Aufgaben:Exercise 3.3Z: Convolution and D-Transformation - Revision history

Guenter at 16:47, 10 November 2022

2022-11-10T16:47:19Z

Guenter at 16:28, 10 November 2022

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Noah at 16:44, 26 September 2022

2022-09-26T16:44:36Z

Noah at 21:03, 23 September 2022

2022-09-23T21:03:38Z

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Guenter at 09:14, 12 September 2022

2022-09-12T09:14:05Z

Guenter: Guenter moved page Aufgaben:Aufgabe 3.3Z: Faltung und D–Transformation to Aufgaben:Exercise 3.3Z: Convolution and D-Transformation

2022-09-12T09:04:57Z

Guenter moved page Aufgabe 3.3Z: Faltung und D–Transformation to Exercise 3.3Z: Convolution and D-Transformation

Javier: Text replacement - "”" to """

2021-05-28T14:32:10Z

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Javier: Text replacement - "Category:Aufgaben zu Kanalcodierung" to "Category:Channel Coding: Exercises"

2021-03-23T12:50:41Z

Text replacement - "Category:Aufgaben zu Kanalcodierung" to "Category:Channel Coding: Exercises"

Javier: Text replacement - "Signal_Representation/Faltungssatz_und_Faltungsoperation" to "Signal_Representation/The_Convolution_Theorem_and_Operation"

2020-09-01T09:53:12Z

Text replacement - "Signal_Representation/Faltungssatz_und_Faltungsoperation" to "Signal_Representation/The_Convolution_Theorem_and_Operation"

