Aufgaben:Exercise 4.5Z: Impulse Response once again - Revision history

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	[[Category:Linear and Time-Invariant Systems: Exercises\|^4.2 Coaxial Cable^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^4.2 Coaxial Cable^]]
	[[de:~~Zusatzaufgaben~~:4.~~5_Nochmals_Impulsantwort~~]]		[[de:Aufgaben:Aufgabe 4.5Z: Nochmals Impulsantwort]]

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 A &nbsp;"standard coaxial cable"&nbsp; $\text{(2.6 mm}$&nbsp; core diameter,&nbsp; $\text{9.5 mm}$&nbsp; outer diameter$)$&nbsp; of length &nbsp;$l = 1 \ \rm km$&nbsp; with the following frequency response is used as transmission medium:
-:$$H_{\rm K}(f)     =  {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} \beta_1 \hspace{0.05cm}\cdot \hspace{0.05cm}  f
+:$$H_{\rm K}(f)     =  {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} \beta_1 \hspace{0.05cm}\cdot \hspace{0.05cm}  f\hspace{0.05cm}\cdot \hspace{0.05cm}l}\cdot {\rm e}^{- \alpha_2  \hspace{0.01cm}\sqrt{f}\hspace{0.05cm}l}\cdot {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2  \hspace{0.05cm}\cdot \hspace{0.05cm}\sqrt{f}\hspace{0.05cm}\cdot \hspace{0.05cm}l}=   H_1(f) \cdot H_2(f) \cdot H_3(f)$$The partial frequency responses &nbsp;$H_1(f)$, &nbsp;$H_2(f)$&nbsp; and &nbsp;$H_3(f)$&nbsp; are used here only as abbreviations.&nbsp; The line parameters are::$$\beta_1 = 21.78\, \frac{\rm rad}{\rm km \cdot MHz}\hspace{0.05cm}, $$:$$ \alpha_2 = 0.2722\, \frac{\rm Np}{\rm km \cdot \sqrt{MHz}}\hspace{0.05cm},$$:$$ \beta_2 = 0.2722\, \frac{\rm rad}{\rm km \cdot \sqrt{MHz}}\hspace{0.05cm}.$$The graph shows the resulting impulse response &nbsp;$h_{\rm K}(t\hspace{0.05cm}')$, where &nbsp;$t\hspace{0.05cm}' = t/T$&nbsp; represents the normalized time. Without considering the (normalized) phase running time &nbsp;$\tau\hspace{0.05cm}' = \tau/T$&nbsp;, &nbsp;$h_{\rm K}(t\hspace{0.05cm}')$&nbsp; can be written as follows::$$h_{\rm K}(t\hspace{0.05cm}')  = \frac {1}{T} \cdot \frac {a_\rm \star/\pi}{ \sqrt{2\hspace{0.05cm}t\hspace{0.05cm}'^3}}\cdot {\rm e}^{ -{a_\rm \star^2}/( {2\pi\hspace{0.05cm}t\hspace{0.05cm}')} } \hspace{0.05cm}, \hspace{0.2cm} \hspace{0.15cm}{\rm with}\hspace{0.15cm}{a}_{\rm \star}\hspace{0.15cm}{\rm in}\hspace{0.15cm}{\rm Neper}\hspace{0.05cm}.$$
 *This equation gives the Fourier retransform of the product &nbsp;$H_2(f) \cdot H_3(f)$&nbsp;.
 *The characteristic cable attenuation &nbsp;${a}_{\rm \star} =  \alpha_2 \cdot \sqrt {R/2} \cdot l \hspace{0.05cm}$ is used here.

