Aufgaben:Exercise 4.07: Decision Boundaries once again - Revision history

Guenter at 16:03, 13 March 2023

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Hwang at 11:13, 1 September 2022

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Guenter at 14:35, 28 July 2022

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Guenter at 14:18, 28 July 2022

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Guenter at 13:09, 28 July 2022

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Hwang at 14:57, 5 July 2022

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Guenter at 10:09, 23 June 2022

2022-06-23T10:09:29Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.07: Nochmals Entscheidungsgrenzen to Aufgaben:Exercise 4.07: Decision Boundaries once again

2022-06-23T09:59:32Z

Guenter moved page Aufgabe 4.07: Nochmals Entscheidungsgrenzen to Exercise 4.07: Decision Boundaries once again

Javier: Text replacement - "”" to """

2021-05-28T14:42:26Z

Text replacement - "”" to """

@@ Line 124: / Line 124: @@
 *With threshold&nbsp; $G = 0$,&nbsp; we get&nbsp; $p_{\rm S} = {\rm Q}(0.5) \ \underline {\approx 0.309}$.
 *The parameter&nbsp; $\gamma = 1.4$&nbsp; now has four times the value compared to subtask&nbsp; '''(5)''', <br>so that the optimal decision boundary is now&nbsp; $G_{\rm opt} = \underline {1.4 \cdot s_0}$.
-*Thus,&nbsp; the&nbsp; (undisturbed)&nbsp; '''KORREKTUR: (without noise)''' signal value&nbsp; $s_0$&nbsp; does not belong to the decision region&nbsp; $I_0$,&nbsp; but to&nbsp; $I_1$,&nbsp; characterized by&nbsp; $\rho < G_{\rm opt}$.
+*Thus,&nbsp; the&nbsp;  (noiseless)&nbsp; signal value&nbsp; $s_0$&nbsp; does not belong to the decision region&nbsp; $I_0$,&nbsp; but to&nbsp; $I_1$,&nbsp; characterized by&nbsp; $\rho < G_{\rm opt}$.
 *Only with a&nbsp; (positive)&nbsp; noise component:&nbsp; $I_0 (\rho > G_{\rm opt})$&nbsp; is possible at all.&nbsp; For the error probability,&nbsp; with&nbsp; $G_{\rm opt} = 1.4 \cdot s_0$:
 :$$p_{\rm S}  \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  {2}/{3} \cdot {\rm Q} (0.5 \cdot (1 + 1.4)) + {1}/{3} \cdot {\rm Q} (0.5 \cdot (1 - 1.4)) =

@@ Line 124: / Line 124: @@
 *With threshold&nbsp; $G = 0$,&nbsp; we get&nbsp; $p_{\rm S} = {\rm Q}(0.5) \ \underline {\approx 0.309}$.
 *The parameter&nbsp; $\gamma = 1.4$&nbsp; now has four times the value compared to subtask&nbsp; '''(5)''', <br>so that the optimal decision boundary is now&nbsp; $G_{\rm opt} = \underline {1.4 \cdot s_0}$.
-*Thus,&nbsp; the&nbsp; (undisturbed)&nbsp; signal value&nbsp; $s_0$&nbsp; does not belong to the decision region&nbsp; $I_0$,&nbsp; but to&nbsp; $I_1$,&nbsp; characterized by&nbsp; $\rho < G_{\rm opt}$.
+*Thus,&nbsp; the&nbsp; (undisturbed)&nbsp; '''KORREKTUR: (without noise)''' signal value&nbsp; $s_0$&nbsp; does not belong to the decision region&nbsp; $I_0$,&nbsp; but to&nbsp; $I_1$,&nbsp; characterized by&nbsp; $\rho < G_{\rm opt}$.
 *Only with a&nbsp; (positive)&nbsp; noise component:&nbsp; $I_0 (\rho > G_{\rm opt})$&nbsp; is possible at all.&nbsp; For the error probability,&nbsp; with&nbsp; $G_{\rm opt} = 1.4 \cdot s_0$:
 :$$p_{\rm S}  \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  {2}/{3} \cdot {\rm Q} (0.5 \cdot (1 + 1.4)) + {1}/{3} \cdot {\rm Q} (0.5 \cdot (1 - 1.4)) =

