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

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2026-03-16T13:47:34Z

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2026-03-16T12:27:03Z

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	[[Category:Mobile Communications: Exercises\|^1.4 Fading with Direct Path Component^]]		[[Category:Mobile Communications: Exercises\|^1.4 Fading with Direct Path Component^]]
			[[de:Exercises:Exercise_1.6Z:_Comparison_of_Rayleigh_and_Rice]]

Hwang at 12:19, 18 January 2023

2023-01-18T12:19:13Z

← Older revision		Revision as of 14:19, 18 January 2023
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	'''(2)'''  For the given Rayleigh fading, we have  $\sigma^2 = 0.5$.  For the ~~quadratic mean~~ of the multiplicative factor  $z(t)$, this results in:		'''(2)'''  For the given Rayleigh fading, we have  $\sigma^2 = 0.5$.  For the second moment of the multiplicative factor  $z(t)$, this results in:
	:$${\rm E}\left [ \|z(t)\|^2 \right ] = 2 \sigma^2 = 1		:$${\rm E}\left [ \|z(t)\|^2 \right ] = 2 \sigma^2 = 1
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

Guenter at 14:44, 17 February 2022

2022-02-17T14:44:54Z

← Older revision		Revision as of 16:44, 17 February 2022
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	:$$f_a(a) = {a}/{\sigma^2} \cdot {\rm e}^{ -{a^2 }/({2\sigma^2})} \hspace{0.05cm}.$$		:$$f_a(a) = {a}/{\sigma^2} \cdot {\rm e}^{ -{a^2 }/({2\sigma^2})} \hspace{0.05cm}.$$

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

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

2022-02-17T11:29:04Z

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

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

	{{ML-Fuß}}		{{ML-Fuß}}

Javier: Text replacement - "„" to """

2021-05-28T14:46:40Z

Text replacement - "„" to """

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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).

Javier: Text replacement - "”" to """

2021-05-28T14:40:08Z

Text replacement - "”" to """

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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).

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

2021-05-03T15:13:10Z

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

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	In this task "Rayleigh fading" and "Rice fading" are to be compared with each other.		In this task "Rayleigh fading" and "Rice fading" are to be compared with each other.

	The graph shows the complex factor  $z(t) = x(t) + {\rm j} \cdot y(t)$   in the complex plane.  For the low-pass transmission signal  $s(t) = 1$  $($which with respect to a band pass system corresponds to a cosine oscillation with amplitude  $1)$,  the low-pass reception signal is  $r(t)=z(t)$.		The graph shows the complex factor  $z(t) = x(t) + {\rm j} \cdot y(t)$   in the complex plane.  For the low-pass transmission signal  $s(t) = 1$  $($which with respect to a band-pass system corresponds to a cosine oscillation with amplitude  $1)$,  the low-pass reception signal is  $r(t)=z(t)$.

	*The upper diagram describes  [[Mobile_Communications/Probability_Density_of_Rayleigh_Fading\| Rayleigh fading]], where the component signals  $x(t)$  and  $y(t)$  are each Gaussian distributed with variance  $\sigma^2$.  For  $a ≥ 0$, the probability density function of the magnitude  $a(t) = \|z(t)\|$  is:		*The upper diagram describes  [[Mobile_Communications/Probability_Density_of_Rayleigh_Fading\| Rayleigh fading]], where the component signals  $x(t)$  and  $y(t)$  are each Gaussian distributed with variance  $\sigma^2$.  For  $a ≥ 0$, the probability density function of the magnitude  $a(t) = \|z(t)\|$  is:
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*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:		*On the other hand, solution 2 applies to all possible signals:
	:$$\|z(t)\| = x_0 = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} r(t) = s(t)		:$$\|z(t)\| = x_0 = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} r(t) = s(t)

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"

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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^]]

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

2020-12-15T14:19:22Z

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

← Older revision		Revision as of 16:19, 15 December 2020
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	In this task "Rayleigh fading" and "Rice fading" are to be compared with each other.		In this task "Rayleigh fading" and "Rice fading" are to be compared with each other.

