Difference between revisions of "Aufgaben:Exercise 3.5: Recursive Filters for GF(2)"

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{{quiz-Header|Buchseite=Kanalcodierung/Algebraische und polynomische Beschreibung}}
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{{quiz-Header|Buchseite=Channel_Coding/Algebraic_and_Polynomial_Description}}
  
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[[File:P_ID2647__KC_A_3_5.png|right|frame|Allgemeines rekursives Filter und betrachtete Realisierung]]
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[[File:P_ID2647__KC_A_3_5.png|right|frame|General recursive filter and considered realization]]
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Die obere der beiden dargestellten Schaltungen zeigt ein rekursives Filter zweiter Ordnung in allgemeiner Form. Mit
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The upper of the two circuits shown shows a second order recursive filter in general form. With
 
:$$A(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}  a_0 + a_1 \cdot D + a_2 \cdot D^2  \hspace{0.05cm},$$
 
:$$A(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}  a_0 + a_1 \cdot D + a_2 \cdot D^2  \hspace{0.05cm},$$
 
:$$B(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}  1 + b_1 \cdot D + b_2 \cdot D^2 $$
 
:$$B(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}  1 + b_1 \cdot D + b_2 \cdot D^2 $$
  
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erhält man für die Übertragungsfunktion
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one obtains for the transfer function
 
:$$G(D) = \frac{A(D)}{B(D)} = \frac{a_0 + a_1 \cdot D + a_2 \cdot D^2}{1 + b_1 \cdot D + b_2 \cdot D^2} \hspace{0.05cm}.$$
 
:$$G(D) = \frac{A(D)}{B(D)} = \frac{a_0 + a_1 \cdot D + a_2 \cdot D^2}{1 + b_1 \cdot D + b_2 \cdot D^2} \hspace{0.05cm}.$$
  
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Zu beachten ist, dass sich alle Rechenoperationen auf  ${\rm GF(2)}$  beziehen. Damit sind auch die Filterkoeffizienten  $a_0, \ a_1, \ a_2, \ b_1$  und  $b_2$  binär $($entweder  $0$  oder  $1)$.
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It should be noted that all arithmetic operations refer to  ${\rm GF(2)}$ . Thus, the filter coefficients  $a_0, \ a_1, \ a_2, \ b_1$  and  $b_2$  are also binary $($entweder  $0$  or  $1)$.
  
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Die untere Grafik zeigt das für die vorliegende Aufgabe spezifische Filter:  
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The bottom graph shows the filter specific to the exercise at hand:  
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*Ein Filterkoeffizient ergibt sich zu  $a_i = 1$, wenn die Verbindung durchgeschaltet ist  $(0 ≤ i ≤ 2)$.  
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*A filter coefficient results in  $a_i = 1$ if the connection is through  $(0 ≤ i ≤ 2)$.  
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*Andernfalls ist  $a_i = 0$. Die gleiche Systematik gilt für die Koeffizienten  $b_1$  und  $b_2$.  
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*Otherwise,  $a_i = 0$. The same system applies to the coefficients  $b_1$  and  $b_2$.  
  
  
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In den Teilaufgaben '''(1)''', ... , '''(3)''' sollen Sie für verschiedene Eingangssequenzen
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In the subtasks '''(1)''', ... , '''(3)''' you are to determine the respective output sequence  $\underline{x}$  for different input sequences
 
* $\underline{u} = (1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$,
 
* $\underline{u} = (1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$,
 
* $\underline{u} = (0, \, 1, \, 0, \, 1, \, 0, \, 0, \, 1, \, 1, \, \text{...}\hspace{0.05cm})$,
 