@@ Line 53: / Line 53: @@
 {What is the error probability for&nbsp; '''optimal threshold'''&nbsp; &nbsp;$G = G_{\rm opt}$?
 |type="{}"}
-$G = G_{\rm opt} \text{:} \hspace{0.2cm} p_{\rm S}\ = \ $ { 12.6 3% } $\ \%$
+$G = G_{\rm opt} \text{:} \hspace{0.2cm} p_{\rm S}\ = \ $ { 0.126 3% } $\ \%$
 {What error probabilities are obtained with the noise variance&nbsp; $\sigma_n^2 = E_{\rm S}$?

@@ Line 53: / Line 53: @@
 {What is the error probability for&nbsp; '''optimal threshold'''&nbsp; &nbsp;$G = G_{\rm opt}$?
 |type="{}"}
-$G = G_{\rm opt} \text{:} \hspace{0.2cm} p_{\rm S}\ = \ $ { 0.126 3% } $\ \%$
+$G = G_{\rm opt} \text{:} \hspace{0.2cm} p_{\rm S}\ = \ $ { 12.6 3% } $\ \%$
 {What error probabilities are obtained with the noise variance&nbsp; $\sigma_n^2 = E_{\rm S}$?

@@ Line 73: / Line 73: @@
 :$${\rm Pr}(m_1) = 2 \cdot {\rm Pr}(m_0) \hspace{0.05cm},\hspace{0.2cm} {\rm Pr}(m_0) + {\rm Pr}(m_1) = 1$$
-it follows directly&nbsp; ${\rm Pr}(m_0) = 1/3 \ \underline {\approx 0.333}$&nbsp; and&nbsp; ${\rm Pr}(m_1) = 2/3 \ \underline {\approx 0.667}$.
+:it follows directly&nbsp; ${\rm Pr}(m_0) = 1/3 \ \underline {\approx 0.333}$&nbsp; and&nbsp; ${\rm Pr}(m_1) = 2/3 \ \underline {\approx 0.667}$.
-'''(2)'''&nbsp; With the decision boundary $G = 0$, independently of the occurrence probabilities:
+'''(2)'''&nbsp; With the decision boundary&nbsp; $G = 0$,&nbsp; independently of the occurrence probabilities:
 :$$p_{\rm S}  =   {\rm Q} \left ( \frac{d/2}{\sigma_n} \right ) \hspace{0.05cm}.$$
-With $d = 2 \cdot \sqrt{E_{\rm S}}$ and $\sigma_n = \sqrt{E_{\rm S}}/3$, this gives: &nbsp; $p_{\rm S}  =   {\rm Q} (3) \hspace{0.1cm} \hspace{0.15cm}\underline {\approx 0.135 \%} \hspace{0.05cm}.$
+*With&nbsp; $d = 2 \cdot \sqrt{E_{\rm S}}$&nbsp; and&nbsp; $\sigma_n = \sqrt{E_{\rm S}}/3$,&nbsp; this gives: &nbsp;
-'''(3)'''&nbsp; According to the specification, for the "normalized threshold":
+'''(3)'''&nbsp; According to the specification,&nbsp; for the&nbsp; "normalized threshold":
 :$$\gamma = \frac{G_{\rm opt}}{E_{\rm S}^{1/2}} = 2 \cdot  \frac{   \sigma_n^2}{d^2} \cdot {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}( m_1)}{{\rm Pr}( m_0)}
 = \frac{  2 \cdot E_{\rm S}/9}{4 \cdot E_{\rm S}} \cdot {\rm ln} \hspace{0.15cm} \frac{2/3}{1/3}  \hspace{0.1cm}\hspace{0.15cm}\underline {\approx 0.04}
 \hspace{0.05cm}.$$
-*Thus, $G_{\rm opt} = \gamma \cdot \sqrt{E_{\rm S}} = 0.04 \cdot \sqrt{E_{\rm S}}$.
+*Thus,&nbsp; $G_{\rm opt} = \gamma \cdot \sqrt{E_{\rm S}} = 0.04 \cdot \sqrt{E_{\rm S}}$.
