Aufgaben:Exercise 4.09: Recursive Systematic Convolutional Codes - Revision history

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@@ Line 94: / Line 94: @@
-[[File:EN_KC_A_4_9b_v1.png|right|frame|Clarification of&nbsp;  $\underline{p} = (1, \, 1, \, 0, \, 1)^{\rm T} \cdot \mathbf{G}$ '''Korrektur''']]
+[[File:EN_KC_A_4_9b_v1.png|right|frame|Clarification of&nbsp;  $\underline{p} = (1, \, 1, \, 0, \, 1)^{\rm T} \cdot \mathbf{G}$]]
 '''(2)'''&nbsp; The graph shows the solution of this exercise according to the equation&nbsp; $\underline{p} = \underline{u}^{\rm T} \cdot \mathbf{G}$.
 * Here the generator matrix&nbsp; $\mathbf{G}$&nbsp; is infinitely extended downward and to the right.

@@ Line 94: / Line 94: @@
-[[File:P_ID3054__KC_A_4_9b_v1.png|right|frame|Clarification of&nbsp;  $\underline{p} = (1, \, 1, \, 0, \, 1)^{\rm T} \cdot \mathbf{G}$ '''Korrektur''']]
+[[File:EN_KC_A_4_9b_v1.png|right|frame|Clarification of&nbsp;  $\underline{p} = (1, \, 1, \, 0, \, 1)^{\rm T} \cdot \mathbf{G}$ '''Korrektur''']]
 '''(2)'''&nbsp; The graph shows the solution of this exercise according to the equation&nbsp; $\underline{p} = \underline{u}^{\rm T} \cdot \mathbf{G}$.
 * Here the generator matrix&nbsp; $\mathbf{G}$&nbsp; is infinitely extended downward and to the right.

@@ Line 105: / Line 105: @@
 *Between the impulse response&nbsp; $\underline{g}$&nbsp; and the&nbsp; $D$&ndash;transfer function&nbsp; $\mathbf{G}(D)$&nbsp; there is the relation according to the first proposed solution:
 :$$\underline{g}= (\hspace{-0.05cm}1\hspace{-0.05cm},
-\hspace{-0.05cm}1\hspace{-0.05cm},
+\hspace{-0.07cm}1\hspace{-0.07cm},
-\hspace{-0.05cm}1\hspace{-0.05cm},
+\hspace{-0.07cm}1\hspace{-0.07cm},
-\hspace{-0.05cm}0\hspace{-0.05cm},
+\hspace{-0.07cm}0\hspace{-0.07cm},
-\hspace{-0.05cm}1\hspace{-0.05cm},
+\hspace{-0.07cm}1\hspace{-0.07cm},
-\hspace{-0.05cm}1\hspace{-0.05cm},
+\hspace{-0.07cm}1\hspace{-0.07cm},
-\hspace{-0.05cm}0\hspace{-0.05cm},
+\hspace{-0.07cm}0\hspace{-0.07cm},
-\hspace{-0.05cm}1\hspace{-0.05cm},
+\hspace{-0.07cm}1\hspace{-0.07cm},
-\hspace{-0.05cm}1\hspace{-0.05cm},
+\hspace{-0.07cm}1\hspace{-0.07cm},
 ... ) \hspace{0.3cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet \hspace{0.3cm}
 G(D) =  1\hspace{-0.05cm}+\hspace{-0.05cm} D\hspace{-0.05cm} +\hspace{-0.05cm} D^2\hspace{-0.05cm} +\hspace{-0.05cm} D^4 \hspace{-0.05cm}+\hspace{-0.05cm} D^5 \hspace{-0.05cm}+\hspace{-0.05cm} D^7 \hspace{-0.05cm}+\hspace{-0.05cm} D^8 +   \text{...} .$$
@@ Line 128: / Line 128: @@
 cm}+ D^2} \hspace{0.05cm}.$$
-*But since equation (2) is true, the last equation must be false.
+*But since equation&nbsp; '''(2)'''&nbsp; is true,&nbsp; the last equation must be false.