* $\underline{u} = (0, \, 1, \, 0, \, 1, \, 0, \, 0, \, 1, \, 1, \, \text{...}\hspace{0.05cm})$,
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die jeweilige Ausgangssequenz  $\underline{x}$  anhand der vorgegebenen Schaltung ermitteln. Es ist zu berücksichtigen:  
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using the given circuit. It should be taken into account:  
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* Besteht die Eingangssequenz  $\underline{u}$  aus einer Eins gefolgt von lauter Nullen, so bezeichnet man diese spezifische Ausgangssequenz  $\underline{x}$  als die  [[Linear_and_Time_Invariant_Systems/Systembeschreibung_im_Zeitbereich#Impulsantwort|Impulsantwort]]  $\underline{g}$, und es gilt:
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* If the input sequence  $\underline{u}$  consists of a one followed by all zeros, this specific output sequence  $\underline{x}$  is called the  [[Linear_and_Time_Invariant_Systems/System_Description_in_Time_Domain#Impulse_response|"Impulse Response"]]  $\underline{g}$, and it holds:
 
:$$\underline{g} \hspace{0.15cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.15cm}{G}(D)\hspace{0.05cm}. $$
 
:$$\underline{g} \hspace{0.15cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.15cm}{G}(D)\hspace{0.05cm}. $$
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* Andernfalls ergibt sich die Ausgangssequenz als das  [[Signal_Representation/The_Convolution_Theorem_and_Operation| Faltungsprodukt]]  zwischen Eingangssequenz und Impulsantwort:
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* Otherwise, the output sequence results as the  [[Signal_Representation/The_Convolution_Theorem_and_Operation|"Convolution Product"]]  between input sequence and impulse response:
 
:$$\underline{x} = \underline{u} * \underline{g} \hspace{0.05cm}.$$
 
:$$\underline{x} = \underline{u} * \underline{g} \hspace{0.05cm}.$$
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* Die Faltungsoperation lässt sich mit dem Umweg über die  [[Channel_Coding/Algebraische_und_polynomische_Beschreibung#GF.282.29.E2.80.93Beschreibungsformen_eines_Digitalen_Filters| $D$–Transformation]]  umgehen.  
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* The convolution operation can be bypassed by  [[Channel_Coding/Algebraic_and_Polynomial_Description#GF.282.29_description_forms_of_a_digital_filter| "$D$ transform"]] .  
  
  
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''Hinweise:''
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Hints:
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* Die Aufgabe gehört zum  Kapitel  [[Channel_Coding/Algebraische_und_polynomische_Beschreibung| Algebraische und polynomische Beschreibung]].
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* This exercise belongs to the chapter  [[Channel_Coding/Algebraic_and_Polynomial_Description| "Algebraic and Polynomial Description"]].
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* Bezug genommen wird insbesondere auf die Seite  [[Channel_Coding/Algebraische_und_polynomische_Beschreibung#Filterstruktur_bei_gebrochen.E2.80.93rationaler_.C3.9Cbertragungsfunktion| Filterstruktur bei gebrochen–rationaler Übertragungsfunktion]]
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* Reference is made in particular to the page  [[Channel_Coding/Algebraic_and_Polynomial_Description#Filter_structure_with_fractional.E2.80.93rational_transfer_function| "Filter structure with fractional–rational transfer function"]]
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===Fragebogen===
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===Solution===
 
<quiz display=simple>
 
<quiz display=simple>
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{Welche Aussagen gelten für die Impulsantwort&nbsp; $\underline{g}$&nbsp; des rekursiven Filters?
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{What statements hold for the impulse response&nbsp; $\underline{g}$&nbsp; of the recursive filter?
 
|type="[]"}
 
|type="[]"}
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- Es gilt&nbsp; $\underline{g} = (0, \, 1, \, 1, \, 0, \, 1, \, 0, \, 0, \, 1, \, \text{...}\hspace{0.05cm})$.
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- It holds&nbsp; $\underline{g} = (0, \, 1, \, 1, \, 0, \, 1, \, 0, \, 0, \, 1, \, \text{...}\hspace{0.05cm})$.
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+ Es gilt&nbsp; $\underline{g} = (1, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, \text{...}\hspace{0.05cm})$.
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+ It holds&nbsp; $\underline{g} = (1, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, \text{...}\hspace{0.05cm})$.
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+ Die Impulsantwort&nbsp; $\underline{g}$&nbsp; ist unendlich weit ausgedehnt.
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+ The impulse response&nbsp; $\underline{g}$&nbsp; is infinitely extended.
  