-*Thus, the optimal decision boundary is shifted to the right (towards the more improbable symbol $s_0$), because&nbsp; ${\rm Pr}(m_0) < {\rm Pr}(m_1)$.
-'''(4)'''&nbsp; With this optimal decision boundary, the error probability is slightly smaller compared to subtask '''(2)''':
+'''(4)'''&nbsp; With this optimal decision boundary,&nbsp; the error probability is slightly smaller compared to subtask&nbsp; '''(2)''':
 :$$p_{\rm S}  \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  {2}/{3} \cdot {\rm Q} (3 \cdot 1.04) + {1}/{3} \cdot {\rm Q} (3 \cdot 0.96) = {2}/{3} \cdot 0.090 \cdot 10^{-2} + {1}/{3} \cdot 0.199 \cdot 10^{-2}
 \hspace{0.1cm}\hspace{0.15cm}\underline {\approx 0.126 \%} \hspace{0.05cm}.$$
-'''(5)'''&nbsp; With the decision boundary in the middle between the symbols ($G = 0$), the result is analogous to subtask '''(2)''' with the noise variance now larger:
+'''(5)'''&nbsp; With the decision boundary in the middle between the symbols&nbsp; $(G = 0)$,&nbsp; the result is analogous to subtask&nbsp; '''(2)'''&nbsp; with the noise variance now larger:
-[[File:P_ID2035__Dig_A_4_7e.png|right|frame|Density functions with &nbsp;$\sigma_n^2 = E_{\rm S}$]]
+[[File:P_ID2035__Dig_A_4_7e.png|right|frame|Probability density functions with &nbsp;$\sigma_n^2 = E_{\rm S}$]]
 :$$p_{\rm S}  =   {\rm Q} \left ( \frac{d/2}{\sigma_n} \right ) =   {\rm Q} \left ( \frac{\sqrt{E_{\rm S}}}{\sqrt{E_{\rm S}}} \right )=
 {\rm Q} (1)\hspace{0.1cm} \hspace{0.15cm}\underline {\approx 15.9 \%} \hspace{0.05cm}.$$
-The parameter $\gamma$ (normalized best possible displacement of the decision boundary) is
+*The parameter&nbsp; $\gamma$&nbsp; (normalized best possible displacement of the decision boundary)&nbsp; is
 :$$\gamma = 2 \cdot \frac{   \sigma_n^2}{d^2} \cdot {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}( m_1)}{{\rm Pr}( m_0)}
 = \frac{  2 \cdot E_{\rm S}}{4 \cdot E_{\rm S}} \cdot {\rm ln} \hspace{0.15cm} \frac{2/3}{1/3} =  \frac{{\rm ln} \hspace{0.15cm} 2}{2} \approx 0.35 $$
@@ Line 111: / Line 113: @@
 \hspace{0.05cm}.$$
-The more frequent symbol is now less frequently falsified &#8658; the mean error probability becomes smaller:
+*The more frequent symbol is now less frequently falsified &#8658; the mean error probability becomes smaller:
 :$$p_{\rm S}  =   {2}/{3} \cdot {\rm Q} (1.35) + {1}/{3} \cdot {\rm Q} (0.65) =  {2}/{3} \cdot 0.0885 +{1}/{3} \cdot 0.258