@@ Line 54: / Line 54: @@
-Hints:
+<u>Hints:</u>
 * The exercise belongs to the chapter&nbsp; [[Channel_Coding/Algebraic_and_Polynomial_Description| "Algebraic and Polynomial Description"]].
@@ Line 103: / Line 103: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The only two filter coefficients different from $0$ are $g_0 \ \underline{= 1}$ and $g_1 \ \underline{= 1}$.
+'''(1)'''&nbsp; The only two filter coefficients different from&nbsp; $0$&nbsp; are&nbsp; $g_0 \ \underline{= 1}$&nbsp; and&nbsp; $g_1 \ \underline{= 1}$.
-*From this follows $g_2 \ \underline{= 0}$ and for the $D$ transform of the impulse response:
+*From this follows&nbsp; $g_2 \ \underline{= 0}$&nbsp; and for the D&ndash;transform of the impulse response:
 :$$\underline{g} = (1\hspace{0.05cm},\hspace{0.05cm} 1)   \hspace{0.25cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.25cm}
 {G}(D) = 1+ D \hspace{0.05cm}.$$
@@ Line 110: / Line 110: @@
-'''(2)'''&nbsp; The impulse response of the considered filter is $\underline{g} = (1, \, 1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$.
+'''(2)'''&nbsp; The impulse response of the considered filter is&nbsp; $\underline{g} = (1, \, 1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$.
-*For the initial sequence, therefore, we obtain the convolution product
+*For the output sequence,&nbsp; therefore,&nbsp; we obtain the convolution product
 :$$\underline{x} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} \underline{u}* \underline{g} =
 (1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0,\hspace{0.05cm} ...\hspace{0.05cm}) *
@@ Line 117: / Line 117: @@
   \hspace{0.05cm}.$$
-*The same result is obtained using the $D$ transforms $U(D) = 1 + D^3$ and $G(D) = 1 + D$:
+*The same result is obtained using the D&ndash;transforms&nbsp; $U(D) = 1 + D^3$&nbsp; and&nbsp; $G(D) = 1 + D$:
 :$${X}(D) = U(D) \cdot G(D) = ( 1+D^3) \cdot (1+D) = 1 +D + D^3 +D^4
 \hspace{0.05cm}.$$
-*The back transformation leads again to the result $\underline{x} = (1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$ &nbsp;&#8658;&nbsp; <u>Proposed solution 3</u>.
+*The inverse transformation leads again to the result&nbsp; $\underline{x} = (1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$ &nbsp; &#8658; &nbsp; <u>Proposed solution 3</u>.
-'''(3)'''&nbsp; Here we immediately use the path over the $D$ transforms:
+'''(3)'''&nbsp; Here we immediately use the path over the D&ndash;transforms:
 :$${X}(D) =  ( 1+D+D^2) \cdot (1+D) = 1 +D + D +D^2 +D^2 +D^3 = 1+ D^3\hspace{0.3cm}
 \Rightarrow \hspace{0.3cm} \underline{x}  =
@@ Line 131: / Line 131: @@
   \hspace{0.05cm}.$$
-*The result corresponds to the <u>proposed solution 2</u>. The following calculation is to illustrate the path in the time domain:
+*The result corresponds to the&nbsp; <u>proposed solution 2</u>.&nbsp; The following calculation is to illustrate the path in the time domain:
 :$$(1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}\text{...}\hspace{0.05cm}) *
 (1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} \text{...} \hspace{0.05cm})\hspace{-0.15cm} \ = \ \hspace{-0.15cm}(1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0,\hspace{0.05cm} \text{...}\hspace{0.05cm}) \hspace{0.05cm},$$
@@ Line 139: / Line 139: @@
 (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} ... \hspace{0.05cm})\hspace{-0.15cm} \ = \ \hspace{-0.15cm}(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},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0,\hspace{0.05cm} \text{...}\hspace{0.05cm}) \hspace{0.05cm}.$$
-*Since convolution is a linear operation, in the Galois field ${\rm GF}(2)$ results from summation:
+*Since the convolution is a linear operation,&nbsp; in the Galois field&nbsp; ${\rm GF}(2)$&nbsp; results from summation:
 :$$(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},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}\text{...}\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{0.08cm}=\hspace{0.08cm}(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},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0,\hspace{0.05cm} \text{...}\hspace{0.05cm}) \hspace{0.05cm}.$$
-*If the convolution had been performed not in ${\rm GF}(2)$ but for real numbers, we would have obtained the result $\underline{x} = (1, \, 2, \, 2, \, 1, \, 0, \, 0, \, \text{...})$.
+*If the convolution had been performed not in&nbsp; ${\rm GF}(2)$&nbsp; but for real numbers,&nbsp; we would have obtained the result&nbsp; $\underline{x} = (1, \, 2, \, 2, \, 1, \, 0, \, 0, \, \text{...})$.
-'''(4)'''&nbsp; The sample solution to the subtask '''(3)''' already suggests that the <u>proposed solution 1</u> is correct here.
+'''(4)'''&nbsp; The sample solution to the subtask&nbsp; '''(3)'''&nbsp; already suggests that the&nbsp; <u>proposed solution 1</u>&nbsp; is correct here.
-*The way over the $D$ transforms confirms this result:
+*The way over the D&ndash;transforms confirms this result:
 :$$\underline{u} = (1\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}\text{...}\hspace{0.05cm})  \hspace{0.25cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.25cm}
 {U}(D)= 1+ D + D^2+ D^3 + \text{...}\hspace{0.15cm}.$$
-*Using the equation valid for calculations in ${\rm GF}(2)$
+*Using the equation valid for calculations in&nbsp; ${\rm GF}(2)$
 :$$1 + D + D^2 + D^3  + \hspace{0.05cm}\text{...} \hspace{0.1cm}= \frac{1}{1+D}$$
@@ Line 162: / Line 162: @@
-'''(5)'''&nbsp; The path over the $D$ transforms leads to <u>solution 2</u>.
+'''(5)'''&nbsp; The path over the D&ndash;transforms leads to&nbsp; <u>solution 2</u>.
-*For this alternating sequence $\underline{u}$, starting with 1, one obtains:
+*For this alternating sequence&nbsp; $\underline{u}$,&nbsp; starting with&nbsp; $1$,&nbsp; one obtains:
 :$${X}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1 \cdot (1+D)  + D^2 \cdot (1+D) + D^4 \cdot (1+D) + \text{...} = 1 + D + D^2 + D^3 + D^4 + D^5 +\hspace{0.05cm} ...
 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}
@@ Line 169: / Line 169: @@
 \hspace{0.05cm}.$$
-*This result can also be read by direct application of the convolution as in subtask (2).
+*This result can also be read by direct application of the convolution as in subtask&nbsp; '''(2)'''.
 *With $\underline{u} = (0, \, 1, \, 0, \, 1, \, 0, \, 1, \, \text{...})$ you get $\underline{x} = (0, \, 1, \, 1, \, 1, \, 1, \, 1, \,\text{...})$.
-*This differs from the "sequence of ones" only in the first bit. It is then $x_1 = 0$ instead of $x_1 = 1$.
+*With&nbsp; $\underline{u} = (0, \, 1, \, 0, \, 1, \, 0, \, 1, \, \text{...})$&nbsp; you get&nbsp; $\underline{x} = (0, \, 1, \, 1, \, 1, \, 1, \, 1, \,\text{...})$.
 {{ML-Fuß}}
 [[Category:Channel Coding: Exercises|^3.2 Polynomial Description^]]

