Aufgaben:Exercise 2.2: Multi-Level Signals - Revision history

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	[[Category:Theory of Stochastic Signals: Exercises\|^2.2 Moments of Discrete Random Variables^]]		[[Category:Theory of Stochastic Signals: Exercises\|^2.2 Moments of Discrete Random Variables^]]
	[[de:Aufgaben:2.~~2_Mehrstufensignale~~]]		[[de:Aufgaben:Aufgabe 2.2: Mehrstufensignale]]

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	[[Category:Theory of Stochastic Signals: Exercises\|^2.2 Moments of Discrete Random Variables^]]		[[Category:Theory of Stochastic Signals: Exercises\|^2.2 Moments of Discrete Random Variables^]]
			[[de:Aufgaben:2.2_Mehrstufensignale]]

Hwang at 12:20, 18 January 2023

2023-01-18T12:20:44Z

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	'''(2)'''  Analogously to  '''(1)'''  one obtains for the ~~quadratic mean~~  (second order moment):		'''(2)'''  Analogously to  '''(1)'''  one obtains for the second moment  (second order moment):
	: $$m_{\rm 2\it x}= \rm \sum_{\mu=0}^{\it M -\rm 1}\it p_\mu\cdot x_{\mu}^{\rm 2}=\frac{\rm 1}{\it M}\cdot \sum_{\mu=\rm 0}^{\rm M- 1}\mu^{\rm 2} = \frac{\rm 1}{\it M}\cdot\frac{(\it M-\rm 1)\cdot \it M\cdot(\rm 2\it M-\rm 1)}{\rm 6} = \frac{(\it M-\rm 1)\cdot(\rm 2\it M-\rm 1)}{\rm 6}. $$		: $$m_{\rm 2\it x}= \rm \sum_{\mu=0}^{\it M -\rm 1}\it p_\mu\cdot x_{\mu}^{\rm 2}=\frac{\rm 1}{\it M}\cdot \sum_{\mu=\rm 0}^{\rm M- 1}\mu^{\rm 2} = \frac{\rm 1}{\it M}\cdot\frac{(\it M-\rm 1)\cdot \it M\cdot(\rm 2\it M-\rm 1)}{\rm 6} = \frac{(\it M-\rm 1)\cdot(\rm 2\it M-\rm 1)}{\rm 6}. $$

	*In the special case $M= 5$  the ~~quadratic mean~~  $m_{2x} {=6}$.		*In the special case $M= 5$  the second moment  $m_{2x} {=6}$.
	*From this, the variance can be calculated using Steiner's theorem:		*From this, the variance can be calculated using Steiner's theorem:
	:$$\sigma_x^{\rm 2}=m_{\rm 2\it x}-m_x^{\rm 2}=\frac{(\it M-\rm 1)\cdot(\rm 2\it M-\rm 1)}{\rm 6}-\frac{(\it M-\rm 1)^{\rm 2}}{\rm 4}=\frac{\it M^{\rm 2}-\rm 1}{\rm 12}.$$		:$$\sigma_x^{\rm 2}=m_{\rm 2\it x}-m_x^{\rm 2}=\frac{(\it M-\rm 1)\cdot(\rm 2\it M-\rm 1)}{\rm 6}-\frac{(\it M-\rm 1)^{\rm 2}}{\rm 4}=\frac{\it M^{\rm 2}-\rm 1}{\rm 12}.$$

Guenter at 13:02, 7 December 2021

2021-12-07T13:02:13Z

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	'''(2)'''  Analogously to  '''(1)'''  one obtains for the ~~root~~ mean ~~square~~:		'''(2)'''  Analogously to  '''(1)'''  one obtains for the quadratic mean  (second order moment):
	: $$m_{\rm 2\it x}= \rm \sum_{\mu=0}^{\it M -\rm 1}\it p_\mu\cdot x_{\mu}^{\rm 2}=\frac{\rm 1}{\it M}\cdot \sum_{\mu=\rm 0}^{\rm M- 1}\mu^{\rm 2} = \frac{\rm 1}{\it M}\cdot\frac{(\it M-\rm 1)\cdot \it M\cdot(\rm 2\it M-\rm 1)}{\rm 6} = \frac{(\it M-\rm 1)\cdot(\rm 2\it M-\rm 1)}{\rm 6}. $$		: $$m_{\rm 2\it x}= \rm \sum_{\mu=0}^{\it M -\rm 1}\it p_\mu\cdot x_{\mu}^{\rm 2}=\frac{\rm 1}{\it M}\cdot \sum_{\mu=\rm 0}^{\rm M- 1}\mu^{\rm 2} = \frac{\rm 1}{\it M}\cdot\frac{(\it M-\rm 1)\cdot \it M\cdot(\rm 2\it M-\rm 1)}{\rm 6} = \frac{(\it M-\rm 1)\cdot(\rm 2\it M-\rm 1)}{\rm 6}. $$