	The graph shows the complex factor  $z(t) = x(t) + {\rm j} \cdot y(t)$   in the complex plane.  For the low pass transmission signal  $s(t) = 1$  $($which with respect to a band pass system corresponds to a cosine oscillation with amplitude  $1)$,  the low pass reception signal is  $r(t)=z(t)$.		The graph shows the complex factor  $z(t) = x(t) + {\rm j} \cdot y(t)$   in the complex plane.  For the low-pass transmission signal  $s(t) = 1$  $($which with respect to a band pass system corresponds to a cosine oscillation with amplitude  $1)$,  the low-pass reception signal is  $r(t)=z(t)$.

	*The upper diagram describes  [[Mobile_Communications/Probability_Density_of_Rayleigh_Fading\| Rayleigh fading]], where the component signals  $x(t)$  and  $y(t)$  are each Gaussian distributed with variance  $\sigma^2$.  For  $a ≥ 0$, the probability density function of the magnitude  $a(t) = \|z(t)\|$  is:		*The upper diagram describes  [[Mobile_Communications/Probability_Density_of_Rayleigh_Fading\| Rayleigh fading]], where the component signals  $x(t)$  and  $y(t)$  are each Gaussian distributed with variance  $\sigma^2$.  For  $a ≥ 0$, the probability density function of the magnitude  $a(t) = \|z(t)\|$  is:
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	For the subtasks  '''(1)'''  to  '''(4)''',  let  $s(t) = 1$.  For the subtasks  '''(5)'''  and  '''(6)''', a BPSK signal is assumed .  The low pass signal  $s(t)$  thus has a rectangular shape with the possible values  $±1$.  The duration of each rectangular impulse is  $T = 10 \ \rm ms$.		For the subtasks  '''(1)'''  to  '''(4)''',  let  $s(t) = 1$.  For the subtasks  '''(5)'''  and  '''(6)''', a BPSK signal is assumed .  The low-pass signal  $s(t)$  thus has a rectangular shape with the possible values  $±1$.  The duration of each rectangular impulse is  $T = 10 \ \rm ms$.


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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*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:		*On the other hand, solution 2 applies to all possible signals:
	:$$\|z(t)\| = x_0 = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} r(t) = s(t)		:$$\|z(t)\| = x_0 = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} r(t) = s(t)

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	:The noncentral second moment  ("power")  of  $z(t)$  is  $1$:		:The noncentral second moment  ("power")  of  $z(t)$  is  $1$:
	:$${\rm E}\left [ \|z(t)\|^2 \right ] = 2 \sigma^2 = 1 \hspace{0.3cm}		:$${\rm E}\left [ \|z(t)\|^2 \right ] = 2 \sigma^2 = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\sigma = {1}/{\sqrt{2}} \approx 0.707\hspace{0.05cm}.$$
	\Rightarrow \hspace{0.3cm}
	\sigma = {1}/{\sqrt{2}} \approx 0.707
	\hspace{0.05cm}.$$

	*The lower phase diagram is the result of  [[Mobile_Communications/Non-~~Frequency_Selective_Fading_With_Direct_Component~~#Example_of_signal_behaviour_with_Rice_fading\|Rice fading]].  Here, too,   $x(t)$  and  $y(t)$  are Gaussian distributed with variance  $\sigma^2$, but now with mean value  $x_0$  and  $y_0$.  The PDF for  $a ≥ 0$   is		*The lower phase diagram is the result of  [[Mobile_Communications/Non-Frequency-Selective_Fading_With_Direct_Component#Example_of_signal_behaviour_with_Rice_fading\|Rice fading]].  Here, too,   $x(t)$  and  $y(t)$  are Gaussian distributed with variance  $\sigma^2$, but now with mean value  $x_0$  and  $y_0$.  The PDF for  $a ≥ 0$   is
	:$$f_a(a) ={a}/{\sigma^2} \cdot {\rm e}^{ -{(a^2+ \|z_0\|^2) }/({2\sigma^2})}\cdot {\rm I}_0 \left [ {a \cdot \|z_0\|}/{\sigma^2} \right ]\hspace{0.05cm}.$$		:$$f_a(a) ={a}/{\sigma^2} \cdot {\rm e}^{ -{(a^2+ \|z_0\|^2) }/({2\sigma^2})}\cdot {\rm I}_0 \left [ {a \cdot \|z_0\|}/{\sigma^2} \right ]\hspace{0.05cm}.$$
	:where  ${\rm I}_0$  is the modified Bessel function of the first kind.  The power now includes the contribution of the direct component  $z_0 = x_0 + {\rm j} \cdot y_0$:		:where  ${\rm I}_0$  is the modified Bessel function of the first kind.  The power now includes the contribution of the direct component  $z_0 = x_0 + {\rm j} \cdot y_0$:
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	'''(1)'''  <u>Solution 2</u> is correct:		'''(1)'''  <u>Solution 2</u> is correct:
	*The first suggestion corresponds to a Rayleigh fading model.  With the last setting, we would have		*The first suggestion corresponds to a Rayleigh fading model.  With the last setting, we would have
	:$$\|z(t)\| = {\rm j} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} r(t) = s(t) \cdot z(t) = {\rm j} \cdot s(t)		:$$\|z(t)\| = {\rm j} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} r(t) = s(t) \cdot z(t) = {\rm j} \cdot s(t)\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*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:		*On the other hand, solution 2 applies to all possible signals:
	:$$\|z(t)\| = x_0 = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} r(t) = s(t)		:$$\|z(t)\| = x_0 = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} r(t) = s(t)\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$


