Aufgaben:Exercise 4.06: Optimal Decision Boundaries - Revision history

Guenter at 14:57, 1 October 2022

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Guenter at 11:53, 3 September 2022

2022-09-03T11:53:02Z

Hwang at 10:55, 1 September 2022

2022-09-01T10:55:09Z

Guenter at 15:47, 27 July 2022

2022-07-27T15:47:42Z

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Hwang at 13:04, 5 July 2022

2022-07-05T13:04:43Z

Hwang at 12:54, 5 July 2022

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Guenter at 10:07, 23 June 2022

2022-06-23T10:07:46Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.06: Optimale Entscheidungsgrenzen to Aufgaben:Exercise 4.06: Optimal Decision Boundaries

2022-06-23T09:51:05Z

Guenter moved page Aufgabe 4.06: Optimale Entscheidungsgrenzen to Exercise 4.06: Optimal Decision Boundaries

Javier: Text replacement - "„" to """

2021-05-28T14:39:25Z

Text replacement - "„" to """

@@ Line 86: / Line 86: @@
 \hspace{0.05cm}.$$
-[[File:P_ID2033__Dig_A_4_6a.png|right|frame|Decision line and decision regions for&nbsp; $K=0$]]
+[[File:P_ID2033__Dig_A_4_6a.png|right|frame|Decision regions for&nbsp; $K=0$]]
 *The decision line lies in the middle between&nbsp; $s_0$&nbsp; and&nbsp; $s_1$&nbsp; and is rotated by&nbsp; $90^\circ$&nbsp; compared to the connecting line between the two symbols.&nbsp; It goes through the point $(2.5, \ \,  3)$.&nbsp; So the&nbsp; <u>first solution</u>&nbsp; is correct.

@@ Line 39: / Line 39: @@
 <quiz display=simple>
-{Where lies the optimal decision limit '''KORREKTUR: boundary''' for equally probable symbols?&nbsp; At
+{Where lies the optimal decision  boundary for equally probable symbols?&nbsp; At
 |type="[]"}
 + $\rho_2 = 3/4 \cdot \rho_1 + 9/8$,
@@ Line 81: / Line 81: @@
 \boldsymbol{ s }_1 - \boldsymbol{ s }_0 = (3,\hspace{0.1cm}-4) \hspace{0.05cm}$$
-:one obtains the following equation for the decision limits '''KORREKTUR: boundaries''':
+:one obtains the following equation for the decision boundaries:
 :$$3 \cdot \rho_1 - 4 \cdot \rho_2  =  ({17-26})/{2} = -  {9}/{2}
 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\rho_2  = 3/4 \cdot \rho_1 + 9/8

@@ Line 39: / Line 39: @@
 <quiz display=simple>
-{Where lies the optimal decision limit for equally probable symbols?&nbsp; At
+{Where lies the optimal decision limit '''KORREKTUR: boundary''' for equally probable symbols?&nbsp; At
 |type="[]"}
 + $\rho_2 = 3/4 \cdot \rho_1 + 9/8$,
@@ Line 81: / Line 81: @@
 \boldsymbol{ s }_1 - \boldsymbol{ s }_0 = (3,\hspace{0.1cm}-4) \hspace{0.05cm}$$
-:one obtains the following equation for the decision limits:
+:one obtains the following equation for the decision limits '''KORREKTUR: boundaries''':
 :$$3 \cdot \rho_1 - 4 \cdot \rho_2  =  ({17-26})/{2} = -  {9}/{2}
 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\rho_2  = 3/4 \cdot \rho_1 + 9/8