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{Es sei nun&nbsp; $\underline{u} = (0, \, 1, \, 0, \, 1, \, 0, \, 0, \, 1, \, 1)$. Welche Aussagen treffen zu?
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{Let now&nbsp; $\underline{u} = (0, \, 1, \, 0, \, 1, \, 0, \, 0, \, 1, \, 1)$. Which statements are true?
 
|type="[]"}
 
|type="[]"}
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+ Die Ausgangssequenz lautet:&nbsp; $\underline{x} = (0, \, 1, \, 1, \, 0, \, 1, \, 0, \, 0, \, 1, \, \text{...}\hspace{0.05cm})$.
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+ The output sequence is:&nbsp; $\underline{x} = (0, \, 1, \, 1, \, 0, \, 1, \, 0, \, 0, \, 1, \, \text{...}\hspace{0.05cm})$.
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- Die Ausgangssequenz lautet:&nbsp; $\underline{x} = (1, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, \text{...}\hspace{0.05cm})$.
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- The output sequence is:&nbsp; $\underline{x} = (1, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, \text{...}\hspace{0.05cm})$.
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+ Die Ausgangssequenz&nbsp; $\underline{x}$&nbsp; reicht bis ins Unendliche.
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+ The output sequence&nbsp; $\underline{x}$&nbsp; extends to infinity.
  
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{Nun gelte&nbsp; $\underline{u} = (1, \, 1, \, 1)$. Welche Aussagen treffen zu?
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{Now apply&nbsp; $\underline{u} = (1, \, 1, \, 1)$. Which statements are true?
 
|type="[]"}
 
|type="[]"}
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+ Die Ausgangssequenz&nbsp; $\underline{x}$&nbsp; beginnt mit&nbsp; $(1, \, 0, \, 1)$.
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+ The output sequence&nbsp; $\underline{x}$&nbsp; starts with&nbsp; $(1, \, 0, \, 1)$.
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- Die Ausgangssequenz&nbsp; $\underline{x}$&nbsp; beginnt mit&nbsp; $(1, \, 1, \, 1)$.
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- The output sequence&nbsp; $\underline{x}$&nbsp; begins with&nbsp; $(1, \, 1, \, 1)$.
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- Die Ausgangssequenz&nbsp; $\underline{x}$&nbsp; reicht bis ins Unendliche.
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- The output sequence&nbsp; $\underline{x}$&nbsp; extends to infinity.
  
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{Welche Aussagen gelten für die Übertragungsfunktion&nbsp; $G(D)$?
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{What statements are valid for the transfer function&nbsp; $G(D)$?
 
|type="[]"}
 
|type="[]"}
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+ Es gilt&nbsp; $G(D) = (1 + D^2)/(1 + D + D^2)$.
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+ It holds&nbsp; $G(D) = (1 + D^2)/(1 + D + D^2)$.
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- Es gilt&nbsp; $G(D) = (1 + D + D^2)/(1 + D^2)$.
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- It holds&nbsp; $G(D) = (1 + D + D^2)/(1 + D^2)$.
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+ Es gilt&nbsp; $G(D) = 1 + D + D^2 + D^4 + D^5 + D^7 + D^8 + \hspace{0.05cm} \text{...}\hspace{0.05cm} $ .
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+ It holds&nbsp; $G(D) = 1 + D + D^2 + D^4 + D^5 + D^7 + D^8 + \hspace{0.05cm} \text{...}\hspace{0.05cm} $ .
 
</quiz>
 
</quiz>
  
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===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
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'''(1)'''&nbsp; Richtig sind die  <u>Lösungsvorschläge 2 und  3</u>:
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'''(1)'''&nbsp; Correct are the <u>proposed solutions 2 and 3</u>:
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[[File:P_ID2643__KC_A_3_5a.png|right|frame|Zur Berechnung der Impulsantwort $\underline{g}$]]
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[[File:P_ID2643__KC_A_3_5a.png|right|frame|To calculate the impulse response $\underline{g}$]]
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*Die Impulsantwort $\underline{g}$ ist gleich der Sequenz $\underline{x}$ für die Eingangssequenz $\underline{u} = (1, \, 0, \, 0, \, \text{...})$. Anhand der Filterstruktur ergibt sich mit $w_0 = w_{-1} = 0$ und den Gleichungen
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*The impulse response $\underline{g}$ is equal to the sequence $\underline{x}$ for the input sequence $\underline{u} = (1, \, 0, \, 0, \, \text{...})$. Based on the filter structure, $w_0 = w_{-1} = 0$ and the equations
 
