Aufgaben:Exercise 2.3: Algebraic Sum of Binary Numbers - Revision history

Guenter at 13:58, 16 February 2022

2022-02-16T13:58:36Z

Guenter at 14:00, 18 December 2021

2021-12-18T14:00:14Z

Guenter at 12:43, 11 December 2021

2021-12-11T12:43:30Z

Guenter at 12:37, 11 December 2021

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Guenter at 11:48, 11 December 2021

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Guenter at 15:47, 10 December 2021

2021-12-10T15:47:57Z

Noah at 12:06, 10 December 2021

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Noah at 12:03, 10 December 2021

2021-12-10T12:03:13Z

Guenter at 16:43, 30 November 2021

2021-11-30T16:43:18Z

Guenter: Guenter moved page Aufgaben:Aufgabe 2.3: Summe von Binärzahlen to Aufgaben:Exercise 2.3: Algebraic Sum of Binary Numbers

2021-11-30T16:40:48Z

Guenter moved page Aufgabe 2.3: Summe von Binärzahlen to Exercise 2.3: Algebraic Sum of Binary Numbers

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-{Find the rms value of the random variable&nbsp; $y$.
+{Find the standard deviation of the random variable&nbsp; $y$.
 |type="{}"}
 $\sigma_y \ = \ $ { 1.061 3% }
@@ Line 83: / Line 83: @@
-'''(4)'''&nbsp; Accordingly,&nbsp; for the rms value of the binomial distribution:
+'''(4)'''&nbsp; Accordingly,&nbsp; for the standard deviation of the binomial distribution:
 :$$\sigma_y=\sqrt{ I \cdot p \cdot( 1- p)} \hspace{0.15cm} \underline{= \rm 1.061}.$$

← Older revision		Revision as of 14:00, 18 December 2021
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	*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Binomial_Distribution\|Binomial Distribution]].		*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Binomial_Distribution\|Binomial Distribution]].

−	*To check your results you can use the interactive applet  [[Applets:Binomial_and_Poisson_Distribution_(Applet)\|Binomial and Poisson distribution]]~~ ~~.	+	*To check your results you can use the interactive HTML5/JavaScript applet  [[Applets:Binomial_and_Poisson_Distribution_(Applet)\|Binomial and Poisson distribution]].

@@ Line 50: / Line 50: @@
 {Are the random numbers&nbsp; $y_\nu$&nbsp; statistically independent?&nbsp; Justify your result.
-|type="[]"}
+|type="()"}
 - The random numbers are statistically independent.
 + The random numbers are statistically dependent.