-'''(4)'''&nbsp; Correct is only the <u>proposed solution 1</u>:
+'''(4)'''&nbsp; Correct is only the&nbsp; <u>proposed solution 1</u>:
-*From $\underline{u} = (1, \, 1, \, 1)$ follows $U(D) = 1 + D + D^2$. Thus also holds:
+*From &nbsp; $\underline{u} = (1, \, 1, \, 1)$ &nbsp; follows &nbsp; $U(D) = 1 + D + D^2$.&nbsp; Thus also holds:
 :$$P(D) = U(D) \cdot G(D) = (1+D+D^2) \cdot \frac{1+D^2}{1+D+D^2}= 1+D^2\hspace{0.3cm}
 \Rightarrow\hspace{0.3cm}  \underline{p}= (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\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}. $$
-* If the variables&nbsp; $u_i$&nbsp; and&nbsp; $g_i$&nbsp; were real-valued, the (discrete) convolution&nbsp; $\underline{p} = \underline{u} * \underline{g}$&nbsp;  would always lead to a broadening&nbsp; &#8658; &nbsp; $\underline{p}$&nbsp; in this case would be broader than&nbsp; $\underline{u}$&nbsp; and also broader than&nbsp; $\underline{g}$.
+* If the variables &nbsp; $u_i$ &nbsp; and &nbsp; $g_i$ &nbsp; were real-valued,&nbsp; the&nbsp; $($discrete$)$&nbsp; convolution &nbsp; $\underline{p} = \underline{u} * \underline{g} $&nbsp;  would always lead to a broadening.
-* On the other hand, for&nbsp; $u_i &#8712; {\rm GF}(2)$&nbsp; and&nbsp; $g_i &#8712; {\rm GF}(2)$&nbsp; it may (but need not) happen that even with unbounded&nbsp;  $\underline{u}$&nbsp; or for unbounded&nbsp; $\underline{g}$&nbsp; the convolution product&nbsp; $\underline{p} = \underline{u} * \underline{g}$&nbsp; is bounded.
+[[File:EN_KC_A_4_9d_v1.png|right|frame|Clarification of&nbsp;  $\underline{p} = (1, \, 1, \, 1, \, 0, \, ...)^{\rm T} \cdot \mathbf{G}$]]
-[[File:EN_KC_A_4_9d_v1.png|right|frame|Verdeutlichung von&nbsp;  $\underline{p} = (1, \, 1, \, 1, \, 0, \, ...)^{\rm T} \cdot \mathbf{G}$]]
+*In this case &nbsp; $\underline{p}$&nbsp; would be broader than &nbsp; $\underline{u}$ &nbsp; and also broader than &nbsp; $\underline{g}$.
-'''(5)'''&nbsp; Following a similar procedure as in [[Aufgaben:Exercise_4.08:_Repetition_to_the_Convolutional_Codes|"Exercise A4.8, (4)"]], one can see:
-*The free distance is again determined by the path&nbsp; $S_0 &#8594; S_0 &#8594; S_1 &#8594; S_2 &#8594; S_0 &#8594; S_0 &#8594; \hspace{0.05cm}\text{...}\hspace{0.05cm}$.
-*But the associated code sequence&nbsp; $\underline{x}$&nbsp; is now &nbsp;" $00 \ 11 \ 10 \ 11 \ 00 \ ... $".
+'''(5)'''&nbsp; Following a similar procedure as in&nbsp; [[Aufgaben:Exercise_4.08:_Repetition_to_the_Convolutional_Codes|$\text{Exercise A4.8}$,&nbsp; (4)]],&nbsp; one can see:
-*This gives the free distance to&nbsp; $d_{\rm F} \ \underline{= 5}$.
+*The free distance is again determined by the path&nbsp;
-*In contrast, in the non-recursive code (exercise 4.8), the free distance was&nbsp; $d_{\rm F} = 3$.
