Aufgaben:Exercise 2.5Z: Multi-Path Scenario - Versionsgeschichte

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2026-03-16T12:38:35Z

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← Nächstältere Version		Version vom 16. März 2026, 14:38 Uhr
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	''Notes:''		''Notes:''
	* This task belongs to chapter  [[Mobile_Kommunikation/Das_GWSSUS%E2%80%93Kanalmodell\| Das GWSSUS–Kanalmodell]].		* This task belongs to chapter  [[Mobile_Kommunikation/Das_GWSSUS%E2%80%93Kanalmodell\| Das GWSSUS–Kanalmodell]].
	*We focus especially on the   [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung#Gebr.C3.A4uchliches_Pfadverlustmodell\| path-loss model]]  and the   [[~~Mobile_Kommunikation~~/~~Statistische_Bindungen_innerhalb_des_Rayleigh-Prozesses~~#~~Dopplerfrequenz_und_deren_Verteilung~~\| Doppler effect]].		*We focus especially on the   [[Mobile_Kommunikation/Distanzabh%C3%A4ngige_D%C3%A4mpfung_und_Abschattung#Gebr.C3.A4uchliches_Pfadverlustmodell\| path-loss model]]  and the   [[Mobile_Communications/Statistical_Bindings_within_the_Rayleigh_Process#Doppler_frequency_and_its_distribution\| Doppler effect]].


Zeile 134:		Zeile 134:

	[[Category:Exercises for Mobile Communications\|^2.3 The GWSSUS Channel Model^]]		[[Category:Exercises for Mobile Communications\|^2.3 The GWSSUS Channel Model^]]
			[[en:Aufgaben:Exercise_2.5Z:_Multi-Path_Scenario]]

Maintenance script: Join multi-line formulas for MW 1.43 parser compat

2026-02-24T13:40:19Z

Join multi-line formulas for MW 1.43 parser compat

← Nächstältere Version		Version vom 24. Februar 2026, 15:40 Uhr
Zeile 5:		Zeile 5:
	In  [[Aufgaben:Exercise_2.5:_Scatter_Function\| Exercise 2.5]], a delay–Doppler function (or scatter function) was given. From this, we will calculate and interpret the other system functions. The given scatter function  $s(\tau_0, f_{\rm D})$  was		In  [[Aufgaben:Exercise_2.5:_Scatter_Function\| Exercise 2.5]], a delay–Doppler function (or scatter function) was given. From this, we will calculate and interpret the other system functions. The given scatter function  $s(\tau_0, f_{\rm D})$  was
	:$$s(\tau_0, f_{\rm D}) =\frac{1}{\sqrt{2}} \cdot \delta (\tau_0) \cdot \delta (f_{\rm D} - 100\,{\rm Hz}) \ - \ $$		:$$s(\tau_0, f_{\rm D}) =\frac{1}{\sqrt{2}} \cdot \delta (\tau_0) \cdot \delta (f_{\rm D} - 100\,{\rm Hz}) \ - \ $$
	:$$\hspace{1.5cm} \ - \ \hspace{-0.2cm} \frac{1}{2} \cdot \delta (\tau_0 \hspace{-0.05cm}- \hspace{-0.05cm}1\,{\rm \mu s}) \cdot \delta (f_{\rm D} \hspace{-0.05cm}- \hspace{-0.05cm}50\,{\rm Hz}) \ - \frac{1}{2} \cdot \delta (\tau_0 \hspace{-0.05cm}- \hspace{-0.05cm}1\,{\rm \mu s}) \cdot \delta (f_{\rm D}\hspace{-0.05cm} + \hspace{-0.05cm}50\,{\rm Hz})		:$$\hspace{1.5cm} \ - \ \hspace{-0.2cm} \frac{1}{2} \cdot \delta (\tau_0 \hspace{-0.05cm}- \hspace{-0.05cm}1\,{\rm \mu s}) \cdot \delta (f_{\rm D} \hspace{-0.05cm}- \hspace{-0.05cm}50\,{\rm Hz}) \ - \frac{1}{2} \cdot \delta (\tau_0 \hspace{-0.05cm}- \hspace{-0.05cm}1\,{\rm \mu s}) \cdot \delta (f_{\rm D}\hspace{-0.05cm} + \hspace{-0.05cm}50\,{\rm Hz}) \hspace{0.05cm}.$$
	\hspace{0.05cm}.$$