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	\sqrt{f}		\sqrt{f}
	\hspace{0.05cm}\cdot \hspace{0.05cm}l}		\hspace{0.05cm}\cdot \hspace{0.05cm}l}
	= H_1(f) \cdot H_2(f) \cdot H_3(f)$$		= H_1(f) \cdot H_2(f) \cdot H_3(f)$$The partial frequency responses  $H_1(f)$,  $H_2(f)$  and  $H_3(f)$  are used here only as abbreviations.  The line parameters are::$$\beta_1 = 21.78\, \frac{\rm rad}{\rm km \cdot MHz}\hspace{0.05cm}, $$:$$ \alpha_2 = 0.2722\, \frac{\rm Np}{\rm km \cdot \sqrt{MHz}}\hspace{0.05cm},$$:$$ \beta_2 = 0.2722\, \frac{\rm rad}{\rm km \cdot \sqrt{MHz}}
			\hspace{0.05cm}.$$The graph shows the resulting impulse response  $h_{\rm K}(t\hspace{0.05cm}')$, where  $t\hspace{0.05cm}' = t/T$  represents the normalized time. Without considering the (normalized) phase running time  $\tau\hspace{0.05cm}' = \tau/T$ ,  $h_{\rm K}(t\hspace{0.05cm}')$  can be written as follows::$$h_{\rm K}(t\hspace{0.05cm}') = \frac {1}{T} \cdot \frac {a_\rm \star/\pi}{ \sqrt{2
	The partial frequency responses  $H_1(f)$,  $H_2(f)$  and  $H_3(f)$  are used here only as abbreviations.  The line parameters are:
	:$$\beta_1 = 21.78\, \frac{\rm rad}{\rm km \cdot MHz}\hspace{0.05cm}, $$
	:$$ \alpha_2 = 0.2722\, \frac{\rm Np}{\rm km \cdot \sqrt{MHz}}\hspace{0.05cm},$$
	:$$ \beta_2 = 0.2722\, \frac{\rm rad}{\rm km \cdot \sqrt{MHz}}
	\hspace{0.05cm}.$$
	The graph shows the resulting impulse response  $h_{\rm K}(t\hspace{0.05cm}')$, where  $t\hspace{0.05cm}' = t/T$  represents the normalized time. Without considering the (normalized) phase running time  $\tau\hspace{0.05cm}' = \tau/T$ ,  $h_{\rm K}(t\hspace{0.05cm}')$  can be written as follows:
	:$$h_{\rm K}(t\hspace{0.05cm}') = \frac {1}{T} \cdot \frac {a_\rm \star/\pi}{ \sqrt{2
	\hspace{0.05cm}t\hspace{0.05cm}'^3}}\cdot {\rm e}^{ -{a_\rm \star^2}/( {2\pi		\hspace{0.05cm}t\hspace{0.05cm}'^3}}\cdot {\rm e}^{ -{a_\rm \star^2}/( {2\pi
	\hspace{0.05cm}t\hspace{0.05cm}')} } \hspace{0.05cm}, \hspace{0.2cm} \hspace{0.15cm}		\hspace{0.05cm}t\hspace{0.05cm}')} } \hspace{0.05cm}, \hspace{0.2cm} \hspace{0.15cm}
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	*You can use the  (German language)  interactive SWF applet  [[Applets:Zeitverhalten_von_Kupferkabeln\|"Zeitverhalten von Kupferkabeln"]]   ⇒   "Time behavior of copper cables"  to check your results.		*You can use the  (German language)  interactive SWF applet  [[Applets:Zeitverhalten_von_Kupferkabeln\|"Zeitverhalten von Kupferkabeln"]]   ⇒   "Time behavior of copper cables"  to check your results.
	*In  [[Aufgaben:Exercise_4.5:_Coaxial_Cable_-_Impulse_Response\|Exercise 4.5]]  the maximum value of the normalized impulse response was calculated as follows:		*In  [[Aufgaben:Exercise_4.5:_Coaxial_Cable_-_Impulse_Response\|Exercise 4.5]]  the maximum value of the normalized impulse response was calculated as follows:
	:$${\rm Max}\, \big[T \cdot h_{\rm K}(t)\big ] = \frac {\sqrt{13.5 \pi} \cdot {\rm e}^{-1.5} }{{a}_{\rm \star}^2} \approx		:$${\rm Max}\, \big[T \cdot h_{\rm K}(t)\big ] = \frac {\sqrt{13.5 \pi} \cdot {\rm e}^{-1.5} }{{a}_{\rm \star}^2} \approx\frac {1.453 }{{a}_{\rm \star}^2} \hspace{0.05cm}, \hspace{0.2cm} \hspace{0.15cm}{\rm with}\hspace{0.15cm}{a}_{\rm \star}\hspace{0.15cm}{\rm in}\hspace{0.15cm}{\rm Neper}\hspace{0.05cm}.$$
	\frac {1.453 }{{a}_{\rm \star}^2} \hspace{0.05cm}, \hspace{0.2cm} \hspace{0.15cm}
	{\rm with}\hspace{0.15cm}{a}_{\rm \star}\hspace{0.15cm}
	{\rm in}\hspace{0.15cm}
	{\rm Neper}\hspace{0.05cm}.$$