:$$w_i \hspace{-0.2cm} \ = \ \hspace{-0.2cm} u_i + w_{i-1} + w_{i-2}  \hspace{0.05cm},$$
 
:$$w_i \hspace{-0.2cm} \ = \ \hspace{-0.2cm} u_i + w_{i-1} + w_{i-2}  \hspace{0.05cm},$$
 
:$$x_i \hspace{-0.2cm} \ = \ \hspace{-0.2cm} w_i + w_{i-2} $$
 
:$$x_i \hspace{-0.2cm} \ = \ \hspace{-0.2cm} w_i + w_{i-2} $$
  
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:das Ergebnis $\underline{g} = \underline{x} = (1, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, \text{...})$ entsprechend <u>Lösungsvorschlag 2</u>, wie nebenstehende Berechnung zeigt.
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:the result $\underline{g} = \underline{x} = (1, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, \text{...})$ corresponding to <u>proposed solution 2</u>, as shown in the adjacent calculation.
  
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*Richtig ist aber zusätzlich auch noch der <u>Lösungsvorschlag 3</u>, da man aus diesem Berechnungsschema weiter folgende Periodizitäten der Impulsantwort $\underline{g}$ (bis ins Unendliche) wegen jeweils gleicher Registerbelegung erkennt:
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*But additionally the <u>proposed solution 3</u> is correct, because one recognizes from this calculation scheme further following periodicities of the impulse response $\underline{g}$ (up to infinity) because of in each case same register assignment:
 
   
 
   
 
:$$g_3 \hspace{-0.2cm} \ = \ \hspace{-0.2cm} g_6 = g_9 = \hspace{0.05cm}\text{...} \hspace{0.05cm}= 1 \hspace{0.05cm},$$
 
:$$g_3 \hspace{-0.2cm} \ = \ \hspace{-0.2cm} g_6 = g_9 = \hspace{0.05cm}\text{...} \hspace{0.05cm}= 1 \hspace{0.05cm},$$
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<br><br><br><br><br>
 
<br><br><br><br><br>
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'''(2)'''&nbsp;Nach ähnlichen Berechnungen wie in Teilaufgabe (1) erkennt man die Richtigkeit der <u>Lösungsvorschläge 1 und 3</u>:
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'''(2)'''&nbsp;After similar calculations as in subtask (1) one recognizes the correctness of <u>solutions 1 and 3</u>:
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*Auch die Ausgangssequenz $\underline{x}$ reicht bis ins Unendliche.
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*The initial sequence $\underline{x}$ also extends to infinity.
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*Es zeigen sich auch wieder Periodizitäten.
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*Periodicities show up again.
  
  
  
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Zum gleichen Ergebnis gelangt man, wenn man die um eine, drei, sechs bzw. sieben Positionen (nach rechts) verschobenen Impulsantworten $\underline{g} = (1, \, 0, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, \text{...} \hspace{0.05cm})$ im Galoisfeld ${\rm GF(2)}$ addiert:
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The same result is obtained by adding the impulse responses $\underline{g} = (1, \, 0, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, \text{...} \hspace{0.05cm})$ n the Galois field ${\rm GF(2)}$ shifted by one, three, six, and seven positions (to the right, respectively):
 
:$$\underline{x} = (0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}  \text{...}\hspace{0.05cm}) \hspace{0.05cm} + \hspace{0.05cm} (0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}  \text{...}\hspace{0.05cm})  \hspace{0.05cm} + \hspace{0.05cm}  (0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}  \text{...}\hspace{0.05cm})  \hspace{0.05cm} + \hspace{0.05cm} (0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}  \text{...}\hspace{0.05cm}) $$
 
