Aufgaben:Exercise 1.7: PDF of Rice Fading - Revision history

Javier at 08:19, 7 July 2021

2021-07-07T08:19:42Z

Javier: Text replacement - "Category:Exercises for Mobile Communications" to "Category:Mobile Communications: Exercises"

2021-03-23T12:37:11Z

Text replacement - "Category:Exercises for Mobile Communications" to "Category:Mobile Communications: Exercises"

Javier: Text replacement - "===Solutions===" to "===Solution==="

2020-12-15T14:11:11Z

Text replacement - "===Solutions===" to "===Solution==="

Javier: Text replacement - "Mobile Kommunikation/Nichtfrequenzselektives Fading mit Direktkomponente" to "Mobile_Communications/Non-Frequency_Selective_Fading_With_Direct_Component"

2020-12-15T13:58:40Z

Text replacement - "Mobile Kommunikation/Nichtfrequenzselektives Fading mit Direktkomponente" to "Mobile_Communications/Non-Frequency_Selective_Fading_With_Direct_Component"

Guenter at 15:26, 8 December 2020

2020-12-08T15:26:31Z

Guenter at 15:18, 8 December 2020

2020-12-08T15:18:13Z

Javier: Text replacement - "PSD" to "PDS"

2020-08-25T09:57:02Z

Text replacement - "PSD" to "PDS"

Guenter at 10:00, 13 August 2020

2020-08-13T10:00:57Z

Guenter at 09:49, 13 August 2020

2020-08-13T09:49:13Z

Guenter at 09:44, 13 August 2020

2020-08-13T09:44:46Z

@@ Line 37: / Line 37: @@
 ===Questions===
 <quiz display=simple>
-{Calculate some PDF values for&nbsp; $|z_0| = 2$&nbsp; and&nbsp; $\sigma = 2$:
+{Calculate some PDF values for&nbsp; $|z_0| = 2$&nbsp; and&nbsp; $\sigma = 1$:
 |type="{}"}
 $f_a(a = 1) \ = \ ${ 0.187 3% }
@@ Line 62: / Line 62: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; With&nbsp; $|z_0| = 2$&nbsp; and&nbsp; $\sigma = 2$&nbsp; the Rice PDF is
+'''(1)'''&nbsp; With&nbsp; $|z_0| = 2$&nbsp; and&nbsp; $\sigma = 1$&nbsp; the Rice PDF is
 :$$f_a(a) = a \cdot {\rm exp} [ -\frac{a^2 + 4}{2}] \cdot {\rm I}_0 (2a)\hspace{0.05cm}.$$

← Older revision		Revision as of 12:37, 23 March 2021
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−	[[Category:~~Exercises for~~ Mobile Communications\|^1.4 Fading with Direct Path Component^]]	+	[[Category:Mobile Communications: Exercises\|^1.4 Fading with Direct Path Component^]]

@@ Line 60: / Line 60: @@
 </quiz>
-===Solutions===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; With&nbsp; $|z_0| = 2$&nbsp; and&nbsp; $\sigma = 2$&nbsp; the Rice PDF is

← Older revision		Revision as of 13:58, 15 December 2020
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−	{{quiz-Header\|Buchseite=~~Mobile Kommunikation~~/~~Nichtfrequenzselektives Fading mit Direktkomponente~~}}	+	{{quiz-Header\|Buchseite=Mobile_Communications/Non-Frequency_Selective_Fading_With_Direct_Component}}

	[[File:P_ID2133__Mob_A_1_7.png\|right\|frame\| Rice fading for different values of  $\|z_0\|^2$]]		[[File:P_ID2133__Mob_A_1_7.png\|right\|frame\| Rice fading for different values of  $\|z_0\|^2$]]

