Aufgaben:Exercise 2.2: Simple Two-Path Channel Model - Versionsgeschichte

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	[[Category:Exercises for Mobile Communications\|^2.2 Multi-Path Reception in Wireless Systems^]]		[[Category:Exercises for Mobile Communications\|^2.2 Multi-Path Reception in Wireless Systems^]]
			[[en:Aufgaben:Exercise_2.2:_Simple_Two-Path_Channel_Model]]

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← Nächstältere Version		Version vom 24. Februar 2026, 15:40 Uhr
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	By Fourier transformation you get the frequency response		By Fourier transformation you get the frequency response
	:$$H_{\rm 0}(f) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1 + k_0 \cdot {\rm e}^{- {\rm j}\cdot 2 \pi f \tau_0}=1 + k_0 \cdot {\cos}( 2 \pi f \tau_0) + {\rm j}\cdot k_0 \cdot {\sin		:$$H_{\rm 0}(f) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1 + k_0 \cdot {\rm e}^{- {\rm j}\cdot 2 \pi f \tau_0}=1 + k_0 \cdot {\cos}( 2 \pi f \tau_0) + {\rm j}\cdot k_0 \cdot {\sin}( 2 \pi f \tau_0)\hspace{0.05cm},$$
	}( 2 \pi f \tau_0)
	\hspace{0.05cm},$$

	and thus to the following magnitude of the frequency response:		and thus to the following magnitude of the frequency response:
	$$\|H_{\rm 0}(f)\| = \sqrt{ \left [ 1 + k_0 \cdot {\cos}( 2 \pi f \tau_0)\right ]^2 + k_0^2 \cdot {\sin^2 }( 2 \pi f \tau_0)}\hspace{0.3cm}		$$\|H_{\rm 0}(f)\| = \sqrt{ \left [ 1 + k_0 \cdot {\cos}( 2 \pi f \tau_0)\right ]^2 + k_0^2 \cdot {\sin^2 }( 2 \pi f \tau_0)}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\|H_{\rm 0}(f = 0)\| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1+ k_0 \hspace{0.1cm} \underline {=1.2} \hspace{0.05cm},$$
	\Rightarrow \hspace{0.3cm}\|H_{\rm 0}(f = 0)\| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1+ k_0 \hspace{0.1cm} \underline {=1.2} \hspace{0.05cm},$$
	$$\|H_{\rm 0}(f = 250\,{\rm kHz})\| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{ \left [ 1 + k_0 \cdot {\cos}( \pi/2)\right ]^2 + k_0^2 \cdot {\sin^2 }( \pi/2)} = \sqrt{1+ k_0^2} \hspace{0.1cm} \underline {\approx 1.02} \hspace{0.05cm},$$		$$\|H_{\rm 0}(f = 250\,{\rm kHz})\| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{ \left [ 1 + k_0 \cdot {\cos}( \pi/2)\right ]^2 + k_0^2 \cdot {\sin^2 }( \pi/2)} = \sqrt{1+ k_0^2} \hspace{0.1cm} \underline {\approx 1.02} \hspace{0.05cm},$$
	$$\|H_{\rm 0}(f = 500\,{\rm kHz})\| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{ \left [ 1 + k_0 \cdot {\cos}( \pi)\right ]^2 + k_0^2 \cdot {\sin^2 }( \pi)} = {1- k_0} \hspace{0.1cm} \underline {= 0.8} \hspace{0.05cm}.$$		$$\|H_{\rm 0}(f = 500\,{\rm kHz})\| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{ \left [ 1 + k_0 \cdot {\cos}( \pi)\right ]^2 + k_0^2 \cdot {\sin^2 }( \pi)} = {1- k_0} \hspace{0.1cm} \underline {= 0.8} \hspace{0.05cm}.$$

Javier am 20. Mai 2020 um 15:41 Uhr

2020-05-20T15:41:21Z

← Nächstältere Version		Version vom 20. Mai 2020, 17:41 Uhr
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	[[Datei:EN_Mob_A_2_2.png\|right\|frame\|Two equivalent models <br>for the two-path channel]]		[[Datei:EN_Mob_A_2_2.png\|right\|frame\|Two equivalent models <br>for the two-path channel]]
	Here we consider a two-~~way–~~channel for mobile radio according to the adjacent graph, characterized by the model parameters		Here we consider a two-path channel for mobile radio according to the adjacent graph, characterized by the model parameters
	:$$k_1 = 10^{-4}\hspace{0.05cm}, \hspace{0.2cm} \tau_{1} = 10\,{\rm µ s}\hspace{0.05cm}, \hspace{0.2cm}\tau_{2} = 11\,{\rm µ s} \hspace{0.05cm}.$$		:$$k_1 = 10^{-4}\hspace{0.05cm}, \hspace{0.2cm} \tau_{1} = 10\,{\rm µ s}\hspace{0.05cm}, \hspace{0.2cm}\tau_{2} = 11\,{\rm µ s} \hspace{0.05cm}.$$