Zeile 19:		Zeile 18:
	* ${\rm S}_2$  and  ${\rm S}_3$  are to be understood here as fictitious transmitters from whose position the angles of incidence  $\alpha_2$  and  $\alpha_3$  of the secondary paths can be determined.		* ${\rm S}_2$  and  ${\rm S}_3$  are to be understood here as fictitious transmitters from whose position the angles of incidence  $\alpha_2$  and  $\alpha_3$  of the secondary paths can be determined.
	* Let the signal frequency be   $f_{\rm S}$, the angle of incidence   $\alpha$, the velocity  $v$  and the velocity of light  $c = 3 \cdot 10^8 \ \rm m/s$. Then, the Doppler frequency is		* Let the signal frequency be   $f_{\rm S}$, the angle of incidence   $\alpha$, the velocity  $v$  and the velocity of light  $c = 3 \cdot 10^8 \ \rm m/s$. Then, the Doppler frequency is
	:$$f_{\rm D}= {v}/{c} \cdot f_{\rm S} \cdot \cos(\alpha)		:$$f_{\rm D}= {v}/{c} \cdot f_{\rm S} \cdot \cos(\alpha)\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$
	* The damping factors  $k_1$,  $k_2$  and  $k_3$  are inversely proportional to the path lengths  $d_1$,  $d_2$  and  $d_3$. This corresponds to the path loss exponent  $\gamma = 2$.		* The damping factors  $k_1$,  $k_2$  and  $k_3$  are inversely proportional to the path lengths  $d_1$,  $d_2$  and  $d_3$. This corresponds to the path loss exponent  $\gamma = 2$.
	*This means:   The signal power decreases quadratically with distance  $d$  and accordingly the signal amplitude decreases linearly with  $d$.		*This means:   The signal power decreases quadratically with distance  $d$  and accordingly the signal amplitude decreases linearly with  $d$.
Zeile 88:		Zeile 86:

	'''(2)'''  The equation for the Doppler frequency is		'''(2)'''  The equation for the Doppler frequency is
	:$$f_{\rm D}= \frac{v}{c} \cdot f_{\rm S} \cdot \cos(\alpha)		:$$f_{\rm D}= \frac{v}{c} \cdot f_{\rm S} \cdot \cos(\alpha)\hspace{0.05cm},$$
	\hspace{0.05cm},$$
	If the angle of incidence is $\alpha=0$, the Doppler frequency is		If the angle of incidence is $\alpha=0$, the Doppler frequency is
	:$$f_d=\frac{v}{c}\cdot f_S$$		:$$f_d=\frac{v}{c}\cdot f_S$$

	*The speed of the receiver is then		*The speed of the receiver is then
	:$$v = \frac{f_{\rm D}}{f_{\rm S}} \cdot c = \frac{10^2\,{\rm Hz}}{2 \cdot 10^9\,{\rm Hz}} \cdot 3 \cdot 10^8\,{\rm m/s} = 15\,{\rm m/s}		:$$v = \frac{f_{\rm D}}{f_{\rm S}} \cdot c = \frac{10^2\,{\rm Hz}}{2 \cdot 10^9\,{\rm Hz}} \cdot 3 \cdot 10^8\,{\rm m/s} = 15\,{\rm m/s}\hspace{0.1cm} \underline {= 54 \,{\rm km/h}} \hspace{0.05cm}.$$
	\hspace{0.1cm} \underline {= 54 \,{\rm km/h}}
	\hspace{0.05cm}.$$