@@ Line 13: / Line 13: @@
 At each clock instant,&nbsp; the contents of this shift register are shifted one place to the right and the algebraic sum&nbsp; $y_\nu$&nbsp; of the shift register contents is formed in each case:
 :$$y_{\nu}=\sum\limits_{i=0}^{5}x_{\nu-i}=x_{\nu}+x_{\nu-1}+\ \text{...} \ +x_{\nu-5}.$$
@@ Line 67: / Line 64: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Each cell can contain a&nbsp; $0$&nbsp; or a&nbsp; $1$&nbsp;.&nbsp; Therefore, the sum can take all integer values between&nbsp; $0$&nbsp; ánd&nbsp; $6$&nbsp;:
+'''(1)'''&nbsp; Each cell can contain a&nbsp; $0$&nbsp; or a&nbsp; $1$&nbsp;.&nbsp; Therefore,&nbsp; the sum can take all integer values between&nbsp; $0$&nbsp; ánd&nbsp; $6$&nbsp;:
 :$$y_{\nu}\in\{0,1,\ \text{...} \ ,6\}\hspace{0.3cm}\Rightarrow\hspace{0.3cm}
 y_{\rm max} \hspace{0.15cm} \underline{= 6}.$$
@@ Line 73: / Line 70: @@
-'''(2)'''&nbsp; There is a binomial distribution.&nbsp; Therefore, with&nbsp; $p = 0.25$:
+'''(2)'''&nbsp; There is a binomial distribution.&nbsp; Therefore,&nbsp; with&nbsp; $p = 0.25$:
 :$${\rm Pr}(y =0)=(1-p)^{\it I}=0.75^6=0.178,$$
 :$${\rm Pr}(y=1)=\left({ I \atop {1}}\right)\cdot (1-p)^{I-1}\cdot p= \rm 6\cdot 0.75^5\cdot 0.25=0.356,$$
@@ Line 81: / Line 78: @@
-'''(3)'''&nbsp; According to the general equation, the mean of the binomial distribution is:
+'''(3)'''&nbsp; According to the general equation,&nbsp; the mean of the binomial distribution is:
 :$$m_y= I\cdot p\hspace{0.15cm} \underline{=\rm 1.5}.$$
-'''(4)'''&nbsp; Accordingly, for the dispersion of the binomial distribution:
+'''(4)'''&nbsp; Accordingly,&nbsp; for the rms value of the binomial distribution:
 :$$\sigma_y=\sqrt{ I \cdot p \cdot( 1- p)} \hspace{0.15cm} \underline{= \rm 1.061}.$$
-'''(5)'''&nbsp; Correct is the <u>proposed solution 2</u>:
+'''(5)'''&nbsp; Correct is the&nbsp; <u>proposed solution 2</u>:
-*If &nbsp; $y_\nu = 0$, then only the values&nbsp; $0$&nbsp; and&nbsp; $1$&nbsp; can follow at the next time point, but not&nbsp; $2$, ... , $6$.
+*If &nbsp; $y_\nu = 0$, then only the values&nbsp; $0$&nbsp; and&nbsp; $1$&nbsp; can follow at the next time point,&nbsp; but not&nbsp; $2$, ... , $6$.
 *That is: &nbsp; The sequence&nbsp; $ \langle y_\nu \rangle$&nbsp; has (strong) statistical bindings.
-'''(6)'''&nbsp; The probability we are looking for is identical to the probability that the new binary symbol is equal to the symbol dropped out of the shift register. It follows that:
+'''(6)'''&nbsp; The probability we are looking for is identical to the probability that the new binary symbol is equal to the symbol dropped out of the shift register.&nbsp; It follows that:
 :$${\rm Pr} (y_{\nu} = \mu\hspace{0.05cm}| \hspace{0.05cm} y_{\nu-{1}} = \mu) = {\rm Pr}(x_{\nu}= x_{\nu-6}). $$
-*Since the symbols&nbsp; $x_\nu$&nbsp; are statistically independent of each other, we can also write for this:
+*Since the symbols&nbsp; $x_\nu$&nbsp; are statistically independent of each other,&nbsp; we can also write for this:
 :$${\rm Pr}(x_{\nu} = x_{\nu-6}) = {\rm Pr}\big[(x_{\nu}= 1)\hspace{0.05cm}\cap\hspace{0.05cm}(x_{\nu-6}= 1)\hspace{0.05cm}\cup \hspace{0.05cm}(x_\nu=0)\hspace{0.05cm}\cap\hspace{0.05cm}(x_{\nu-6} =0)\big]= p^{2}+(1- p)^{2}=\rm 0.25^2 + 0.75^2\hspace{0.15cm} \underline{ = 0.625}. $$

