Aufgaben:Exercise 4.4: Coaxial Cable - Frequency Response - 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:Aufgaben:4.~~4_Koaxialkabel_–_Frequenzgang~~]]		[[de:Aufgaben:Aufgabe 4.4: Koaxialkabel – Frequenzgang]]

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	has the following frequency response:		has the following frequency response:
	:$$H_{\rm K}(f) = {\rm e}^{- \alpha_0 \hspace{0.05cm} \cdot \hspace{0.05cm} l} \cdot		:$$H_{\rm K}(f) = {\rm e}^{- \alpha_0 \hspace{0.05cm} \cdot \hspace{0.05cm} l} \cdot{\rm e}^{- \alpha_1 \hspace{0.05cm}\cdot \hspace{0.05cm}l \hspace{0.05cm}\cdot \hspace{0.05cm}f} \cdot{\rm e}^{- \alpha_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot\sqrt{f}} \cdot{\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_1 \hspace{0.05cm}\cdot \hspace{0.05cm} l \hspace{0.05cm}\cdot \hspace{0.05cm}f} \cdot{\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot\sqrt{f}} \hspace{0.05cm}.$$
	{\rm e}^{- \alpha_1 \hspace{0.05cm}\cdot \hspace{0.05cm}l \hspace{0.05cm}\cdot \hspace{0.05cm}f} \cdot
	{\rm e}^{- \alpha_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot
	\sqrt{f}} \cdot
	{\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_1 \hspace{0.05cm}\cdot \hspace{0.05cm} l \hspace{0.05cm}\cdot \hspace{0.05cm}f} \cdot
	{\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot
	\sqrt{f}} \hspace{0.05cm}.$$
	The attenuation parameters  $\alpha_0$,  $\alpha_1$ and  $\alpha_2$ are to be used in  "Neper per kilometer"  (Np/km)  and the phase parameters  $\beta_1$ and  $\beta_2$ in  "Radian per kilometer"  (rad/km).  The following numerical values apply:		The attenuation parameters  $\alpha_0$,  $\alpha_1$ and  $\alpha_2$ are to be used in  "Neper per kilometer"  (Np/km)  and the phase parameters  $\beta_1$ and  $\beta_2$ in  "Radian per kilometer"  (rad/km).  The following numerical values apply:
	:$$\alpha_0 = 0.00162 \hspace{0.15cm}{\rm Np}/{\rm km} \hspace{0.05cm},$$		:$$\alpha_0 = 0.00162 \hspace{0.15cm}{\rm Np}/{\rm km} \hspace{0.05cm},$$

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	'''(5)'''  For any positive quantity  $x$  the following holds:		'''(5)'''  For any positive quantity  $x$  the following holds:
	:$$x_{\rm Np} = {\rm ln} \hspace{0.10cm} x = \frac{{\rm lg} \hspace{0.10cm} x}{{\rm lg} \hspace{0.10cm} {\rm e}}= \frac{1}{{20 \cdot \rm lg} \hspace{0.10cm} {\rm e}} \cdot(20 \cdot {\rm lg} \hspace{0.10cm} x) = 0.1151 \cdot x_{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x_{\rm dB} = 8.6859 \cdot x_{\~~rmNp~~}\hspace{0.05cm}.$$		:$$x_{\rm Np} = {\rm ln} \hspace{0.10cm} x = \frac{{\rm lg} \hspace{0.10cm} x}{{\rm lg} \hspace{0.10cm} {\rm e}}= \frac{1}{{20 \cdot \rm lg} \hspace{0.10cm} {\rm e}} \cdot(20 \cdot {\rm lg} \hspace{0.10cm} x) = 0.1151 \cdot x_{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x_{\rm dB} = 8.6859 \cdot x_{\rm Np}\hspace{0.05cm}.$$
	The attenuation value  $4.555 \ {\rm Np}$  is thus identical to  ${a}_{\rm K}(f = 70\ \rm MHz)\hspace{0.15cm}\underline{=39.56 \ \rm dB}$.		The attenuation value  $4.555 \ {\rm Np}$  is thus identical to  ${a}_{\rm K}(f = 70\ \rm MHz)\hspace{0.15cm}\underline{=39.56 \ \rm dB}$.