Zeile 102:		Zeile 97:
	*The Doppler frequency $f_{\rm D} = 50 \ \rm Hz$ comes from the blue path, because the receiver moves towards the virtual transmitter ${\rm S}_2$ (i.e., towards the reflection point), although not directly. In other words, the movement of the receiver <b>reduces</b> the blue path's length.		*The Doppler frequency $f_{\rm D} = 50 \ \rm Hz$ comes from the blue path, because the receiver moves towards the virtual transmitter ${\rm S}_2$ (i.e., towards the reflection point), although not directly. In other words, the movement of the receiver <b>reduces</b> the blue path's length.
	*The angle $\alpha_2$ between the direction of movement and the connecting line ${\rm S_2 – E}$ is $60^\circ$:		*The angle $\alpha_2$ between the direction of movement and the connecting line ${\rm S_2 – E}$ is $60^\circ$:
	:$$\cos(\alpha_2) = \frac{f_{\rm D}}{f_{\rm S}} \cdot \frac{c}{v} = \frac{50 \,{\rm Hz}\cdot 3 \cdot 10^8\,{\rm m/s}}{2 \cdot 10^9\,{\rm Hz}\cdot 15\,{\rm m/s}} = 0.5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \alpha_2		:$$\cos(\alpha_2) = \frac{f_{\rm D}}{f_{\rm S}} \cdot \frac{c}{v} = \frac{50 \,{\rm Hz}\cdot 3 \cdot 10^8\,{\rm m/s}}{2 \cdot 10^9\,{\rm Hz}\cdot 15\,{\rm m/s}} = 0.5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \alpha_2\hspace{0.1cm} \underline {= 60^{\circ} } \hspace{0.05cm}.$$
	\hspace{0.1cm} \underline {= 60^{\circ} }
	\hspace{0.05cm}.$$


Zeile 129:		Zeile 122:
	*The minus sign takes into account the $180^\circ$ phase rotation on the secondary paths.		*The minus sign takes into account the $180^\circ$ phase rotation on the secondary paths.
	*From the weights of the Dirac functions one can read $k_1 = \sqrt{0.5}$ and $k_2 = -0.5$. From this follows:		*From the weights of the Dirac functions one can read $k_1 = \sqrt{0.5}$ and $k_2 = -0.5$. From this follows:
	:$$\frac{d_2}{d_1} = \frac{k_1}{-k_2} = \frac{1/\sqrt{2}}{0.5} = \sqrt{2}		:$$\frac{d_2}{d_1} = \frac{k_1}{-k_2} = \frac{1/\sqrt{2}}{0.5} = \sqrt{2}\hspace{0.15cm} \underline {= 1.414} \hspace{0.05cm}.$$
	\hspace{0.15cm} \underline {= 1.414}
	\hspace{0.05cm}.$$

	*The constant $K$ is only an auxiliary variable that does not need to be considered further.		*The constant $K$ is only an auxiliary variable that does not need to be considered further.
Zeile 138:		Zeile 129:

	'''(8)'''  From   $d_2/d_1 = 2^{-0.5}$  and  $\Delta d = d_2 \, - d_1 = 300 \ \rm m$  finally follows:		'''(8)'''  From   $d_2/d_1 = 2^{-0.5}$  and  $\Delta d = d_2 \, - d_1 = 300 \ \rm m$  finally follows:
	:$$\sqrt{2} \cdot d_1 - d_1 = 300\,{\rm m} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}		:$$\sqrt{2} \cdot d_1 - d_1 = 300\,{\rm m} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}d_1 = \frac{300\,{\rm m}}{\sqrt{2} - 1} \hspace{0.15cm} \underline {= 724\,{\rm m}} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} d_2 = \sqrt{2} \cdot d_1 \hspace{0.15cm} \underline {= 1024\,{\rm m}}\hspace{0.05cm}. $$
	d_1 = \frac{300\,{\rm m}}{\sqrt{2} - 1} \hspace{0.15cm} \underline {= 724\,{\rm m}}
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm} d_2 = \sqrt{2} \cdot d_1 \hspace{0.15cm} \underline {= 1024\,{\rm m}}
	\hspace{0.05cm}. $$
	{{ML-Fuß}}		{{ML-Fuß}}


	[[Category:Exercises for Mobile Communications\|^2.3 The GWSSUS Channel Model^]]		[[Category:Exercises for Mobile Communications\|^2.3 The GWSSUS Channel Model^]]

Guenter am 11. Mai 2020 um 14:12 Uhr

2020-05-11T14:12:48Z

← Nächstältere Version		Version vom 11. Mai 2020, 16:12 Uhr
Zeile 2:		Zeile 2:
	{{quiz-Header\|Buchseite=Mobile Kommunikation/Das GWSSUS–Kanalmodell}}		{{quiz-Header\|Buchseite=Mobile Kommunikation/Das GWSSUS–Kanalmodell}}

