Exercise 2.6Z: PN Generator of Length 3 - Revision history

Guenter at 17:19, 28 December 2021

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Guenter at 16:18, 28 December 2021

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Guenter at 16:17, 28 December 2021

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Noah at 08:43, 21 December 2021

2021-12-21T08:43:19Z

Noah at 08:42, 21 December 2021

2021-12-21T08:42:24Z

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Guenter at 17:01, 10 December 2021

2021-12-10T17:01:31Z

Guenter at 15:49, 10 December 2021

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Guenter at 14:46, 9 December 2021

2021-12-09T14:46:24Z

Guenter at 10:15, 1 December 2021

2021-12-01T10:15:57Z

Guenter: Guenter moved page Aufgabe 2.6Z: PN-Generator der Länge 3 to Exercise 2.6Z: PN Generator of Length 3

2021-12-01T10:10:47Z

Guenter moved page Aufgabe 2.6Z: PN-Generator der Länge 3 to Exercise 2.6Z: PN Generator of Length 3

@@ Line 63: / Line 63: @@
 ===Solution===
 {{ML-Kopf}}
-[[File:EN_Sto_Z_2_6b.png|right|frame|PN&ndash;generator with octal identifier $15$]]
+[[File:EN_Sto_Z_2_6b.png|right|frame|PN generator with octal identifier&nbsp; $15$]]
 '''(1)'''&nbsp; It is an M-sequence with&nbsp; $L= 3$.&nbsp; It follows that $P= 2^L - 1 \hspace{0.15cm}\underline{= 7}$.
@@ Line 74: / Line 74: @@
-The result is entered in the first row of the above table (marked in red):
+The result is entered in the first row of the above table&nbsp; (marked in red):
 *At the clock time&nbsp; $\nu = 7$&nbsp; results in the same memory usage as at the time&nbsp; $\nu = 0$.
-*From this follows&nbsp; $ {P = 7}$&nbsp; and the sequence is from&nbsp; $\nu = 1$&nbsp; corresponding to <u>solution 3</u> :
+*From this follows&nbsp; $ {P = 7}$&nbsp; and the sequence is from&nbsp; $\nu = 1$&nbsp; corresponding to&nbsp; <u>solution 3</u>:
 :$$\langle z_\nu \rangle = 1 \ 1 \ 0 \ 0 \ 1 \ 1 \ 0 \ 0 \ 1 \ 0 \ 1 \ 1 \text{...}$$
-*In contrast, proposal 1 describes the M sequence of the PN generator with length&nbsp; $L=4$&nbsp; and identifier&nbsp; $(31)$ &nbsp; &rArr; &nbsp; period length is&nbsp; $P= 15$.
+*In contrast,&nbsp; proposal 1 describes the M sequence of the PN generator with length&nbsp; $L=4$&nbsp; and identifier&nbsp; $(31)$ &nbsp; &rArr; &nbsp; period length is&nbsp; $P= 15$.
-*In proposal 2, the period length&nbsp; $P= 4$&nbsp; is too short.
+*In proposal 2,&nbsp; the period length&nbsp; $P= 4$&nbsp; is too short.
-*Finally, the last proposal would have the desired period length&nbsp; $P= 7$, but from the modulo 2 addition of&nbsp; $S_2= 0$&nbsp; and&nbsp; $S_3= 1$&nbsp; $($f&uuml;r&nbsp; $\nu = 0)$&nbsp; it necessarily follows at the next time&nbsp; $(\nu = 1)$&nbsp; : &nbsp; $S_1= 1$. &nbsp; This property is not exhibited by sequence 4.