@@ Line 103: / Line 103: @@
 :$${\rm Pr}(a \le 1) \approx {\rm Pr}(g \le -3.37) = {\rm Q}(3.37) { \approx 0.04\,\%} \hspace{0.05cm}.$$
-*If one assumes the equal power Gaussian distributio&nbsp;n (see the note to the last subtask), the mean value is&nbsp; $m_g = \sqrt{21}\approx 4.58$,&nbsp; and the probability would then be
+*If one assumes the equal power Gaussian distribution&nbsp; (see the note to the last subtask), the mean value is&nbsp; $m_g = \sqrt{21}\approx 4.58$,&nbsp; and the probability would then be
 :$${\rm Pr}(a \le 1) \approx {\rm Q}(3.58) \hspace{0.15cm} \underline{ \approx 0.02\,\%} \hspace{0.05cm}.$$

@@ Line 30: / Line 30: @@
 ''Notes:''
 * This task belongs to chapter&nbsp; [[Mobile_Communications/Non-frequency_selective_fading_with_direct_component| Non-frequency selective fading with direct component]].
-* For the numerical solutions of the last subtasks, we recommend the interaction module&nbsp; [[Applets:Complementary_Gaussian_Error_Functions|Complementary Gaussian Error Functions]].
+* For the numerical solutions of the last subtasks, we recommend the applet&nbsp; [[Applets:Complementary_Gaussian_Error_Functions|Complementary Gaussian Error Functions]].

@@ Line 3: / Line 3: @@
 [[File:P_ID2133__Mob_A_1_7.png|right|frame| Rice fading for different values of&nbsp; $|z_0|^2$]]
-As you can see in the diagram, we consider the same scenario as in&nbsp; [[Aufgaben:Exercise_1.6:_Autocorrelation_Function_and_PSD_with_Rice_Fading| Exercise 1.6]]:
+As you can see in the diagram, we consider the same scenario as in&nbsp; [[Aufgaben:Exercise_1.6:_Autocorrelation_Function_and_PDS_with_Rice_Fading| Exercise 1.6]]:
 * Rice fading&nbsp; with variance of the Gaussian processes &nbsp; $\sigma^2 = 1$&nbsp; and parameter&nbsp; $|z_0|$&nbsp; for the direct path.
 * Regarding direct path, we are interested in the parameter values&nbsp; $|z_0|^2 = 0, \ 2, \ 4, \ 10, \ 20$&nbsp; (see graph).