:$$\underline{x} = (0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}  \text{...}\hspace{0.05cm}) \hspace{0.05cm} + \hspace{0.05cm} (0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}  \text{...}\hspace{0.05cm})  \hspace{0.05cm} + \hspace{0.05cm}  (0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}  \text{...}\hspace{0.05cm})  \hspace{0.05cm} + \hspace{0.05cm} (0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}  \text{...}\hspace{0.05cm}) $$
 
:$$\Rightarrow \hspace{0.3cm}\underline{x} = (0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}    \text{...}\hspace{0.05cm})  \hspace{0.05cm}. $$
 
:$$\Rightarrow \hspace{0.3cm}\underline{x} = (0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}    \text{...}\hspace{0.05cm})  \hspace{0.05cm}. $$
  
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Aufgrund der Linearität des betrachteten Systems ist dies erlaubt.
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Due to the linearity of the system under consideration, this is allowed.
  
  
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'''(3)'''&nbsp; Hier wählen wir den Weg über die $D$&ndash;Transformierten:
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'''(3)'''&nbsp; Here we choose the way over the $D$ transforms:
 
:$$\underline{u}= (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quad
 
:$$\underline{u}= (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quad
 
U(D) =  1+ D + D^2 \hspace{0.05cm}.$$
 
U(D) =  1+ D + D^2 \hspace{0.05cm}.$$
  
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*Mit der Übertragungsfunktion $G(D) = (1 + D^2)/(1 + D + D^2)$ erhält man somit für die $D$&ndash;Transformierte der Ausgangssequenz:
+
*With the transfer function $G(D) = (1 + D^2)/(1 + D + D^2)$ one thus obtains for the $D$ transform of the output sequence:
 
:$$X(D) = {U(D)} \cdot G(D) = {1+D+D^2} \cdot \frac{1+D^2}{1+D+D^2} = 1+D^2 \hspace{0.05cm}\hspace{0.3cm}
 
:$$X(D) = {U(D)} \cdot G(D) = {1+D+D^2} \cdot \frac{1+D^2}{1+D+D^2} = 1+D^2 \hspace{0.05cm}\hspace{0.3cm}
 
\Rightarrow \hspace{0.3cm}\underline{x} = (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} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.01cm}  \text{...}\hspace{0.05cm} \hspace{0.05cm} ) \hspace{0.05cm}.$$
 
\Rightarrow \hspace{0.3cm}\underline{x} = (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} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.01cm}  \text{...}\hspace{0.05cm} \hspace{0.05cm} ) \hspace{0.05cm}.$$
  
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*Richtig ist hier nur der <u>Lösungsvorschlag 1</u>: Trotz unendlich langer Impulsantwort $\underline{g}$ ist bei dieser Eingangssequenz $\underline{u}$ die Ausgangssequenz $\underline{x}$ auf drei Bit begrenzt.  
+
*Only the <u>proposed solution 1</u> is correct here: despite infinitely long impulse response $\underline{g}$, for this input sequence $\underline{u}$ the output sequence $\underline{x}$ is limited to three bits.  
  
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*Zum gleichen Ergebnis kommt man wieder durch Addition verschobener Impulsantworten:
+
*The same result is again obtained by adding shifted impulse responses:
 
:$$\underline{x} = (1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}  \text{...}\hspace{0.05cm}) + (0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}  \text{...}\hspace{0.05cm}) +  (0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}    \text{...}\hspace{0.05cm}) =  (1\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm}    \text{...}\hspace{0.05cm})  \hspace{0.05cm}. $$
 
:$$\underline{x} = (1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}  \text{...}\hspace{0.05cm}) + (0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}  \text{...}\hspace{0.05cm}) +  (0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}    \text{...}\hspace{0.05cm}) =  (1\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm}    \text{...}\hspace{0.05cm})  \hspace{0.05cm}. $$
  
  
−
[[File:P_ID2645__KC_A_3_5e.png|right|frame|$\rm GF(2)$–Polynomdivision $(1 + D^2)/(1 + D + D^2)$]]  
+
[[File:P_ID2645__KC_A_3_5e.png|right|frame|$\rm GF(2)$ polynomial division $(1 + D^2)/(1 + D + D^2)$]]  
−
'''(4)'''&nbsp; Richtig sind die  <u>Lösungsvorschläge 1 und  3</u>:
+
'''(4)'''&nbsp; Correct are the <u>proposed solutions 1 and 3</u>:
−
*Auf dem Angabenblatt ist die allgemeine Übertragungsfunktion eines rekursiven Filters zweiter Ordnung wie folgt gegeben.  
+
*On the data sheet, the general transfer function of a second&ndash;order recursive filter is given as follows.  
 