	*In the special case $M= 5$  the ~~root~~ mean ~~square results in~~  $m_{2x} {=6}$.		*In the special case $M= 5$  the quadratic mean  $m_{2x} {=6}$.
	*From this, the variance can be calculated using Steiner's theorem:		*From this, the variance can be calculated using Steiner's theorem:
	:$$\sigma_x^{\rm 2}=m_{\rm 2\it x}-m_x^{\rm 2}=\frac{(\it M-\rm 1)\cdot(\rm 2\it M-\rm 1)}{\rm 6}-\frac{(\it M-\rm 1)^{\rm 2}}{\rm 4}=\frac{\it M^{\rm 2}-\rm 1}{\rm 12}.$$		:$$\sigma_x^{\rm 2}=m_{\rm 2\it x}-m_x^{\rm 2}=\frac{(\it M-\rm 1)\cdot(\rm 2\it M-\rm 1)}{\rm 6}-\frac{(\it M-\rm 1)^{\rm 2}}{\rm 4}=\frac{\it M^{\rm 2}-\rm 1}{\rm 12}.$$

Guenter at 16:34, 6 December 2021

2021-12-06T16:34:35Z

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	[[File:P_ID85__Sto_A_2_2.png\|right\|frame\|Two similar multi-~~stepped~~ signals]]		[[File:P_ID85__Sto_A_2_2.png\|right\|frame\|Two similar multi-level signals]]
	Let the rectangular signal  $x(t)$  be dimensionless and can only have the current values  $0, \ 1, \ 2, \ \text{...} \ , \ M-2, \ M-1$  with equal probability. The upper graph shows this signal for the special case  $M = 5$.		Let the rectangular signal  $x(t)$  be dimensionless and can only have the current values  $0, \ 1, \ 2, \ \text{...} \ , \ M-2, \ M-1$  with equal probability. The upper graph shows this signal for the special case  $M = 5$.

Guenter at 16:33, 6 December 2021

2021-12-06T16:33:57Z

@@ Line 45: / Line 45: @@
-{What is the variance&nbsp; $\sigma_y^2$&nbsp; of the random variable&nbsp; $y$&nbsp; in general and for&nbsp; $M= 5$??&nbsp; Consider the result from&nbsp; '''(2)'''.
+{What is the variance&nbsp; $\sigma_y^2$&nbsp; of the random variable&nbsp; $y$&nbsp; in general and for&nbsp; $M= 5$?&nbsp; Consider the result from&nbsp; '''(2)'''.
 |type="{}"}
 $\sigma_y^2\ = \ $ { 2 3% } $\ \rm V^2$
@@ Line 64: / Line 64: @@
-'''(2)'''&nbsp; Analogously,'''(1)'''&nbsp; Man erh&auml;lt durch Mittelung über alle möglichen Signalwerte für den linearen Mittelwert:
+'''(2)'''&nbsp; Analogously to&nbsp; '''(1)'''&nbsp; one obtains for the root mean square:
 : $$m_{\rm 2\it x}= \rm \sum_{\mu=0}^{\it M -\rm 1}\it p_\mu\cdot x_{\mu}^{\rm 2}=\frac{\rm 1}{\it M}\cdot \sum_{\mu=\rm 0}^{\rm M- 1}\mu^{\rm 2} = \frac{\rm 1}{\it M}\cdot\frac{(\it M-\rm 1)\cdot \it M\cdot(\rm 2\it M-\rm 1)}{\rm 6} = \frac{(\it M-\rm 1)\cdot(\rm 2\it M-\rm 1)}{\rm 6}. $$
@@ Line 108: / Line 75: @@
-'''(3)'''&nbsp; Because of the symmetry of&nbsp; $y$&nbsp;, holds independently of&nbsp; $M$:
+'''(3)'''&nbsp; Because of the symmetry of&nbsp; $y$,&nbsp; it holds independently of&nbsp; $M$:
-:$$m_x \;\underline{= 2}.$$
+:$$m_y \;\underline{= 0}.$$
-'''(4)'''&nbsp; Between&nbsp; $x(t)$&nbsp; and&nbsp; $y(t)$&nbsp; the following relation holds:
+'''(4)'''&nbsp; The following relation holds between&nbsp; $x(t)$&nbsp; and&nbsp; $y(t)$:
 :$$y(t)=\frac{2\cdot y_{\rm 0}}{M-\rm 1}\cdot \big[x(t)-m_x\big].$$