	[[Datei:~~P_ID2169__Mob_Z_2_5~~.png\|right\|frame\|~~Mobilfunk–Szenario mit drei Pfaden~~]]		[[Datei:EN_Mob_A_2_5Z.png\|right\|frame\|Mobile radio scenario with three paths]]
	In  [[Aufgaben:Exercise_2.5:_Scatter_Function\| Exercise 2.5]], a delay–Doppler function (or scatter function) was given. From this, we will calculate and interpret the other system functions. The given scatter function  $s(\tau_0, f_{\rm D})$  was		In  [[Aufgaben:Exercise_2.5:_Scatter_Function\| Exercise 2.5]], a delay–Doppler function (or scatter function) was given. From this, we will calculate and interpret the other system functions. The given scatter function  $s(\tau_0, f_{\rm D})$  was
	:$$s(\tau_0, f_{\rm D}) =\frac{1}{\sqrt{2}} \cdot \delta (\tau_0) \cdot \delta (f_{\rm D} - 100\,{\rm Hz}) \ - \ $$		:$$s(\tau_0, f_{\rm D}) =\frac{1}{\sqrt{2}} \cdot \delta (\tau_0) \cdot \delta (f_{\rm D} - 100\,{\rm Hz}) \ - \ $$

Javier: Javier verschob die Seite Exercises:Exercise 2.5Z: Multi-Path Scenario nach Aufgaben:Exercise 2.5Z: Multi-Path Scenario

2020-04-22T12:40:47Z

Javier verschob die Seite Exercises:Exercise 2.5Z: Multi-Path Scenario nach Aufgaben:Exercise 2.5Z: Multi-Path Scenario

← Nächstältere Version		Version vom 22. April 2020, 14:40 Uhr
(kein Unterschied)

Javier am 22. April 2020 um 12:40 Uhr

2020-04-22T12:40:35Z

← Nächstältere Version		Version vom 22. April 2020, 14:40 Uhr
Zeile 76:		Zeile 76:
	$d_2/d_1 \ = \ ${ 1,414 3% }		$d_2/d_1 \ = \ ${ 1,414 3% }

	{~~Indicate~~ the distances  $d_1$  and  $d_2$ .		{Find the distances  $d_1$  and  $d_2$ .
	\|type="{}"}		\|type="{}"}
	$d_1 \ = \ ${ 724 3% } $\ \rm m$		$d_1 \ = \ ${ 724 3% } $\ \rm m$

Javier am 22. April 2020 um 12:39 Uhr

2020-04-22T12:39:48Z

← Nächstältere Version		Version vom 22. April 2020, 14:39 Uhr
Zeile 78:		Zeile 78:
	{Indicate the distances  $d_1$  and  $d_2$ .		{Indicate the distances  $d_1$  and  $d_2$ .
	\|type="{}"}		\|type="{}"}
	$d_1 \ = \ ${724 3% } $\ \ \rm m$		$d_1 \ = \ ${ 724 3% } $\ \rm m$
	$d_2 \ = \ ${ 1024 3% } $\ \ \rm m$		$d_2 \ = \ ${ 1024 3% } $\ \rm m$
	</quiz>		</quiz>

Javier am 22. April 2020 um 12:39 Uhr

2020-04-22T12:39:07Z

← Nächstältere Version		Version vom 22. April 2020, 14:39 Uhr
Zeile 129:		Zeile 129:
	*The minus sign takes into account the $180^\circ$ phase rotation on the secondary paths.		*The minus sign takes into account the $180^\circ$ phase rotation on the secondary paths.
	*From the weights of the Dirac functions one can read $k_1 = \sqrt{0.5}$ and $k_2 = -0.5$. From this follows:		*From the weights of the Dirac functions one can read $k_1 = \sqrt{0.5}$ and $k_2 = -0.5$. From this follows:
	:$$\frac{\2}{d_1} = \frac{k_1}{-k_2} = \frac{1/\sqrt{2}}{0.5} = \sqrt{2}		:$$\frac{d_2}{d_1} = \frac{k_1}{-k_2} = \frac{1/\sqrt{2}}{0.5} = \sqrt{2}
	\hspace{0.15cm} \underline {= 1,414}		\hspace{0.15cm} \underline {= 1.414}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