@@ Line 62: / Line 62: @@
 ===Solutions===
 {{ML-Kopf}}
-'''(1)'''&nbsp; With $|z_0| = 2$ and $\sigma = 2$ the Rice PDF is
+'''(1)'''&nbsp; With&nbsp; $|z_0| = 2$&nbsp; and&nbsp; $\sigma = 2$&nbsp; the Rice PDF is
-$$f_a(a) = a \cdot {\rm exp} [ -\frac{a^2 + 4}{2}] \cdot {\rm I}_0 (2a)\hspace{0.05cm}.$$
+:$$f_a(a) = a \cdot {\rm exp} [ -\frac{a^2 + 4}{2}] \cdot {\rm I}_0 (2a)\hspace{0.05cm}.$$
 *This gives the desired values:
@@ Line 74: / Line 74: @@
-'''(2)'''&nbsp; With the result of the subtask '''(1)''' &nbsp; &rArr; &nbsp; $f_a(a = 1) = 0.187$ the triangle approximation gives
+'''(2)'''&nbsp; With the result of the subtask&nbsp; '''(1)''' &nbsp; &rArr; &nbsp; $f_a(a = 1) = 0.187$&nbsp; the triangle approximation gives
-$${\rm Pr}(a \le 1) = {1}/{2} \cdot 0.187 \cdot 1\hspace{0.15cm} \underline{ \approx 9.4\,\%} \hspace{0.05cm}.$$
+:$${\rm Pr}(a \le 1) = {1}/{2} \cdot 0.187 \cdot 1\hspace{0.15cm} \underline{ \approx 9.4\,\%} \hspace{0.05cm}.$$
-*This result will be a bit too large, because the blue curve is below the connecting line from $(0, 0)$ to $(1, 0.187)$ &nbsp; &rArr; &nbsp; convex curve.
+*This result will be a bit too large, because the blue curve is below the connecting line from&nbsp; $(0, 0)$&nbsp; to&nbsp; $(1, 0.187)$ &nbsp; &rArr; &nbsp; convex curve.
-*The actual probability value will be slightly larger because the red curve is concave in the range between $0$ and $1$.
+'''(3)'''&nbsp; For the red curve the PDF value&nbsp; $f_a(a = 1) \approx 0.35$&nbsp; can be read from the graph:
-*Here, $g$ denotes a Gaussian distributed random variable with mean $0$ and standard deviation $\sigma = 1$.
+'''(4)'''&nbsp; The Gaussian approximation states that one can approximate the Rice distribution by a Gaussian distribution with mean&nbsp; $|z_0| = \sqrt{10} = 3.16$&nbsp; and standard deviation&nbsp; $\sigma = 1$&nbsp; if the quotient&nbsp; $|z_0|/\sigma$ &nbsp; is sufficiently large.&nbsp; Then we have
-*The numerical value was determined with the specified interactive [[Applets:QFunction|Applet]].
+:$${\rm Pr}(a \le 1) \approx {\rm Pr}(g \le -2.16) = {\rm Q}(2.16) \hspace{0.15cm} \underline{ \approx 1.5\,\%} \hspace{0.05cm}.$$
 <i>Note:</i> &nbsp; The Gaussian approximation is certainly associated with a certain error here:
-*From the graph you can see that the average value of the green curve is not $a = 3.16$ , but rather $3.31$.
+*From the graph you can see that the average value of the green curve is not&nbsp; $a = 3.16$,&nbsp; but rather&nbsp; $3.31$.
-*Then the power of the Gaussian approximation $(3.31^2 + 1^2 = 12)$ is exactly the same as that of the Rice distribution:
+*Then the power of the Gaussian approximation&nbsp; $(3.31^2 + 1^2 = 12)$&nbsp; is exactly the same as that of the Rice distribution:
 :$$|z_0|^2 + 2 \sigma^2= 10 + 2 =12\hspace{0.05cm}.$$
-'''(5)'''&nbsp; Using the same calculation method, replace the Rice PDF with a Gaussian PDF with mean value $\sqrt{20} \approx $4.47 and standard deviation $\sigma = $1 and you get
+'''(5)'''&nbsp; Using the same calculation method, replace the Rice PDF with a Gaussian PDF with mean value&nbsp; $\sqrt{20} \approx 4.47$&nbsp; and standard deviation&nbsp; $\sigma = 1$&nbsp; and you get
-$${\rm Pr}(a \le 1) \approx {\rm Pr}(g \le -3.37) = {\rm Q}(3.37) { \approx 0.04\,\%} \hspace{0.05cm}.$$
+:$${\rm Pr}(a \le 1) \approx {\rm Pr}(g \le -3.37) = {\rm Q}(3.37) { \approx 0.04\,\%} \hspace{0.05cm}.$$
-*If one assumes the equal power Gaussian distribution (see the note to the last subtask), the mean value is $m_g = \sqrt{21}\approx 4.58$, and the probability would then be
+*If one assumes the equal power Gaussian distributio&nbsp;n (see the note to the last subtask), the mean value is&nbsp; $m_g = \sqrt{21}\approx 4.58$,&nbsp; and the probability would then be
-$${\rm Pr}(a \le 1) \approx {\rm Q}(3.58) \hspace{0.15cm} \underline{ \approx 0.02\,\%} \hspace{0.05cm}.$$
+:$${\rm Pr}(a \le 1) \approx {\rm Q}(3.58) \hspace{0.15cm} \underline{ \approx 0.02\,\%} \hspace{0.05cm}.$$
 {{ML-Fuß}}