+*Finally,&nbsp; the last proposal would have the desired period length&nbsp; $P= 7$,&nbsp; but from the modulo 2 addition of&nbsp; $S_2= 0$&nbsp; and&nbsp; $S_3= 1$&nbsp; $($for&nbsp; $\nu = 0)$&nbsp; it necessarily follows at the next time&nbsp; $(\nu = 1)$: &nbsp; $S_1= 1$. &nbsp; This property is not exhibited by sequence 4.
 '''(3)'''&nbsp; Correct are <u>solutions 2, 3, and 4</u>:
-*The maximum number of consecutive ones is&nbsp; $L$&nbsp; (namely if there is a one in all&nbsp; $L$&nbsp; memory cells).
+*The maximum number of consecutive&nbsp; "ones"&nbsp; is&nbsp; $L$&nbsp; (namely if there is a&nbsp; "one"&nbsp; in all&nbsp; $L$&nbsp; memory cells).
-*On the other hand, it is not possible that all memory cells are filled with zeros. Therefore, there is always one more than zeros.
+*On the other hand,&nbsp; it is not possible that all memory cells are filled with&nbsp; "zeros".&nbsp; Therefore,&nbsp; there is always one more&nbsp; "one"&nbsp; than&nbsp; "zeros".
-*The period length of the last sequence is&nbsp; $P = 2$.&nbsp; On the other hand, for an M-sequence&nbsp; $P= 2^L - 1.$&nbsp; For no value of&nbsp; $L$&nbsp; is&nbsp; $P = 2$&nbsp; possible.
+*The period length of the last sequence is&nbsp; $P = 2$.&nbsp; On the other hand,&nbsp; for an M-sequence&nbsp; $P= 2^L - 1.$&nbsp; For no value of&nbsp; $L$:&nbsp; &nbsp; $P = 2$&nbsp; is possible.
-[[File: EN_Sto_Z_2_6d.png|right|frame|PN&ndash;generator with octal identifier&nbsp; $13$]]
+[[File: EN_Sto_Z_2_6d.png|right|frame|PN generator with octal identifier&nbsp; $13$]]
-'''(4)'''&nbsp; In the adjacent table the emergence of the PN&ndash;sequence at the reciprocal polynomial&nbsp; $G_{\rm R}(D)$&nbsp; is entered.
+'''(4)'''&nbsp; In the adjacent table the emergence of the PN sequence at the reciprocal polynomial&nbsp; $G_{\rm R}(D)$&nbsp; is entered.&nbsp; It can be seen that the&nbsp; <u>proposed solution 2</u> applies:
-It can be seen that the <u>proposed solution 2</u> applies:
+*Also for the reciprocal arrangement,&nbsp; the period length&nbsp; $P = 7$&nbsp; must hold,&nbsp; so that proposition 1&nbsp; $($with&nbsp; $P = 15)$&nbsp; is eliminated.
-*Also for the reciprocal arrangement, the period length&nbsp; $P = 7$&nbsp; must hold, so that proposition 1&nbsp; $($with&nbsp; $P = 15)$&nbsp; is eliminated.
+*Proposal 3 is just a version of the output sequence of&nbsp; $(15)$ shifted by two clocks.
-*Proposal 3 is just a version of the initial sequence of&nbsp; $(15)$ shifted by two clocks.
+*In contrast,&nbsp; in the (correct) second proposal,&nbsp; the inverse of ...&nbsp;$ 1 \ 1 \ 0 \ 0 \ 1 \ 0 \ 1$&nbsp;... &ndash; thus the sequence ...&nbsp;$ 1 \ 0 \ 1 \ 0 \ 0 \ 1 \ 1$&nbsp;... &ndash; are included,&nbsp; albeit with a phase shift.