@@ Line 4: / Line 4: @@
 In this exercise, we use a simple example to describe
-* the finite&nbsp; <b>impulse response</b> &nbsp;of a filter:
+* the finite&nbsp; &raquo;<b>impulse response</b>&laquo; &nbsp;of a filter:
 :$$\underline{g} = \left (g_0, g_1, \hspace{0.05cm}\text{...}\hspace{0.1cm}, g_l, \hspace{0.05cm}\text{...}\hspace{0.1cm}, g_m \right )
 \hspace{0.05cm},\hspace{0.2cm}g_l \in {\rm GF(2) } = \{ 0, 1 \}\hspace{0.05cm}, $$
 * the&nbsp; <b>input sequence</b>&nbsp; of the filter:
 :$$\underline{u} = \left (u_0, u_1, \hspace{0.05cm}\text{...}\hspace{0.1cm}, u_i, \hspace{0.05cm}\text{...}\hspace{0.1cm} \right )
 \hspace{0.05cm},\hspace{0.2cm}u_i \in {\rm GF(2) } = \{ 0, 1 \}\hspace{0.05cm}, $$
-* the&nbsp; <b>output sequence</b>&nbsp; of the filter:
+* the&nbsp; &raquo;<b>output sequence</b>&laquo;&nbsp; of the filter:
 :$$\underline{x} = \left (x_0, x_1, \hspace{0.05cm}\text{...}\hspace{0.1cm}, x_i, \hspace{0.05cm}\text{...}\hspace{0.1cm} \right )
 \hspace{0.05cm},\hspace{0.2cm}x_i \in {\rm GF(2) } = \{ 0, 1 \}\hspace{0.05cm}. $$
-We have adapted the nomenclature for this (digital) filter description to the book "Introduction to Channel Coding". In other&nbsp; $\rm LNTww$ books often&nbsp; $\underline{x}$&nbsp; denotes the filter input,&nbsp; $\underline{y}$&nbsp; the filter output, and the impulse response is called&nbsp; $h$&nbsp;.
+We have adapted the nomenclature for this&nbsp; $($digital$)$&nbsp; filter description to the book&nbsp; "Introduction to Channel Coding".&nbsp; In other&nbsp; $\rm LNTww$ books often&nbsp;
-In general, for the output sequence corresponding to the&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_in_the_time_domain| "convolution"]]&nbsp;:
+In general,&nbsp; for the output sequence corresponding to the&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation#Convolution_in_the_time_domain| $\text{convolution}$]]&nbsp;:
-:$$\underline{x} = \underline{u}* \underline{g} = \left (x_0, x_1, \hspace{0.05cm}\text{...}\hspace{0.1cm}, x_i, \hspace{0.05cm}\text{...}\hspace{0.1cm} \right )\hspace{0.05cm},\hspace{0.1cm} {\rm mit} \hspace{0.2cm} x_i = \sum_{l = 0}^{m} g_l \cdot u_{i-l}\hspace{0.05cm}.$$
+:$$\underline{x} = \underline{u}* \underline{g} = \left (x_0, x_1, \hspace{0.05cm}\text{...}\hspace{0.1cm}, x_i, \hspace{0.05cm}\text{...}\hspace{0.1cm} \right )\hspace{0.05cm},\hspace{0.1cm} {\rm with} \hspace{0.2cm} x_i = \sum_{l = 0}^{m} g_l \cdot u_{i-l}\hspace{0.05cm}.$$
-We now represent the time functions&nbsp; $\underline{g}, \ \underline{u}$&nbsp; and&nbsp; $\underline{x}$&nbsp; by polynomials in a dummy&ndash;variable&nbsp; $D$&nbsp; and call these the&nbsp; $D$ transforms:
+*We now represent the time functions &nbsp; $\underline{g}, \ \underline{u}$ &nbsp; and &nbsp; $\underline{x}$&nbsp; by polynomials in a dummy variable&nbsp; $D$&nbsp; and call these the&nbsp; D&ndash;transforms:
 :$$\underline{g}   \hspace{0.25cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.25cm}
 {G}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}  \sum_{l = 0}^{m} g_l \cdot D\hspace{0.03cm}^l = g_0 + g_1 \cdot D + g_2 \cdot D^2 + \text{...} + g_m \cdot D\hspace{0.03cm}^m\hspace{0.05cm},$$
@@ Line 27: / Line 32: @@
 {X}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}  \sum_{i = 0}^{\infty} x_i \cdot D\hspace{0.03cm}^i = x_0 + x_1 \cdot D + x_2 \cdot D^2 + \text{...} \hspace{0.05cm}.$$
