Aufgaben:Exercise 3.1Z: Frequency Response of the Coaxial Cable - Revision history

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	[[Category:Digital Signal Transmission: Exercises\|^3.1 Intersymbol Interference^]]		[[Category:Digital Signal Transmission: Exercises\|^3.1 Intersymbol Interference^]]
	[[de:3.~~1Z_Frequenzgang_des_Koaxialkabels~~]]		[[de:Aufgaben:Aufgabe 3.1Z: Frequenzgang des Koaxialkabels]]

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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 entered in  "Neper"  $(\rm Np)$,  the phase parameters  $\beta_1$  and  $\beta_2$  in  "radian"  $(\rm rad)$.		The attenuation parameters  $\alpha_0$,  $\alpha_1$  and  $\alpha_2$  are to be entered in  "Neper"  $(\rm Np)$,  the phase parameters  $\beta_1$  and  $\beta_2$  in  "radian"  $(\rm rad)$.

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	'''(5)'''  For any positive quantity  $x$  holds:		'''(5)'''  For any positive quantity  $x$  holds:
	:$$x_{\rm Np} = {\rm ln} \hspace{0.10cm} x = \frac{{\rm lg} \hspace{0.10cm} x}{{\rm lg} \hspace{0.10cm} {\rm e}}= \frac{1}{{20 \cdot \rm lg} \hspace{0.10cm} {\rm e}} \cdot(20 \cdot {\rm lg} \hspace{0.10cm} x) = 0.1151 \cdot x_{\rm dB}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} x_{\rm dB} = 8.686 \cdot x_{\~~rmNp~~}\hspace{0.05cm}.$$		:$$x_{\rm Np} = {\rm ln} \hspace{0.10cm} x = \frac{{\rm lg} \hspace{0.10cm} x}{{\rm lg} \hspace{0.10cm} {\rm e}}= \frac{1}{{20 \cdot \rm lg} \hspace{0.10cm} {\rm e}} \cdot(20 \cdot {\rm lg} \hspace{0.10cm} x) = 0.1151 \cdot x_{\rm dB}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} x_{\rm dB} = 8.686 \cdot x_{\rm Np}\hspace{0.05cm}.$$
	*Thus,  the attenuation value  $4.555\,{\rm Np}$  is identical to  $\underline{39.57\,{\rm dB} }$.		*Thus,  the attenuation value  $4.555\,{\rm Np}$  is identical to  $\underline{39.57\,{\rm dB} }$.

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	'''(6)'''  <u>Solutions 1, 4 and 5</u>  are correct:		'''(6)'''  <u>Solutions 1, 4 and 5</u>  are correct:
	*With the restriction to the attenuation term with  $\alpha_2$,  the following holds for the frequency response:		*With the restriction to the attenuation term with  $\alpha_2$,  the following holds for 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 we omit the  $\beta_1$  phase term,  nothing changes with respect to the distortions.  Only the phase and group delay would become smaller  (both equal)  by the value  $\tau_1 = (\beta_1 \cdot l)/2\pi$.		*If we omit the  $\beta_1$  phase term,  nothing changes with respect to the distortions.  Only the phase and group delay would become smaller  (both equal)  by the value  $\tau_1 = (\beta_1 \cdot l)/2\pi$.
	*On the other hand,  if the  $\beta_2$–term is omitted,  completely different ratios result:		*On the other hand,  if the  $\beta_2$–term is omitted,  completely different ratios result:

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	The attenuation parameters  $\alpha_0$,  $\alpha_1$  and  $\alpha_2$  are to be entered in  "Neper"  $(\rm Np)$,  the phase parameters  $\beta_1$  and  $\beta_2$  in  "radian"  $(\rm rad)$.		The attenuation parameters  $\alpha_0$,  $\alpha_1$  and  $\alpha_2$  are to be entered in  "Neper"  $(\rm Np)$,  the phase parameters  $\beta_1$  and  $\beta_2$  in  "radian"  $(\rm rad)$.
	The following numerical values apply:		The following numerical values apply:
	:$$\alpha_0 = 0.00162 \hspace{0.15cm}\frac{\rm Np}{\rm km} \hspace{0.05cm},\hspace{0.2cm}		:$$\alpha_0 = 0.00162 \hspace{0.15cm}\frac{\rm Np}{\rm km} \hspace{0.05cm},\hspace{0.2cm}\alpha_1 = 0.000435 \hspace{0.15cm}\frac{\rm Np}{\rm km\cdot{\rm MHz}} \hspace{0.05cm},\hspace{0.2cm}\alpha_2 = 0.2722 \hspace{0.15cm}\frac{\rm Np}{\rm km\cdot\sqrt{\rm MHz}} \hspace{0.05cm}.$$
	\alpha_1 = 0.000435 \hspace{0.15cm}\frac{\rm Np}{\rm km\cdot{\rm MHz}} \hspace{0.05cm},
	\hspace{0.2cm}
	\alpha_2 = 0.2722 \hspace{0.15cm}\frac{\rm Np}{\rm km\cdot\sqrt{\rm MHz}} \hspace{0.05cm}.$$

	Often,  to describe a  [[Linear_and_Time_Invariant_Systems/Systembeschreibung_im_Frequenzbereich\|"linear time-invariant system"]]  $\rm (LTI)$  in terms of system theory,  one uses		Often,  to describe a  [[Linear_and_Time_Invariant_Systems/Systembeschreibung_im_Frequenzbereich\|"linear time-invariant system"]]  $\rm (LTI)$  in terms of system theory,  one uses
	* the attenuation function  $($in  $\rm Np$  or  $\rm dB)$:		* the attenuation function  $($in  $\rm Np$  or  $\rm dB)$:
	:$$a_{\rm K}(f) = - {\rm ln} \hspace{0.10cm}\|H_{\rm K}(f)\|= - 20 \cdot {\rm lg} \hspace{0.10cm}\|H_{\rm K}(f)\|		:$$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  $\rm rad$  or  $\rm degrees)$:		* the phase function  $($in  $\rm rad$  or  $\rm degrees)$:
	:$$b_{\rm K}(f) = - {\rm arc} \hspace{0.10cm}H_{\rm K}(f)		:$$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{0.2cm}b_{\rm K}(f) = a_{\rm K}(f) \cdot\frac{\rm rad}{\rm Np}\hspace{0.05cm}.$$
	\sqrt{f}} \cdot
	{\rm e}^{- {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l\hspace{0.05cm}\hspace{0.05cm}\cdot
	\sqrt{f}}$$
	:$$\Rightarrow \hspace{0.3cm} a_{\rm K}(f) = \alpha_2 \cdot l \cdot
	\sqrt{f}, \hspace{0.2cm}b_{\rm K}(f) = a_{\rm K}(f) \cdot
	\frac{\rm rad}{\rm Np}\hspace{0.05cm}.$$

	This is allowed because  $\alpha_2$  and  $\beta_2$  have exactly the same numerical value – just different pseudo units.		This is allowed because  $\alpha_2$  and  $\beta_2$  have exactly the same numerical value – just different pseudo units.
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	'''(2)'''  With  $a_0 = \alpha_0 \cdot l$,  the following equation must be satisfied:		'''(2)'''  With  $a_0 = \alpha_0 \cdot l$,  the following equation must be satisfied:
	:$${\rm e}^{- a_0 } \ge 0.99		:$${\rm e}^{- a_0 } \ge 0.99\hspace{0.3cm} \Rightarrow \hspace{0.3cm}a_0 < {\rm ln}\hspace{0.10cm}\frac{1}{0.99}\approx 0.01\,\,{\rm (Np)}\hspace{0.05cm}.$$
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm}a_0 < {\rm ln}
	\hspace{0.10cm}\frac{1}{0.99}\approx 0.01\,\,{\rm (Np)}
	\hspace{0.05cm}.$$