Javier am 20. Mai 2020 um 15:41 Uhr

2020-05-20T15:41:05Z

← Nächstältere Version		Version vom 20. Mai 2020, 17:41 Uhr
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	{{quiz-Header\|Buchseite=Mobile Kommunikation/Mehrwegeempfang beim Mobilfunk}}		{{quiz-Header\|Buchseite=Mobile Kommunikation/Mehrwegeempfang beim Mobilfunk}}

	[[Datei:EN_Mob_A_2_2.png\|right\|frame\|~~Zwei äquivalente Modelle~~ <br>~~für den Zweiwege~~-~~Kanal~~]]		[[Datei:EN_Mob_A_2_2.png\|right\|frame\|Two equivalent models <br>for the two-path channel]]
	Here we consider a two-way–channel for mobile radio according to the adjacent graph, characterized by the model parameters		Here we consider a two-way–channel for mobile radio according to the adjacent graph, characterized by the model parameters
	:$$k_1 = 10^{-4}\hspace{0.05cm}, \hspace{0.2cm} \tau_{1} = 10\,{\rm µ s}\hspace{0.05cm}, \hspace{0.2cm}\tau_{2} = 11\,{\rm µ s} \hspace{0.05cm}.$$		:$$k_1 = 10^{-4}\hspace{0.05cm}, \hspace{0.2cm} \tau_{1} = 10\,{\rm µ s}\hspace{0.05cm}, \hspace{0.2cm}\tau_{2} = 11\,{\rm µ s} \hspace{0.05cm}.$$

Guenter am 11. Mai 2020 um 14:05 Uhr

2020-05-11T14:05:23Z

← Nächstältere Version		Version vom 11. Mai 2020, 16:05 Uhr
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	{{quiz-Header\|Buchseite=Mobile Kommunikation/Mehrwegeempfang beim Mobilfunk}}		{{quiz-Header\|Buchseite=Mobile Kommunikation/Mehrwegeempfang beim Mobilfunk}}

	[[Datei:~~P_ID2157__Mob_A_2_2~~.png\|right\|frame\|Zwei äquivalente Modelle <br>für den Zweiwege-Kanal]]		[[Datei:EN_Mob_A_2_2.png\|right\|frame\|Zwei äquivalente Modelle <br>für den Zweiwege-Kanal]]
	Here we consider a two-way–channel for mobile radio according to the adjacent graph, characterized by the model parameters		Here we consider a two-way–channel for mobile radio according to the adjacent graph, characterized by the model parameters
	:$$k_1 = 10^{-4}\hspace{0.05cm}, \hspace{0.2cm} \tau_{1} = 10\,{\rm µ s}\hspace{0.05cm}, \hspace{0.2cm}\tau_{2} = 11\,{\rm µ s} \hspace{0.05cm}.$$		:$$k_1 = 10^{-4}\hspace{0.05cm}, \hspace{0.2cm} \tau_{1} = 10\,{\rm µ s}\hspace{0.05cm}, \hspace{0.2cm}\tau_{2} = 11\,{\rm µ s} \hspace{0.05cm}.$$

Javier am 15. April 2020 um 13:02 Uhr

2020-04-15T13:02:51Z

← Nächstältere Version		Version vom 15. April 2020, 15:02 Uhr
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	*The result is independent from $k_2$. It applies to $k_2 = 2 \cdot 10^{-5} \Rightarrow k_0 = 0.2$ and $k_2 = 10^{-4} \Rightarrow k_0 = $1 in the same way.		*The result is independent from $k_2$. It applies to $k_2 = 2 \cdot 10^{-5} \Rightarrow k_0 = 0.2$ and $k_2 = 10^{-4} \Rightarrow k_0 = $1 in the same way.
	*This approximation $B_{\rm K}\hspace{0.01cm}'$ of the coherence bandwidth si shown in the graph.		*This approximation $B_{\rm K}\hspace{0.01cm}'$ of the coherence bandwidth is shown in the graph.