-Thus the (more complicated) convolution becomes a multiplication:
+*Thus the&nbsp; $($more complicated$)$&nbsp; convolution becomes a multiplication:
 :$$\underline{x} = \underline{u}* \underline{g}   \hspace{0.25cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.25cm}
 {X}(D) = U(D) \cdot G(D) \hspace{0.05cm}.$$
-Formally, this relationship can be demonstrated as follows:
+*Formally,&nbsp; this relationship can be demonstrated as follows:
 :$${X}(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}  \sum_{i = 0}^{\infty} x_i \cdot D\hspace{0.03cm}^i = \sum_{i = 0}^{\infty} \sum_{l = 0}^{m}\hspace{0.1cm}
   g_l \cdot u_{i-l} \cdot D\hspace{0.03cm}^{i} =  \sum_{l = 0}^{m} \hspace{0.1cm} g_l \cdot \sum_{j = -l}^{\infty} \hspace{0.1cm}
@@ Line 39: / Line 44: @@
   \hspace{0.05cm}.$$
-Here it was considered that all&nbsp; $u_j$&nbsp; for&nbsp; $j < 0$&nbsp; do not exist and can be set to zero.
+*Here it was considered that all &nbsp; $u_j$ &nbsp; for&nbsp; $j < 0$&nbsp; do not exist and can be set to zero.
-Both procedures for computing the initial sequence&nbsp; $\underline{x}$, viz.
+*Both procedures for computing the initial sequence&nbsp; $\underline{x}$&nbsp; shall be demonstrated for the digital filter outlined above,&nbsp; viz.
-* using the convolution
+# &nbsp; using the convolution,
-* by means of the $D$ transformation,
+# &nbsp; by means of the D&ndash;transformation.
@@ Line 52: / Line 54: @@
-Hints:
+* In the solution,&nbsp; consider the following identity for calculations in&nbsp; $\rm GF(2)$:
 :$$1 + D + D^2 + D^3  + \hspace{0.05cm}\text{...}\hspace{0.1cm}= \frac{1}{1+D} \hspace{0.05cm}.$$
@@ Line 71: / Line 73: @@
 $g_2 \ = \ ${ 0. }
-{The sequence&nbsp; $\underline{u} = (1, \, 0, \, 0, \, 1)$&nbsp; let be finite. What is the initial sequence?
+{The sequence&nbsp; $\underline{u} = (1, \, 0, \, 0, \, 1)$&nbsp; let be finite.&nbsp; What is the output sequence?
 |type="()"}
 - $\underline{x} = (1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$.
@@ Line 78: / Line 80: @@
 - $\underline{x} = (1, \, 1, \, 1, \, 1, \, \text{...}\hspace{0.05cm})$ &nbsp; &#8658; &nbsp; "sequence of ones".
-{The sequence&nbsp; $\underline{u} = (1, \, 1, \, 1)$&nbsp; let be finite. What is the initial sequence?
+{The sequence&nbsp; $\underline{u} = (1, \, 1, \, 1)$&nbsp; let be finite.&nbsp; What is the output sequence?
 |type="()"}
 - $\underline{x} = (1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$.
@@ Line 95: / Line 97: @@
 |type="()"}
 - $\underline{u} = (1, \, 1, \, 1, \, 1, \, \text{...}\hspace{0.05cm})$ &nbsp; &#8658; &nbsp; "sequence of ones"
-+ $\underline{u} = (1, \, 0, \, 1, \, 0, \, 1, \, 0, \,\text{...}\hspace{0.05cm})$ &nbsp; &#8658; &nbsp; alternating sequence, starting with $1$.
++ $\underline{u} = (1, \, 0, \, 1, \, 0, \, 1, \, 0, \,\text{...}\hspace{0.05cm})$ &nbsp; &#8658; &nbsp; alternating sequence, starting with&nbsp; "$1$".
-- $\underline{u} = (0, \, 1, \, 0, \, 1, \, 0, \, 1, \, \text{...}\hspace{0.05cm})$ &nbsp; &#8658; &nbsp; alternating sequence, starting with $0$.
+- $\underline{u} = (0, \, 1, \, 0, \, 1, \, 0, \, 1, \, \text{...}\hspace{0.05cm})$ &nbsp; &#8658; &nbsp; alternating sequence, starting with&nbsp; "$0$".
 </quiz>