Zeile 139:		Zeile 139:
	'''(8)'''  From   $d_2/d_1 = 2^{-0.5}$  and  $\Delta d = d_2 \, - d_1 = 300 \ \rm m$  finally follows:		'''(8)'''  From   $d_2/d_1 = 2^{-0.5}$  and  $\Delta d = d_2 \, - d_1 = 300 \ \rm m$  finally follows:
	:$$\sqrt{2} \cdot d_1 - d_1 = 300\,{\rm m} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}		:$$\sqrt{2} \cdot d_1 - d_1 = 300\,{\rm m} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}
	d_1 = \frac{300\,{\rm m}}}{\sqrt{2} - 1} \hspace{0.15cm} \underline {= 724\,{\rm m}}		d_1 = \frac{300\,{\rm m}}{\sqrt{2} - 1} \hspace{0.15cm} \underline {= 724\,{\rm m}}
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm} d_2 = \sqrt{2} \cdot d_1 \hspace{0.15cm} \underline {= 1024\,{\rm m}}		\hspace{0.3cm} \Rightarrow \hspace{0.3cm} d_2 = \sqrt{2} \cdot d_1 \hspace{0.15cm} \underline {= 1024\,{\rm m}}
	\hspace{0.05cm}. $$		\hspace{0.05cm}. $$

Javier am 22. April 2020 um 12:37 Uhr

2020-04-22T12:37:52Z

Javier am 22. April 2020 um 12:31 Uhr

2020-04-22T12:31:03Z

← Nächstältere Version		Version vom 22. April 2020, 14:31 Uhr
Zeile 70:		Zeile 70:
	{What is the difference in time  $\Delta d = d_2 - d_1$?		{What is the difference in time  $\Delta d = d_2 - d_1$?
	\|type="{}"}		\|type="{}"}
	$\ Delta d \ = \ ${ 300 3% } $\ \ \rm m$		$\Delta d \ = \ ${ 300 3% } $\ \ \rm m$

	{What is the relationship between  $d_2$  and  $d_1$?		{What is the relationship between  $d_2$  and  $d_1$?

Javier am 22. April 2020 um 12:30 Uhr

2020-04-22T12:30:48Z

← Nächstältere Version		Version vom 22. April 2020, 14:30 Uhr
Zeile 58:		Zeile 58:
	{What statements apply to the green path?		{What statements apply to the green path?
	\|type="[]"}		\|type="[]"}
	+ ~~For this, ~~ $\tau_0 = 1 \ \ \rm µ s$  and  $f_{\rm D} = \, –50 \ \ \rm Hz$.		+ We have $\tau_0 = 1 \ \ \rm µ s$  and  $f_{\rm D} = \, –50 \ \ \rm Hz$.
	- The angle  $\alpha_3$  (see ~~graphic~~) is  $60^\circ$.		- The angle  $\alpha_3$  (see graph) is  $60^\circ$.
	+ The angle  $\alpha_3$  is  $240^\circ$.		+ The angle  $\alpha_3$  is  $240^\circ$.

← Nächstältere Version		Version vom 22. April 2020, 14:37 Uhr
Zeile 99:		Zeile 99:


	'''(3)'''  ~~Correct are the~~ <u>~~solutions~~ 1 and 4</u>:		'''(3)'''  <u>Solutions 1 and 4</u> are correct:
	*The Doppler frequency $f_{\rm D} = 50 \ \rm Hz$ comes from the blue path, because the receiver ~~somehow~~ moves towards the virtual transmitter ${\rm S}_2$ (at the reflection point), although not ~~in a direct direction~~.		*The Doppler frequency $f_{\rm D} = 50 \ \rm Hz$ comes from the blue path, because the receiver moves towards the virtual transmitter ${\rm S}_2$ (i.e., towards the reflection point), although not directly. In other words, the movement of the receiver <b>reduces</b> the blue path's length.
	*The angle $\alpha_2$ between the direction of movement and the connecting line ${\rm S_2 – E}$ is $60^\circ$:		*The angle $\alpha_2$ between the direction of movement and the connecting line ${\rm S_2 – E}$ is $60^\circ$:
	:$$\cos(\alpha_2) = \frac{f_{\rm D}}}{f_{\rm S}} \cdot \frac{c}{v} = \frac{50 \,{\rm Hz}\cdot 3 \cdot 10^8\,{\rm m/s}}}{2 \cdot 10^9\,{\rm Hz}\cdot 15\,{\rm m/s}} = 0.5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \alpha_2		:$$\cos(\alpha_2) = \frac{f_{\rm D}}{f_{\rm S}} \cdot \frac{c}{v} = \frac{50 \,{\rm Hz}\cdot 3 \cdot 10^8\,{\rm m/s}}{2 \cdot 10^9\,{\rm Hz}\cdot 15\,{\rm m/s}} = 0.5 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \alpha_2
	\hspace{0.1cm} \underline {= 60^{\circ} }		\hspace{0.1cm} \underline {= 60^{\circ} }
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$