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	* the attenuation function  (in Np or dB):		* the attenuation function  (in Np or dB):
	:$${ a}_{\rm K}(f) = - {\rm ln} \hspace{0.10cm}\|H_{\rm K}(f)\|= - 20 \cdot {\rm lg} \hspace{0.10cm}\|H_{\rm K}(f)\|		:$${ a}_{\rm K}(f) = - {\rm ln} \hspace{0.10cm}\|H_{\rm K}(f)\|= - 20 \cdot {\rm lg} \hspace{0.10cm}\|H_{\rm K}(f)\|\hspace{0.05cm},$$
	\hspace{0.05cm},$$

	* the phase function  (in rad or degree):		* the phase function  (in rad or degree):
	:$$b_{\rm K}(f) = - {\rm arc} \hspace{0.10cm}H_{\rm K}(f)		:$$b_{\rm K}(f) = - {\rm arc} \hspace{0.10cm}H_{\rm K}(f)\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	In practice,  one often uses the approximation		In practice,  one often uses the approximation
	:$$H_{\rm K}(f) =		:$$H_{\rm K}(f) ={\rm e}^{- \alpha_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot\sqrt{f}} \cdot{\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot\sqrt{f}}$$
	{\rm e}^{- \alpha_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot		:$$\Rightarrow \hspace{0.3cm} a_{\rm K}(f) = \alpha_2 \cdot l \cdot\sqrt{f}, \hspace{0.8cm}b_{\rm K}(f) = a_{\rm K}(f) \cdot{\rm rad}/{\rm Np}\hspace{0.05cm}.$$
	\sqrt{f}} \cdot
	{\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot
	\sqrt{f}}$$
	:$$\Rightarrow \hspace{0.3cm} a_{\rm K}(f) = \alpha_2 \cdot l \cdot
	\sqrt{f}, \hspace{0.8cm}b_{\rm K}(f) = a_{\rm K}(f) \cdot
	{\rm rad}/{\rm Np}\hspace{0.05cm}.$$
	This is allowed because  $\alpha_2$  and  $\beta_2$  have exactly the same numerical value and differ only by different pseudo units.		This is allowed because  $\alpha_2$  and  $\beta_2$  have exactly the same numerical value and differ only by different pseudo units.