 \hspace{0.1cm} \hspace{0.15cm}\underline {\approx 14.5 \%} \hspace{0.05cm}.$$
-The figure makes it clear that the optimal decision boundary can also be determined graphically as the intersection of the two (weighted) probability density functions:
+*The sketch makes it clear that the optimal decision boundary can also be determined graphically as intersection of the two&nbsp; (weighted)&nbsp; probability density functions:
-'''(6)'''&nbsp; <u>All solutions</u> of this rather academic subtask <u>are correct</u>:
+'''(6)'''&nbsp; <u>All solutions</u>&nbsp; of this rather academic subtask&nbsp; <u>are correct</u>:
-[[File:P_ID2036__Dig_A_4_7f_version2.png|right|frame|Density functions with &nbsp;$\sigma_n^2 = 4 \cdot E_{\rm S}$]]
+[[File:P_ID2036__Dig_A_4_7f_version2.png|right|frame|Probability density functions with &nbsp;$\sigma_n^2 = 4 \cdot E_{\rm S}$]]
-*With threshold $G = 0$, we get $p_{\rm S} = {\rm Q}(0.5) \ \underline {\approx 0.309}$.
+*With threshold&nbsp; $G = 0$,&nbsp; we get&nbsp; $p_{\rm S} = {\rm Q}(0.5) \ \underline {\approx 0.309}$.
-*The parameter $\gamma = 1.4$ now has four times the value compared to subtask '''(5)''', <br>so that the optimal decision boundary is now $G_{\rm opt} = \underline {1.4 \cdot s_0}$.
+*The parameter&nbsp; $\gamma = 1.4$&nbsp; now has four times the value compared to subtask&nbsp; '''(5)''', <br>so that the optimal decision boundary is now&nbsp; $G_{\rm opt} = \underline {1.4 \cdot s_0}$.
-*Thus, the (undisturbed) signal value $s_0$ does not belong to the decision region $I_0$, but to $I_1$, characterized by $\rho < G_{\rm opt}$.
+*Thus,&nbsp; the&nbsp; (undisturbed)&nbsp; signal value&nbsp; $s_0$&nbsp; does not belong to the decision region&nbsp; $I_0$,&nbsp; but to&nbsp; $I_1$,&nbsp; characterized by&nbsp; $\rho < G_{\rm opt}$.
-*Only with a (positive) noise component $I_0 (\rho > G_{\rm opt})$ is possible at all. For the error probability, with $G_{\rm opt} = 1.4 \cdot s_0$:
+*Only with a&nbsp; (positive)&nbsp; noise component:&nbsp; $I_0 (\rho > G_{\rm opt})$&nbsp; is possible at all.&nbsp; For the error probability,&nbsp; with&nbsp; $G_{\rm opt} = 1.4 \cdot s_0$:
 :$$p_{\rm S}  \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  {2}/{3} \cdot {\rm Q} (0.5 \cdot (1 + 1.4)) + {1}/{3} \cdot {\rm Q} (0.5 \cdot (1 - 1.4)) =
 \hspace{0.15cm}\underline {\approx 27\%}  \hspace{0.05cm}.$$
-The accompanying graph illustrates the statements made here.
+*The accompanying graph illustrates the statements made here.