	*This gives the following for the maximum cable length:		*This gives the following for the maximum cable length:
	:$$l_{\rm max} = \frac{a_0 }{\alpha_0 } = \frac{0.01\,\,{\rm Np}}{0.00162\,\,{\rm Np/km}}\hspace{0.15cm}\underline {\approx 6.173\,\,{\rm km}}		:$$l_{\rm max} = \frac{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)'''  For the attenuation curve,  considering all terms:		'''(3)'''  For the attenuation curve,  considering all terms:
	:$$a_{\rm K}(f) \ = \ \big[\alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot		:$$a_{\rm K}(f) \ = \ \big[\alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot\sqrt{f}\hspace{0.05cm}\big] \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} = \hspace{0.15cm}\underline {= 4.619\, {\rm Np}}\hspace{0.05cm}.$$
	\sqrt{f}\hspace{0.05cm}\big] \cdot l = \big [0.00162 + 0.000435 \cdot 70 + 0.2722 \cdot \sqrt{70}\hspace{0.05cm}\big]\, \frac{\rm Np}{\rm km} \cdot 2\,{\rm km} = \hspace{0.15cm}\underline {= 4.619\, {\rm Np}}\hspace{0.05cm}.$$


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	'''(5)'''  For any positive quantity  $x$  holds:		'''(5)'''  For any positive quantity  $x$  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} = 8.686 \cdot x_{\rmNp}\hspace{0.05cm}.$$
	= \frac{1}{{20 \cdot \rm lg} \hspace{0.10cm} {\rm e}} \cdot
	(20 \cdot {\rm lg} \hspace{0.10cm} x) = 0.1151 \cdot x_{\rm dB}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} x_{\rm dB} = 8.686 \cdot x_{\rm
	Np}\hspace{0.05cm}.$$
	*Thus,  the attenuation value  $4.555\,{\rm Np}$  is identical to  $\underline{39.57\,{\rm dB} }$.		*Thus,  the attenuation value  $4.555\,{\rm Np}$  is identical to  $\underline{39.57\,{\rm dB} }$.

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	# The impulse response  $h_{\rm K}(t)$  is symmetrical about  $t = 0$  with real frequency response,  which does not correspond to the conditions.		# The impulse response  $h_{\rm K}(t)$  is symmetrical about  $t = 0$  with real frequency response,  which does not correspond to the conditions.
	*Therefore as an approximation for the coaxial cable frequency response is allowed:		*Therefore as an approximation for the coaxial cable frequency response is allowed:
	:$$a_{\rm K}(f) = \alpha_2 \cdot l \cdot		:$$a_{\rm K}(f) = \alpha_2 \cdot l \cdot\sqrt{f}, \hspace{0.2cm}b_{\rm K}(f) = a_{\rm K}(f) \cdot\frac{\rm rad}{\rm Np}\hspace{0.05cm}.$$
	\sqrt{f}, \hspace{0.2cm}b_{\rm K}(f) = a_{\rm K}(f) \cdot
	\frac{\rm rad}{\rm Np}\hspace{0.05cm}.$$
	*That means:  $a_{\rm K}(f)$  and  $b_{\rm K}(f)$  of a coaxial cable are identical in form and differ only in their units.		*That means:  $a_{\rm K}(f)$  and  $b_{\rm K}(f)$  of a coaxial cable are identical in form and differ only in their units.
	For a digital system with bit rate  $R_{\rm B} = 140\,{\rm Mbit/s}$   ⇒   $R_{\rm B}/2 = 70\,{\rm Mbit/s}$  and cable length  $l = 2\,{\rm km}$:  <br>       $a_ ≈ 40\,{\rm dB}$ is indeed valid – see solution to subtask  '''(5)'''.		For a digital system with bit rate  $R_{\rm B} = 140\,{\rm Mbit/s}$   ⇒   $R_{\rm B}/2 = 70\,{\rm Mbit/s}$  and cable length  $l = 2\,{\rm km}$:  <br>       $a_ ≈ 40\,{\rm dB}$ is indeed valid – see solution to subtask  '''(5)'''.