Javier am 15. April 2020 um 13:02 Uhr

2020-04-15T13:02:11Z

← Nächstältere Version		Version vom 15. April 2020, 15:02 Uhr
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	*On the other hand:		*On the other hand:
	:$$\|H_{\rm 0}(f = 750\,{\rm kHz})\| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \|H_{\rm 0}(f = 250\,{\rm kHz})\| \approx 1.02 > 1\hspace{0.05cm},$$		:$$\|H_{\rm 0}(f = 750\,{\rm kHz})\| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \|H_{\rm 0}(f = 250\,{\rm kHz})\| \approx 1.02 > 1\hspace{0.05cm},$$
	:$$\|H_{\rm 0}(f = 1\,{\rm MHz})\| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \|H_{\rm 0}(f = 0)\| = 1.2 > 1 \hspace{0.05cm}.$		:$$\|H_{\rm 0}(f = 1\,{\rm MHz})\| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \|H_{\rm 0}(f = 0)\| = 1.2 > 1 \hspace{0.05cm}.$$

Javier am 15. April 2020 um 13:01 Uhr

2020-04-15T13:01:50Z

← Nächstältere Version		Version vom 15. April 2020, 15:01 Uhr
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	'''(5)'''  The difference $\tau_{\rm max} \ – \tau_{\rm min}$ between the delays in the two paths is equal to $\tau_0 = 1 \ \ \rm µ s$.		'''(5)'''  The difference $\tau_{\rm max} \ – \tau_{\rm min}$ between the delays in the two paths is equal to $\tau_0 = 1 \ \ \rm µ s$.
	*So the coherence bandwidth is		*So the coherence bandwidth is
	:$$B_{\rm K}\hspace{0.01cm}' = {1}/{\tau_{\rm 0} \hspace{0.1cm} \underline {=1\,{\rm MHz} \hspace{0.05cm}.$$		:$$B_{\rm K}\hspace{0.01cm}' = {1}/{\tau_{\rm 0} } \hspace{0.1cm} \underline {=1\,{\rm MHz}} \hspace{0.05cm}.$$

	*The result is independent from $k_2$. It applies to $k_2 = 2 \cdot 10^{-5} \Rightarrow k_0 = 0.2$ and $k_2 = 10^{-4} \Rightarrow k_0 = $1 in the same way.		*The result is independent from $k_2$. It applies to $k_2 = 2 \cdot 10^{-5} \Rightarrow k_0 = 0.2$ and $k_2 = 10^{-4} \Rightarrow k_0 = $1 in the same way.
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	*The channel is non-frequency-selective if the coherence bandwidth $B_{\rm K}$ is significantly larger than the signal bandwidth $B_{\rm S}$.		*The channel is non-frequency-selective if the coherence bandwidth $B_{\rm K}$ is significantly larger than the signal bandwidth $B_{\rm S}$.
	*For the given channel, this is true for GSM, but not for UMTS. For UMTS, this is frequency-selective channel.		*For the given channel, this is true for GSM, but not for UMTS. For UMTS, this is frequency-selective channel.
	{{{ML-Fuß}}		{{ML-Fuß}}

	[[Category:Exercises for Mobile Communications\|^2.2 Multi-Path Reception in Wireless Systems^]]		[[Category:Exercises for Mobile Communications\|^2.2 Multi-Path Reception in Wireless Systems^]]

Javier am 15. April 2020 um 13:01 Uhr

2020-04-15T13:01:04Z

← Nächstältere Version		Version vom 15. April 2020, 15:01 Uhr
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	\hspace{0.05cm},$$		\hspace{0.05cm},$$

	and thus to the following magnitude frequency response:		and thus to the following magnitude of the frequency response:
	$$\|H_{\rm 0}(f)\| = \sqrt{ \left [ 1 + k_0 \cdot {\cos}( 2 \pi f \tau_0)\right ]^2 + k_0^2 \cdot {\sin^2 }( 2 \pi f \tau_0)}\hspace{0.3cm}		$$\|H_{\rm 0}(f)\| = \sqrt{ \left [ 1 + k_0 \cdot {\cos}( 2 \pi f \tau_0)\right ]^2 + k_0^2 \cdot {\sin^2 }( 2 \pi f \tau_0)}\hspace{0.3cm}
	\Rightarrow \hspace{0.3cm}\|H_{\rm 0}(f = 0)\| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1+ k_0 \hspace{0.1cm} \underline {=1.2} \hspace{0.05cm},$$		\Rightarrow \hspace{0.3cm}\|H_{\rm 0}(f = 0)\| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 1+ k_0 \hspace{0.1cm} \underline {=1.2} \hspace{0.05cm},$$
Zeile 94:		Zeile 94:
	$$\|H_{\rm 0}(f = 500\,{\rm kHz})\| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{ \left [ 1 + k_0 \cdot {\cos}( \pi)\right ]^2 + k_0^2 \cdot {\sin^2 }( \pi)} = {1- k_0} \hspace{0.1cm} \underline {= 0.8} \hspace{0.05cm}.$$		$$\|H_{\rm 0}(f = 500\,{\rm kHz})\| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{ \left [ 1 + k_0 \cdot {\cos}( \pi)\right ]^2 + k_0^2 \cdot {\sin^2 }( \pi)} = {1- k_0} \hspace{0.1cm} \underline {= 0.8} \hspace{0.05cm}.$$