@@ Line 19: / Line 19: @@
 For AWGN noise with variance&nbsp; $\sigma_n^2$,&nbsp; the decision boundary is the solution of the following vectorial equation with respect to the vector&nbsp; $\boldsymbol{ \rho } =(\rho_1, \rho_2)$:
 :$$|| \boldsymbol{ s }_1||^2 - || \boldsymbol{ s }_0||^2 + 2 \cdot \sigma_n^2 \cdot  {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}( m_0)}{{\rm Pr}( m_1)} =
-\cdot \boldsymbol{ \rho }^{\rm T} \cdot (\boldsymbol{ s }_1 - \boldsymbol{ s }_0)\hspace{0.05cm},\hspace{0.2cm}
+\cdot \boldsymbol{ \rho }^{\rm T} \cdot (\boldsymbol{ s }_1 - \boldsymbol{ s }_0)\hspace{0.05cm}.$$
 In addition,&nbsp; two received values
@@ Line 73: / Line 72: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; With ${\rm Pr}(m_0) = {\rm Pr}(m_1) = 0.5$, the equation of the boundary line between the decision areas $I_0$ and $I_1$ reads:
+'''(1)'''&nbsp; With&nbsp; ${\rm Pr}(m_0) = {\rm Pr}(m_1) = 0.5$,&nbsp; the equation of the boundary line between the decision areas&nbsp; $I_0$&nbsp; and&nbsp; $I_1$&nbsp; reads:
 :$$|| \boldsymbol{ s }_1||^2 - || \boldsymbol{ s }_0||^2  =
 \cdot \boldsymbol{ \rho }^{\rm T} \cdot (\boldsymbol{ s }_1 - \boldsymbol{ s }_0)\hspace{0.05cm}.$$
-With the given vector values, i.e. the numerical values
+*With the given vector values,&nbsp; i.e. the numerical values
 :$$|| \boldsymbol{ s }_1||^2  = 4^2 + 1^2 = 17\hspace{0.05cm}, \hspace{0.2cm}
   || \boldsymbol{ s }_0||^2  =  1^2 + 5^2 = 26\hspace{0.05cm}, \hspace{0.2cm}
 \boldsymbol{ s }_1 - \boldsymbol{ s }_0 = (3,\hspace{0.1cm}-4) \hspace{0.05cm}$$
-one obtains the following equation for the decision limits:
+:one obtains the following equation for the decision limits:
 :$$3 \cdot \rho_1 - 4 \cdot \rho_2  =  ({17-26})/{2} = -  {9}/{2}
 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\rho_2  = 3/4 \cdot \rho_1 + 9/8
 \hspace{0.05cm}.$$
-[[File:P_ID2033__Dig_A_4_6a.png|right|frame|Decision line and decision regions for $K=0$]]
+[[File:P_ID2033__Dig_A_4_6a.png|right|frame|Decision line and decision regions for&nbsp; $K=0$]]
-The decision limit lies in the middle between $s_0$ and $s_1$ and is rotated by $90^\circ$ compared to the connecting line between the two symbols. It goes through the point $(2.5, \ \,  3)$. So the <u>first solution</u> is correct.
+*The decision line lies in the middle between&nbsp; $s_0$&nbsp; and&nbsp; $s_1$&nbsp; and is rotated by&nbsp; $90^\circ$&nbsp; compared to the connecting line between the two symbols.&nbsp; It goes through the point $(2.5, \ \,  3)$.&nbsp; So the&nbsp; <u>first solution</u>&nbsp; is correct.
-'''(3)'''&nbsp; The decision line also goes through the point $(3, \ \,  3.375)$. $B = (3, \ \, 3.5)$ lies above and therefore belongs to the decision region $I_0$ according to <u>solution 1</u>.
+'''(3)'''&nbsp; The decision line also goes through the point&nbsp; $(3, \ \,  3.375)$.
-'''(4)'''&nbsp; According to the equation in the information section and the calculations for subtask (1), the following now applies:
+'''(4)'''&nbsp; According to the equation in the information section and the calculations for subtask&nbsp; '''(1)''',&nbsp; the following now applies:
-[[File:P_ID2034__Dig_A_4_6c.png|right|frame|Decision areas for different $K$ values]]
+[[File:P_ID2034__Dig_A_4_6c.png|right|frame|Decision regions for different&nbsp; $K$]]
 :$$|| \boldsymbol{ s }_1||^2 - || \boldsymbol{ s }_0||^2 + 2 \cdot \sigma_n^2 \cdot  {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}( m_0)}{{\rm Pr}( m_1)} =
 \cdot \boldsymbol{ \rho }^{\rm T} \cdot (\boldsymbol{ s }_1 - \boldsymbol{ s }_0)\hspace{0.05cm}.$$
-With $|| \boldsymbol{ s }_1||^2 = 17$, $|| \boldsymbol{ s }_0||^2 = 26$, $ \boldsymbol{ s }_1 \, &ndash;\boldsymbol{ s }_0 = (3, \ \, &ndash;4)$ we obtain:
+*With&nbsp; $|| \boldsymbol{ s }_1||^2 = 17$,&nbsp; $|| \boldsymbol{ s }_0||^2 = 26$,&nbsp; $ \boldsymbol{ s }_1 \, &ndash;\boldsymbol{ s }_0 = (3, \ \, &ndash;4)$&nbsp; we obtain:
 :$$\rho_2  = 3/4 \cdot \rho_1 + 9/8 - K /8
