Aufgaben:Exercise 4.08: Decision Regions at Three Symbols - Revision history

Guenter at 11:56, 3 September 2022

2022-09-03T11:56:30Z

Hwang at 11:19, 1 September 2022

2022-09-01T11:19:00Z

Hwang at 11:18, 1 September 2022

2022-09-01T11:18:20Z

Hwang at 11:16, 1 September 2022

2022-09-01T11:16:31Z

Guenter at 15:19, 28 July 2022

2022-07-28T15:19:00Z

Guenter at 15:01, 28 July 2022

2022-07-28T15:01:17Z

Hwang at 14:41, 11 July 2022

2022-07-11T14:41:40Z

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Guenter at 10:09, 23 June 2022

2022-06-23T10:09:58Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.08: Entscheidungsregionen bei drei Symbolen to Aufgaben:Exercise 4.08: Decision Regions at Three Symbols

2022-06-23T10:00:47Z

Guenter moved page Aufgabe 4.08: Entscheidungsregionen bei drei Symbolen to Exercise 4.08: Decision Regions at Three Symbols

Javier: Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

2021-03-23T12:52:08Z

Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

← Older revision		Revision as of 11:56, 3 September 2022
Line 23:		Line 23:
	\boldsymbol{ C } = (0.75, \hspace{0.1cm}0.50)$$		\boldsymbol{ C } = (0.75, \hspace{0.1cm}0.50)$$

−	~~'''KORREKTUR:~~ drawn,~~'''~~ each of which is to be assigned to a region $I_i$  in subtask  '''(5)'''.	+	are drawn,nbsp; each of which is to be assigned to a region $I_i$  in subtask  '''(5)'''.

@@ Line 93: / Line 93: @@
-'''(4)'''&nbsp; The graph already shows that the correct answer &nbsp; &rArr; &nbsp;  <u>YES</u>.
+'''(4)'''&nbsp; The graph already shows the correct answer &nbsp; &rArr; &nbsp;  <u>YES</u>.
 *The intersection of&nbsp; $G_{\rm 01}$&nbsp; and&nbsp; $G_{\rm 12}$&nbsp; (white circle)&nbsp; is at&nbsp; $(9/14,\ 3/14)$,&nbsp; because of
 :$${3}/{2} - 2 x = {x}/{3} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}

@@ Line 79: / Line 79: @@
 '''(2)'''&nbsp; <u>Solution 3</u>&nbsp; is correct:
-*The connecting line between&nbsp; $\boldsymbol{s}_0 = (&ndash;1,\ 1)$&nbsp; and&nbsp; $\boldsymbol{s}_2 = (2,\ 1)$&nbsp; has the gradient&nbsp; $&ndash;2/3$&nbsp; and intersects the decission boundary&nbsp; $G_{\rm 02}$&nbsp; $($with gradient $3/2)$&nbsp; at $(0.5,\ 0)$.
+*The connecting line between&nbsp; $\boldsymbol{s}_0 = (&ndash;1,\ 1)$&nbsp; and&nbsp; $\boldsymbol{s}_2 = (2,\ 1)$&nbsp; has the gradient&nbsp; $&ndash;2/3$&nbsp; and intersects the decision boundary&nbsp; $G_{\rm 02}$&nbsp; $($with gradient $3/2)$&nbsp; at $(0.5,\ 0)$.
 *From this follows: &nbsp; $y = {3}/{2} \left ( x - {1}/{2} \right )
     = -{3}/{4} + {3}/{2} \cdot x\hspace{0.05cm}.$

← Older revision		Revision as of 11:16, 1 September 2022
Line 23:		Line 23:
	\boldsymbol{ C } = (0.75, \hspace{0.1cm}0.50)$$		\boldsymbol{ C } = (0.75, \hspace{0.1cm}0.50)$$

−	each of which is to be assigned to a region $I_i$  in subtask  '''(5)'''.	+	'''KORREKTUR: drawn,''' each of which is to be assigned to a region $I_i$  in subtask  '''(5)'''.