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	[[Category:Digital Signal Transmission: Exercises\|^3.1 Intersymbol Interference^]]		[[Category:Digital Signal Transmission: Exercises\|^3.1 Intersymbol Interference^]]
			[[de:3.1Z_Frequenzgang_des_Koaxialkabels]]

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	$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  $\rm dB$&~~nbsp~~; value for the attenuation calculated in subtask  '''(4)'''?		{What is the formula for the conversion between  $\rm Np$  and  $\rm dB$?  What is the  $\rm dB$–value for the attenuation calculated in subtask  '''(4)'''?
	\|type="{}"}		\|type="{}"}
	$a_{\rm K}(f = 70\,{\rm MHz})\ = \ $ { 39.57 3% } $\ {\rm dB} $		$a_{\rm K}(f = 70\,{\rm MHz})\ = \ $ { 39.57 3% } $\ {\rm dB} $
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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 frequency-independent attenuation and the $\beta_1$–term (linear phase) causes frequency-independent delay.		*The  $\alpha_0$–term causes only frequency-independent attenuation and the  $\beta_1$–term  (linear phase)  causes frequency-independent delay.
	*All other terms contribute to the (linear) distortions.		*All other terms contribute to the  (linear)  distortions.



	'''(2)'''  With $a_0 = \alpha_0 \cdot l$, the following equation must be satisfied:		'''(2)'''  With  $a_0 = \alpha_0 \cdot l$,  the following equation must be satisfied:
	:$${\rm e}^{- a_0 } \ge 0.99		:$${\rm e}^{- a_0 } \ge 0.99
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm}a_0 < {\rm ln}		\hspace{0.3cm} \Rightarrow \hspace{0.3cm}a_0 < {\rm ln}
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	This gives the following for the maximum cable length		*This gives the following for the maximum cable length:
	:$$l_{\rm max} = \frac{a_0 }{\alpha_0 } = \frac{0.01\,\,{\rm Np}}{0.00162\,\,{\rm Np/km}}\hspace{0.15cm}\underline {\approx 6.173\,\,{\rm km}}		:$$l_{\rm max} = \frac{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)'''  For the attenuation curve, considering all terms:		'''(3)'''  For the attenuation curve,  considering all terms:
	:$$a_{\rm K}(f) \ = \ \big[\alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot		:$$a_{\rm K}(f) \ = \ \big[\alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot
	\sqrt{f}\hspace{0.05cm}\big] \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} = \hspace{0.15cm}\underline {= 4.619\, {\rm Np}}\hspace{0.05cm}.$$		\sqrt{f}\hspace{0.05cm}\big] \cdot l = \big [0.00162 + 0.000435 \cdot 70 + 0.2722 \cdot \sqrt{70}\hspace{0.05cm}\big]\, \frac{\rm Np}{\rm km} \cdot 2\,{\rm km} = \hspace{0.15cm}\underline {= 4.619\, {\rm Np}}\hspace{0.05cm}.$$


	'''(4)'''  According to the calculation at point (3), the attenuation value $\underline {4.555\,{\rm Np}}$ is obtained here.		'''(4)'''  According to the calculation at point  '''(3)''',  the attenuation value  $\underline {4.555\,{\rm Np}}$  is obtained here.


	'''(5)'''  For any positive quantity $x$ holds:		'''(5)'''  For any positive quantity  $x$  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
	(20 \cdot {\rm lg} \hspace{0.10cm} x) = 0.1151 \cdot x_{\rm dB}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} x_{\rm dB} = 8.686 \cdot x_{\rm		(20 \cdot {\rm lg} \hspace{0.10cm} x) = 0.1151 \cdot x_{\rm dB}\hspace{0.3cm} \Rightarrow \hspace{0.3cm} x_{\rm dB} = 8.686 \cdot x_{\rm
	Np}\hspace{0.05cm}.$$		Np}\hspace{0.05cm}.$$
	Thus, the attenuation value $4.555\,{\rm Np}$ is identical to $\underline{39.57\,{\rm dB} }$.		*Thus,  the attenuation value  $4.555\,{\rm Np}$  is identical to  $\underline{39.57\,{\rm dB} }$.


