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

Javier am 20. Mai 2020 um 15:41 Uhr

2020-05-20T15:41:21Z

Javier am 20. Mai 2020 um 15:41 Uhr

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Guenter am 11. Mai 2020 um 14:05 Uhr

2020-05-11T14:05:23Z

Javier am 15. April 2020 um 13:02 Uhr

2020-04-15T13:02:51Z

Javier am 15. April 2020 um 13:02 Uhr

2020-04-15T13:02:11Z

Javier am 15. April 2020 um 13:01 Uhr

2020-04-15T13:01:50Z

Javier am 15. April 2020 um 13:01 Uhr

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Javier am 15. April 2020 um 12:58 Uhr

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Javier am 15. April 2020 um 12:57 Uhr

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Javier am 15. April 2020 um 12:56 Uhr

2020-04-15T12:56:30Z

@@ Zeile 3: / Zeile 3: @@
 [[Datei:EN_Mob_A_2_2.png|right|frame|Two equivalent models <br>for the two-path channel]]
-Here we consider a two-way&ndash;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
 :$$k_1 = 10^{-4}\hspace{0.05cm}, \hspace{0.2cm} \tau_{1} = 10\,{\rm &micro; s}\hspace{0.05cm}, \hspace{0.2cm}\tau_{2} = 11\,{\rm &micro; s} \hspace{0.05cm}.$$

@@ Zeile 2: / Zeile 2: @@
 {{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&ndash;channel for mobile radio according to the adjacent graph, characterized by the model parameters
 :$$k_1 = 10^{-4}\hspace{0.05cm}, \hspace{0.2cm} \tau_{1} = 10\,{\rm &micro; s}\hspace{0.05cm}, \hspace{0.2cm}\tau_{2} = 11\,{\rm &micro; s} \hspace{0.05cm}.$$

@@ Zeile 2: / Zeile 2: @@
 {{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&ndash;channel for mobile radio according to the adjacent graph, characterized by the model parameters
 :$$k_1 = 10^{-4}\hspace{0.05cm}, \hspace{0.2cm} \tau_{1} = 10\,{\rm &micro; s}\hspace{0.05cm}, \hspace{0.2cm}\tau_{2} = 11\,{\rm &micro; s} \hspace{0.05cm}.$$

← Nächstältere Version		Version vom 15. April 2020, 13:02 Uhr
Zeile 113:		Zeile 113:

	*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.

@@ Zeile 105: / Zeile 105: @@
 *On the other hand:
 :$$|H_{\rm 0}(f = 750\,{\rm kHz})| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} |H_{\rm 0}(f = 250\,{\rm kHz})| \approx 1.02 > 1\hspace{0.05cm},$$
-:$$|H_{\rm 0}(f = 1\,{\rm MHz})| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} |H_{\rm 0}(f = 0)| = 1.2 > 1 \hspace{0.05cm}.$
+:$$|H_{\rm 0}(f = 1\,{\rm MHz})| \hspace{-0.1cm} \ = \ \hspace{-0.1cm} |H_{\rm 0}(f = 0)| = 1.2 > 1 \hspace{0.05cm}.$$

@@ Zeile 110: / Zeile 110: @@
 '''(5)'''&nbsp; The difference $\tau_{\rm max} \ &ndash; \tau_{\rm min}$ between the delays in the two paths is equal to $\tau_0 = 1 \ \ \rm &micro; s$.
 *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.
@@ Zeile 119: / Zeile 119: @@
 *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.
-{{{ML-Fuß}}
+{{ML-Fuß}}
 [[Category:Exercises for Mobile Communications|^2.2 Multi-Path Reception in Wireless Systems^]]

@@ Zeile 88: / Zeile 88: @@
    \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}
 \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}.$$
-[[File:P_ID2158__Mob_A_2_2c.png|right|frame|Amount frequency response of a two-way channel]]
+[[File:P_ID2158__Mob_A_2_2c.png|right|frame|Magnitude of the frequency response of a two-way channel]]
 The graphic (red curve) shows the function $|H_0(f)|$.
 *The values you are looking for are marked by the yellow dots.
@@ Zeile 101: / Zeile 101: @@
-'''(4)'''&nbsp; Correct is the <u>solution 1</u>:
+'''(4)'''&nbsp; <u>Solution 1</u> is correct:
-*Destructive overlays exist for $|H_0(f)| < 1$, for example for $f = 500 \ \rm kHz$.
+*Destructive interference occurs for $|H_0(f)| < 1$, for example for $f = 500 \ \rm kHz$.
 *On the other hand:
 :$$|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},$$
@@ Zeile 108: / Zeile 108: @@
-'''(5)'''&nbsp; The difference $\tau_{\rm max} \ &ndash; \tau_{\rm min}$ of the delay times in the two paths is equal to $\tau_0 = 1 \ \ \rm &micro; s$.
+'''(5)'''&nbsp; The difference $\tau_{\rm max} \ &ndash; \tau_{\rm min}$ between the delays in the two paths is equal to $\tau_0 = 1 \ \ \rm &micro; s$.
-*So the coherence bandwidth
+*So the coherence bandwidth is
 :$$B_{\rm K}\hspace{0.01cm}' = {1}/{\tau_{\rm 0} \hspace{0.1cm} \underline {=1\,{\rm MHz} \hspace{0.05cm}.$$
 *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)'''&nbsp; Correct is the <u>solution 2</u>:
+'''(6)'''&nbsp; <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}}
 [[Category:Exercises for Mobile Communications|^2.2 Multi-Path Reception in Wireless Systems^]]

@@ Zeile 81: / Zeile 81: @@
 '''(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

@@ Zeile 74: / Zeile 74: @@
 ===Sample solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; The following applies $\tau_1 = d_1/c$ &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 $\tau_1 = d_1/c$ &nbsp;&#8658;&nbsp; $ d_1 = \tau_1 \cdot c = 10^{-5} \rm s \cdot 3 \cdot 10^8 \ m/s \ \ \underline {= 3 \ km}$.

@@ Zeile 63: / Zeile 63: @@
 {What is the approximate coherence bandwidth for&nbsp; $k_2 = 2 \cdot 10^{-5}$&nbsp; or.&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?