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

	The Rice fading should have exactly the same performance.  That is, it should apply:		The Rice fading should have exactly the same performance.  That is, it should apply:
	:$$\|z_0\|^2 + 2 \sigma^2 = 1		:$$\|z_0\|^2 + 2 \sigma^2 = 1\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	Furthermore, it was requested:		Furthermore, it was requested:
	* The ratio of the power of the deterministic part  $(\|z_0\|^2)$  and the stochastic part  $(2\sigma^2)$  is  $4$.  It follows from this:		* The ratio of the power of the deterministic part  $(\|z_0\|^2)$  and the stochastic part  $(2\sigma^2)$  is  $4$.  It follows from this:
	:$$2 \sigma^2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \sigma^2 = 0.1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \sigma = \frac{1}{\sqrt{10}} \hspace{0.25cm} \underline{ \approx 0.316} \hspace{0.05cm},$$		:$$2 \sigma^2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \sigma^2 = 0.1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \sigma = \frac{1}{\sqrt{10}} \hspace{0.25cm} \underline{ \approx 0.316} \hspace{0.05cm},$$
	:$$\|z_0\|^2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.8 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \|z_0\| = 0.894		:$$\|z_0\|^2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.8 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \|z_0\| = 0.894\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$
	* The phase of  $z_0 = x_0 + {\rm j} \cdot y_0$  results from the graph.  You can see that  $y_0 = x_0$  $($center of the cloud in the first quadrant below $45^\circ)$:		* The phase of  $z_0 = x_0 + {\rm j} \cdot y_0$  results from the graph.  You can see that  $y_0 = x_0$  $($center of the cloud in the first quadrant below $45^\circ)$:
	:$$x_0 = y_0 = \frac{\|z_0\|}{\sqrt{2}} = \frac{0.894}{\sqrt{2}} \hspace{0.25cm} \underline{ \approx 0.632} \hspace{0.05cm}.$$		:$$x_0 = y_0 = \frac{\|z_0\|}{\sqrt{2}} = \frac{0.894}{\sqrt{2}} \hspace{0.25cm} \underline{ \approx 0.632} \hspace{0.05cm}.$$
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	'''(4)'''  Correct is <u>solution 3</u>:		'''(4)'''  Correct is <u>solution 3</u>:
	*For the Rayleigh distribution, the requested probability is		*For the Rayleigh distribution, the requested probability is
	:$${\rm Pr}(20 \cdot {\rm lg}\hspace{0.15cm} a(t) \le 0\,\,{\rm dB}) = {\rm Pr}(a(t) \le 1) = 1 - {\rm e}^{-1} \approx 63\%\,\,\		:$${\rm Pr}(20 \cdot {\rm lg}\hspace{0.15cm} a(t) \le 0\,\,{\rm dB}) = {\rm Pr}(a(t) \le 1) = 1 - {\rm e}^{-1} \approx 63\%\,\,\\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*From the graph you can see that the probability for the Rice version (with the selected parameters) is about the same as the probability of falling below.		*From the graph you can see that the probability for the Rice version (with the selected parameters) is about the same as the probability of falling below.