+:$$S_0 &#8594; S_0 &#8594; S_1 &#8594; S_2 &#8594; S_0 &#8594; S_0 &#8594; \hspace{0.05cm}\text{...}\hspace{0.05cm}.$$
 {{ML-Fuß}}

@@ Line 104: / Line 104: @@
 '''(3)'''&nbsp; Correct are&nbsp; <u>the proposed solutions 1 and 2</u>:
 *Between the impulse response&nbsp; $\underline{g}$&nbsp; and the&nbsp; $D$&ndash;transfer function&nbsp; $\mathbf{G}(D)$&nbsp; there is the relation according to the first proposed solution:
-$$\underline{g}= (\hspace{-0.05cm}1\hspace{-0.05cm},
+:$$\underline{g}= (\hspace{-0.05cm}1\hspace{-0.05cm},
 \hspace{-0.05cm}1\hspace{-0.05cm},
 \hspace{-0.05cm}1\hspace{-0.05cm},
@@ Line 113: / Line 113: @@
 \hspace{-0.05cm}1\hspace{-0.05cm},
 \hspace{-0.05cm}1\hspace{-0.05cm},
-... ) \hspace{0.3cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\\hspace{0.3cm}
+... ) \hspace{0.3cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet \hspace{0.3cm}
-G(D) =  1\hspace{-0.05cm}+\hspace{-0.05cm} D\hspace{-0.05cm} +\hspace{-0.05cm} D^2\hspace{-0.05cm} +\hspace{-0.05cm} D^4 \hspace{-0.05cm}+\hspace{-0.05cm} D^5 \hspace{-0.05cm}+\hspace{-0.05cm} D^7 \hspace{-0.05cm}+\hspace{-0.05cm} D^8 +  \hspace{0.05cm} \text{...} \hspace{0.05cm}.$$
+G(D) =  1\hspace{-0.05cm}+\hspace{-0.05cm} D\hspace{-0.05cm} +\hspace{-0.05cm} D^2\hspace{-0.05cm} +\hspace{-0.05cm} D^4 \hspace{-0.05cm}+\hspace{-0.05cm} D^5 \hspace{-0.05cm}+\hspace{-0.05cm} D^7 \hspace{-0.05cm}+\hspace{-0.05cm} D^8 +   \text{...} .$$
 *Let us now examine the second proposal:

@@ Line 73: / Line 73: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Tracing the transitions in the state diagram for the sequence&nbsp;  $\underline{u} = (1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0)$&nbsp;  at the input, we get the path
+'''(1)'''&nbsp; Tracing the transitions in the state diagram for the sequence &nbsp;  $\underline{u} = (1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0)$ &nbsp;  at the input,&nbsp; we get the path
 :$$S_0 &#8594; S_1 &#8594; S_3 &#8594; S_2 &#8594; S_1 &#8594; S_3 &#8594; S_2 &#8594; S_1 &#8594; S_3 &#8594; \hspace{0.05cm}\text{...}  \hspace{0.05cm}$$
-*For each transition, the first code symbol $x_i^{(1)}$ is equal to the information bit $u_i$ and the code symbol $x_i^{(2)}$ indicates the parity-check bit $p_i$.
+*For each transition,&nbsp; the first code symbol&nbsp; $x_i^{(1)}$&nbsp; is equal to the information bit&nbsp; $u_i$&nbsp; and the code symbol&nbsp; $x_i^{(2)}$&nbsp; indicates the parity-check bit&nbsp; $p_i$.
 *This gives the result corresponding to <u>solution proposition 1</u>:
 :$$\underline{p}= (\hspace{0.05cm}1\hspace{0.05cm},
@@ Line 89: / Line 90: @@
 = \underline{g}\hspace{0.05cm}.$$
-*For any RSC code, the impulse response $\underline{g}$ is infinite in length and becomes periodic at some point, in this example with period&nbsp; $P = 3$&nbsp; and&nbsp; "$0, \, 1, \, 1$".
+*For any RSC code,&nbsp; the impulse response&nbsp; $\underline{g}$&nbsp; is infinite in length and becomes periodic at some point,&nbsp; in this example with period&nbsp; $P = 3$&nbsp; and&nbsp; "$0, \, 1, \, 1$".