@@ Line 168: / Line 168: @@
 *This result can also be read by direct application of the convolution as in subtask (2).
-*Mit $\underline{u} = (0, \, 1, \, 0, \, 1, \, 0, \, 1, \, \text{...})$ erhält man dagegen $\underline{x} = (0, \, 1, \, 1, \, 1, \, 1, \, 1, \,\text{...})$.
+*With $\underline{u} = (0, \, 1, \, 0, \, 1, \, 0, \, 1, \, \text{...})$ you get $\underline{x} = (0, \, 1, \, 1, \, 1, \, 1, \, 1, \,\text{...})$.
 *This differs from the "sequence of ones" only in the first bit. It is then $x_1 = 0$ instead of $x_1 = 1$.
 {{ML-Fuß}}

← Older revision		Revision as of 09:14, 12 September 2022
Line 173:		Line 173:


−	[[Category:Channel Coding: Exercises\|^3.2 ~~Polynomische Beschreibung~~^]]	+	[[Category:Channel Coding: Exercises\|^3.2 Polynomial Description^]]

@@ Line 14: / Line 14: @@
 \hspace{0.05cm},\hspace{0.2cm}x_i \in {\rm GF(2) } = \{ 0, 1 \}\hspace{0.05cm}. $$
-Die Nomenklatur für diese (digitale) Filterbeschreibung haben wir an das Buch "Einführung in die Kanalcodierung&rdquo; angepasst. In anderen&nbsp; $\rm LNTwww$&ndash;Büchern bezeichnet oft&nbsp; $\underline{x}$&nbsp; den Filtereingang,&nbsp; $\underline{y}$&nbsp; den Filterausgang, und die Impulsantwort wird&nbsp; $h$&nbsp; genannt.
+Die Nomenklatur für diese (digitale) Filterbeschreibung haben wir an das Buch "Einführung in die Kanalcodierung" angepasst. In anderen&nbsp; $\rm LNTwww$&ndash;Büchern bezeichnet oft&nbsp; $\underline{x}$&nbsp; den Filtereingang,&nbsp; $\underline{y}$&nbsp; den Filterausgang, und die Impulsantwort wird&nbsp; $h$&nbsp; genannt.
 Allgemein gilt für die Ausgangssequenz entsprechend der&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation#Faltung_im_Zeitbereich| Faltung]]&nbsp; (englisch: &nbsp;<i>Convolution</i>&nbsp;):
@@ Line 76: / Line 76: @@
 - $\underline{x} = (1, \, 0, \, 0, \, 1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$.
 + $\underline{x} = (1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$.
-- $\underline{x} = (1, \, 1, \, 1, \, 1, \, \text{...}\hspace{0.05cm})$ &nbsp; &#8658; &nbsp; "Dauer&ndash;Einsfolge&rdquo;.
+- $\underline{x} = (1, \, 1, \, 1, \, 1, \, \text{...}\hspace{0.05cm})$ &nbsp; &#8658; &nbsp; "Dauer&ndash;Einsfolge".
 {Die Sequenz&nbsp; $\underline{u} = (1, \, 1, \, 1)$&nbsp; sei endlich. Wie lautet die Ausgangssequenz?
@@ Line 83: / Line 83: @@
 + $\underline{x} = (1, \, 0, \, 0, \, 1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$.
 - $\underline{x} = (1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$.
-- $\underline{x} = (1, \, 1, \, 1, \, 1, \, \text{...}\hspace{0.05cm})$ &nbsp; &#8658; &nbsp; "Dauer&ndash;Einsfolge&rdquo;.
+- $\underline{x} = (1, \, 1, \, 1, \, 1, \, \text{...}\hspace{0.05cm})$ &nbsp; &#8658; &nbsp; "Dauer&ndash;Einsfolge".
-{Wie lautet die Ausgangssequenz für&nbsp; $\underline{u} = (1, \, 1, \, 1, \, 1, \, \text{...}\hspace{0.05cm}.)$ &nbsp; &#8658; &nbsp; "Dauer&ndash;Einsfolge&rdquo;?
+{Wie lautet die Ausgangssequenz für&nbsp; $\underline{u} = (1, \, 1, \, 1, \, 1, \, \text{...}\hspace{0.05cm}.)$ &nbsp; &#8658; &nbsp; "Dauer&ndash;Einsfolge"?
 |type="()"}
 + $\underline{x} = (1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$.
 - $\underline{x} = (1, \, 0, \, 0, \, 1, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$.
 - $\underline{x} = (1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 0, \,\text{...}\hspace{0.05cm})$.
-- $\underline{x} = (1, \, 1, \, 1, \, 1, \, \text{...}\hspace{0.05cm})$ &nbsp; &#8658; &nbsp; "Dauer&ndash;Einsfolge&rdquo;.
+- $\underline{x} = (1, \, 1, \, 1, \, 1, \, \text{...}\hspace{0.05cm})$ &nbsp; &#8658; &nbsp; "Dauer&ndash;Einsfolge".
 {Für welchen Vektor&nbsp; $\underline{u}$&nbsp; tritt am Ausgang die Folge&nbsp; $\underline{x} = (1, \, 1, \, 1, \, 1, \ \text{...}\hspace{0.05cm})$&nbsp; auf?
 |type="()"}
-- $\underline{u} = (1, \, 1, \, 1, \, 1, \, \text{...}\hspace{0.05cm})$ &nbsp; &#8658; &nbsp; "Dauer&ndash;Einsfolge&rdquo;
+- $\underline{u} = (1, \, 1, \, 1, \, 1, \, \text{...}\hspace{0.05cm})$ &nbsp; &#8658; &nbsp; "Dauer&ndash;Einsfolge"
 + $\underline{u} = (1, \, 0, \, 1, \, 0, \, 1, \, 0, \,\text{...}\hspace{0.05cm})$ &nbsp; &#8658; &nbsp; alternierende Folge, beginnend mit $1$.
 - $\underline{u} = (0, \, 1, \, 0, \, 1, \, 0, \, 1, \, \text{...}\hspace{0.05cm})$ &nbsp; &#8658; &nbsp; alternierende Folge, beginnend mit $0$.
@@ Line 169: / Line 169: @@
 *Auch bei direkter Anwendung der Faltung wie in Teilaufgabe (2) kann man dieses Ergebnis ablesen.
 *Mit $\underline{u} = (0, \, 1, \, 0, \, 1, \, 0, \, 1, \, \text{...})$ erhält man dagegen $\underline{x} = (0, \, 1, \, 1, \, 1, \, 1, \, 1, \,\text{...})$.
-*Diese unterscheidet sich von der "Dauer&ndash;Einsfolge&rdquo; nur im ersten Bit. Es ist dann $x_1 = 0$ statt $x_1 = 1$.
+*Diese unterscheidet sich von der "Dauer&ndash;Einsfolge" nur im ersten Bit. Es ist dann $x_1 = 0$ statt $x_1 = 1$.
 {{ML-Fuß}}
 [[Category:Channel Coding: Exercises|^3.2 Polynomische Beschreibung^]]