	[[File:P_ID2158__Mob_A_2_2c.png\|right\|frame\|~~Amount~~ frequency response of a two-way channel]]		[[File:P_ID2158__Mob_A_2_2c.png\|right\|frame\|Magnitude of the frequency response of a two-way channel]]
	The graphic (red curve) shows the function $\|H_0(f)\|$.		The graphic (red curve) shows the function $\|H_0(f)\|$.
	*The values you are looking for are marked by the yellow dots.		*The values you are looking for are marked by the yellow dots.
Zeile 101:		Zeile 101:


	'''(4)'''  ~~Correct is the~~ <u>~~solution~~ 1</u>:		'''(4)'''  <u>Solution 1</u> is correct:
	*Destructive ~~overlays exist~~ for $\|H_0(f)\| < 1$, for example for $f = 500 \ \rm kHz$.		*Destructive interference occurs for $\|H_0(f)\| < 1$, for example for $f = 500 \ \rm kHz$.
	*On the other hand:		*On the other hand:
	:$$\|H_{\rm 0}(f = 750\,{\rm kHz})\| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \|H_{\rm 0}(f = 250\,{\rm kHz})\| \approx 1.02 > 1\hspace{0.05cm},$$		:$$\|H_{\rm 0}(f = 750\,{\rm kHz})\| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \|H_{\rm 0}(f = 250\,{\rm kHz})\| \approx 1.02 > 1\hspace{0.05cm},$$
Zeile 108:		Zeile 108:


	'''(5)'''  The difference $\tau_{\rm max} \ – \tau_{\rm min}$ of the ~~delay times~~ in the two paths is equal to $\tau_0 = 1 \ \ \rm µ s$.		'''(5)'''  The difference $\tau_{\rm max} \ – \tau_{\rm min}$ between the delays in the two paths is equal to $\tau_0 = 1 \ \ \rm µ s$.
	*So the coherence bandwidth		*So the coherence bandwidth is
	:$$B_{\rm K}\hspace{0.01cm}' = {1}/{\tau_{\rm 0} \hspace{0.1cm} \underline {=1\,{\rm MHz} \hspace{0.05cm}.$$		:$$B_{\rm K}\hspace{0.01cm}' = {1}/{\tau_{\rm 0} \hspace{0.1cm} \underline {=1\,{\rm MHz} \hspace{0.05cm}.$$

	*The result is independent from $k_2$. It applies to $k_2 = 2 \cdot 10^{-5} \Rightarrow k_0 = 0.2$ and $k_2 = 10^{-4} \Rightarrow k_0 = $1 in the same way.		*The result is independent from $k_2$. It applies to $k_2 = 2 \cdot 10^{-5} \Rightarrow k_0 = 0.2$ and $k_2 = 10^{-4} \Rightarrow k_0 = $1 in the same way.
	*~~In the graph this~~ approximation $B_{\rm K}\hspace{0.01cm}'$ ~~is drawn for~~ the coherence bandwidth.		*This approximation $B_{\rm K}\hspace{0.01cm}'$ of the coherence bandwidth si shown in the graph.


	'''(6)'''  ~~Correct is the~~ <u>~~solution~~ 2</u>:		'''(6)'''  <u>Solution 2</u> is correct:
	*The channel is ~~not~~ frequency selective if the coherence bandwidth $B_{\rm K}$ is significantly larger than the signal bandwidth $B_{\rm S}$.		*The channel is non-frequency-selective if the coherence bandwidth $B_{\rm K}$ is significantly larger than the signal bandwidth $B_{\rm S}$.
	*For the given channel, this is true for GSM, but not for UMTS. For UMTS ~~there~~ is a frequency-selective channel.		*For the given channel, this is true for GSM, but not for UMTS. For UMTS, this is frequency-selective channel.
	{{{ML-Fuß}}		{{{ML-Fuß}}

	[[Category:Exercises for Mobile Communications\|^2.2 Multi-Path Reception in Wireless Systems^]]		[[Category:Exercises for Mobile Communications\|^2.2 Multi-Path Reception in Wireless Systems^]]

Javier am 15. April 2020 um 12:58 Uhr

2020-04-15T12:58:00Z

← Nächstältere Version		Version vom 15. April 2020, 14:58 Uhr
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	'''(3)'''  The impulse location is		'''(3)'''  The impulse location is
	$$h_{\rm 0}(\tau) = \delta(\tau) + k_0 \cdot \delta(\tau - \tau_0) \hspace{0.05cm}.$		$$h_{\rm 0}(\tau) = \delta(\tau) + k_0 \cdot \delta(\tau - \tau_0) \hspace{0.05cm}.$$

	By Fourier transformation you get the frequency response		By Fourier transformation you get the frequency response