-[[File:P_ID3054__KC_A_4_9b_v1.png|right|frame|Clarification of&nbsp;  $\underline{p} = (1, \, 1, \, 0, \, 1)^{\rm T} \cdot \mathbf{G}$]]
+[[File:P_ID3054__KC_A_4_9b_v1.png|right|frame|Clarification of&nbsp;  $\underline{p} = (1, \, 1, \, 0, \, 1)^{\rm T} \cdot \mathbf{G}$ '''Korrektur''']]
 '''(2)'''&nbsp; The graph shows the solution of this exercise according to the equation&nbsp; $\underline{p} = \underline{u}^{\rm T} \cdot \mathbf{G}$.
 * Here the generator matrix&nbsp; $\mathbf{G}$&nbsp; is infinitely extended downward and to the right.
 *The correct solution is <u>proposal 2</u>.
-<br clear=all>
+*The correct solution is&nbsp; <u>proposal 2</u>.
 '''(3)'''&nbsp; Correct are <u>the proposed solutions 1 and 2</u>:
 *Between the impulse response $\underline{g}$ and the $D$&ndash;transfer function $\mathbf{G}(D)$ there is the relation according to the first proposed solution:
-:$$\underline{g}= (\hspace{-0.05cm}1\hspace{-0.05cm},
 \hspace{-0.05cm}1\hspace{-0.05cm},
 \hspace{-0.05cm}1\hspace{-0.05cm},
@@ Line 109: / Line 113: @@
 \hspace{-0.05cm}1\hspace{-0.05cm},
 \hspace{-0.05cm}1\hspace{-0.05cm},
-... ) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quad
+... ) \hspace{0.3cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\\hspace{0.3cm}
 G(D) =  1\hspace{-0.05cm}+\hspace{-0.05cm} D\hspace{-0.05cm} +\hspace{-0.05cm} D^2\hspace{-0.05cm} +\hspace{-0.05cm} D^4 \hspace{-0.05cm}+\hspace{-0.05cm} D^5 \hspace{-0.05cm}+\hspace{-0.05cm} D^7 \hspace{-0.05cm}+\hspace{-0.05cm} D^8 +  \hspace{0.05cm} \text{...} \hspace{0.05cm}.$$

@@ Line 34: / Line 34: @@
 * The following vectorial quantities are used in the questions for this exercise:
-** the parity-check sequence:&nbsp; $\hspace{0.2cm} \underline{p} = (p_1, \, p_2, \text{...} \hspace{0.05cm})$,
+:* the information sequence:&nbsp; $\hspace{0.2cm} \underline{u} = (u_1, \, u_2, \text{...} \hspace{0.05cm} )$,
+:* the parity-check sequence:&nbsp; $\hspace{0.2cm} \underline{p} = (p_1, \, p_2, \text{...} \hspace{0.05cm})$,
-** the impulse response:&nbsp; $\hspace{0.2cm} \underline{g} = (g_1, \, g_2, \text{...} \hspace{0.05cm} ); \hspace{0.2cm}$ this is equal to the parity sequence $\underline{p}$&nbsp; for&nbsp; $\underline{u} = (1, \, 0, \, 0, \text{...} \hspace{0.05cm} )$.
+:* the impulse response:&nbsp; $\hspace{0.2cm} \underline{g} = (g_1, \, g_2, \text{...} \hspace{0.05cm} ); \hspace{0.2cm}$ this is equal to the parity sequence&nbsp; $\underline{p}$&nbsp; for&nbsp; $\underline{u} = (1, \, 0, \, 0, \text{...} \hspace{0.05cm} )$.
@@ Line 46: / Line 45: @@
 {What is the impulse response&nbsp; $\underline{g}$&nbsp;?
 |type="()"}
-+&nbsp;  $\underline{g} = (1, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, 1, \text{...}  \hspace{0.05cm})$ holds.
++&nbsp;  $\underline{g} = (1, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, 1, \text{...}  \hspace{0.05cm})$.
--&nbsp;  $\underline{g} = (1, \, 0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \text{...}  \hspace{0.05cm})$ holds.
+-&nbsp;  $\underline{g} = (1, \, 0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \text{...}  \hspace{0.05cm})$.
-{It holds&nbsp; $\underline{u} = (1, \, 1, \, 0, \, 1)$. Which statements hold for the parity sequence&nbsp; $\underline{p}$&nbsp;?