	'''(4)'''  ~~Correct are the~~ <u>~~statements~~ 1 and 3</u>:		'''(4)'''  <u>Statements 1 and 3</u> are correct:
	*From $f_{\rm D} = \, –50 \ \rm Hz$ follows $\alpha_3 = \alpha_2 ± \pi$, so $\alpha_3 \ \underline {= 240^\circ}$.		*From $f_{\rm D} = \, –50 \ \rm Hz$ follows $\alpha_3 = \alpha_2 ± \pi$, so $\alpha_3 \ \underline {= 240^\circ}$.

Zeile 113:		Zeile 113:

	'''(5)'''  <u>All statements are correct</u>:		'''(5)'''  <u>All statements are correct</u>:
	*The two Dirac functions at $± 50 \ \ \rm Hz$ have the same ~~running time~~. ~~For both durations~~ $\tau_3 = \tau_2 = \tau_1 + \tau_0$ ~~is valid~~.		*The two Dirac functions at $± 50 \ \ \rm Hz$ have the same delay. We have $\tau_3 = \tau_2 = \tau_1 + \tau_0$.
	*From the ~~same transit time~~, however, also follows  $d_3 = d_2$~~  and with~~ the same length ~~also the same~~ damping factors.		*From the equality of the delays, however, also follows that  $d_3 = d_2$. As both paths have the same length, their damping factors are also equal.



	'''(6)'''  The ~~runtime~~ difference is $\tau_0 = 1 \ \rm µ s$, as shown in the equation for $s(\tau_0, f_{\rm D})$.		'''(6)'''  The delay difference is $\tau_0 = 1 \ \rm µ s$, as shown in the equation for $s(\tau_0, f_{\rm D})$.
	* This gives the difference in length:		* This gives the difference in length:
	:$$\Delta d = \tau_0 \cdot c = 10^{–6} {\rm s} \cdot 3 \cdot 10^8 \ \rm m/s \ \ \underline {= 300 \ \ \rm m}.$$		:$$\Delta d = \tau_0 \cdot c = 10^{–6} {\rm s} \cdot 3 \cdot 10^8 \ \rm m/s \ \ \underline {= 300 \ \ \rm m}.$$
Zeile 127:		Zeile 127:
	*Then $k_1 = K/d_1$ and $k_2 = K/d_2$.		*Then $k_1 = K/d_1$ and $k_2 = K/d_2$.

	*The minus sign takes into account the $180^\circ$~~–~~phase rotation on the secondary paths.		*The minus sign takes into account the $180^\circ$ phase rotation on the secondary paths.
	*From the weights of the Dirac functions one can read $k_1 = \sqrt{0.5}$ and $k_2 = -0.5$. From this follows:		*From the weights of the Dirac functions one can read $k_1 = \sqrt{0.5}$ and $k_2 = -0.5$. From this follows:
	:$$\frac{\2}{d_1} = \frac{k_1}{-k_2} = \frac{1/\sqrt{2}}{0.5} = \sqrt{2}		:$$\frac{\2}{d_1} = \frac{k_1}{-k_2} = \frac{1/\sqrt{2}}{0.5} = \sqrt{2}
Zeile 137:		Zeile 137:


	'''(8)'''  ~~Aus~~  $d_2/d_1 = 2^{-0.5}$  and  $\Delta d = d_2 \, - d_1 = 300 \ \rm m$  finally follows:		'''(8)'''  From   $d_2/d_1 = 2^{-0.5}$  and  $\Delta d = d_2 \, - d_1 = 300 \ \rm m$  finally follows:
	:$$\sqrt{2} \cdot d_1 - d_1 = 300\,{\rm m} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}		:$$\sqrt{2} \cdot d_1 - d_1 = 300\,{\rm m} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}
	d_1 = \frac{300\,{\rm m}}}{\sqrt{2} - 1} \hspace{0.15cm} \underline {= 724\,{\rm m}}		d_1 = \frac{300\,{\rm m}}}{\sqrt{2} - 1} \hspace{0.15cm} \underline {= 724\,{\rm m}}