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	'''(2)'''  According to the information provided, the following applies:		'''(2)'''  According to the information provided, the following applies:
	:$$2\pi \cdot f \cdot \tau = \beta_1 \cdot f \cdot l \Rightarrow \hspace{0.3cm}\tau= \frac {\beta_1 \cdot l}{2\pi} =		:$$2\pi \cdot f \cdot \tau = \beta_1 \cdot f \cdot l \Rightarrow \hspace{0.3cm}\tau= \frac {\beta_1 \cdot l}{2\pi} =\frac {21.78\, {\rm rad}/{({\rm km \cdot MHz})}\cdot 10\,{\rm km}}{2\pi} =34.7\,{\rm µ s}$$
	\frac {21.78\, {\rm rad}/{({\rm km \cdot MHz})}\cdot 10\,{\rm km}}{2\pi} =		:$$\Rightarrow \hspace{0.3cm}\tau '= {\tau}/{T} = 694 \Rightarrow \hspace{0.3cm}T = \frac {34.7\,{\rm µ s}}{700} \approx0.05\,{\rm µ s}\hspace{0.05cm}.$$
	34.7\,{\rm µ s}$$
	:$$\Rightarrow \hspace{0.3cm}\tau '= {\tau}/{T} = 694 \Rightarrow \hspace{0.3cm}
	T = \frac {34.7\,{\rm µ s}}{700} \~~approx~~
	0.05\,{\rm µ s}\hspace{0.05cm}.$$
	*The bit rate is equal to the reciprocal of the symbol duration:		*The bit rate is equal to the reciprocal of the symbol duration:
	:$$\underline{R = 20 \ \rm Mbit/s}.$$		:$$\underline{R = 20 \ \rm Mbit/s}.$$
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	'''(3)'''  For the characteristic cable attenuation one obtains:		'''(3)'''  For the characteristic cable attenuation one obtains:
	:$${a}_{\rm \star} = \alpha_2 \cdot \sqrt {R/2} \cdot l =		:$${a}_{\rm \star} = \alpha_2 \cdot \sqrt {R/2} \cdot l =0.2722\, \frac{\rm Np}{\rm km \cdot \sqrt{MHz}} \cdot \sqrt {10\,{\rm MHz}} \cdot 10\,{\rm km} \hspace{0.15cm}\underline{\approx8.6\,{\rm Np}}\hspace{0.05cm}.$$
	0.2722\, \frac{\rm Np}{\rm km \cdot \sqrt{MHz}} \cdot \sqrt {10\,{\rm MHz}} \cdot 10\,{\rm km} \hspace{0.15cm}\underline{\~~approx~~
	8.6\,{\rm Np}}\hspace{0.05cm}.$$
	*The corresponding dB value is  ${a}_{\rm \star} = 75 \ \rm dB$.		*The corresponding dB value is  ${a}_{\rm \star} = 75 \ \rm dB$.

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	'''(4)'''  Using the given equation and the result of subtask  '''(3)''' , we obtain:		'''(4)'''  Using the given equation and the result of subtask  '''(3)''' , we obtain:
	:$${\rm Max}\, \big[T \cdot h_{\rm K}(t)\big] \approx		:$${\rm Max}\, \big[T \cdot h_{\rm K}(t)\big] \approx\frac {1.453 }{{a}_{\rm \star}^2} = \frac {1.453 }{8.6^2}\hspace{0.15cm}\underline{ \approx 0.02}\hspace{0.05cm}.$$
	\frac {1.453 }{{a}_{\rm \star}^2} = \frac {1.453 }{8.6^2}
	\hspace{0.15cm}\underline{ \approx 0.02}\hspace{0.05cm}.$$

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← Older revision		Revision as of 14:29, 16 March 2026
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	[[Category:Linear and Time-Invariant Systems: Exercises\|^4.2 Coaxial Cable^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^4.2 Coaxial Cable^]]
			[[de:Zusatzaufgaben:4.5_Nochmals_Impulsantwort]]

Hwang at 09:14, 18 November 2021

2021-11-18T09:14:44Z

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Guenter at 14:52, 29 October 2021

2021-10-29T14:52:01Z

← Older revision		Revision as of 16:52, 29 October 2021
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	[[Category:Linear and Time-Invariant Systems: Exercises\|^4.2 ~~Koaxialkabel~~^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^4.2 Coaxial Cable^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 4.5Z: Nochmals Impulsantwort to Aufgaben:Exercise 4.5Z: Impulse Response once again