+{It holds &nbsp; $\underline{u} = (1, \, 1, \, 0, \, 1)$.&nbsp; Which statements hold for the parity sequence&nbsp; $\underline{p}$&nbsp;?
 |type="[]"}
--&nbsp;  $\underline{p} = (1, \, 0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \text{...}  \hspace{0.05cm})$ holds.
+-&nbsp;  $\underline{p} = (1, \, 0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \text{...}  \hspace{0.05cm})$.
-+&nbsp;  $\underline{p} = (1, \, 0, \, 0, \, 0, \, 0, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \text{...}  \hspace{0.05cm})$ holds.
++&nbsp;  $\underline{p} = (1, \, 0, \, 0, \, 0, \, 0, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \text{...}  \hspace{0.05cm})$.
-- With limited information sequence&nbsp; $\underline{u}$&nbsp; is always also&nbsp; $\underline{p}$&nbsp; limited.
+- With limited information sequence&nbsp; $\underline{u}$ &nbsp; &rArr; &nbsp; $\underline{p}$&nbsp; is also limited.
 {What is the&nbsp; $D$&ndash;transfer function&nbsp; $G(D)$?
 |type="[]"}
-+&nbsp;  $G(D) = 1 + D + D^2 + D^4 + D^5 + D^7 + D^8 + \text{...}  \hspace{0.05cm}$ holds.
++&nbsp;  $G(D) = 1 + D + D^2 + D^4 + D^5 + D^7 + D^8 + \text{...}  \hspace{0.05cm}$.
-+&nbsp;  $G(D) = (1 + D^2)/(1 + D + D^2)$ holds.
++&nbsp;  $G(D) = (1 + D^2)/(1 + D + D^2)$.
--&nbsp;  $G(D) = (1 + D + D^2)/(1 + D^2)$ holds.
+-&nbsp;  $G(D) = (1 + D + D^2)/(1 + D^2)$.
-{Now let&nbsp; $\underline{u} = (1, \, 1, \, 1)$ hold. Which statements hold for the parity sequence&nbsp; $\underline{p}$?
+{Now let&nbsp; $\underline{u} = (1, \, 1, \, 1)$.&nbsp; Which statements hold for the parity sequence&nbsp; $\underline{p}$?
 |type="[]"}
-+&nbsp;  $\underline{p} = (1, \, 0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \text{...}  \hspace{0.05cm})$ holds.
++&nbsp;  $\underline{p} = (1, \, 0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \text{...}  \hspace{0.05cm})$.
--&nbsp;  $\underline{p} = (1, \, 0, \, 0, \, 0, \, 0, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \text{...}  \hspace{0.05cm})$ holds.
+-&nbsp;  $\underline{p} = (1, \, 0, \, 0, \, 0, \, 0, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \text{...}  \hspace{0.05cm})$.
-- With unlimited impulse response&nbsp; $\underline{g}$&nbsp; is always also&nbsp; $\underline{p}$&nbsp; unlimited.
+- With limited information sequence&nbsp; $\underline{u}$ &nbsp; &rArr; &nbsp; $\underline{p}$&nbsp; is also limited.
 {What is the free distance&nbsp; $d_{\rm F}$&nbsp; of this RSC encoder?

@@ Line 30: / Line 30: @@
-Hints:
+<u>Hints:</u>
 * The exercise refers to the chapter&nbsp; [[Channel_Coding/The_Basics_of_Turbo_Codes|"Basics of Turbo Codes"]].
 * The following vectorial quantities are used in the questions for this exercise:
 ** the information sequence:&nbsp; $\hspace{0.2cm} \underline{u} = (u_1, \, u_2, \text{...} \hspace{0.05cm} )$,
 ** the parity-check sequence:&nbsp; $\hspace{0.2cm} \underline{p} = (p_1, \, p_2, \text{...} \hspace{0.05cm})$,
 ** the impulse response:&nbsp; $\hspace{0.2cm} \underline{g} = (g_1, \, g_2, \text{...} \hspace{0.05cm} ); \hspace{0.2cm}$ this is equal to the parity sequence $\underline{p}$&nbsp; for&nbsp; $\underline{u} = (1, \, 0, \, 0, \text{...} \hspace{0.05cm} )$.