:$$G(D) =  \frac{a_0 + a_1 \cdot D + a_2 \cdot D^2}{1 + b_1 \cdot D + b_2 \cdot D^2} \hspace{0.05cm}.$$
 
:$$G(D) =  \frac{a_0 + a_1 \cdot D + a_2 \cdot D^2}{1 + b_1 \cdot D + b_2 \cdot D^2} \hspace{0.05cm}.$$
  
 
*Das hier betrachtete Filter ist durch die Koeffizienten $a_0 = a_2 = b_1 = b_2 = 1$ und $a_1 = 0$ bestimmt. Somit erhält man das Ergebnis entsprechend dem <u>Lösungsvorschlag 1</u>:
 
*Das hier betrachtete Filter ist durch die Koeffizienten $a_0 = a_2 = b_1 = b_2 = 1$ und $a_1 = 0$ bestimmt. Somit erhält man das Ergebnis entsprechend dem <u>Lösungsvorschlag 1</u>:
 
:$$G(D) =  \frac{1 +  D^2}{1 +  D + D^2} \hspace{0.05cm}. $$
 
:$$G(D) =  \frac{1 +  D^2}{1 +  D + D^2} \hspace{0.05cm}. $$
−
*Gleichzeitig ist aber $G(D)$ auch die $D$&ndash;Transformierte der Impulsantwort:
+
*At the same time $G(D)$ is also the $D$ transformed of the impulse response:
 
:$$\underline{g}= (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} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}0 ,\hspace{0.05cm} \text{ ...}\hspace{0.05cm}) \hspace{0.15cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.15cm}
 
:$$\underline{g}= (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} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}0 ,\hspace{0.05cm} \text{ ...}\hspace{0.05cm}) \hspace{0.15cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.15cm}
 
{G}(D)$$
 
{G}(D)$$
Line 126: Line 126:
 
{G}(D)= 1 + D + D^2 + D^4+ D^5  +\text{...}  
 
{G}(D)= 1 + D + D^2 + D^4+ D^5  +\text{...}  
 
\hspace{0.1cm}. $$
 
\hspace{0.1cm}. $$
−
*Das bedeutet: Richtig ist auch der <u>Lösungsvorschlag 3</u>.  
+
*This means: The <u>proposed solution 3</u> is also correct.  
−
*Zum gleichen Ergebnis wäre man durch Division der beiden Polynome $1 + D^2$ und $1 + D + D^2$ gekommen, wie die nebenstehende Rechnung zeigt.
+
*The same result would have been obtained by dividing the two polynomials $1 + D^2$ and $1 + D + D^2$, as the calculation opposite shows.
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  
  
 
[[Category:Channel Coding: Exercises|^3.2 Polynomial Description^]]
 
[[Category:Channel Coding: Exercises|^3.2 Polynomial Description^]]

Revision as of 21:40, 26 September 2022

General recursive filter and considered realization

The upper of the two circuits shown shows a second order recursive filter in general form. With

$$A(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} a_0 + a_1 \cdot D + a_2 \cdot D^2 \hspace{0.05cm},$$
$$B(D) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1 + b_1 \cdot D + b_2 \cdot D^2 $$

one obtains for the transfer function

$$G(D) = \frac{A(D)}{B(D)} = \frac{a_0 + a_1 \cdot D + a_2 \cdot D^2}{1 + b_1 \cdot D + b_2 \cdot D^2} \hspace{0.05cm}.$$

It should be noted that all arithmetic operations refer to  ${\rm GF(2)}$ . Thus, the filter coefficients  $a_0, \ a_1, \ a_2, \ b_1$  and  $b_2$  are also binary $($entweder  $0$  or  $1)$.