	'''(6)'''  <u>Solutions 1, 4 and 5</u> are correct:		'''(6)'''  <u>Solutions 1, 4 and 5</u>  are correct:
	*With the restriction to the attenuation term with $\alpha_2$, the following holds for the frequency response:		*With the restriction to the attenuation term with  $\alpha_2$,  the following holds for 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 we omit the $\beta_1$ phase term, nothing changes with respect to the distortions. Only the phase and group delay would become smaller (both equal) by the value $\tau_1 = (\beta_1 \cdot l)/2\pi$.		*If we omit the  $\beta_1$  phase term,  nothing changes with respect to the distortions.  Only the phase and group delay would become smaller  (both equal)  by the value  $\tau_1 = (\beta_1 \cdot l)/2\pi$.
	*On the other hand, if the $\beta_2$–term is omitted, completely different ratios result:		*On the other hand,  if the  $\beta_2$–term is omitted,  completely different ratios result:
	:: The frequency response $H_{\rm K}(f)$ now no longer satisfies the requirement of a causal system; for such a system $H_{\rm K}(f)$ must be minimum-phase.		# The frequency response  $H_{\rm K}(f)$  now no longer satisfies the requirement of a causal system;  for such a system  $H_{\rm K}(f)$  must be minimum-phase.
	:: The impulse response $h_{\rm K}(t)$ is symmetrical about $t = 0$ with real frequency response, which does not correspond to the conditions.		# The impulse response  $h_{\rm K}(t)$  is symmetrical about  $t = 0$  with real frequency response,  which does not correspond to the conditions.
	*Therefore as an approximation for the coaxial cable frequency response is allowed:		*Therefore as an approximation for the coaxial cable frequency response is allowed:
	:$$a_{\rm K}(f) = \alpha_2 \cdot l \cdot		:$$a_{\rm K}(f) = \alpha_2 \cdot l \cdot
	\sqrt{f}, \hspace{0.2cm}b_{\rm K}(f) = a_{\rm K}(f) \cdot		\sqrt{f}, \hspace{0.2cm}b_{\rm K}(f) = a_{\rm K}(f) \cdot
	\frac{\rm rad}{\rm Np}\hspace{0.05cm}.$$		\frac{\rm rad}{\rm Np}\hspace{0.05cm}.$$
	*That means: $a_{\rm K}(f)$ and $b_{\rm K}(f)$ of a coaxial cable are identical in form and differ only in their units.		*That means:  $a_{\rm K}(f)$  and  $b_{\rm K}(f)$  of a coaxial cable are identical in form and differ only in their units.
	For a digital system with bit rate $R_{\rm B} = 140\,{\rm Mbit/s}$ ⇒ $R_{\rm B}/2 = 70\,{\rm Mbit/s}$ and cable length $l = 2\,{\rm km}$, $a_ ≈ 40\,{\rm dB}$ is indeed valid – see solution to '''(5)'''.		For a digital system with bit rate  $R_{\rm B} = 140\,{\rm Mbit/s}$   ⇒   $R_{\rm B}/2 = 70\,{\rm Mbit/s}$  and cable length  $l = 2\,{\rm km}$:  <br>       $a_ ≈ 40\,{\rm dB}$ is indeed valid – see solution to subtask  '''(5)'''.
	*A system with four times the bit rate ($R_{\rm B}/2 = 280\,{\rm Mbit/s}$) and half the length ($l = 1\,{\rm km}$) results in the same characteristic cable attenuation.		*A system with four times the bit rate  $(R_{\rm B}/2 = 280\,{\rm Mbit/s})$  and half the length  $(l = 1\,{\rm km})$  results in the same characteristic cable attenuation.
	*In contrast, for a system with $R_{\rm B}/2 = 35\,{\rm Mbit/s}$ and $l = 2\,{\rm km}$ to		*In contrast,  for a system with  $R_{\rm B}/2 = 35\,{\rm Mbit/s}$  and  $l = 2\,{\rm km}$  holds:
	:$$a_{\rm dB} = 0.2722 \frac{\rm Np}{\rm km\cdot \sqrt{\rm MHz}} \cdot 2 \ \rm km \cdot \sqrt{35 \ \rm MHz} \cdot 8.6859 \frac{\rm dB}{\rm Np} ≈ 28 \ \rm dB.$$		:$$a_{\rm dB} = 0.2722 \frac{\rm Np}{\rm km\cdot \sqrt{\rm MHz}} \cdot 2 \ \rm km \cdot \sqrt{35 \ \rm MHz} \cdot 8.6859 \frac{\rm dB}{\rm Np} ≈ 28 \ \rm dB.$$

