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

Hwang at 12:20, 18 January 2023

2023-01-18T12:20:44Z

Guenter at 13:02, 7 December 2021

2021-12-07T13:02:13Z

Guenter at 16:34, 6 December 2021

2021-12-06T16:34:35Z

Guenter at 16:33, 6 December 2021

2021-12-06T16:33:57Z

Guenter at 16:26, 6 December 2021

2021-12-06T16:26:49Z

Guenter at 16:21, 6 December 2021

2021-12-06T16:21:30Z

Noah at 20:33, 3 December 2021

2021-12-03T20:33:51Z

Noah at 20:10, 3 December 2021

2021-12-03T20:10:48Z

Noah at 18:41, 3 December 2021

2021-12-03T18:41:41Z

Noah at 18:39, 3 December 2021

2021-12-03T18:39:27Z

@@ Line 64: / Line 64: @@
-'''(2)'''&nbsp; Analogously to&nbsp; '''(1)'''&nbsp; one obtains for the quadratic mean&nbsp; (second order moment):
+'''(2)'''&nbsp; Analogously to&nbsp; '''(1)'''&nbsp; one obtains for the second moment&nbsp; (second order moment):
 : $$m_{\rm 2\it x}= \rm \sum_{\mu=0}^{\it M -\rm 1}\it p_\mu\cdot x_{\mu}^{\rm 2}=\frac{\rm 1}{\it M}\cdot \sum_{\mu=\rm 0}^{\rm M- 1}\mu^{\rm 2} = \frac{\rm 1}{\it M}\cdot\frac{(\it M-\rm 1)\cdot \it M\cdot(\rm 2\it M-\rm 1)}{\rm 6} = \frac{(\it M-\rm 1)\cdot(\rm 2\it M-\rm 1)}{\rm 6}. $$
-*In the special case $M= 5$&nbsp; the quadratic mean&nbsp; $m_{2x} {=6}$.
+*In the special case $M= 5$&nbsp; the second moment&nbsp; $m_{2x} {=6}$.
 *From this, the variance can be calculated using Steiner's theorem:
 :$$\sigma_x^{\rm 2}=m_{\rm 2\it x}-m_x^{\rm 2}=\frac{(\it M-\rm 1)\cdot(\rm 2\it M-\rm 1)}{\rm 6}-\frac{(\it M-\rm 1)^{\rm 2}}{\rm 4}=\frac{\it M^{\rm 2}-\rm 1}{\rm 12}.$$

@@ Line 64: / Line 64: @@
-'''(2)'''&nbsp; Analogously to&nbsp; '''(1)'''&nbsp; one obtains for the root mean square:
+'''(2)'''&nbsp; Analogously to&nbsp; '''(1)'''&nbsp; one obtains for the quadratic mean&nbsp; (second order moment):
 : $$m_{\rm 2\it x}= \rm \sum_{\mu=0}^{\it M -\rm 1}\it p_\mu\cdot x_{\mu}^{\rm 2}=\frac{\rm 1}{\it M}\cdot \sum_{\mu=\rm 0}^{\rm M- 1}\mu^{\rm 2} = \frac{\rm 1}{\it M}\cdot\frac{(\it M-\rm 1)\cdot \it M\cdot(\rm 2\it M-\rm 1)}{\rm 6} = \frac{(\it M-\rm 1)\cdot(\rm 2\it M-\rm 1)}{\rm 6}. $$
-*In the special case $M= 5$&nbsp; the root mean square results in&nbsp; $m_{2x} {=6}$.
+*In the special case $M= 5$&nbsp; the quadratic mean&nbsp; $m_{2x} {=6}$.
 *From this, the variance can be calculated using Steiner's theorem:
 :$$\sigma_x^{\rm 2}=m_{\rm 2\it x}-m_x^{\rm 2}=\frac{(\it M-\rm 1)\cdot(\rm 2\it M-\rm 1)}{\rm 6}-\frac{(\it M-\rm 1)^{\rm 2}}{\rm 4}=\frac{\it M^{\rm 2}-\rm 1}{\rm 12}.$$

← Older revision		Revision as of 16:34, 6 December 2021
Line 4:		Line 4:


−	[[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  $M = 5$.

@@ 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; $y(t)$:
 :$$y(t)=\frac{2\cdot y_{\rm 0}}{M-\rm 1}\cdot \big[x(t)-m_x\big].$$