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	'''(2)'''  With  ${\rm a}_0 = \alpha_0 \cdot l$  the following equation must be satisfied:		'''(2)'''  With  ${\rm a}_0 = \alpha_0 \cdot l$  the following equation must be satisfied:
	:$${\rm e}^{- {\rm a}_0 } \ge 0.99		:$${\rm e}^{- {\rm a}_0 } \ge 0.99\hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\rm a}_0 < {\rm ln}\hspace{0.10cm}\frac{1}{0.99}\approx 0.01\,\,{\rm (Np)}\hspace{0.05cm}.$$
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\rm a}_0 < {\rm ln}
	\hspace{0.10cm}\frac{1}{0.99}\approx 0.01\,\,{\rm (Np)}
	\hspace{0.05cm}.$$
	*This gives the maximum cable length:		*This gives the maximum cable length:
	:$$l_{\rm max} = \frac{{\rm a}_0 }{\alpha_0 } = \frac{0.01\,\,{\rm Np}}{0.00162\,\,{\rm Np/km}}\hspace{0.15cm}\underline{\approx 6.173\,\,{\rm km}}		:$$l_{\rm max} = \frac{{\rm a}_0 }{\alpha_0 } = \frac{0.01\,\,{\rm Np}}{0.00162\,\,{\rm Np/km}}\hspace{0.15cm}\underline{\approx 6.173\,\,{\rm km}}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	'''(3)'''  The following applies to the attenuation curve  when all terms are taken into account:		'''(3)'''  The following applies to the attenuation curve  when all terms are taken into account:
	:$${a}_{\rm K}(f) = [\alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot		:$${a}_{\rm K}(f) = [\alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot\sqrt{f}\hspace{0.05cm}] \cdot l= \big[0.00162 + 0.000435 \cdot 70 + 0.2722 \cdot \sqrt{70}\hspace{0.05cm}\big]\, \frac{\rm Np}{\rm km} \cdot 2\,{\rm km} $$
	\sqrt{f}\hspace{0.05cm}] \cdot l
	= \big[0.00162 + 0.000435 \cdot 70 + 0.2722 \cdot \sqrt{70}\hspace{0.05cm}\big]\, \frac{\rm Np}{\rm km} \cdot 2\,{\rm km} $$
	:$$ \Rightarrow \hspace{0.3cm}{a}_{\rm K}(f = 70\ \rm MHz) = \big[0.003 + 0.061 + 4.555 \hspace{0.05cm}\big]\, {\rm Np}\hspace{0.15cm}\underline{= 4.619\, {\rm Np}}\hspace{0.05cm}.$$		:$$ \Rightarrow \hspace{0.3cm}{a}_{\rm K}(f = 70\ \rm MHz) = \big[0.003 + 0.061 + 4.555 \hspace{0.05cm}\big]\, {\rm Np}\hspace{0.15cm}\underline{= 4.619\, {\rm Np}}\hspace{0.05cm}.$$

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	'''(5)'''  For any positive quantity  $x$  the following holds:		'''(5)'''  For any positive quantity  $x$  the following holds:
	:$$x_{\rm Np} = {\rm ln} \hspace{0.10cm} x = \frac{{\rm lg} \hspace{0.10cm} x}{{\rm lg} \hspace{0.10cm} {\rm e}}		:$$x_{\rm Np} = {\rm ln} \hspace{0.10cm} x = \frac{{\rm lg} \hspace{0.10cm} x}{{\rm lg} \hspace{0.10cm} {\rm e}}= \frac{1}{{20 \cdot \rm lg} \hspace{0.10cm} {\rm e}} \cdot(20 \cdot {\rm lg} \hspace{0.10cm} x) = 0.1151 \cdot x_{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x_{\rm dB} = 8.6859 \cdot x_{\rmNp}\hspace{0.05cm}.$$
	= \frac{1}{{20 \cdot \rm lg} \hspace{0.10cm} {\rm e}} \cdot
	(20 \cdot {\rm lg} \hspace{0.10cm} x) = 0.1151 \cdot x_{\rm dB} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} x_{\rm dB} = 8.6859 \cdot x_{\rm
	Np}\hspace{0.05cm}.$$
	The attenuation value  $4.555 \ {\rm Np}$  is thus identical to  ${a}_{\rm K}(f = 70\ \rm MHz)\hspace{0.15cm}\underline{=39.56 \ \rm dB}$.		The attenuation value  $4.555 \ {\rm Np}$  is thus identical to  ${a}_{\rm K}(f = 70\ \rm MHz)\hspace{0.15cm}\underline{=39.56 \ \rm dB}$.