Guenter at 13:11, 31 May 2022

2022-05-31T13:11:28Z

← Older revision		Revision as of 15:11, 31 May 2022
Line 3:		Line 3:

	[[File:P_ID1371__Dig_Z_3_1.png\|right\|frame\|Some coaxial cable types]]		[[File:P_ID1371__Dig_Z_3_1.png\|right\|frame\|Some coaxial cable types]]
	A so-called standard coaxial cable with core diameter  $2.6 \ \rm mm$, outer diameter  $9.5 \ \rm mm$  and length  $l$  has the following frequency response:		A so-called  "standard coaxial cable"
			*with core diameter  $2.6 \ \rm mm$,
			*outer diameter  $9.5 \ \rm mm$  and
			*length  $l$


			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_1 \hspace{0.05cm}\cdot \hspace{0.05cm}l \hspace{0.05cm}\cdot \hspace{0.05cm}f} \cdot
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	\sqrt{f}} \hspace{0.05cm}.$$		\sqrt{f}} \hspace{0.05cm}.$$

	The attenuation parameters  $\alpha_0$,  $\alpha_1$  and  $\alpha_2$  are to be entered in Neper  $(\rm Np)$, the phase parameters  $\beta_1$  and  $\beta_2$  in ~~Radian~~  $(\rm rad)$.		The attenuation parameters  $\alpha_0$,  $\alpha_1$  and  $\alpha_2$  are to be entered in  "Neper"  $(\rm Np)$,  the phase parameters  $\beta_1$  and  $\beta_2$  in  "radian"  $(\rm rad)$.
	The following numerical values apply:		The following numerical values apply:
	:$$\alpha_0 = 0.00162 \hspace{0.15cm}\frac{\rm Np}{\rm km} \hspace{0.05cm},\hspace{0.2cm}		:$$\alpha_0 = 0.00162 \hspace{0.15cm}\frac{\rm Np}{\rm km} \hspace{0.05cm},\hspace{0.2cm}
Line 19:		Line 25:
	\alpha_2 = 0.2722 \hspace{0.15cm}\frac{\rm Np}{\rm km\cdot\sqrt{\rm MHz}} \hspace{0.05cm}.$$		\alpha_2 = 0.2722 \hspace{0.15cm}\frac{\rm Np}{\rm km\cdot\sqrt{\rm MHz}} \hspace{0.05cm}.$$