@@ Line 43: / Line 43: @@
 $f_a(a = 3) \ = \ ${ 0.303 3% }
-{Let &nbsp; $|z_0| = 2$ &nbsp; &rArr; &nbsp; $|z_0|^2 = 4$&nbsp; ('''blue curve'''). How big is&nbsp; ${\rm Pr}(a &#8804; 1)$? Use the&nbsp; '''triangular approximation'''.
+{Let &nbsp; $|z_0| = 2$ &nbsp; &rArr; &nbsp; $|z_0|^2 = 4$&nbsp; '''(blue curve)'''.&nbsp; How big is&nbsp; ${\rm Pr}(a &#8804; 1)$?&nbsp; Use the&nbsp; '''triangular approximation'''.
 |type="{}"}
 ${\rm Pr}(a &#8804; 1)\ = \ ${ 9.4 3% } $\ \%$
-{Let &nbsp; $|z_0|^2 = 2$&nbsp; ('''red curve'''). How big is&nbsp; ${\rm Pr}(a &#8804; 1)$? Use the &nbsp;'''triangular approximation'''.
+{Let &nbsp; $|z_0|^2 = 2$&nbsp; '''(red curve)'''.&nbsp; How big is&nbsp; ${\rm Pr}(a &#8804; 1)$?&nbsp; Use the &nbsp;'''triangular approximation'''.
 |type="{}"}
 ${\rm Pr}(a &#8804; 1) \ = \ ${ 17.5 3% } $\ \%$
-{Let&nbsp; $|z_0|^2 = 10$&nbsp; ('''green curve'''). How big is&nbsp; ${\rm Pr}(a &#8804; 1)$? Use the&nbsp; '''Gaussian approximation'''.
+{Let&nbsp; $|z_0|^2 = 10$&nbsp; '''(green curve)'''.&nbsp; How big is&nbsp; ${\rm Pr}(a &#8804; 1)$?&nbsp; Use the&nbsp; '''Gaussian approximation'''.
 |type="{}"}
 ${\rm Pr}(a &#8804; 1) \ = \ ${ 1.5 3% } $\ \%$
-{Let&nbsp; $|z_0|^2 = 20$&nbsp; ('''purple curve'''). How big is&nbsp; ${\rm Pr}(a &#8804; 1)$? Use the &nbsp;'''Gaussian approximation'''.
+{Let&nbsp; $|z_0|^2 = 20$&nbsp; '''(purple curve)'''.&nbsp; How big is&nbsp; ${\rm Pr}(a &#8804; 1)$?&nbsp; Use the &nbsp;'''Gaussian approximation'''.
 |type="{}"}
 ${\rm Pr}(a &#8804; 1) \ = \ ${ 0.02 3% } $\ \%$