 \hspace{0.05cm}.$$
-The following abbreviation was used here:
+*The following abbreviation was used here:
 :$$K =  2 \cdot \sigma_n^2 \cdot  {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}( m_0)}{{\rm Pr}( m_1)} =
 \cdot 1^2 \cdot 1.5 = 3 \hspace{0.05cm}.$$
-From this it follows:
+*From this it follows:
 :$$\rho_2  = 3/4 \cdot \rho_1 + 9/8 - 3 /8 = 3/4 \cdot \rho_1 + 3/4
 \hspace{0.05cm}.$$
-The decision line is shifted down by $3/8$ (black curve, labeled "$K = 3$" in the graphic). So <u>solution 2</u> is correct.
+*The decision line is shifted down by&nbsp; $3/8$&nbsp; $($black curve,&nbsp; labeled&nbsp; "$K = 3$"&nbsp; in the graphic$)$.&nbsp; So,&nbsp; <u>solution 2</u>&nbsp; is correct.
-*The first equation describes the optimal decision boundary for equally probable symbols ($K = 0$, dashed gray).
+#The first equation describes the optimal decision line for equally probable symbols&nbsp; $(K = 0$,&nbsp; dashed gray$)$.
-*The third equation is valid for $K = \, &ndash;3$. This results with $\sigma_n^2 = 1$ for the symbol probabilities ${\rm Pr}(m_1) \approx 0.817$ and ${\rm Pr}(m_0) \approx 0.138$ (green curve).
+#The third equation is valid for&nbsp; $K = \, &ndash;3$.&nbsp; This results with&nbsp; $\sigma_n^2 = 1$&nbsp; for the symbol probabilities&nbsp; ${\rm Pr}(m_1) \approx 0.817$&nbsp; and&nbsp; ${\rm Pr}(m_0) \approx 0.138$&nbsp; (green curve).
-*The violet straight line results with $K = 9$, i.e. with the same probabilities as for the black curve, but now with the variance $\sigma_n^2 = 3$.
+#The violet straight line results with&nbsp; $K = 9$,&nbsp; i.e. with the same probabilities as for the black curve,&nbsp; but now with the variance $\sigma_n^2 = 3$.
-'''(5)'''&nbsp; The graphic above already shows that both $A$ and $B$ now belong to the decision regio $I_0$. <u>Solutions 1 and 3</u> are correct.
+'''(5)'''&nbsp; The graphic above already shows that both&nbsp; $A$&nbsp; and&nbsp; $B$&nbsp; now belong to the decision region&nbsp; $I_0$.&nbsp; <u>Solutions 1 and 3</u>&nbsp; are correct.
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 [[File:P_ID2015__Dig_A_4_6.png|right|frame|Signal space constellation with<br> $N = 2, \ M = 2$]]
-We consider a binary message system &nbsp;$(M = 2)$ that is defined by the drawn 2D signal space constellation &nbsp;$(N = 2)$.&nbsp; The following applies to the two possible transmitted vectors that are directly coupled to the messages&nbsp; $m_0$&nbsp; and&nbsp; $m_1$:&nbsp;
+We consider a binary transmission system &nbsp;$(M = 2)$&nbsp; that is defined by the drawn two-dimensional signal space constellation &nbsp;$(N = 2)$.&nbsp; The following applies to the two possible transmitted vectors that are directly coupled to the messages&nbsp; $m_0$&nbsp; and&nbsp; $m_1$:&nbsp;
 :$$\boldsymbol{ s }_0 \hspace{-0.1cm} \ =\ \hspace{-0.1cm}  \sqrt {E} \cdot (1,\hspace{0.1cm} 5) \hspace{0.2cm} \Leftrightarrow \hspace{0.2cm} m_0   \hspace{0.05cm},$$
 :$$ \boldsymbol{ s }_1 \hspace{-0.1cm} \ =\ \hspace{-0.1cm}  \sqrt {E} \cdot (4, \hspace{0.1cm}1) \hspace{0.2cm} \Leftrightarrow \hspace{0.2cm} m_1   \hspace{0.05cm}.$$
-The optimal decision boundary between the regions&nbsp; $I_0 &#8660; m_0$&nbsp; and&nbsp; $I_1 &#8660; m_1$ is sought, whereby the following assumptions are made:
+The optimal decision boundary between the regions&nbsp; $I_0 &#8660; m_0$&nbsp; and&nbsp; $I_1 &#8660; m_1$ is sought.&nbsp; The following assumptions are made:
-* The following applies to subtasks '''(1)''' to '''(3)'''
+* It applies to subtasks&nbsp; '''(1)'''&nbsp; to&nbsp; '''(3)''':
 :$${\rm Pr}(m_0 ) = {\rm Pr}(m_1 ) = 0.5
   \hspace{0.05cm}. $$
-* For subtasks '''(4)''' and '''(5)''', on the other hand, the following should apply:
+* For subtasks&nbsp; '''(4)'''&nbsp; and&nbsp; '''(5)'''&nbsp; should apply:
 :$${\rm Pr}(m_0 ) = 0.817 \hspace{0.05cm},\hspace{0.2cm} {\rm Pr}(m_1 ) = 0.183\hspace{0.3cm}
   \Rightarrow \hspace{0.3cm} {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}( m_0)}{{\rm Pr}( m_1)} =