@@ Line 16: / Line 16: @@
 #[[Channel_Coding/Code_Description_with_State_and_Trellis_Diagram#Definition_of_the_free_distance|"Definition of the free distance"]]
 #[[Channel_Coding/Algebraic_and_Polynomial_Description#GF.282.29_description_forms_of_a_digital_filter|"GF(2) description forms of a digital filter"]]
-#[[Channel_Coding/Algebraic_and_Polynomial_Description#Application_of_the_D.E2.80.93transform_to_rate_.7F.27.22.60UNIQ-MathJax160-QINU.60.22.27.7F_convolution_encoders| "Application of the  D–transform  to rate  1/n convolution encoders"]]
+#[[Channel_Coding/Algebraic_and_Polynomial_Description#Application_of_the_D.E2.80.93transform_to_rate_.7F.27.22.60UNIQ-MathJax158-QINU.60.22.27.7F_convolution_encoders| "Application of the&nbsp;  $D$–transform  to rate&nbsp;  $1/n$&nbsp; convolution encoders"]]
 #[[Channel_Coding/Algebraic_and_Polynomial_Description#Filter_structure_with_fractional.E2.80.93rational_transfer_function|"Filter structure with fractional–rational transfer function"]]

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Channel_Coding/The_Basics_of_Turbo_Codes}}
-[[File:P_ID3040__KC_A_4_9_v1.png|right|frame|State transition diagram of an RSC code]]
+[[File:P_ID3040__KC_A_4_9_v1.png|right|frame|State transition diagram <br>of an RSC code]]
-In the&nbsp; [[Aufgaben:Exercise_4.08:_Repetition_to_the_Convolutional_Codes|"Exercise 4.8"]]&nbsp; important properties of convolutional codes have already been derived from the state transition diagram, assuming a non-recursive filter structure.
+In&nbsp; [[Aufgaben:Exercise_4.08:_Repetition_to_the_Convolutional_Codes|$\text{Exercise 4.8}$]]&nbsp; important properties of convolutional codes have already been derived from the state transition diagram,&nbsp; assuming a non-recursive filter structure.
-Now a rate $1/2$ RSC code is treated in a similar manner. Here "RSC" stands for "Recursive Systematic Convolutional".
+Now a rate&nbsp; $1/2$&nbsp; RSC code is treated in a similar manner.&nbsp; Here&nbsp; $\rm RSC$&nbsp; stands for&nbsp; "recursive systematic convolutional".
 The transfer function matrix of an RSC convolutional code can be specified as follows:
@@ Line 10: / Line 10: @@
   \hspace{0.05cm}.$$
-Otherwise, exactly the same conditions apply here as in exercise 4.8. We refer again to the following theory pages:
+Otherwise,&nbsp; exactly the same conditions apply here as in Exercise 4.8.&nbsp; We refer again to the following theory pages:
 #[[Channel_Coding/Algebraic_and_Polynomial_Description#Systematic_convolutional_codes|"Systematic convolutional codes"]]
@@ Line 20: / Line 20: @@
-In the state transition diagram, the state&nbsp; $S_0$&nbsp; is always assumed. Two arrows go from each state. The label is "$u_i \hspace{0.05cm}| \hspace{0.05cm} x_i^{(1)}x_i^{(2)}$". For a systematic code, this involves:
+In the state transition diagram,&nbsp; two arrows go from each state.&nbsp; The label is&nbsp; "$u_i \hspace{0.05cm}| \hspace{0.05cm} x_i^{(1)}x_i^{(2)}$".&nbsp; For a systematic code,&nbsp; this involves:
 * The first code bit is identical to the information bit: &nbsp; $\hspace{0.2cm} x_i^{(1)} = u_i &#8712; \{0, \, 1\}$.
 * The second code bit is the parity-check bit: &nbsp; $\hspace{0.2cm} x_i^{(2)} = p_i &#8712; \{0, \, 1\}$.