Aufgaben:Exercise 2.2: Simple Two-Path Channel Model - Revision history

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

2021-03-23T12:37:43Z

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

Guenter at 14:55, 11 January 2021

2021-01-11T14:55:32Z

Guenter at 14:41, 11 January 2021

2021-01-11T14:41:31Z

Guenter at 14:39, 11 January 2021

2021-01-11T14:39:19Z

Javier: Text replacement - "===Sample solution===" to "===Solution==="

2020-12-16T13:16:30Z

Text replacement - "===Sample solution===" to "===Solution==="

Javier: Text replacement - "Mobile Kommunikation/Mehrwegeempfang beim Mobilfunk" to "Mobile_Communications/Multipath_Reception_in_Mobile_Communications"

2020-12-15T14:00:22Z

Text replacement - "Mobile Kommunikation/Mehrwegeempfang beim Mobilfunk" to "Mobile_Communications/Multipath_Reception_in_Mobile_Communications"

Javier: Text replacement - "Mobile_Kommunikation/" to "Mobile_Communications/"

2020-07-09T09:13:38Z

Text replacement - "Mobile_Kommunikation/" to "Mobile_Communications/"

Javier at 15:41, 20 May 2020

2020-05-20T15:41:21Z

Javier at 15:41, 20 May 2020

2020-05-20T15:41:05Z

Guenter at 14:05, 11 May 2020

2020-05-11T14:05:23Z

← Older revision		Revision as of 12:37, 23 March 2021
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	{{ML-Fuß}}		{{ML-Fuß}}

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

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 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; We have $\tau_1 = d_1/c$ &nbsp;&#8658;&nbsp; $ d_1 = \tau_1 \cdot c = 10^{-5} \rm s \cdot 3 \cdot 10^8 \ m/s \ \ \underline {= 3 \ km}$.
+'''(1)'''&nbsp; We have&nbsp; $\tau_1 = d_1/c$ &nbsp;&#8658;&nbsp; $ d_1 = \tau_1 \cdot c = 10^{-5} \rm s \cdot 3 \cdot 10^8 \ m/s \ \ \underline {= 3 \ km}$.
-'''(2)'''&nbsp; The damping factor is $k_0 = k_2/k_1 \ \ \underline {= 0.2}$ and the delay time $\tau_0 = \tau_2 \ &ndash; \tau_1 \ \underline {= 1 \ \ \rm &micro; s}$.
+'''(2)'''&nbsp; The damping factor is&nbsp; $k_0 = k_2/k_1 \ \ \underline {= 0.2}$&nbsp; and the delay time&nbsp; $\tau_0 = \tau_2 \ - \tau_1 \ \underline {= 1 \ \ \rm &micro; s}$.
-*The path loss effective for both paths is thus $k_1 = 10^{-4}$ and the basic delay time is $\tau_1 = 10 \ \ \rm &micro; s$.
+*The effective path loss for both paths is thus&nbsp; $k_1 = 10^{-4}$&nbsp; and the basic delay time is&nbsp; $\tau_1 = 10 \ \ \rm &micro; s$.
 '''(3)'''&nbsp; 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
@@ Line 89: / Line 90: @@
 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},$$
-$$|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}.$$
 [[File:P_ID2158__Mob_A_2_2c.png|right|frame|Magnitude of the frequency response of a two-way channel]]
-The graphic (red curve) shows the function $|H_0(f)|$.
+The graphic (red curve) shows the function&nbsp;  $|H_0(f)|$.
 *The values you are looking for are marked by the yellow dots.
-*The blue curve refers to the subtask (5) with $k_0 = 1 \ \Rightarrow \ k_2 = k_0 \cdot k_1 = 10^{&ndash;4}$.
+*The blue curve refers to&nbsp; '''(5)'''&nbsp; with&nbsp; $k_0 = 1 \ \Rightarrow \ k_2 = k_0 \cdot k_1 = 10^{-4}$.
 '''(4)'''&nbsp; <u>Solution 1</u> is correct:
-*Destructive interference occurs for $|H_0(f)| < 1$, for example for $f = 500 \ \rm kHz$.
+*Destructive interference occurs for $|H_0(f)| < 1$, e.g. for $f = 500 \ \rm kHz$.
 *On the other hand:
 :$$|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},$$
@@ Line 108: / Line 109: @@
-'''(5)'''&nbsp; The difference $\tau_{\rm max} \ &ndash; \tau_{\rm min}$ between the delays in the two paths is equal to $\tau_0 = 1 \ \ \rm &micro; s$.
+'''(5)'''&nbsp; The difference&nbsp; $\tau_{\rm max} - \tau_{\rm min}$&nbsp; between the delays in the two paths is equal to&nbsp; $\tau_0 = 1 \ \ \rm &micro; s$.
 *So the coherence bandwidth is
 :$$B_{\rm K}\hspace{0.01cm}' = {1}/{\tau_{\rm 0} } \hspace{0.1cm} \underline {=1\,{\rm MHz}} \hspace{0.05cm}.$$
-*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&nbsp; $k_2$.&nbsp; It applies to&nbsp; $k_2 = 2 \cdot 10^{-5} \Rightarrow \ k_0 = 0.2$&nbsp; and&nbsp; $k_2 = 10^{-4} \Rightarrow \ k_0 = 1$&nbsp; in the same way.
-*This approximation $B_{\rm K}\hspace{0.01cm}'$ of the coherence bandwidth is shown in the graph.
+*This approximation&nbsp; $B_{\rm K}\hspace{0.01cm}'$&nbsp; of the coherence bandwidth is shown in the graph.
 '''(6)'''&nbsp; <u>Solution 2</u> is correct:
-*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&ndash;frequency&ndash;selective if the coherence bandwidth&nbsp; $B_{\rm K}$&nbsp; is significantly larger than the signal bandwidth&nbsp; $B_{\rm S}$.
-*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 a frequency&nbsp;selective channel.
 {{ML-Fuß}}
 [[Category:Exercises for Mobile Communications|^2.2 Multi-Path Reception in Wireless Systems^]]