@@ Line 71: / Line 71: @@
 {{ML-Kopf}}
 [[File:P_ID2038__Dig_A_4_8d.png|right|frame|Decision regions]]
-'''(1)'''&nbsp; <u>Solution 1</u> is correct:
+'''(1)'''&nbsp; <u>Solution 1</u>&nbsp; is correct:
-*The connecting line between the signal points $\boldsymbol{s}_0 = (&ndash;1, 1)$ and $\boldsymbol{s}_1 = (1, 2)$ has the gradient $1/2$ (see diagram).
+*The connecting line between the signal points&nbsp; $\boldsymbol{s}_0 = (&ndash;1,\ 1)$&nbsp; and&nbsp; $\boldsymbol{s}_1 = (1,\ 2)$&nbsp; has the gradient&nbsp; $1/2$&nbsp; (see diagram).
-*The decision boundary intersects the connecting line at $(\boldsymbol{s}_0 + \boldsymbol{s}_1)/2 = (0, 1.5)$ and has the slope $2$ (rotation of the connecting line by $90^\circ$).
+*The decision boundary intersects the connecting line at&nbsp; $(\boldsymbol{s}_0 + \boldsymbol{s}_1)/2 = (0,\ 1.5)$&nbsp; and has the gradient&nbsp; $2$&nbsp; $($rotation of the connecting line by&nbsp; $90^\circ)$.
 *From this follows: &nbsp; $y  = 1.5 - 2 x \hspace{0.05cm}.$
-'''(2)'''&nbsp; <u>Solution 3</u> is correct:
+'''(2)'''&nbsp; <u>Solution 3</u>&nbsp; is correct:
-*The connecting line between $\boldsymbol{s}_0 = (&ndash;1, 1)$ and $\boldsymbol{s}_2 = (2, 1)$ has the slope $&ndash;2/3$ and intersects the decider boundary $G_{\rm 02}$ (with the slope $3/2$) at $(0.5, 0)$.
+*The connecting line between&nbsp; $\boldsymbol{s}_0 = (&ndash;1,\ 1)$&nbsp; and&nbsp; $\boldsymbol{s}_2 = (2,\ 1)$&nbsp; has the gradient&nbsp; $&ndash;2/3$&nbsp; and intersects the decission boundary&nbsp; $G_{\rm 02}$&nbsp; $($with gradient $3/2)$&nbsp; at $(0.5,\ 0)$.
 *From this follows: &nbsp; $y = {3}/{2} \left ( x - {1}/{2} \right )
     = -{3}/{4} + {3}/{2} \cdot x\hspace{0.05cm}.$
@@ Line 85: / Line 85: @@
-'''(3)'''&nbsp; Here <u>solution 2</u> is applicable:
+'''(3)'''&nbsp; Here&nbsp; <u>solution 2</u>&nbsp; is applicable:
-*The line connecting $\boldsymbol{s}_1 = (1, 2)$ and $\boldsymbol{s}_2 = (2, \, &ndash;1)$ intersects the decision boundary $G_{\rm 12}$ at $(1.5, 0.5)$ and has slope $&ndash;3$.
+*The line connecting&nbsp; $\boldsymbol{s}_1 = (1,\ 2)$&nbsp; and&nbsp; $\boldsymbol{s}_2 = (2, \, &ndash;1)$&nbsp; intersects the decision boundary&nbsp; $G_{\rm 12}$&nbsp; at&nbsp; $(1.5,\ 0.5)$&nbsp; and has gradient&nbsp; $&ndash;3$.
-*Consequently, the slope of $G_{\rm 12} = 1/3$ and the equation of the decision boundary $G_{\rm 12}$ is:
+*Consequently,&nbsp; the gradient of&nbsp; $G_{\rm 12} = 1/3$&nbsp; and the equation of the decision boundary $G_{\rm 12}$ is:
 :$$y - {1}/{2} = {1}/{3} \cdot \left ( x - {3}/{2} \right )
      = {x}/{3} - {1}/{2}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}y =   {x}/{3}
@@ Line 93: / Line 93: @@
-'''(4)'''&nbsp; The graph already shows that the correct answer is <u>YES</u>.
+'''(4)'''&nbsp; The graph already shows that the correct answer &nbsp; &rArr; &nbsp;  <u>YES</u>.
-*The intersection of $G_{\rm 01}$ and $G_{\rm 12}$ (white circle) is at $(9/14, 3/14)$, because of
+*The intersection of&nbsp; $G_{\rm 01}$&nbsp; and&nbsp; $G_{\rm 12}$&nbsp; (white circle)&nbsp; is at&nbsp; $(9/14,\ 3/14)$,&nbsp; because of
 :$${3}/{2} - 2 x = {x}/{3} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
      {3}/{2} = {7}/{3} \cdot x \hspace{0.3cm}
@@ Line 100: / Line 100: @@
      \hspace{0.05cm}.$$
-*The straight line $G_{\rm 02}$ also passes through this point:
+*The straight line&nbsp; $G_{\rm 02}$&nbsp; also passes through this point:
 :$$y(x = {9}/{14}) =-{3}/{4} + {3}/{2} \cdot x = -{3}/{4} + {3}/{2} \cdot {9}/{14} =\frac{-21+27}{28}= {3}/{14}
      \hspace{0.05cm}.$$
-'''(5)'''&nbsp; According to the graph <u>all statements mentioned are correct</u>.
+'''(5)'''&nbsp; According to the graph:&nbsp; <u>all statements mentioned are correct</u>.
 {{ML-Fuß}}
 [[Category:Digital Signal Transmission: Exercises|^4.3 BER Approximation^]]