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	'''(6)'''  <u>Solutions 1, 4 and 5</u> are correct.  Explanation:		'''(6)'''  <u>Solutions 1, 4 and 5</u> are correct.  Explanation:
	*With the restriction to the attenuation term with  $\alpha_2$,  the following applies to the frequency response:		*With the restriction to the attenuation term with  $\alpha_2$,  the following applies to the frequency response:
	:$$H_{\rm K}(f) =		:$$H_{\rm K}(f) ={\rm e}^{- \alpha_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot\sqrt{f}} \cdot{\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_1 \hspace{0.05cm}\cdot \hspace{0.05cm} l \hspace{0.05cm}\cdot f} \cdot{\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot\sqrt{f}} \hspace{0.05cm}.$$
	{\rm e}^{- \alpha_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot
	\sqrt{f}} \cdot
	{\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_1 \hspace{0.05cm}\cdot \hspace{0.05cm} l \hspace{0.05cm}\cdot f} \cdot
	{\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot
	\sqrt{f}} \hspace{0.05cm}.$$
	*If the  $\beta_1$–phase term is omitted,  nothing changes with respect to the distortions.   Only the phase and group delay would be  (both equal)  smaller by the value  $\tau_1 = (\beta_1 \cdot l)/(2\pi)$.		*If the  $\beta_1$–phase term is omitted,  nothing changes with respect to the distortions.   Only the phase and group delay would be  (both equal)  smaller by the value  $\tau_1 = (\beta_1 \cdot l)/(2\pi)$.

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	::'''(b)''' The impulse response  $h_{\rm K}(t)$  is symmetrical at  $t = 0$  with real frequency response,  which does not correspond to the conditions.		::'''(b)''' The impulse response  $h_{\rm K}(t)$  is symmetrical at  $t = 0$  with real frequency response,  which does not correspond to the conditions.
	*Therefore,  as an approximation for the coaxial cable frequency response,  the following is allowed:		*Therefore,  as an approximation for the coaxial cable frequency response,  the following is allowed:
	:$${a}_{\rm K}(f) = \alpha_2 \cdot l \cdot		:$${a}_{\rm K}(f) = \alpha_2 \cdot l \cdot\sqrt{f},$$
	\sqrt{f},$$		:$$ b_{\rm K}(f) = a_{\rm K}(f) \cdot{\rm rad}/{\rm Np}\hspace{0.05cm}.$$
	:$$ b_{\rm K}(f) = a_{\rm K}(f) \cdot
	{\rm rad}/{\rm Np}\hspace{0.05cm}.$$
	*That means:  ${a}_{\rm K}(f)$  and  ${b}_{\rm K}(f)$  of a coaxial cable are in first approximation identical in shape and differ only in their units.		*That means:  ${a}_{\rm K}(f)$  and  ${b}_{\rm K}(f)$  of a coaxial cable are in first approximation identical in shape and differ only in their units.

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	*A system with four times the bit rate  $R/2 = 280 \ \rm Mbit/s$  and half the length  $(l = 1 \ \rm km)$  results in the same characteristic cable attenuation.		*A system with four times the bit rate  $R/2 = 280 \ \rm Mbit/s$  and half the length  $(l = 1 \ \rm km)$  results in the same characteristic cable attenuation.
	*In contrast,  the following holds for a system with  $R/2 = 35 \ \rm Mbit/s$  and  $l = 2 \ \rm km$:		*In contrast,  the following holds for a system with  $R/2 = 35 \ \rm Mbit/s$  and  $l = 2 \ \rm km$:
	:$${a}_\star = 0.2722 \hspace{0.15cm}\frac {\rm Np}{\rm km \cdot \sqrt{MHz}} \cdot {\rm 2\,km}\cdot\sqrt{\rm 35\,MHz}		:$${a}_\star = 0.2722 \hspace{0.15cm}\frac {\rm Np}{\rm km \cdot \sqrt{MHz}} \cdot {\rm 2\,km}\cdot\sqrt{\rm 35\,MHz}\cdot 8.6859 \,\frac {\rm dB}{\rm Np} \approx 28\,{\rm dB}\hspace{0.05cm}.$$
	\cdot 8.6859 \,\frac {\rm dB}{\rm Np} \approx 28\,{\rm dB}
	\hspace{0.05cm}.$$

	{{ML-Fuß}}		{{ML-Fuß}}

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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:Aufgaben:4.4_Koaxialkabel_–_Frequenzgang]]

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	}}		}}

	[[File:LZI_A_4_4_vers3.png\|right\|frame\|Various coaxial ~~cables~~]]		[[File:LZI_A_4_4_vers3.png\|right\|frame\|Various coaxial cable types]]
	A so-called normal coaxial cable of length  $l$  with		A so-called normal coaxial cable of length  $l$  with
	*core diameter  $\text{2.6 mm}$,  and		*core diameter  $\text{2.6 mm}$,  and