@@ Line 17: / Line 17: @@
 \hspace{0.05cm}.$$
-For AWGN noise with variance&nbsp; $\sigma_n^2$,&nbsp; the decision limit is the solution of the following vectorial equation with respect to the vector&nbsp; $(\rho_1, \rho_2)$:
+For AWGN noise with variance&nbsp; $\sigma_n^2$,&nbsp; the decision boundary is the solution of the following vectorial equation with respect to the vector&nbsp; $\boldsymbol{ \rho } =(\rho_1, \rho_2)$:
 :$$|| \boldsymbol{ s }_1||^2 - || \boldsymbol{ s }_0||^2 + 2 \cdot \sigma_n^2 \cdot  {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}( m_0)}{{\rm Pr}( m_1)} =
 \cdot \boldsymbol{ \rho }^{\rm T} \cdot (\boldsymbol{ s }_1 - \boldsymbol{ s }_0)\hspace{0.05cm},\hspace{0.2cm}
 \boldsymbol{ \rho } = (\rho_1 , \hspace{0.1cm}\rho_2 )\hspace{0.05cm}.$$
-In addition, two received values
+In addition,&nbsp; two received values
-:$$\boldsymbol{ A }= \sqrt {E} \cdot (1.5, \hspace{0.1cm}2)\hspace{0.05cm}, \hspace{0.2cm} \boldsymbol{ B }= \sqrt {E} \cdot (3, \hspace{0.1cm}3.5)   $$
+:$$\boldsymbol{ A }= \sqrt {E} \cdot (1.5, \hspace{0.1cm}2)\hspace{0.05cm},$$
-are drawn in the graphic. It must be checked whether these should be assigned to the regions&nbsp; $I_0$&nbsp; $($and thus the message $m_0)$&nbsp; or&nbsp; $I_1$&nbsp; $($message $m_1)$&nbsp; given the corresponding boundary conditions.
+are drawn in the graphic.&nbsp; It must be checked whether these should be assigned to the regions&nbsp; $I_0$&nbsp; $($and thus the message $m_0)$&nbsp; or to&nbsp; $I_1$&nbsp; $($message $m_1)$&nbsp; given the corresponding boundary conditions.
 Notes:
 * The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability|"Approximation of the Error Probability"]].
-''Notes:''
-* The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability|"Approximation of the Error Probability"]].
+* For numeric calculations,&nbsp; the energy&nbsp; $E = 1$&nbsp; can be set for simplification.
@@ Line 40: / Line 40: @@
 <quiz display=simple>
-{Where lies the optimal decision limit for equally probable symbols? At
+{Where lies the optimal decision limit for equally probable symbols?&nbsp; At
 |type="[]"}
 + $\rho_2 = 3/4 \cdot \rho_1 + 9/8$,
@@ Line 46: / Line 46: @@
 - $\rho_2 = 3$.
-{To which decision area does the received value&nbsp; $A = (1.5, \ \, 2)$ belong?
+{To which decision area does the received value&nbsp; $A = (1.5, \ \, 2)$&nbsp; belong?
 |type="()"}
 - To decision area&nbsp; $I_0$,
 + to decision area&nbsp; $I_1$.
-{To which decision area does the received value&nbsp; $B = (3, \ \, 3.5)$ belong?
+{To which decision area does the received value&nbsp; $B = (3, \ \, 3.5)$&nbsp; belong?
 |type="()"}
 + To decision area&nbsp; $I_0$,
 - to decision area&nbsp; $I_1$.
-{What is the equation of the decision line for&nbsp; ${\rm Pr}(m_0) = 0.817, \sigma_n = 1$?
+{What is the equation of the decision line for&nbsp; ${\rm Pr}(m_0) = 0.817, \  \sigma_n = 1$?
 |type="()"}
 - $\rho_2 = 3/4 \cdot \rho_1 + 9/8$,

@@ Line 32: / Line 32: @@
 ''Notes:''
-* The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability| Approximation of the Error Probability]].
+* The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability|"Approximation of the Error Probability"]].
 * For numeric calculations, the energy&nbsp; $E = 1$&nbsp; can be set for simplification.

← Older revision		Revision as of 10:07, 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^]]

@@ Line 116: / Line 116: @@
 \hspace{0.05cm}.$$
-Die Entscheidungsgerade ist um $3/8$ nach unten verschoben (schwarze Kurve, mit &bdquo;$K = 3$" bezeichnet in der Grafik). Richtig ist also der <u>Lösungsvorschlag 2</u>.
+Die Entscheidungsgerade ist um $3/8$ nach unten verschoben (schwarze Kurve, mit "$K = 3$" bezeichnet in der Grafik). Richtig ist also der <u>Lösungsvorschlag 2</u>.
 *Die erste Gleichung beschreibt die optimale Entscheidungsgrenze für gleichwahrscheinliche Symbole ($K = 0$, grau gestrichelt).