Guenter at 16:26, 6 December 2021

2021-12-06T16:26:49Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Man erhält durch Mittelung über alle möglichen Signalwerte für den linearen Mittelwert:
			'''(1)'''  One obtains by averaging over all possible signal values for the linear mean:
			:$$m_{\it x}=\rm \sum_{\mu=0}^{\it M-{\rm 1}} \it p_\mu\cdot x_{\mu}=\frac{\rm 1}{\it M} \cdot \sum_{\mu=\rm 0}^{\it M-\rm 1}\mu=\frac{\rm 1}{\it M}\cdot\frac{(\it M-\rm 1)\cdot \it M}{\rm 2}=\frac{\it M-\rm 1}{\rm 2}.$$

			*In the special case  $M= 5$  the linear mean results in  $m_x \;\underline{= 2}$.



			'''(2)'''  Analogously,'''(1)'''  Man erhält durch Mittelung über alle möglichen Signalwerte für den linearen Mittelwert:
	:$$m_{\it x}=\rm \sum_{\mu=0}^{\it M-{\rm 1}} \it p_\mu\cdot x_{\mu}=\frac{\rm 1}{\it M} \cdot \sum_{\mu=\rm 0}^{\it M-\rm 1}\mu=\frac{\rm 1}{\it M}\cdot\frac{(\it M-\rm 1)\cdot \it M}{\rm 2}=\frac{\it M-\rm 1}{\rm 2}.$$		:$$m_{\it x}=\rm \sum_{\mu=0}^{\it M-{\rm 1}} \it p_\mu\cdot x_{\mu}=\frac{\rm 1}{\it M} \cdot \sum_{\mu=\rm 0}^{\it M-\rm 1}\mu=\frac{\rm 1}{\it M}\cdot\frac{(\it M-\rm 1)\cdot \it M}{\rm 2}=\frac{\it M-\rm 1}{\rm 2}.$$

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	{{ML-Fuß}}		{{ML-Fuß}}
			for the root mean square:
	~~'''(1)'''  One obtains by averaging over all possible signal values for the linear mean:~~
	:$$m_{\it x}=\rm \sum_{\mu=0}^{\it M-{\rm 1}} \it p_\mu\cdot x_{\mu}=\frac{\rm 1}{\it M} \cdot \sum_{\mu=\rm 0}^{\it M-\rm 1}\mu=\frac{\rm 1}{\it M}\cdot\frac{(\it M-\rm 1)\cdot \it M}{\rm 2}=\frac{\it M-\rm 1}{\rm 2}.$$

	*In the special case  $M= 5$  the linear mean results in  $m_x \;\underline{= 2}$.



	~~'''(2)'''  Analogously,~~ for the root mean square:
	: $$m_{\rm 2\it x}= \rm \sum_{\mu=0}^{\it M -\rm 1}\it p_\mu\cdot x_{\mu}^{\rm 2}=\frac{\rm 1}{\it M}\cdot \sum_{\mu=\rm 0}^{\rm M- 1}\mu^{\rm 2} = \frac{\rm 1}{\it M}\cdot\frac{(\it M-\rm 1)\cdot \it M\cdot(\rm 2\it M-\rm 1)}{\rm 6} = \frac{(\it M-\rm 1)\cdot(\rm 2\it M-\rm 1)}{\rm 6}. $$		: $$m_{\rm 2\it x}= \rm \sum_{\mu=0}^{\it M -\rm 1}\it p_\mu\cdot x_{\mu}^{\rm 2}=\frac{\rm 1}{\it M}\cdot \sum_{\mu=\rm 0}^{\rm M- 1}\mu^{\rm 2} = \frac{\rm 1}{\it M}\cdot\frac{(\it M-\rm 1)\cdot \it M\cdot(\rm 2\it M-\rm 1)}{\rm 6} = \frac{(\it M-\rm 1)\cdot(\rm 2\it M-\rm 1)}{\rm 6}. $$