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>Solutions 1 and 4</u> are correct:		'''(1)'''  <u>Solutions 1 and 4</u>  are correct:
	*The  $\alpha_0$–term causes only a frequency-independent attenuation.		*The  $\alpha_0$–term causes only a frequency-independent attenuation.
	*The  $\beta_1$–term (linear phase) results in a frequency-independent delay.		*The  $\beta_1$–term  (linear phase)  results in a frequency-independent delay.
	*All other terms contribute to the (linear) distortions.		*All other terms contribute to the  (linear)  distortions.




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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

			'''(3)'''  The following applies to the attenuation curve  when all terms are taken into account:


	'''(3)'''  The following applies to the attenuation curve when all terms are taken into account:
	:$${a}_{\rm K}(f) = [\alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot		:$${a}_{\rm K}(f) = [\alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot
	\sqrt{f}\hspace{0.05cm}] \cdot l		\sqrt{f}\hspace{0.05cm}] \cdot l
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			'''(4)'''  According to the calculation in subtask  '''(3)''',  the attenuation value  ${a}_{\rm K}(f = 70\ \rm MHz)\hspace{0.15cm}\underline{=4.555 \ \rm Np}$ is obtained here.


	'''(4)'''  ~~According to the calculation in subtask  '''(3)''' , the attenuation value~~  $~~{a}_{\rm K}(f = 70\ \rm MHz)\hspace{0.15cm}\underline{=4.555 \ \rm Np}~~$ ~~is obtained here.~~		'''(5)'''  For any positive quantity  $x$  the following holds:




	~~'''(5)'''~~  ~~For any positive quantity $x$~~ the following holds:
	:$$x_{\rm Np} = {\rm ln} \hspace{0.10cm} x = \frac{{\rm lg} \hspace{0.10cm} x}{{\rm lg} \hspace{0.10cm} {\rm e}}		:$$x_{\rm Np} = {\rm ln} \hspace{0.10cm} x = \frac{{\rm lg} \hspace{0.10cm} x}{{\rm lg} \hspace{0.10cm} {\rm e}}
	= \frac{1}{{20 \cdot \rm lg} \hspace{0.10cm} {\rm e}} \cdot		= \frac{1}{{20 \cdot \rm lg} \hspace{0.10cm} {\rm e}} \cdot
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			'''(6)'''  <u>Solutions 1, 4 and 5</u> are correct.  Explanation:
	'''(6)'''  <u>Solutions 1, 4 and 5</u> are correct. Explanation:		*With the restriction to the attenuation term with  $\alpha_2$,  the following applies to the frequency response:
	*With the restriction to the attenuation term with  $\alpha_2$ , the following applies to the frequency response:
	:$$H_{\rm K}(f) =		:$$H_{\rm K}(f) =
	{\rm e}^{- \alpha_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot		{\rm e}^{- \alpha_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot
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	{\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot		{\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot
	\sqrt{f}} \hspace{0.05cm}.$$		\sqrt{f}} \hspace{0.05cm}.$$
	*If the  $\beta_1$–phase term is omitted, nothing changes with respect to the distortions.   Only the phase and group delay would be (both equal) smaller by the value  $\tau_1 = (\beta_1 \cdot l)/(2\pi)$~~ ~~.		*If the  $\beta_1$–phase term is omitted,  nothing changes with respect to the distortions.   Only the phase and group delay would be  (both equal)  smaller by the value  $\tau_1 = (\beta_1 \cdot l)/(2\pi)$.