← Older revision	Revision as of 09:04, 12 September 2022
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−	[[Category:~~Aufgaben zu Kanalcodierung~~\|^3.2 Polynomische Beschreibung^]]	+	[[Category:Channel Coding: Exercises\|^3.2 Polynomische Beschreibung^]]

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 Die Nomenklatur für diese (digitale) Filterbeschreibung haben wir an das Buch &bdquo;Einführung in die Kanalcodierung&rdquo; angepasst. In anderen&nbsp; $\rm LNTwww$&ndash;Büchern bezeichnet oft&nbsp; $\underline{x}$&nbsp; den Filtereingang,&nbsp; $\underline{y}$&nbsp; den Filterausgang, und die Impulsantwort wird&nbsp; $h$&nbsp; genannt.
-Allgemein gilt für die Ausgangssequenz entsprechend der&nbsp; [[Signal_Representation/Faltungssatz_und_Faltungsoperation#Faltung_im_Zeitbereich| Faltung]]&nbsp; (englisch: &nbsp;<i>Convolution</i>&nbsp;):
+Allgemein gilt für die Ausgangssequenz entsprechend der&nbsp; [[Signal_Representation/The_Convolution_Theorem_and_Operation#Faltung_im_Zeitbereich| Faltung]]&nbsp; (englisch: &nbsp;<i>Convolution</i>&nbsp;):
 :$$\underline{x} = \underline{u}* \underline{g} = \left (x_0, x_1, \hspace{0.05cm}\text{...}\hspace{0.1cm}, x_i, \hspace{0.05cm}\text{...}\hspace{0.1cm} \right )\hspace{0.05cm},\hspace{0.1cm} {\rm mit} \hspace{0.2cm} x_i = \sum_{l = 0}^{m} g_l \cdot u_{i-l}\hspace{0.05cm}.$$