&nbsp; $($discrete$)$&nbsp; time functionss&nbsp; $\underline{u}$&nbsp; and&nbsp; $\underline{g}$&nbsp; are interconnected by the&nbsp; &raquo;<b>convolution</b>&laquo;:
 ::<math>\underline{x} = \underline{u} * \underline{g} \quad
@@ Line 287: / Line 287: @@
 X(D) = U(D) \cdot G(D) \hspace{0.05cm}.</math>
-*The&nbsp; $($rather simple$)$&nbsp; $\rm proof$&nbsp; of this important result can be found in the specification for&nbsp; [[Aufgaben:Exercise_3.3Z:_Convolution_and_D-Transformation|$\text{Exercise 3.3Z}$]].
+*The&nbsp; $($rather simple$)$&nbsp; $\rm proof$&nbsp; of this important result can be found in the specification for&nbsp; [[Aufgaben:Exercise_3.3Z:_Convolution_and_D-Transformation|"Exercise 3.3Z"]].
 *As in&nbsp; &raquo;[[Linear_and_Time_Invariant_Systems/System_Description_in_Frequency_Domain#Frequency_response_.E2.80. 93_Transfer_function|$\text{system theory}$]]&laquo;&nbsp; commonly,&nbsp;  the&nbsp; D&ndash;transform&nbsp; $G(D)$&nbsp; of the impulse response&nbsp; $\underline{g}$&nbsp; is also called&nbsp; "transfer function".}}
@@ Line 328: / Line 328: @@
 The following equations apply equally to both filters,&nbsp; setting $j = 1$&nbsp; for the upper filter&nbsp; and $j = 2$&nbsp; for the lower filter:
-*The&nbsp; <b>impulse responses</b>&nbsp; of the two filters result in
+*The&nbsp; &raquo;<b>impulse responses</b>&laquo;&nbsp; of the two filters result in
 ::<math>\underline{g}^{(j)} = (g_0^{(j)}, g_1^{(j)}, \hspace{0.05cm}\text{...}\hspace{0.05cm}, g_m^{(j)}\hspace{0.01cm}) \hspace{0.05cm},\hspace{0.2cm}{\rm with }\hspace{0.15cm} j \in \{1,2\}\hspace{0.05cm}.</math>
-*The two&nbsp; <b>output sequences</b>&nbsp; are as follows,&nbsp; considering that both filters operate on the same input sequence&nbsp; $\underline{u} = (u_0, u_1, u_2, \hspace{0.05cm} \text{...})$&nbsp;:
+*The two&nbsp; &raquo;<b>output sequences</b>&laquo;&nbsp; are as follows,&nbsp; considering that both filters operate on the same input sequence&nbsp; $\underline{u} = (u_0, u_1, u_2, \hspace{0.05cm} \text{...})$&nbsp;:
 ::<math>\underline{x}^{(j)} = (x_0^{(j)}, x_1^{(j)}, x_2^{(j)}, \hspace{0.05cm}\text{...}\hspace{0.05cm}) = \underline{u} \cdot \underline{g}^{(j)} \hspace{0.05cm},\hspace{0.2cm}{\rm with }\hspace{0.15cm} j \in \{1,2\}\hspace{0.05cm}.</math>
-*For the&nbsp; <b>D&ndash;transform</b>&nbsp; of the output sequences:
+*For the&nbsp; &raquo;<b>D&ndash;transform</b>&laquo;&nbsp; of the output sequences:
 ::<math>X^{(j)}(D) = U(D) \cdot G^{(j)}(D) \hspace{0.05cm},\hspace{0.2cm}{\rm with }\hspace{0.15cm} j \in \{1,2\}\hspace{0.05cm}.</math>
@@ Line 382: / Line 382: @@
 ::<math>\Rightarrow  \underline{x}^{(2)} = (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm}, \hspace{0.05cm} 0\hspace{0.05cm}, \hspace{0.05cm} \text{...} \hspace{0.05cm}  \hspace{0.05cm}) \hspace{0.05cm}.</math>
-*We got the same result in&nbsp; [[Aufgaben:Exercise_3.1Z:_Convolution_Codes_of_Rate_1/2|$\text{Exercise 3.1Z}$]]&nbsp; on other way.&nbsp; After multiplexing the two strands,&nbsp; you get again: &nbsp;
+*We got the same result in&nbsp; [[Aufgaben:Exercise_3.1Z:_Convolution_Codes_of_Rate_1/2|"Exercise 3.1Z"]]&nbsp; on other way.&nbsp; After multiplexing the two strands,&nbsp; you get again: &nbsp;
 :$$\underline{x} = (11, 10, 00, 01, 01, 11, 00, 00, \hspace{0.05cm} \text{...} \hspace{0.05cm}).$$}}<br>

@@ Line 13: / Line 13: @@
 # &nbsp; Therefore,&nbsp; the generator matrix&nbsp; $\mathbf{G}$&nbsp; will also be infinitely extended in both directions.<br><br>
-In preparation for the introduction of the generator matrix&nbsp; $\mathbf{G}$&nbsp; on the next page,&nbsp;
+In preparation for the introduction of the generator matrix&nbsp; $\mathbf{G}$&nbsp; in the next section,&nbsp;
 *we define&nbsp; $m + 1$&nbsp; "partial matrices",&nbsp; each with&nbsp; $k$&nbsp; rows and&nbsp; $n$&nbsp; columns, which we denote by&nbsp; $\mathbf{G}_l$&nbsp;
@@ Line 305: / Line 305: @@
 G(D) =  1+ D^2 + D^3 \hspace{0.05cm}.</math>
-*As in&nbsp; $\text{Example 3}$&nbsp; $($on this page above$)$,&nbsp; you get also on this solution path:
+*As in&nbsp; $\text{Example 3}$&nbsp; $($in this section above$)$,&nbsp; you get also on this solution path:
 ::<math>X(D) = U(D) \cdot G(D) =  (1+D) \cdot (1+ D^2 + D^3) </math>
@@ Line 319: / Line 319: @@
 == Application of the&nbsp; D&ndash;transform&nbsp; to rate&nbsp; $1/n$&nbsp; convolution encoders ==
 <br>
-We now apply the results of the last page to a convolutional encoder,&nbsp; restricting ourselves for the moment to the special case&nbsp; $k = 1$.
+We now apply the results of the last section to a convolutional encoder,&nbsp; restricting ourselves for the moment to the special case&nbsp; $k = 1$.
 *Such a&nbsp; $(n, \ k = 1)$&nbsp; convolutional code can be realized with&nbsp; $n$&nbsp; digital filters operating in parallel on the same information sequence&nbsp; $\underline{u}$.