	*If, on the other hand, we omit the  $\beta_2$–term, we obtain completely different ~~ratios~~:		*If,  on the other hand,  we omit the  $\beta_2$–term,  we obtain completely different conditions:
	::'''(a)''' The frequency response  $H_{\rm K}(f)$  no longer fulfills the requirement of a causal system; in such a case,  $H_{\rm K}(f)$  would have to be in minimum phase.		::'''(a)''' The frequency response  $H_{\rm K}(f)$  no longer fulfills the requirement of a causal system;  in such a case,  $H_{\rm K}(f)$  would have to be in minimum phase.
	::'''(b)''' The impulse response  $h_{\rm K}(t)$  is symmetrical at  $t = 0$ with real frequency response, which does not correspond to the conditions.		::'''(b)''' The impulse response  $h_{\rm K}(t)$  is symmetrical at  $t = 0$  with real frequency response,  which does not correspond to the conditions.
	*Therefore, as an approximation for the coaxial cable frequency response, the following is allowed:		*Therefore,  as an approximation for the coaxial cable frequency response,  the following is allowed:
	:$${a}_{\rm K}(f) = \alpha_2 \cdot l \cdot		:$${a}_{\rm K}(f) = \alpha_2 \cdot l \cdot
	\sqrt{f},$$		\sqrt{f},$$
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	*For a digital system with bit rate  $R = 140 \ \rm Mbit/s$   ⇒   $R/2 = 70 \ \rm Mbit/s$  and cable length  $l = 2 \ \rm km$ ,   $a_\star \approx 40 \ \rm dB$  holds (see solution to the last sub-task).		*For a digital system with bit rate  $R = 140 \ \rm Mbit/s$   ⇒   $R/2 = 70 \ \rm Mbit/s$  and cable length  $l = 2 \ \rm km$ ,   $a_\star \approx 40 \ \rm dB$  holds (see solution to the last sub-task).
	*A system with four times the bit rate  $R/2 = 280 \ \rm Mbit/s$  and half the length  $(l = 1 \ \rm km)$  results in the same characteristic cable attenuation.		*A system with four times the bit rate  $R/2 = 280 \ \rm Mbit/s$  and half the length  $(l = 1 \ \rm km)$  results in the same characteristic cable attenuation.
	*In contrast, the following holds for a system with  $R/2 = 35 \ \rm Mbit/s$  and  $l = 2 \ \rm km$:		*In contrast,  the following holds for a system with  $R/2 = 35 \ \rm Mbit/s$  and  $l = 2 \ \rm km$:
	:$${a}_\star = 0.2722 \hspace{0.15cm}\frac {\rm Np}{\rm km \cdot \sqrt{MHz}} \cdot {\rm 2\,km}\cdot\sqrt{\rm 35\,MHz}		:$${a}_\star = 0.2722 \hspace{0.15cm}\frac {\rm Np}{\rm km \cdot \sqrt{MHz}} \cdot {\rm 2\,km}\cdot\sqrt{\rm 35\,MHz}
	\cdot 8.6859 \,\frac {\rm dB}{\rm Np} \approx 28\,{\rm dB}		\cdot 8.6859 \,\frac {\rm dB}{\rm Np} \approx 28\,{\rm dB}

Guenter at 16:10, 12 November 2021

2021-11-12T16:10:06Z

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			Notes:



	''Notes:''
	*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Eigenschaften_von_Koaxialkabeln\|Properties of Coaxial Cables]].		*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Eigenschaften_von_Koaxialkabeln\|Properties of Coaxial Cables]].

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	{What is the attenuation (in Np) at frequency  $f = 70 \ \rm MHz$ when the cable length is  $\underline{l = 2 \ \rm km}$~~ ~~?		{What is the attenuation  (in Np)  at frequency  $f = 70 \ \rm MHz$  when the cable length is  $\underline{l = 2 \ \rm km}$?
	\|type="{}"}		\|type="{}"}
	$a_{\rm K}(f = 70\ \rm MHz) \ = \ $ { 4.619 3% } $\ \rm Np$		$a_{\rm K}(f = 70\ \rm MHz) \ = \ $ { 4.619 3% } $\ \rm Np$