@@ Line 4: / Line 4: @@
 [[File:P_ID2017__Dig_A_4_7.png|right|frame|PDF with unequal symbol probabilities]]
 We consider a transmission system with
-* only one basis function &nbsp;$(N = 1)$,
+# only one basis function &nbsp;$(N = 1)$,
-* two signals &nbsp;$(M = 2)$&nbsp; with&nbsp; $s_0 = \sqrt{E_s}$&nbsp; and &nbsp;$s_1 =  -\sqrt{E_s}$ ,
+# two signals &nbsp;$(M = 2)$&nbsp; with&nbsp; $s_0 = \sqrt{E_s}$&nbsp; and &nbsp;$s_1 =  -\sqrt{E_s}$ ,
-* an AWGN channel with variance&nbsp; $\sigma_n^2 = N_0/2$.
+# an AWGN channel with variance&nbsp; $\sigma_n^2 = N_0/2$.
-Since this exercise deals with the general case&nbsp; ${\rm Pr}(m_0) &ne; {\rm Pr}(m_1)$,&nbsp; it is not sufficient to consider the conditional density functions&nbsp; $p_{\it r\hspace{0.05cm}|\hspace{0.05cm}m_i}(\rho\hspace{0.05cm} |\hspace{0.05cm}m_i)$.&nbsp; Rather, these must still be multiplied by the symbol probabilities &nbsp; ${\rm Pr}(m_i)$.&nbsp; For &nbsp; $i$,&nbsp; the values&nbsp; $0$&nbsp; and&nbsp; $1$&nbsp; have to be used here.
+Since this exercise deals with the general case &nbsp; ${\rm Pr}(m_0) &ne; {\rm Pr}(m_1)$,&nbsp; it is not sufficient to consider the conditional density functions &nbsp; $p_{\it r\hspace{0.05cm}|\hspace{0.05cm}m_i}(\rho\hspace{0.05cm} |\hspace{0.05cm}m_i)$.&nbsp; Rather,&nbsp; these must still be multiplied by the symbol probabilities &nbsp; ${\rm Pr}(m_i)$.&nbsp; For &nbsp; $i$,&nbsp; the values&nbsp; $0$&nbsp; and&nbsp; $1$&nbsp; have to be used here.
-If the decision boundary between the two regions&nbsp; $I_0$&nbsp; and&nbsp; $I_1$&nbsp; is at&nbsp; $G = 0$, i.e., in the middle between&nbsp; $\boldsymbol{s}_0$&nbsp; and&nbsp; $\boldsymbol{s}_1$, the error probability is independent of the occurrence probabilities&nbsp; ${\rm Pr}(m_0)$&nbsp; and&nbsp; ${\rm Pr}(m_1)$:
+If the decision boundary between the two regions&nbsp; $I_0$&nbsp; and&nbsp; $I_1$&nbsp; is at&nbsp; $G = 0$,&nbsp; i.e.,&nbsp; in the middle between&nbsp; $\boldsymbol{s}_0$&nbsp; and&nbsp; $\boldsymbol{s}_1$,&nbsp; the error probability is independent of the occurrence probabilities&nbsp; ${\rm Pr}(m_0)$&nbsp; and&nbsp; ${\rm Pr}(m_1)$:
 :$$p_{\rm S}  = {\rm Pr}({ \cal E} ) =   {\rm Q} \left ( \frac{d/2}{\sigma_n} \right ) \hspace{0.05cm}.$$
-Here&nbsp; $d$&nbsp; indicates the distance between the signal points&nbsp; $s_0$&nbsp; and&nbsp; $s_1$&nbsp; and&nbsp; $d/2$&nbsp; accordingly the respective distance of&nbsp; $s_0$&nbsp; and&nbsp; $s_1$&nbsp; from the decision boundary&nbsp; $G = 0$. The rms value (root of the variance) of the AWGN noise is&nbsp; $\sigma_n$.
+*Here&nbsp; $d$&nbsp; indicates the distance between the signal points&nbsp; $s_0$&nbsp; and&nbsp; $s_1$&nbsp; and&nbsp; $d/2$&nbsp; accordingly the respective distance of&nbsp; $s_0$&nbsp; and&nbsp; $s_1$&nbsp; from the decision boundary&nbsp; $G = 0$.
-If, on the other hand, the occurrence probabilities are different &nbsp; &#8658; &nbsp;${\rm Pr}(m_0) &ne; {\rm Pr}(m_1)$, a smaller error probability can be obtained by shifting the decision boundary&nbsp; $G$:&nbsp;