@@ Line 5: / Line 5: @@
 [[File:EN_Sto_A_2_3_neu.png|right|frame|Considered random generator]]
 A random number generator outputs a binary random number &nbsp; $x_\nu$&nbsp; at each clock time&nbsp; $(\nu)$&nbsp;, which can be&nbsp; $0$&nbsp; or&nbsp; $1$&nbsp;.
-*The value "1" occurs with probability&nbsp; $p = 0.25$&nbsp;.
+*The value&nbsp; "1"&nbsp; occurs with probability&nbsp; $p = 0.25$&nbsp;.
 *The individual values&nbsp;  $x_\nu$&nbsp; are statistically independent of each other.
-The binary numbers are stored in a shift register withnbsp; $I = 6$&nbsp; memory cells.
+The binary numbers are stored in a shift register with&nbsp; $I = 6$&nbsp; memory cells.
-At each clock instant, the contents of this shift register are shifted one place to the right and the algebraic sum&nbsp; $y_\nu$&nbsp; of the shift register contents is formed in each case:
+At each clock instant,&nbsp; the contents of this shift register are shifted one place to the right and the algebraic sum&nbsp; $y_\nu$&nbsp; of the shift register contents is formed in each case:
 :$$y_{\nu}=\sum\limits_{i=0}^{5}x_{\nu-i}=x_{\nu}+x_{\nu-1}+\ \text{...} \ +x_{\nu-5}.$$
@@ Line 21: / Line 21: @@
 Hints:
-*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Binomial_Distribution|binomial distribution]].
+*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Binomial_Distribution|Binomial Distribution]].
-*To check your results you can use the interactive applet&nbsp; [[Applets:Binomial-_und_Poissonverteilung_(Applet)|Binomial and Poisson distribution]]&nbsp;.
+*To check your results you can use the interactive applet&nbsp; [[Applets:Binomial_and_Poisson_Distribution_(Applet)|Binomial and Poisson distribution]]&nbsp;.
@@ Line 37: / Line 37: @@
-{Calculate the probability that&nbsp; $y$&nbsp; is greater than&nbsp; $2$&nbsp;.
+{Calculate the probability that&nbsp; $y$&nbsp; is greater than&nbsp; $2$.
 |type="{}"}
 ${\rm Pr}(y > 2) \ = \ $ { 0.169 3% }
-{What is the mean value of the random variable&nbsp; $y$&nbsp;?
+{What is the mean value of the random variable&nbsp; $y$?
 |type="{}"}
 $m_y \ =$ { 1.5 3% }
-{Find the ***standard deviation*** of the random variable&nbsp; $y$.
+{Find the rms value of the random variable&nbsp; $y$.
 |type="{}"}
 $\sigma_y \ = \ $ { 1.061 3% }
@@ Line 58: / Line 58: @@
-{What is the conditional probability that&nbsp; $y_\nu$&nbsp; equals&nbsp; $\mu$&nbsp; again if&nbsp; $y_{\nu-1} = \mu$&nbsp; occured previously?&nbsp; $(\mu = 0, \ 1, \ \text{...} \ , \ I)$.
+{What is the conditional probability that&nbsp; $y_\nu = \mu$&nbsp;  if&nbsp; $y_{\nu-1} = \mu$&nbsp; occured previously?&nbsp; $(\mu = 0, \ 1, \ \text{...} \ , \ I)$.
 |type="{}"}
 ${\rm Pr}(y_\nu = \mu \hspace{0.05cm} | \hspace{0.05cm} y_{\nu-1} = \mu ) \ = \ $ { 0.625 3% }