@@ Line 4: / Line 4: @@
 [[File:P_ID2133__Mob_A_1_7.png|right|frame| Rice fading for different values of&nbsp; $|z_0|^2$]]
 As you can see in the diagram, we consider the same scenario as in&nbsp; [[Aufgaben:Exercise_1.6:_Autocorrelation_Function_and_PSD_with_Rice_Fading| Exercise 1.6]]:
-* <i>Rice fading</i>&nbsp; with variance of the Gaussian processes &nbsp; $\sigma^2 = 1$&nbsp; and parameter&nbsp; $|z_0|$&nbsp; for the direct path.
+* Rice fading&nbsp; with variance of the Gaussian processes &nbsp; $\sigma^2 = 1$&nbsp; and parameter&nbsp; $|z_0|$&nbsp; for the direct path.
 * Regarding direct path, we are interested in the parameter values&nbsp; $|z_0|^2 = 0, \ 2, \ 4, \ 10, \ 20$&nbsp; (see graph).
 * The PDF of the magnitude&nbsp; $a(t) = |z(t)|$&nbsp; is
-:$$f_a(a) = \frac{a}{\sigma^2} \cdot {\rm exp} [ -\frac{a^2 + |z_0|^2}{2\sigma^2}] \cdot {\rm I}_0 \left [ \frac{a \cdot |z_0|}{\sigma^2} \right ]\hspace{0.05cm}.$$
+:$$f_a(a) ={a}/{\sigma^2} \cdot {\rm e}^{  -{(a^2+ |z_0|^2) }/({2\sigma^2})}\cdot {\rm I}_0 \left [ {a \cdot |z_0|}/{\sigma^2} \right ]\hspace{0.05cm}.$$
 * For example, the modified zeroth order Bessel function returns the following values:
-$${\rm I }_0 (2) = 2.28\hspace{0.05cm},\hspace{0.2cm}{\rm I }_0 (4) = 11.30\hspace{0.05cm},\hspace{0.2cm}{\rm I }_0 (3) = 67.23
+:$${\rm I }_0 (2) = 2.28\hspace{0.05cm},\hspace{0.2cm}{\rm I }_0 (4) = 11.30\hspace{0.05cm},\hspace{0.2cm}{\rm I }_0 (3) = 67.23
   \hspace{0.05cm}.$$
 * The power (noncentral second moment) of the multiplicative factor&nbsp; $|z(t)|$ is
-$${\rm E}\left [ a^2 \right ] = {\rm E}\left [ |z(t)|^2 \right ] = 2 \cdot \sigma^2 + |z_0|^2
+:$${\rm E}\left [ a^2 \right ] = {\rm E}\left [ |z(t)|^2 \right ] = 2 \cdot \sigma^2 + |z_0|^2
     \hspace{0.05cm}.$$
-* With&nbsp; $z_0 = 0$,&nbsp; the <i>Rice fading</i>&nbsp; becomes <i>Rayleigh fading</i>, which is more critical. In this case, the probability that&nbsp; $a$&nbsp; lies in the yellow-shaded area between&nbsp; $0$&nbsp; and&nbsp; $1$&nbsp; is
+* With&nbsp; $z_0 = 0$,&nbsp; the Rice fading&nbsp; becomes Rayleigh fading, which is more critical.&nbsp; In this case, the probability that&nbsp; $a$&nbsp; lies in the yellow-shaded area between&nbsp; $0$&nbsp; and&nbsp; $1$&nbsp; is
-$$ {\rm Pr}(a \le 1) = 1 - {\rm e}^{-0.5/\sigma^2}  \approx 0.4
+:$$ {\rm Pr}(a \le 1) = 1 - {\rm e}^{-0.5/\sigma^2}  \approx 0.4
   \hspace{0.05cm}.$$
 In this task the probability&nbsp; ${\rm Pr}(a &#8804; 1)$&nbsp; for &nbsp;$|z_0| &ne; 0$&nbsp; is to be approximated. There are two ways to do this, namely:
-* the <i>triangular approximation</i>:
+* the triangular approximation: &nbsp; ${\rm Pr}(a \le 1) = {1}/{2} \cdot f_a(a=1)
-$${\rm Pr}(a \le 1) = {1}/{2} \cdot f_a(a=1)
+  \hspace{0.05cm}.$
-  \hspace{0.05cm}.$$
+* the  Gaussian approximation: &nbsp; If &nbsp; $|z_0| \gg \sigma$, then the Rice distribution can be approximated by a Gaussian distribution with mean&nbsp; $|z_0|$&nbsp; and standard deviation&nbsp; $\sigma$&nbsp;.
@@ Line 30: / Line 29: @@
 ''Notes:''
-* This task belongs to chapter&nbsp; [[Mobile_Communications/Nichtfrequenzselektives_Fading_mit_Direktkomponente| Nichtfrequenzselektives Fading mit Direktkomponente]].
+* This task belongs to chapter&nbsp; [[Mobile_Communications/Non-frequency_selective_fading_with_direct_component| Non-frequency selective fading with direct component]].
-* For the numerical solutions of the last subtasks, we recommend the interaction module&nbsp; [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen|Complementary Gaussian Error Functions]].
+* For the numerical solutions of the last subtasks, we recommend the interaction module&nbsp; [[Applets:Complementary_Gaussian_Error_Functions|Complementary Gaussian Error Functions]].