The bottom graph shows the filter specific to the exercise at hand:

  • A filter coefficient results in  $a_i = 1$ if the connection is through  $(0 ≤ i ≤ 2)$.
  • Otherwise,  $a_i = 0$. The same system applies to the coefficients  $b_1$  and  $b_2$.


In the subtasks (1), ... , (3) you are to determine the respective output sequence  $\underline{x}$  for different input sequences

  • $\underline{u} = (1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$,
  • $\underline{u} = (0, \, 1, \, 0, \, 1, \, 0, \, 0, \, 1, \, 1, \, \text{...}\hspace{0.05cm})$,
  • $\underline{u} = (1, \, 1, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, \text{...}\hspace{0.05cm})$


using the given circuit. It should be taken into account:

  • If the input sequence  $\underline{u}$  consists of a one followed by all zeros, this specific output sequence  $\underline{x}$  is called the  "Impulse Response"  $\underline{g}$, and it holds:
$$\underline{g} \hspace{0.15cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.15cm}{G}(D)\hspace{0.05cm}. $$
  • Otherwise, the output sequence results as the  "Convolution Product"  between input sequence and impulse response:
$$\underline{x} = \underline{u} * \underline{g} \hspace{0.05cm}.$$




Hints:


Solution

1

What statements hold for the impulse response  $\underline{g}$  of the recursive filter?

It holds  $\underline{g} = (0, \, 1, \, 1, \, 0, \, 1, \, 0, \, 0, \, 1, \, \text{...}\hspace{0.05cm})$.
It holds  $\underline{g} = (1, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, \text{...}\hspace{0.05cm})$.
The impulse response  $\underline{g}$  is infinitely extended.

2

Let now  $\underline{u} = (0, \, 1, \, 0, \, 1, \, 0, \, 0, \, 1, \, 1)$. Which statements are true?

The output sequence is:  $\underline{x} = (0, \, 1, \, 1, \, 0, \, 1, \, 0, \, 0, \, 1, \, \text{...}\hspace{0.05cm})$.
The output sequence is:  $\underline{x} = (1, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, \text{...}\hspace{0.05cm})$.
The output sequence  $\underline{x}$  extends to infinity.

3

Now apply  $\underline{u} = (1, \, 1, \, 1)$. Which statements are true?

The output sequence  $\underline{x}$  starts with  $(1, \, 0, \, 1)$.
The output sequence  $\underline{x}$  begins with  $(1, \, 1, \, 1)$.
The output sequence  $\underline{x}$  extends to infinity.

4

What statements are valid for the transfer function  $G(D)$?

It holds  $G(D) = (1 + D^2)/(1 + D + D^2)$.
It holds  $G(D) = (1 + D + D^2)/(1 + D^2)$.
It holds  $G(D) = 1 + D + D^2 + D^4 + D^5 + D^7 + D^8 + \hspace{0.05cm} \text{...}\hspace{0.05cm} $ .


Solution

Solution

(1)  Correct are the proposed solutions 2 and 3:

To calculate the impulse response $\underline{g}$
  • The impulse response $\underline{g}$ is equal to the sequence $\underline{x}$ for the input sequence $\underline{u} = (1, \, 0, \, 0, \, \text{...})$. Based on the filter structure, $w_0 = w_{-1} = 0$ and the equations
$$w_i \hspace{-0.2cm} \ = \ \hspace{-0.2cm} u_i + w_{i-1} + w_{i-2} \hspace{0.05cm},$$
$$x_i \hspace{-0.2cm} \ = \ \hspace{-0.2cm} w_i + w_{i-2} $$
the result $\underline{g} = \underline{x} = (1, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, \text{...})$ corresponding to proposed solution 2, as shown in the adjacent calculation.
  • But additionally the proposed solution 3 is correct, because one recognizes from this calculation scheme further following periodicities of the impulse response $\underline{g}$ (up to infinity) because of in each case same register assignment:
$$g_3 \hspace{-0.2cm} \ = \ \hspace{-0.2cm} g_6 = g_9 = \hspace{0.05cm}\text{...} \hspace{0.05cm}= 1 \hspace{0.05cm},$$
$$g_4 \hspace{-0.2cm} \ = \ \hspace{-0.2cm} g_7 = g_{10} =\hspace{0.05cm} \text{...} \hspace{0.05cm}= 0 \hspace{0.05cm},$$
Zur Berechnung der Ausgangssequenz $\underline{x}$
$$g_5 \hspace{-0.2cm} \ = \ \hspace{-0.2cm} g_8 = g_{11} =\hspace{0.05cm} \text{...} \hspace{0.05cm}= 1 \hspace{0.05cm}.$$






