Aufgaben:Exercise 1.6Z: Comparison of Rayleigh and Rice - Revision history

Hwang at 12:19, 18 January 2023

2023-01-18T12:19:13Z

Guenter at 14:44, 17 February 2022

2022-02-17T14:44:54Z

Bene: Text replacement - "autocorrelation" to "auto-correlation"

2022-02-17T11:29:04Z

Text replacement - "autocorrelation" to "auto-correlation"

Javier: Text replacement - "„" to """

2021-05-28T14:46:40Z

Text replacement - "„" to """

Javier: Text replacement - "”" to """

2021-05-28T14:40:08Z

Text replacement - "”" to """

Javier: Text replacement - "band pass" to "band-pass"

2021-05-03T15:13:10Z

Text replacement - "band pass" to "band-pass"

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

2021-03-23T12:38:04Z

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

Javier: Text replacement - "low pass" to "low-pass"

2020-12-15T14:19:22Z

Text replacement - "low pass" to "low-pass"

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"

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-'''(2)'''&nbsp; For the given Rayleigh fading, we have&nbsp; $\sigma^2 = 0.5$.&nbsp; For the quadratic mean of the multiplicative factor&nbsp; $z(t)$, this results in:
+'''(2)'''&nbsp; For the given Rayleigh fading, we have&nbsp; $\sigma^2 = 0.5$.&nbsp; For the second moment of the multiplicative factor&nbsp; $z(t)$, this results in:
 :$${\rm E}\left [ |z(t)|^2 \right ] = 2 \sigma^2 = 1
     \hspace{0.05cm}.$$

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 :$$f_a(a) = {a}/{\sigma^2} \cdot {\rm e}^{  -{a^2 }/({2\sigma^2})} \hspace{0.05cm}.$$
-:The noncentral second moment of&nbsp; $z(t)$&nbsp; is &nbsp;$1$:
+:The noncentral second moment&nbsp; ("power")&nbsp; of&nbsp; $z(t)$&nbsp; is &nbsp;$1$:
 :$${\rm E}\left [ |z(t)|^2 \right ] = 2  \sigma^2 = 1 \hspace{0.3cm}
 \Rightarrow \hspace{0.3cm}
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 *The lower phase diagram is the result of&nbsp; [[Mobile_Communications/Non-Frequency_Selective_Fading_With_Direct_Component#Example_of_signal_behaviour_with_Rice_fading|Rice fading]].&nbsp; Here, too, &nbsp; $x(t)$&nbsp; and&nbsp; $y(t)$&nbsp; are Gaussian distributed with variance&nbsp; $\sigma^2$, but now with mean value&nbsp; $x_0$&nbsp; and&nbsp; $y_0$.&nbsp; The PDF for&nbsp; $a &#8805; 0$ &nbsp; is
 :$$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}.$$
-:where&nbsp; ${\rm I}_0$&nbsp; is the modified Bessel function of the first kind.&nbsp; The root mean square now includes the direct component&nbsp; $z_0 = x_0 + {\rm j} \cdot y_0$:
+:where&nbsp; ${\rm I}_0$&nbsp; is the modified Bessel function of the first kind.&nbsp; The power now includes the contribution of the direct component&nbsp; $z_0 = x_0 + {\rm j} \cdot y_0$:
 :$${\rm E}\left [ |z(t)|^2 \right ] = 2 \cdot \sigma^2 + |z_0|^2 \hspace{0.05cm}.$$

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 *However, if you look at the development of the complex representation of&nbsp; $z(t)$&nbsp; and&nbsp; $r(t)$&nbsp; dynamically, there are indeed differences.
 *Larger jumps occur in the complex representation of&nbsp; $r(t)$, whenever there are phase jumps by&nbsp; $&plusmn;180^\circ$&nbsp; in the transmitted signal&nbsp; $s(t)$, i.e. when changing symbols.
-*Thus,&nbsp; ${\it \Phi}_z(f_{\rm D})$&nbsp; and&nbsp; ${\it \Phi}_r(f_{\rm D})$&nbsp; are different for Rice fading (the latter is wider) and accordingly the corresponding autocorrelation function.
+*Thus,&nbsp; ${\it \Phi}_z(f_{\rm D})$&nbsp; and&nbsp; ${\it \Phi}_r(f_{\rm D})$&nbsp; are different for Rice fading (the latter is wider) and accordingly the corresponding auto-correlation function.
 {{ML-Fuß}}