@@ Line 37: / Line 37: @@
-{Determine the output sequence&nbsp; $〈z_ν\rangle$&nbsp; for the time points&nbsp; $1$, ... , $P$.&nbsp; What are the first&nbsp; $15$&nbsp; binary values of the output sequence? <br>Hint:&nbsp;From left to right,&nbsp; label the cells with&nbsp; $S_1$,&nbsp; $S_2$&nbsp; and&nbsp; $S_3$.&nbsp; Output the value&nbsp; $z_ν$ that is currently&nbsp; (at time&nbsp; $\nu$)&nbsp; entered into the memory cell&nbsp; $S_1$.
+{Determine the output sequence&nbsp; $〈z_ν\rangle$&nbsp; for the time points&nbsp; $1$, ... , $P$.&nbsp; What are the first&nbsp; $15$&nbsp; binary values of the output sequence? <br>Hint:&nbsp;From left to right,&nbsp; label the cells with&nbsp; $S_1$,&nbsp; $S_2$&nbsp; and&nbsp; $S_3$.&nbsp; Output the value&nbsp; $z_ν$&nbsp; that is currently&nbsp; (at time&nbsp; $\nu$)&nbsp; entered into the memory cell&nbsp; $S_1$.
 |type="()"}
 - $1\ 0 \ 0 \ 1 \ 1 \ 0 \ 1 \ 0 \ 1 \ 1 \ 1 \ 1 \ 0 \ 0 \ 0$ . . .

@@ Line 14: / Line 14: @@
 has the octal identifier&nbsp; $(1 \ 0 \ 1 \ 1)_{\rm bin} = (13)_{\rm oct}$.
-*At start time, let the three memory cells be preallocated with the binary values&nbsp; $1$,&nbsp; $0$&nbsp; and&nbsp; $1$&nbsp;.
+*At start time,&nbsp; let the three memory cells be preallocated with the binary values&nbsp; $1$,&nbsp; $0$&nbsp; and&nbsp; $1$&nbsp;.
-*Both arrangements generate an M sequence.
+*Both arrangements generate an&nbsp; "M-sequence".
@@ Line 25: / Line 25: @@
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Generation_of_Discrete_Random_Variables|Generation of Discrete Random Variables]].
-*The topic of this chapter is illustrated with examples in the  (German language)  learning video &nbsp; [[Erläuterung_der_PN–Generatoren_an_einem_Beispiel_(Lernvideo)|Erläuterung  der PN-Generatoren an einem Beispiel]] $\Rightarrow$ Explanation of PN generators by example.
+*The topic of this chapter is illustrated with examples in the&nbsp; (German language)&nbsp;  learning video: <br> &nbsp; &nbsp; [[Erläuterung_der_PN–Generatoren_an_einem_Beispiel_(Lernvideo)|"Erläuterung der PN-Generatoren an einem Beispiel"]] &nbsp; $\Rightarrow$ &nbsp; "Explanation of PN generators using an example".
@@ Line 37: / Line 37: @@
-{Determine the output sequence&nbsp; $〈z_ν\rangle$&nbsp; for the time points&nbsp; $1$, ... , $P$.&nbsp; What are the first&nbsp; $15$&nbsp; binary values of the output sequence? <br>''Hint:'' &nbsp;From left to right, label the cells with&nbsp; $S_1$,&nbsp; $S_2$&nbsp; and&nbsp; $S_3$.&nbsp; Output the value&nbsp; $z_ν$ that is currently&nbsp; $\nu$&nbsp; entered into the memory cell&nbsp; $S_1$&nbsp; .
+{Determine the output sequence&nbsp; $〈z_ν\rangle$&nbsp; for the time points&nbsp; $1$, ... , $P$.&nbsp; What are the first&nbsp; $15$&nbsp; binary values of the output sequence? <br>Hint:&nbsp;From left to right,&nbsp; label the cells with&nbsp; $S_1$,&nbsp; $S_2$&nbsp; and&nbsp; $S_3$.&nbsp; Output the value&nbsp; $z_ν$ that is currently&nbsp; (at time&nbsp; $\nu$)&nbsp; entered into the memory cell&nbsp; $S_1$.
 |type="()"}
 - $1\ 0 \ 0 \ 1 \ 1 \ 0 \ 1 \ 0 \ 1 \ 1 \ 1 \ 1 \ 0 \ 0 \ 0$ . . .
@@ Line 44: / Line 44: @@
 - $0 \ 0 \ 1 \ 1 \ 1 \ 0 \ 1 \ 0 \ 0 \ 1 \ 1 \ 1 \ 0 \ 1 \ 0 $. . .
-{Which of the following statements are true for each M sequence?
+{Which of the following statements are true for each M-sequence?
 |type="[]"}
-- The number of zeros and ones is equal.
+- The number of&nbsp; "zeros"&nbsp; and&nbsp; "ones"&nbsp; is equal.
-+ In each period there is one more one than zeros.
++ In each period there is one more&nbsp; "one"&nbsp; than&nbsp; "zeros".
-+ The maximum number of consecutive ones is&nbsp; $L$.
++ The maximum number of consecutive&nbsp; "ones"&nbsp; is&nbsp; $L$.
 + The sequence&nbsp; $1 \ 0 \ 1 \ 0 \ 1 \ 0 $ ... &nbsp; is not possible.
-{Now consider the reciprocal order&nbsp; $(13)$.&nbsp; What are the first&nbsp; $15$&nbsp; binary values of the initial sequence with the same initial assignment here?
+{Now consider the reciprocal order&nbsp; $(13)$.&nbsp; What are the first&nbsp; $15$&nbsp; binary values of the output sequence with the same initial assignment here?
 |type="()"}
 - $0 \ 0 \ 0 \ 1 \ 1 \ 1 \ 1 \ 0 \ 1 \ 0 \ 1 \ 1 \ 0 \ 0 \ 1 $ . . .