@@ Line 3: / Line 3: @@
 [[File:P_ID2018__Dig_A_4_8.png|right|frame|Signal space constellations with&nbsp; $M = 3$&nbsp; symbols]]
-We consider a signal space constellation in two-dimensional space&nbsp; $(N = 2)$&nbsp; with the signal set:
+We consider a signal space constellation in the two-dimensional space&nbsp; $(N = 2)$&nbsp; with the signal set:
 :$$\boldsymbol{ s }_0 = (-1, 1)\hspace{0.05cm}, \hspace{0.2cm}
     \boldsymbol{ s }_1 = (1, 2)\hspace{0.05cm}, \hspace{0.2cm}
@@ Line 10: / Line 10: @@
 in each case referred to the normalization value&nbsp; $\sqrt {E}$.
-The decision regions&nbsp; $I_0$,&nbsp; $I_1$&nbsp; and&nbsp; $I_2$ are sought, with the following considerations:
+The decision regions&nbsp; $I_0$,&nbsp; $I_1$&nbsp; and&nbsp; $I_2$ are sought,&nbsp; with the following considerations:
-* The region&nbsp; $I_i$&nbsp; should contain the signal space point&nbsp; $\boldsymbol{s}_i$ &nbsp; ($i = 0, 1, 2$).
+# The region&nbsp; $I_i$&nbsp; should contain the signal space point&nbsp; $\boldsymbol{s}_i$ &nbsp; ($i = 0,\ 1,\ 2$).
-* The signals&nbsp; $\boldsymbol{s}_0$,&nbsp; $\boldsymbol{s}_1$&nbsp; and&nbsp; $\boldsymbol{s}_2$&nbsp; are equally probable.
+# The signals&nbsp; $\boldsymbol{s}_0$,&nbsp; $\boldsymbol{s}_1$&nbsp; and&nbsp; $\boldsymbol{s}_2$&nbsp; are equally probable.
-* The regions are to be determined in such a way that the smallest error probability results for the AWGN channel.
+# The regions are to be determined in such a way that the smallest error probability results for the AWGN channel.
-With these preconditions, the decision boundaries&nbsp; $G_{\it ik}$&nbsp; between regions&nbsp; $I_i$&nbsp; and&nbsp; $I_k$&nbsp; are respectively straight lines exactly midway between&nbsp; $\boldsymbol{s}_i$&nbsp; and&nbsp; $\boldsymbol{s}_k$&nbsp; &nbsp;$(i = 0, 1, 2; \ \ k = 0, 1, 2; \ \  i &ne; k)$.
+With these preconditions,&nbsp; the decision boundaries&nbsp; $G_{\it ik}$&nbsp; between regions&nbsp; $I_i$&nbsp; and&nbsp; $I_k$&nbsp; are respectively straight lines exactly midway between&nbsp; $\boldsymbol{s}_i$&nbsp; and&nbsp; $\boldsymbol{s}_k$&nbsp; &nbsp;$(i = 0,\ 1,\ 2; \ \ k = 0,\ 1,\ 2; \ \  i &ne; k)$.
 With crosses in the above graph are three received values
@@ Line 23: / Line 23: @@
     \boldsymbol{ C } = (0.75, \hspace{0.1cm}0.50)$$
-each of which is to be assigned to a region $I_i$&nbsp; in subtask '''(5)'''.
+each of which is to be assigned to a region $I_i$&nbsp; in subtask&nbsp; '''(5)'''.
 Notes:
 * The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability|"Approximation of the Error Probability"]].
@@ Line 59: / Line 56: @@
 - $y = \, -3/4 + 3/2 \cdot x$.
-{Sketch the three decision regions&nbsp; $I_0$,&nbsp; $I_1$&nbsp; and&nbsp; $I_2$. Do the decision boundaries&nbsp; $G_{\rm 01}$,&nbsp; $G_{\rm 02}$&nbsp; and&nbsp; $G_{\rm 12}$&nbsp; intersect at a point?
+{Sketch the three decision regions&nbsp; $I_0$,&nbsp; $I_1$&nbsp; and&nbsp; $I_2$.&nbsp; Do the decision boundaries&nbsp; $G_{\rm 01}$,&nbsp; $G_{\rm 02}$&nbsp; and&nbsp; $G_{\rm 12}$&nbsp; intersect at a point?
 |type="()"}
 + Yes.
@@ Line 66: / Line 63: @@
 {Which of the following decisions are correct?
 |type="[]"}
-+ $\boldsymbol{A} = (0.5, 0.25)$&nbsp; belongs to region &nbsp;$I_0$.
++ $\boldsymbol{A} = (0.5,\ 0.25)$&nbsp; belongs to region &nbsp;$I_0$.
-+ $\boldsymbol{B} = (1, 0)$&nbsp; belongs to region &nbsp;$I_2$.
++ $\boldsymbol{B} = (1,\ 0)$&nbsp; belongs to region &nbsp;$I_2$.
-+ $\boldsymbol{C} = (0.75, 0.5)$&nbsp; belongs to region&nbsp; $I_1$.
++ $\boldsymbol{C} = (0.75,\ 0.5)$&nbsp; belongs to region&nbsp; $I_1$.
 </quiz>

← Older revision		Revision as of 10:09, 23 June 2022
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−	[[Category:Digital Signal Transmission: Exercises\|^4.3 BER-Approximation^]]	+	[[Category:Digital Signal Transmission: Exercises\|^4.3 BER Approximation^]]

← Older revision		Revision as of 12:52, 23 March 2021
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−	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^4.3 BER-Approximation^]]	+	[[Category:Digital Signal Transmission: Exercises\|^4.3 BER-Approximation^]]