← Older revision		Revision as of 14:46, 28 May 2021
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−	The graph shows the result of a system simulation with the program ~~&bdquo;~~Mobile radio channel" from the (former) practical course ~~&bdquo;~~Simulation of digital transmission systems".	+	The graph shows the result of a system simulation with the program "Mobile radio channel" from the (former) practical course "Simulation of digital transmission systems".
	:Both from the signal section as well as from the PDF one can see that the blue curve (Rayleigh) has more parts under the violet curve than the red curve (Rice).		:Both from the signal section as well as from the PDF one can see that the blue curve (Rayleigh) has more parts under the violet curve than the red curve (Rice).

← Older revision		Revision as of 14:40, 28 May 2021
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−	The graph shows the result of a system simulation with the program &bdquo;Mobile radio channel~~”~~ from the (former) practical course &bdquo;Simulation of digital transmission systems~~”~~.	+	The graph shows the result of a system simulation with the program &bdquo;Mobile radio channel" from the (former) practical course &bdquo;Simulation of digital transmission systems".
	:Both from the signal section as well as from the PDF one can see that the blue curve (Rayleigh) has more parts under the violet curve than the red curve (Rice).		:Both from the signal section as well as from the PDF one can see that the blue curve (Rayleigh) has more parts under the violet curve than the red curve (Rice).

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 In this task "Rayleigh fading" and "Rice fading" are to be compared with each other.
-The graph shows the complex factor&nbsp; $z(t) = x(t) + {\rm j} \cdot y(t)$ &nbsp; in the complex plane.&nbsp; For the low-pass transmission signal&nbsp; $s(t) = 1$&nbsp; $($which with respect to a band pass system corresponds to a cosine oscillation with amplitude&nbsp; $1)$,&nbsp; the low-pass reception signal is&nbsp; $r(t)=z(t)$.
+The graph shows the complex factor&nbsp; $z(t) = x(t) + {\rm j} \cdot y(t)$ &nbsp; in the complex plane.&nbsp; For the low-pass transmission signal&nbsp; $s(t) = 1$&nbsp; $($which with respect to a band-pass system corresponds to a cosine oscillation with amplitude&nbsp; $1)$,&nbsp; the low-pass reception signal is&nbsp; $r(t)=z(t)$.
 *The upper diagram describes&nbsp; [[Mobile_Communications/Probability_Density_of_Rayleigh_Fading| Rayleigh fading]], where the component signals&nbsp; $x(t)$&nbsp; and&nbsp; $y(t)$&nbsp; are each Gaussian distributed with variance&nbsp; $\sigma^2$.&nbsp;  For&nbsp; $a &#8805; 0$, the probability density function of the magnitude&nbsp; $a(t) = |z(t)|$&nbsp; is:
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     \hspace{0.05cm}.$$
-*If we take into account that we are in the equivalent low-pass range, then a cosine input would produce a negative sinusoidal band pass signal $r_{\rm BP}(t)$.
+*If we take into account that we are in the equivalent low-pass range, then a cosine input would produce a negative sinusoidal band-pass signal $r_{\rm BP}(t)$.
 *On the other hand, solution 2 applies to all possible signals:
 :$$|z(t)| = x_0 = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} r(t) = s(t)

← Older revision		Revision as of 12:38, 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^]]

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 </quiz>
-===Solutions===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; <u>Solution 2</u> is correct:

← 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_ID2135__Mob_Z_1_6.png\|right\|frame\|Phase diagram of the factor  $z(t) = x(t) + {\rm j} \cdot y(t)$  <br>for Rayleigh and Rice fading]]		[[File:P_ID2135__Mob_Z_1_6.png\|right\|frame\|Phase diagram of the factor  $z(t) = x(t) + {\rm j} \cdot y(t)$  <br>for Rayleigh and Rice fading]]