← Older revision		Revision as of 08:43, 21 December 2021
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−	===~~Fragebogen~~===	+	===Questions===

	<quiz display=simple>		<quiz display=simple>

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-[[File: P_ID2897__Sto_Z_2_6d.png|right|frame|PN&ndash;Generator mit Oktalkennung&nbsp; $13$]]
+[[File: EN_Sto_Z_2_6d.png|right|frame|PN&ndash;Generator mit Oktalkennung&nbsp; $13$]]
 '''(4)'''&nbsp; In nebenstehender Tabelle ist die Entstehung der PN&ndash;Folge beim reziproken Polynom&nbsp; $G_{\rm R}(D)$&nbsp; eingetragen.
 Man erkennt, dass der  <u>Lösungsvorschlag 2</u> zutrifft:

@@ Line 64: / Line 64: @@
 ===Musterlösung===
 {{ML-Kopf}}
-[[File:P_ID107__Sto_Z_2_6b.png|right|frame|PN&ndash;Generator mit Oktalkennung $15$]]
+[[File:EN_Sto_Z_2_6b.png|right|frame|PN&ndash;Generator mit Oktalkennung $15$]]
 '''(1)'''&nbsp; Es handelt sich um eine M-Sequenz mit&nbsp; $L= 3$.&nbsp; Daraus folgt $P= 2^L - 1 \hspace{0.15cm}\underline{= 7}$.

@@ Line 10: / Line 10: @@
 Das zugeh&ouml;rige reziproke Polynom
-$$G_{\rm R}(D) =  D^{\rm 3}\cdot  ( D^{\rm -3} + D^{\rm -2} +  1)  = D^{\rm 3} + D^{\rm 1} + \rm 1$$
+:$$G_{\rm R}(D) =  D^{\rm 3}\cdot  ( D^{\rm -3} + D^{\rm -2} +  1)  = D^{\rm 3} + D^{\rm 1} + \rm 1$$
 hat die Oktalkennung&nbsp; $(1 \ 0  \ 1  \ 1)_{\rm bin} = (13)_{\rm oct}$.

← Older revision		Revision as of 10:15, 1 December 2021
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−	[[Category:Theory of Stochastic Signals: Exercises\|^2.5 ~~Erzeugung diskreter Zufallsgrößen~~^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^2.5 Generation of Discrete Random Variables^]]

← Older revision	Revision as of 10:10, 1 December 2021
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