	{Assuming all other things are equal, what is the attenuation when only the  $\alpha_2$–term is considered?		{Assuming all other things are equal,  what is the attenuation when only the  $\alpha_2$–term is considered?
	\|type="{}"}		\|type="{}"}
	$a_{\rm K}(f = 70\ \rm MHz) \ = \ $ { 4.555 3% } $\ \rm Np$		$a_{\rm K}(f = 70\ \rm MHz) \ = \ $ { 4.555 3% } $\ \rm Np$


	{What is the formula for the conversion between  $\rm Np$ and  $\rm dB$?  What is the $\rm dB$ value that results for the attenuation calculated in  '''(4)'''~~ ~~?		{What is the formula for the conversion between  $\rm Np$  and  $\rm dB$?  What is the $\rm dB$ value that results for the attenuation calculated in  '''(4)'''?
	\|type="{}"}		\|type="{}"}
	$a_{\rm K}(f = 70\ \rm MHz) \ = \ $ { 39.56 3% } $\ \rm dB$		$a_{\rm K}(f = 70\ \rm MHz) \ = \ $ { 39.56 3% } $\ \rm dB$
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	{Which statements are true if we restrict ourselves to the  $\alpha_2$–value with respect to the attenuation function?		{Which statements are true if we restrict ourselves to the  $\alpha_2$–value with respect to the attenuation function?
	\|type="[]"}		\|type="[]"}
	+ One can also do without the phase term with  $\beta_1$~~ ~~.		+ One can also do without the phase term with  $\beta_1$.
	- One can also do without the phase term with  $\beta_2$~~ ~~.		- One can also do without the phase term with  $\beta_2$.
	- $a_\star \approx 40 \ \rm dB$  holds for a system with  $R = 70 \ \rm Mbit/s$  and  $l = 2 \ \rm km$.		- $a_\star \approx 40 \ \rm dB$  holds for a system with  $R = 70 \ \rm Mbit/s$  and  $l = 2 \ \rm km$.
	+ $a_\star \approx 40 \ \rm dB$  holds for a system with  $R = 140 \ \rm Mbit/s$  and  $l = 2 \ \rm km$.		+ $a_\star \approx 40 \ \rm dB$  holds for a system with  $R = 140 \ \rm Mbit/s$  and  $l = 2 \ \rm km$.

Guenter at 16:06, 12 November 2021

2021-11-12T16:06:21Z

← Older revision		Revision as of 18:06, 12 November 2021
Line 57:		Line 57:
	*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Eigenschaften_von_Koaxialkabeln\|Properties of Coaxial Cables]].		*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Eigenschaften_von_Koaxialkabeln\|Properties of Coaxial Cables]].

	*You can use the interactive applet  [[Applets:~~Dämpfung_von_Kupferkabeln~~\|~~Dämpfung von Kupferkabeln~~]]  to check your results.		*You can use the interactive  "HTML 5/JS" applet  [[Applets:Attenuation_of_Copper_Cables\|Applets:Attenuation of Copper Cables]]  to check your results.

Guenter at 16:02, 12 November 2021

2021-11-12T16:02:08Z

← Older revision		Revision as of 18:02, 12 November 2021
Line 5:		Line 5:
	[[File:LZI_A_4_4_vers3.png\|right\|frame\|Various coaxial cables]]		[[File:LZI_A_4_4_vers3.png\|right\|frame\|Various coaxial cables]]
	A so-called normal coaxial cable of length  $l$  with		A so-called normal coaxial cable of length  $l$  with
	*~~the~~ core diameter  $\text{2.6 mm}$,  and		*core diameter  $\text{2.6 mm}$,  and
	*~~the~~ outer diameter  $\text{9.5 mm}$		*outer diameter  $\text{9.5 mm}$