+If,&nbsp; on the other hand,&nbsp; the occurrence probabilities are different &nbsp; &#8658; &nbsp;${\rm Pr}(m_0) &ne; {\rm Pr}(m_1)$,&nbsp; a smaller error probability can be obtained by shifting the decision boundary&nbsp; $G$:&nbsp;
 :$$p_{\rm S}  =  {\rm Pr}(m_1) \cdot {\rm Q} \left ( \frac{d/2}{\sigma_n} \cdot (1 + \gamma) \right )
   + {\rm Pr}(m_0) \cdot {\rm Q} \left ( \frac{d/2}{\sigma_n} \cdot (1 - \gamma) \right )\hspace{0.05cm},$$
@@ Line 26: / Line 29: @@
-''Notes:''
+Notes:
 * The exercise belongs to the chapter&nbsp; [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability|"Approximation of the Error Probability"]].
@@ Line 35: / Line 38: @@
 ===Questions===
 <quiz display=simple>
-{What are the underlying symbol probabilities of the graph, if the blue Gaussian curve is exactly twice as high as the red one?
+{What are the underlying symbol probabilities of the graph,&nbsp; if the blue Gaussian curve is exactly twice as high as the red one?
 |type="{}"}
 ${\rm Pr}(m_0)\ = \ $ { 0.333 3% }
 ${\rm Pr}(m_1)\ = \ $ { 0.667 3% }
-{What is the error probability with noise variance&nbsp; $\sigma_n^2 = E_{\rm S}/9$&nbsp; and the '''given threshold''' &nbsp;$G = 0$?
+{What is the error probability with noise variance&nbsp; $\sigma_n^2 = E_{\rm S}/9$&nbsp; and the&nbsp; '''given threshold'''&nbsp; &nbsp;$G = 0$?
 |type="{}"}
 $G = 0 \text{:} \hspace{0.85cm} p_{\rm S} \ = \ $ { 0.135 3% } $\ \%$
-{What is the optimal threshold for the given probabilities?
+{What is the optimal threshold value for the given probabilities?
 |type="{}"}
 $G_{\rm opt}\ = \ $ { 0.04 3% } $\ \cdot \sqrt{E_s}$
-{What is the probability of error for '''optimal threshold''' &nbsp;$G = G_{\rm opt}$?
+{What is the error probability for&nbsp; '''optimal threshold'''&nbsp; &nbsp;$G = G_{\rm opt}$?
 |type="{}"}
 $G = G_{\rm opt} \text{:} \hspace{0.2cm} p_{\rm S}\ = \ $ { 0.126 3% } $\ \%$
@@ Line 59: / Line 62: @@
 {Which statements are true for the noise variance&nbsp; $\sigma_n^2 = 4 \cdot E_{\rm S}$?
 |type="[]"}
-+ With&nbsp; $G = 0$,&nbsp; the probability of error is greater than&nbsp; $30\%$.
++ With&nbsp; $G = 0$,&nbsp; the error probability  is greater than&nbsp; $30\%$.
 + The optimal decision threshold is to the right of&nbsp; $s_0$.
-+ With optimal threshold, the error probability is about&nbsp; $27\%$.
++ With optimal threshold,&nbsp; the error probability is about&nbsp; $27\%$.
 + The estimated value&nbsp; $m_0$&nbsp; is possible only with sufficiently large noise.
 </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^]]

@@ Line 81: / Line 81: @@
-'''(3)'''&nbsp; Entsprechend der Angabe gilt für den "normierten Schwellenwert&rdquo;:
+'''(3)'''&nbsp; Entsprechend der Angabe gilt für den "normierten Schwellenwert":
 :$$\gamma = \frac{G_{\rm opt}}{E_{\rm S}^{1/2}} = 2 \cdot  \frac{   \sigma_n^2}{d^2} \cdot {\rm ln} \hspace{0.15cm} \frac{{\rm Pr}( m_1)}{{\rm Pr}( m_0)}
 = \frac{  2 \cdot E_{\rm S}/9}{4 \cdot E_{\rm S}} \cdot {\rm ln} \hspace{0.15cm} \frac{2/3}{1/3}  \hspace{0.1cm}\hspace{0.15cm}\underline {\approx 0.04}