@@ Line 56: / Line 56: @@
 ===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; Man erh&auml;lt durch Mittelung über alle möglichen Signalwerte für den linearen Mittelwert:
 :$$m_{\it x}=\rm \sum_{\mu=0}^{\it M-{\rm 1}} \it p_\mu\cdot x_{\mu}=\frac{\rm 1}{\it M} \cdot \sum_{\mu=\rm 0}^{\it M-\rm 1}\mu=\frac{\rm 1}{\it M}\cdot\frac{(\it M-\rm 1)\cdot \it M}{\rm 2}=\frac{\it M-\rm 1}{\rm 2}.$$
@@ Line 89: / Line 97: @@
 {{ML-Fuß}}
+ 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 8: / Line 8: @@
-The rectangular signal&nbsp; $y(t)$&nbsp; is also&nbsp; $M$&ndash;stepped, but zero mean and restricted to the range from&nbsp; $y > -y_0$&nbsp; to&nbsp; $y < +y_0$&nbsp; .
+The rectangular zero mean signal&nbsp; $y(t)$&nbsp; can also assume&nbsp; $M$&nbsp; different values.&nbsp;
+*It is restricted to the range from&nbsp; $ -y_0 \le y \le +y_0$.
+*In the graph below you can see the signal&nbsp; $y(t)$, again for the level number&nbsp; $M = 5$.
@@ Line 23: / Line 19: @@
 Hints:
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Momente_einer_diskreten_Zufallsgröße|Moments of a Discrete Random Variable]].
+*Fur numerical calculations,&nbsp; use&nbsp; $y_0 = \rm 2\hspace{0.05cm}V$.
-**The topic of this chapter is illustrated with examples in the&nbsp;  (German language)&nbsp;  learning video<br> [[Momentenberechnung_bei_diskreten_Zufallsgrößen_(Lernvideo)|Momentenberechnung bei diskreten Zufallsgrößen]]&nbsp; $\Rightarrow$ &nbsp; Calculating Moments for Discrete-Valued Random Variables
+*The topic of this chapter is illustrated with examples in the&nbsp;  (German language)&nbsp;  learning video<br> &nbsp; &nbsp; [[Momentenberechnung_bei_diskreten_Zufallsgrößen_(Lernvideo)|"Momentenberechnung bei diskreten Zufallsgrößen"]] &nbsp; &rArr; &nbsp; "Calculating Moments for Discrete-Valued Random Variables"
-*Fur numerical calculations, use&nbsp; $y_0 = 2\hspace{0.05cm}V$.
@@ Line 49: / Line 45: @@
-{What is the variance&nbsp; $\sigma_y^2$&nbsp; of the random variable&nbsp; $y$?&nbsp; Consider the result from&nbsp; '''(2)'''.&nbsp; What is the value again for&nbsp; $M= 5$?
+{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% } $\ \rm V^2$

← Older revision		Revision as of 20:33, 3 December 2021
Line 1:		Line 1:

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

@@ Line 5: / Line 5: @@
 [[File:P_ID85__Sto_A_2_2.png|right|frame|Two similar multi-stepped signals]]
-Let the square wave signal&nbsp; $x(t)$&nbsp; be dimensionless and can only have the current values&nbsp; $0, \ 1, \ 2, \ \text{...} \ , \ M-2, \ M-1$&nbsp; with equal probability. The upper graph shows this signal for the special case&nbsp; $M = 5$.
+Let the rectangular signal&nbsp; $x(t)$&nbsp; be dimensionless and can only have the current values&nbsp; $0, \ 1, \ 2, \ \text{...} \ , \ M-2, \ M-1$&nbsp; with equal probability. The upper graph shows this signal for the special case&nbsp; $M = 5$.
-The square wave signal&nbsp; $y(t)$&nbsp; is also&nbsp; $M$&ndash;stepped, but zero mean and restricted to the range from&nbsp; $y > -y_0$&nbsp; to&nbsp; $y < +y_0$&nbsp; .
+The rectangular signal&nbsp; $y(t)$&nbsp; is also&nbsp; $M$&ndash;stepped, but zero mean and restricted to the range from&nbsp; $y > -y_0$&nbsp; to&nbsp; $y < +y_0$&nbsp; .

← Older revision		Revision as of 18:41, 3 December 2021
Line 4:		Line 4:


−	[[File:P_ID85__Sto_A_2_2.png\|right\|frame\|Two similar ~~multilevel~~ 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 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  $M = 5$.

@@ Line 39: / Line 39: @@
-{What is the variance&nbsp; $\sigma_x^2$&nbsp; of the random variable&nbsp; $x$&nbsp; in general and f&uuml;r&nbsp; $M= 5$?
+{What is the variance&nbsp; $\sigma_x^2$&nbsp; of the random variable&nbsp; $x$&nbsp; in general and for&nbsp; $M= 5$?
 |type="{}"}
 $\sigma_x^2\ = \ $ { 2 3% }
-{Calculate the mean&nbsp; $m_y$&nbsp; of the random variable&nbsp; $y$&nbsp; f&uuml;r&nbsp; $M= 5$.
+{Calculate the mean&nbsp; $m_y$&nbsp; of the random variable&nbsp; $y$&nbsp; for&nbsp; $M= 5$.
 |type="{}"}
 $m_y \ = \ $ { 0. } $\ \rm V$
@@ Line 58: / Line 58: @@
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; Man erh&auml;lt durch Mittelung über alle möglichen Signalwerte für den linearen Mittelwert:
@@ Line 93: / Line 93: @@
 {{ML-Fuß}}
 [[Category:Theory of Stochastic Signals: Exercises|^2.2 Moments of Discrete Random Variables^]]