Line 17:		Line 17:
	{\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot		{\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot
	\sqrt{f}} \hspace{0.05cm}.$$		\sqrt{f}} \hspace{0.05cm}.$$
	The attenuation parameters  $\alpha_0$,  $\alpha_1$ and  $\alpha_2$ are to be used in "Neper per kilometer" (Np/km) and the phase parameters  $\beta_1$ and  $\beta_2$ in "Radian per kilometer" (rad/km). The following numerical values apply:		The attenuation parameters  $\alpha_0$,  $\alpha_1$ and  $\alpha_2$ are to be used in  "Neper per kilometer"  (Np/km)  and the phase parameters  $\beta_1$ and  $\beta_2$ in  "Radian per kilometer"  (rad/km).  The following numerical values apply:
	:$$\alpha_0 = 0.00162 \hspace{0.15cm}{\rm Np}/{\rm km} \hspace{0.05cm},$$		:$$\alpha_0 = 0.00162 \hspace{0.15cm}{\rm Np}/{\rm km} \hspace{0.05cm},$$
	:$$\alpha_1 = 0.000435 \hspace{0.15cm} {\rm Np}/{{\rm km} \cdot {\rm MHz}} \hspace{0.05cm},$$		:$$\alpha_1 = 0.000435 \hspace{0.15cm} {\rm Np}/{{\rm km} \cdot {\rm MHz}} \hspace{0.05cm},$$
	:$$\alpha_2 = 0.2722 \hspace{0.15cm}{\rm Np}/{{\rm km} \cdot \sqrt{\rm MHz}} \hspace{0.05cm}.$$		:$$\alpha_2 = 0.2722 \hspace{0.15cm}{\rm Np}/{{\rm km} \cdot \sqrt{\rm MHz}} \hspace{0.05cm}.$$

	~~Often one uses for~~ the system-theoretical description of a ~~linear time-invariant system~~		For the system-theoretical description of a coaxial cable  (German:  "Koaxialkabel"   ⇒   subscipt  "K"),  one uses

	* the attenuation function (in Np or dB):		* the attenuation function  (in Np or dB):
	:$${ a}_{\rm K}(f) = - {\rm ln} \hspace{0.10cm}\|H_{\rm K}(f)\|= - 20 \cdot {\rm lg} \hspace{0.10cm}\|H_{\rm K}(f)\|		:$${ a}_{\rm K}(f) = - {\rm ln} \hspace{0.10cm}\|H_{\rm K}(f)\|= - 20 \cdot {\rm lg} \hspace{0.10cm}\|H_{\rm K}(f)\|
	\hspace{0.05cm},$$		\hspace{0.05cm},$$

	* the phase function (in rad or degree):		* the phase function  (in rad or degree):
	:$$b_{\rm K}(f) = - {\rm arc} \hspace{0.10cm}H_{\rm K}(f)		:$$b_{\rm K}(f) = - {\rm arc} \hspace{0.10cm}H_{\rm K}(f)
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	In practice, one often uses the approximation		In practice,  one often uses the approximation
	:$$H_{\rm K}(f) =		:$$H_{\rm K}(f) =
	{\rm e}^{- \alpha_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot		{\rm e}^{- \alpha_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot
Line 43:		Line 43:
	This is allowed because  $\alpha_2$  and  $\beta_2$  have exactly the same numerical value and differ only by different pseudo units.		This is allowed because  $\alpha_2$  and  $\beta_2$  have exactly the same numerical value and differ only by different pseudo units.

	With the definition of the characteristic cable attenuation (in Neper or decibel)		With the definition of the characteristic cable attenuation  (in Neper or decibel)
	:$${a}_{\rm \star(Np)} = {a}_{\rm K}(f = {R}/{2}) = 0.1151 \cdot {a}_{\rm \star(dB)}$$		:$${a}_{\rm \star(Np)} = {a}_{\rm K}(f = {R}/{2}) = 0.1151 \cdot {a}_{\rm \star(dB)}$$
	digital systems with different bit rate  $R$  and cable length  $l$  can be treated uniformly.		digital systems with different bit rate  $R$  and cable length  $l$  can be treated uniformly.