(2) After similar calculations as in subtask (1) one recognizes the correctness of solutions 1 and 3:

  • The initial sequence $\underline{x}$ also extends to infinity.
  • Periodicities show up again.


The same result is obtained by adding the impulse responses $\underline{g} = (1, \, 0, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, \text{...} \hspace{0.05cm})$ n the Galois field ${\rm GF(2)}$ shifted by one, three, six, and seven positions (to the right, respectively):

$$\underline{x} = (0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.05cm}) \hspace{0.05cm} + \hspace{0.05cm} (0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.05cm}) \hspace{0.05cm} + \hspace{0.05cm} (0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.05cm}) \hspace{0.05cm} + \hspace{0.05cm} (0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.05cm}) $$
$$\Rightarrow \hspace{0.3cm}\underline{x} = (0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.05cm}) \hspace{0.05cm}. $$

Due to the linearity of the system under consideration, this is allowed.


(3)  Here we choose the way over the $D$ transforms:

$$\underline{u}= (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quad U(D) = 1+ D + D^2 \hspace{0.05cm}.$$
  • With the transfer function $G(D) = (1 + D^2)/(1 + D + D^2)$ one thus obtains for the $D$ transform of the output sequence:
$$X(D) = {U(D)} \cdot G(D) = {1+D+D^2} \cdot \frac{1+D^2}{1+D+D^2} = 1+D^2 \hspace{0.05cm}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}\underline{x} = (1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.01cm} \text{...}\hspace{0.05cm} \hspace{0.05cm} ) \hspace{0.05cm}.$$
  • Only the proposed solution 1 is correct here: despite infinitely long impulse response $\underline{g}$, for this input sequence $\underline{u}$ the output sequence $\underline{x}$ is limited to three bits.
  • The same result is again obtained by adding shifted impulse responses:
$$\underline{x} = (1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.05cm}) + (0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.05cm}) + (0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.05cm}) = (1\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} \text{...}\hspace{0.05cm}) \hspace{0.05cm}. $$


$\rm GF(2)$ polynomial division $(1 + D^2)/(1 + D + D^2)$

(4)  Correct are the proposed solutions 1 and 3:

  • On the data sheet, the general transfer function of a second–order recursive filter is given as follows.
$$G(D) = \frac{a_0 + a_1 \cdot D + a_2 \cdot D^2}{1 + b_1 \cdot D + b_2 \cdot D^2} \hspace{0.05cm}.$$
  • Das hier betrachtete Filter ist durch die Koeffizienten $a_0 = a_2 = b_1 = b_2 = 1$ und $a_1 = 0$ bestimmt. Somit erhält man das Ergebnis entsprechend dem Lösungsvorschlag 1:
$$G(D) = \frac{1 + D^2}{1 + D + D^2} \hspace{0.05cm}. $$
  • At the same time $G(D)$ is also the $D$ transformed of the impulse response:
$$\underline{g}= (1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0 \hspace{0.05cm}\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}0 ,\hspace{0.05cm} \text{ ...}\hspace{0.05cm}) \hspace{0.15cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\hspace{0.15cm} {G}(D)$$
$$\Rightarrow\hspace{0.3cm} {G}(D)= 1 + D + D^2 + D^4+ D^5 +\text{...} \hspace{0.1cm}. $$
  • This means: The proposed solution 3 is also correct.
  • The same result would have been obtained by dividing the two polynomials $1 + D^2$ and $1 + D + D^2$, as the calculation opposite shows.

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