Guenter at 16:21, 6 December 2021

2021-12-06T16:21:30Z

← Older revision		Revision as of 18:21, 6 December 2021
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	The rectangular signal  $y(t)$  is also  $M$&~~ndash~~;~~stepped, but zero mean and~~ restricted to the range from  $~~y >~~ -y_0~~$  to  $~~y < +y_0$~~ ~~ .		The rectangular zero mean signal  $y(t)$  can also assume  $M$  different values.
			*It is restricted to the range from  $ -y_0 \le y \le +y_0$.
			*In the graph below you can see the signal  $y(t)$, again for the level number  $M = 5$.
	In the graph below you can see the signal  $y(t)$, again for the number ~~of steps~~  $M = 5$.





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	Hints:		Hints:
	*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Momente_einer_diskreten_Zufallsgröße\|Moments of a Discrete Random Variable]].		*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Momente_einer_diskreten_Zufallsgröße\|Moments of a Discrete Random Variable]].
			*Fur numerical calculations,  use  $y_0 = \rm 2\hspace{0.05cm}V$.
	**The topic of this chapter is illustrated with examples in the  (German language)  learning video<br> [[Momentenberechnung_bei_diskreten_Zufallsgrößen_(Lernvideo)\|Momentenberechnung bei diskreten Zufallsgrößen]]  ~~$\Rightarrow$~~   Calculating Moments for Discrete-Valued Random Variables		*The topic of this chapter is illustrated with examples in the  (German language)  learning video<br>     [[Momentenberechnung_bei_diskreten_Zufallsgrößen_(Lernvideo)\|"Momentenberechnung bei diskreten Zufallsgrößen"]]   ⇒   "Calculating Moments for Discrete-Valued Random Variables"
	*Fur numerical calculations, use  $y_0 = 2\hspace{0.05cm}V$.


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	{What is the variance  $\sigma_y^2$  of the random variable  $y$?  Consider the result from  '''(2)'''.~~  What is the value again for  $M= 5$?~~		{What is the variance  $\sigma_y^2$  of the random variable  $y$  in general and for  $M= 5$??  Consider the result from  '''(2)'''.
	\|type="{}"}		\|type="{}"}
	$\sigma_y^2\ = \ $ { 2 3% } $\ \rm V^2$		$\sigma_y^2\ = \ $ { 2 3% } $\ \rm V^2$

Noah at 20:33, 3 December 2021

2021-12-03T20:33:51Z

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	{{quiz-Header\|Buchseite=~~Stochastische Signaltheorie~~/~~Wahrscheinlichkeit und relative Häufigkeit~~		{{quiz-Header\|Buchseite=Theory_of_Stochastic_Signals/From_Random_Experiment_to_Random_Variable
	}}		}}

Noah at 20:10, 3 December 2021

2021-12-03T20:10:48Z

← Older revision		Revision as of 22:10, 3 December 2021
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	[[File:P_ID85__Sto_A_2_2.png\|right\|frame\|Two similar multi-stepped signals]]		[[File:P_ID85__Sto_A_2_2.png\|right\|frame\|Two similar multi-stepped signals]]
	Let the ~~square wave~~ signal  $x(t)$  be dimensionless and can only have the current values  $0, \ 1, \ 2, \ \text{...} \ , \ M-2, \ M-1$  with equal probability. The upper graph shows this signal for the special case  $M = 5$.		Let the rectangular signal  $x(t)$  be dimensionless and can only have the current values  $0, \ 1, \ 2, \ \text{...} \ , \ M-2, \ M-1$  with equal probability. The upper graph shows this signal for the special case  $M = 5$.


	The ~~square wave~~ signal  $y(t)$  is also  $M$–stepped, but zero mean and restricted to the range from  $y > -y_0$  to  $y < +y_0$  .		The rectangular signal  $y(t)$  is also  $M$–stepped, but zero mean and restricted to the range from  $y > -y_0$  to  $y < +y_0$  .