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 {{quiz-Header|Buchseite=Mobile_Communications/Multipath_Reception_in_Mobile_Communications}}
-[[File:EN_Mob_A_2_2.png|right|frame|Two equivalent models <br>for the two-path channel]]
+[[File:EN_Mob_A_2_2.png|right|frame|Two equivalent models for the two-path channel]]
 Here we consider a two-path channel for mobile radio according to the adjacent graph, characterized by the model parameters
 :$$k_1 = 10^{-4}\hspace{0.05cm}, \hspace{0.2cm} \tau_{1} = 10\,{\rm &micro; s}\hspace{0.05cm}, \hspace{0.2cm}\tau_{2} = 11\,{\rm &micro; s} \hspace{0.05cm}.$$

@@ Line 11: / Line 11: @@
-An equivalent channel model is shown below, with only the part highlighted in green being considered further. This means: &nbsp;
+An equivalent channel model is shown below, with only the part highlighted in green being considered further. This means:
 *The basic attenuation (path loss) and the basic propagation time are not taken into account here.
-*The frequency response of this&nbsp; $(k_0, \tau_0$)&ndash;model is designated&nbsp; $H_0(f)$&nbsp;.
+*The frequency response of this&nbsp; $(k_0, \tau_0$)&ndash;model is designated&nbsp; $H_0(f)$.
-An important descriptive parameter of any mobile radio system is the coherence bandwidth&nbsp; $B_{\rm K}$, which is defined in the chapter&nbsp; [[Mobile_Communications/Das_GWSSUS%E2%80%93Kanalmodell| GWSSUS&ndash;Kanalmodell]]&nbsp;. The coherence bandwidth indicates whether the system can be approximated as non-frequency-selective:
+An important descriptive parameter of any mobile radio system is the coherence bandwidth&nbsp; $B_{\rm K}$, which is defined in the chapter&nbsp; [[Mobile_Communications/The_GWSSUS_Channel_Model| The GWSSUS Channel Model]]&nbsp;. The coherence bandwidth indicates whether the system can be approximated as non&ndash;frequency&ndash;selective:
 *This is justified if the signal bandwidth&nbsp; $B_{\rm S}$&nbsp; is significantly smaller than the coherence bandwidth&nbsp; $B_{\rm K}$.
-*Otherwise, the mobile radio system is frequency-selective, which requires a more complicated description.
+*Otherwise, the mobile radio system is frequency&ndash;selective, which requires a more complicated description.
-As a simple approximation for the coherence bandwidth, the reciprocal value of pulse broadening is often used in the literature (marked by an apostrophe in our learning tutorial):
+As a simple approximation for the coherence bandwidth, the reciprocal value of pulse broadening is often used in the literature (marked by an apostrophe in our tutorial):
 :$$B_{\rm K}\hspace{0.01cm}' = \frac{1}{\tau_{\rm max} - \tau_{\rm min}} \hspace{0.05cm}.$$
@@ Line 31: / Line 31: @@
 ''Notes:''
-* This task belongs to the topic of the chapter&nbsp; [[Mobile_Communications/Mehrwegeempfang_beim_Mobilfunk| Mehrwegeempfang beim Mobilfunk]].
+* This task belongs to the topic of the chapter&nbsp; [[Mobile_Communications/Multi-Path_Reception_in_Mobile_Communications| Multi&ndash;Path Reception in Mobile Communications]].
 * For the solution you also need the speed of light&nbsp; $c = 3 \cdot 10^8 \ \rm m/s$.
-* For&nbsp; $k_2$&nbsp; only positive values are used here. However, as you may remember,if the secondary path is created by reflection on a wall, a phase change by&nbsp; $\pi$&nbsp; occurs, resulting in a negative value of $k_2$.