2021-10-29T14:50:15Z

Guenter moved page Aufgaben:Aufgabe 4.5Z: Nochmals Impulsantwort to Aufgaben:Exercise 4.5Z: Impulse Response once again

← Older revision		Revision as of 16:50, 29 October 2021
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Javier: Text replacement - "”" to """

2021-05-28T14:37:50Z

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← Older revision		Revision as of 16:37, 28 May 2021
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	Wir betrachten wie in  [[Aufgaben:4.5_Koaxialkabel_–_Impulsantwort\|Aufgabe 4.5]]  ein binäres Übertragungssystem mit der Bitrate  $R$  ⇒   Symboldauer  $T= 1/R$.		Wir betrachten wie in  [[Aufgaben:4.5_Koaxialkabel_–_Impulsantwort\|Aufgabe 4.5]]  ein binäres Übertragungssystem mit der Bitrate  $R$  ⇒   Symboldauer  $T= 1/R$.

	Als Übertragungsmedium wird ein "Normalkoaxialkabel~~”~~  $\text{(2.6 mm}$  Kerndurchmesser,  $\text{9.5 mm}$  Außendurchmesser$)$  der Länge  $l = 1 \ \rm km$  mit folgendem Frequenzgang verwendet:		Als Übertragungsmedium wird ein "Normalkoaxialkabel"  $\text{(2.6 mm}$  Kerndurchmesser,  $\text{9.5 mm}$  Außendurchmesser$)$  der Länge  $l = 1 \ \rm km$  mit folgendem Frequenzgang verwendet:
	:$$H_{\rm K}(f) = {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} \beta_1 \hspace{0.05cm}\cdot \hspace{0.05cm} f		:$$H_{\rm K}(f) = {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} \beta_1 \hspace{0.05cm}\cdot \hspace{0.05cm} f
	\hspace{0.05cm}\cdot \hspace{0.05cm}l}		\hspace{0.05cm}\cdot \hspace{0.05cm}l}

Javier: Text replacement - "„" to """

2021-05-28T14:33:57Z

Text replacement - "„" to """

← Older revision		Revision as of 16:33, 28 May 2021
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	Wir betrachten wie in  [[Aufgaben:4.5_Koaxialkabel_–_Impulsantwort\|Aufgabe 4.5]]  ein binäres Übertragungssystem mit der Bitrate  $R$  ⇒   Symboldauer  $T= 1/R$.		Wir betrachten wie in  [[Aufgaben:4.5_Koaxialkabel_–_Impulsantwort\|Aufgabe 4.5]]  ein binäres Übertragungssystem mit der Bitrate  $R$  ⇒   Symboldauer  $T= 1/R$.

	Als Übertragungsmedium wird ein ~~&bdquo;~~Normalkoaxialkabel”  $\text{(2.6 mm}$  Kerndurchmesser,  $\text{9.5 mm}$  Außendurchmesser$)$  der Länge  $l = 1 \ \rm km$  mit folgendem Frequenzgang verwendet:		Als Übertragungsmedium wird ein "Normalkoaxialkabel”  $\text{(2.6 mm}$  Kerndurchmesser,  $\text{9.5 mm}$  Außendurchmesser$)$  der Länge  $l = 1 \ \rm km$  mit folgendem Frequenzgang verwendet:
	:$$H_{\rm K}(f) = {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} \beta_1 \hspace{0.05cm}\cdot \hspace{0.05cm} f		:$$H_{\rm K}(f) = {\rm e}^{-{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm} \beta_1 \hspace{0.05cm}\cdot \hspace{0.05cm} f
	\hspace{0.05cm}\cdot \hspace{0.05cm}l}		\hspace{0.05cm}\cdot \hspace{0.05cm}l}

Javier: Text replacement - "Category:Exercises for Linear and Time-Invariant Systems" to "Category:Linear and Time-Invariant Systems: Exercises"

2021-03-23T12:40:33Z

Text replacement - "Category:Exercises for Linear and Time-Invariant Systems" to "Category:Linear and Time-Invariant Systems: Exercises"

← Older revision		Revision as of 14:40, 23 March 2021
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	[[Category:~~Exercises for~~ Linear and Time-Invariant Systems\|^4.2 Koaxialkabel^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^4.2 Koaxialkabel^]]