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Stochastische Signaltheorie/Binomialverteilung
+{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Binomial_Distribution
 }}
-[[File:EN_Sto_A_2_3.png|right|frame|Betrachteter Zufallsgenerator]]
+[[File:EN_Sto_A_2_3.png|right|frame|Considered random generator]]
-Ein Zufallsgenerator gibt zu jedem Taktzeitpunkt&nbsp; $(\nu)$&nbsp; eine bin&auml;re Zufallszahl&nbsp; $x_\nu$&nbsp; ab,&nbsp; die&nbsp; $0$&nbsp; oder&nbsp; $1$&nbsp; sein kann.
+A random number generator outputs a binary random number &nbsp; $x_\nu$&nbsp; at each clock time&nbsp; $(\nu)$&nbsp;, which can be&nbsp; $0$&nbsp; or&nbsp; $1$&nbsp;.
-*Der Wert "1" tritt mit Wahrscheinlichkeit&nbsp; $p = 0.25$&nbsp; auf.
+*The value "1" occurs with probability&nbsp; $p = 0.25$&nbsp;.
-*Die einzelnen Werte&nbsp;  $x_\nu$&nbsp; seien statistisch voneinander unabh&auml;ngig.
+*The individual values&nbsp;  $x_\nu$&nbsp; are statistically independent of each other.
-Die Bin&auml;rzahlen werden in ein Schieberegister mit&nbsp; $I = 6$&nbsp; Speicherzellen abgelegt.
+The binary numbers are stored in a shift register withnbsp; $I = 6$&nbsp; memory cells.
-Zu jedem Taktzeitpunkt wird der Inhalt dieses Schieberegisters um eine Stelle nach rechts verschoben und jeweils die algebraische Summe&nbsp; $y_\nu$&nbsp; der Schieberegisterinhalte gebildet:
+At each clock instant, the contents of this shift register are shifted one place to the right and the algebraic sum&nbsp; $y_\nu$&nbsp; of the shift register contents is formed in each case:
 :$$y_{\nu}=\sum\limits_{i=0}^{5}x_{\nu-i}=x_{\nu}+x_{\nu-1}+\ \text{...} \ +x_{\nu-5}.$$
@@ Line 20: / Line 20: @@
-''Hinweise:''
+Hints:
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Binomialverteilung|Binomialverteilung]].
+*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Binomial_Distribution|binomial distribution]].
-*Zur Kontrolle Ihrer Ergebnisse können Sie das interaktive Applet&nbsp; [[Applets:Binomial-_und_Poissonverteilung_(Applet)|Binomial&ndash; und Poissonverteilung]]&nbsp; benutzen.
+*To check your results you can use the interactive applet&nbsp; [[Applets:Binomial-_und_Poissonverteilung_(Applet)|Binomial and Poisson distribution]]&nbsp;.
@@ Line 29: / Line 29: @@
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Welche Werte kann die Summe&nbsp; $y$&nbsp; annehmen?&nbsp; Was ist der gr&ouml;&szlig;tm&ouml;gliche Wert?
+{What values can the sum&nbsp; $y$&nbsp; take?&nbsp; What is the largest possible value?
 |type="{}"}
 $y_\max \ = \ $  { 6 3% }
-{Berechnen Sie die Wahrscheinlichkeit, dass&nbsp; $y$&nbsp; gr&ouml;&szlig;er als&nbsp; $2$&nbsp; ist.
+{Calculate the probability that&nbsp; $y$&nbsp; is greater than&nbsp; $2$&nbsp;.
 |type="{}"}
 ${\rm Pr}(y > 2) \ = \ $ { 0.169 3% }
-{Wie gro&szlig; ist der Mittelwert der Zufallsgr&ouml;&szlig;e&nbsp; $y$&nbsp;?
+{What is the mean value of the random variable&nbsp; $y$&nbsp;?
 |type="{}"}
 $m_y \ =$ { 1.5 3% }
-{Ermitteln Sie die Streuung der Zufallsgr&ouml;&szlig;e&nbsp; $y$.
+{Find the ***standard deviation*** of the random variable&nbsp; $y$.
 |type="{}"}
 $\sigma_y \ = \ $ { 1.061 3% }
-{Sind die Zufallszahlen&nbsp; $y_\nu$&nbsp; statistisch unabh&auml;ngig?&nbsp; Begr&uuml;nden Sie Ihr Ergebnis.
+{Are the random numbers&nbsp; $y_\nu$&nbsp; statistically independent?&nbsp; Justify your result.
 |type="[]"}
-- Die Zufallszahlen sind statistisch unabh&auml;ngig.
+- The random numbers are statistically independent.
-+ Die Zufallszahlen sind statistisch abh&auml;ngig.
++ The random numbers are statistically dependent.
-{Wie groß ist die bedingte Wahrscheinlichkeit, dass&nbsp; $y_\nu$&nbsp; wieder gleich&nbsp; $\mu$&nbsp; ist, wenn vorher&nbsp; $y_{\nu-1} = \mu$&nbsp; aufgetreten ist?&nbsp; $(\mu = 0, \ 1, \ \text{...} \ , \ I)$.
+{What is the conditional probability that&nbsp; $y_\nu$&nbsp; equals&nbsp; $\mu$&nbsp; again if&nbsp; $y_{\nu-1} = \mu$&nbsp; occured previously?&nbsp; $(\mu = 0, \ 1, \ \text{...} \ , \ I)$.
 |type="{}"}
 ${\rm Pr}(y_\nu = \mu \hspace{0.05cm} | \hspace{0.05cm} y_{\nu-1} = \mu ) \ = \ $ { 0.625 3% }
@@ Line 65: / Line 65: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; In jeder Zelle kann eine&nbsp; $0$&nbsp; oder eine&nbsp; $1$&nbsp; stehen.&nbsp; Deshalb kann die Summe alle ganzzahligen Werte zwischen&nbsp; $0$&nbsp; und&nbsp; $6$&nbsp; annehmen:

← Older revision		Revision as of 16:43, 30 November 2021
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−	[[Category:Theory of Stochastic Signals: Exercises\|^2.3 ~~Binomialverteilung~~^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^2.3 Binomial Distribution^]]

← Older revision	Revision as of 16:40, 30 November 2021
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