+* For&nbsp; $k_2$&nbsp; only positive values are used here. However, as you may remember, if the secondary path is created by reflection on a wall, a phase change by&nbsp; $\pi$&nbsp; occurs, resulting in a negative value of&nbsp; $k_2$.
@@ Line 44: / Line 44: @@
 $d_1 \ = \ ${ 3 3% } $\ \ \rm km$
-{What are the parameters of the simplified model for&nbsp; $k_2 = 2 \cdot 10^{-5}$?
+{What are the parameters of the simplified model for&nbsp; $k_2 = 2 \cdot 10^{-5}$&nbsp;?
 |type="{}"}
 $k_0 \ = \ ${ 0.2 3% }
 $\tau_0 \ = \ ${ 1 3% } $\ \ \rm &micro; s$
-{Calculate the magnitude of the frequency response &nbsp; &#8658; &nbsp; $|H_0(f)|$&nbsp; of the simplified model for the frequencies&nbsp; $f = 0$,&nbsp; $f = 250 \ \rm kHz$&nbsp; and&nbsp; $f = 500 \ \rm kHz$
+{Calculate the magnitude of the frequency response &nbsp; &#8658; &nbsp; $|H_0(f)|$&nbsp; of the simplified model for the frequencies&nbsp; $f = 0$,&nbsp; $f = 250 \ \rm kHz$&nbsp; and&nbsp; $f = 500 \ \rm kHz$&nbsp;?
 |type="{}"}
 $|H_0(f = 0)| \ = \ ${ 1.2 3% }
@@ Line 61: / Line 61: @@
 - $f_{\rm S} = 1 \ \rm MHz$.
-{What is the approximate coherence bandwidth for&nbsp; $k_2 = 2 \cdot 10^{-5}$&nbsp; or.&nbsp; $k_2 = 10^{-4}$&nbsp;?
+{What is the approximate coherence bandwidth for&nbsp; $k_2 = 2 \cdot 10^{-5}$&nbsp; and&nbsp; $k_2 = 10^{-4}$&nbsp;?
 |type="{}"}
-$k_2 = 2 \cdot 10^{-5} \text{:} \ \hspace{0.4cm} B_{\rm K}\hspace{0.01cm}' = \ ${ 1 3% } $\ \rm MHz
+$k_2 = 2 \cdot 10^{-5} \text{:} \ \hspace{0.4cm} B_{\rm K}\hspace{0.01cm}' = \ ${ 1 3% } $\ \rm MHz$
-$k_2 = 10^{-4} \text{:} \ \hspace{0.4cm} B_{\rm K}\hspace{0.01cm}' = \ ${ 1 3% } $\ \rm MHz
+$k_2 = 10^{-4} \text{:} \ \hspace{0.4cm} B_{\rm K}\hspace{0.01cm}' = \ ${ 1 3% } $\ \rm MHz$
 {Which statements are correct regarding frequency selectivity if&nbsp; $B_{\rm S}$&nbsp; denotes the signal bandwidth?
 |type="[]"}
-- For GSM: &nbsp; $(B_{\rm S} = 200 \ \rm kHz)$&nbsp; the channel is frequency selective.
+- For GSM&nbsp; $(B_{\rm S} = 200 \ \rm kHz)$:&nbsp; the channel is frequency selective.
-+ For UMTS: &nbsp; $(B_{\rm S} = 5 \ \rm MHz)$&nbsp; the channel is frequency selective.
++ For UMTS&nbsp; $(B_{\rm S} = 5 \ \rm MHz)$:&nbsp; the channel is frequency selective.
 </quiz>

@@ Line 72: / Line 72: @@
 </quiz>
-===Sample solution===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; We have $\tau_1 = d_1/c$ &nbsp;&#8658;&nbsp; $ d_1 = \tau_1 \cdot c = 10^{-5} \rm s \cdot 3 \cdot 10^8 \ m/s \ \ \underline {= 3 \ km}$.

← Older revision		Revision as of 14:00, 15 December 2020
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−	{{quiz-Header\|Buchseite=~~Mobile Kommunikation~~/~~Mehrwegeempfang beim Mobilfunk~~}}	+	{{quiz-Header\|Buchseite=Mobile_Communications/Multipath_Reception_in_Mobile_Communications}}

	[[File:EN_Mob_A_2_2.png\|right\|frame\|Two equivalent models <br>for the two-path channel]]		[[File:EN_Mob_A_2_2.png\|right\|frame\|Two equivalent models <br>for the two-path channel]]

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Mobile Kommunikation/Mehrwegeempfang beim Mobilfunk}}
-[[File:P_ID2157__Mob_A_2_2.png|right|frame|Zwei äquivalente Modelle <br>für den Zweiwege-Kanal]]
+[[File:EN_Mob_A_2_2.png|right|frame|Zwei äquivalente Modelle <br>für den Zweiwege-Kanal]]
 Here we consider a two-way&ndash;channel for mobile radio according to the adjacent graph, characterized by the model parameters
 :$$k_1 = 10^{-4}\hspace{0.05cm}, \hspace{0.2cm} \tau_{1} = 10\,{\rm &micro; s}\hspace{0.05cm}, \hspace{0.2cm}\tau_{2} = 11\,{\rm &micro; s} \hspace{0.05cm}.$$