	Often, to describe a  [[Linear_and_Time_Invariant_Systems/Systembeschreibung_im_Frequenzbereich\|"linear time-invariant system"]]  (~~LZI~~) in terms of system theory, one uses		Often,  to describe a  [[Linear_and_Time_Invariant_Systems/Systembeschreibung_im_Frequenzbereich\|"linear time-invariant system"]]  $\rm (LTI)$  in terms of system theory,  one uses
	* the attenuation function $($in  $\rm Np$  or  $\rm dB)$:		* the attenuation function  $($in  $\rm Np$  or  $\rm dB)$:
	:$$a_{\rm K}(f) = - {\rm ln} \hspace{0.10cm}\|H_{\rm K}(f)\|= - 20 \cdot {\rm lg} \hspace{0.10cm}\|H_{\rm K}(f)\|		:$$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  $\rm rad$ or $\rm degrees)$:		* the phase function  $($in  $\rm rad$  or  $\rm degrees)$:
	:$$b_{\rm K}(f) = - {\rm arc} \hspace{0.10cm}H_{\rm K}(f)		:$$b_{\rm K}(f) = - {\rm arc} \hspace{0.10cm}H_{\rm K}(f)
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
Line 32:		Line 38:
	\sqrt{f}} \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		{\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{0.3cm}~~\Rightarrow \hspace{0.3cm} a_{\rm K}(f) = \alpha_2 \cdot l \cdot		\sqrt{f}}$$
			:$$\Rightarrow \hspace{0.3cm} a_{\rm K}(f) = \alpha_2 \cdot l \cdot
	\sqrt{f}, \hspace{0.2cm}b_{\rm K}(f) = a_{\rm K}(f) \cdot		\sqrt{f}, \hspace{0.2cm}b_{\rm K}(f) = a_{\rm K}(f) \cdot
	\frac{\rm rad}{\rm Np}\hspace{0.05cm}.$$		\frac{\rm rad}{\rm Np}\hspace{0.05cm}.$$

	This is allowed because  $\alpha_2$  and  $\beta_2$  have exactly the same numerical value – just different pseudo units. Using the definition of the  '''characteristic cable attenuation'''  (in Neper or decibels)		This is allowed because  $\alpha_2$  and  $\beta_2$  have exactly the same numerical value – just different pseudo units.

			Using the definition of the  '''characteristic cable attenuation'''  (in Neper or decibels)
	:$$a_{\rm * (Np)} = a_{\rm K}(f = {R_{\rm B}}/{2}) = 0.1151 \cdot a_{\rm * (dB)}$$		:$$a_{\rm * (Np)} = a_{\rm K}(f = {R_{\rm B}}/{2}) = 0.1151 \cdot a_{\rm * (dB)}$$

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			Notes:

	''Notes:''
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference\|"Causes and Effects of Intersymbol Interference"]].		*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference\|"Causes and Effects of Intersymbol Interference"]].
	*Reference is made in particular to the section  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces\|"Signals, Basis Functions and Vector Spaces"]].		*Reference is made in particular to the section  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces\|"Signals, Basis Functions and Vector Spaces"]].
Line 56:		Line 63:

	<quiz display=simple>		<quiz display=simple>
	{Which terms of  $H_{\rm K}(f)$  do not lead to distortions? The		{Which terms of  $H_{\rm K}(f)$  do not lead to distortions?  The
	\|type="[]"}		\|type="[]"}
	+ $\alpha_0$–term,		+ $\alpha_0$–term,
Line 68:		Line 75:
	$l_{\rm max} \ = \ $ { 6.173 3% } $\ {\rm km} $		$l_{\rm max} \ = \ $ { 6.173 3% } $\ {\rm km} $

	{What is the attenuation $($in  $\rm Np)$  at the frequency  $f = 70\,{\rm MHz}$, if the cable length is  $l = 2\,{\rm km}$?		{What is the attenuation  $($in  $\rm Np)$  at the frequency  $f = 70\,{\rm MHz}$,  if the cable length is  $l = 2\,{\rm km}$?
	\|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} $

	{All other things being equal, what attenuation results when only the  $\alpha_2$–term is considered?		{All other things being equal,  what attenuation results 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 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 for the attenuation calculated in subtask  '''(4)'''?
	\|type="{}"}		\|type="{}"}
	$a_{\rm K}(f = 70\,{\rm MHz})\ = \ $ { 39.57 3% } $\ {\rm dB} $		$a_{\rm K}(f = 70\,{\rm MHz})\ = \ $ { 39.57 3% } $\ {\rm dB} $
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	{Which of the statements are true provided that one restricts oneself to the  $\alpha_2$–value with respect to the attenuation function?		{Which of the statements are true provided that one restricts oneself to the  $\alpha_2$–value with respect to the attenuation function?
	\|type="[]"}		\|type="[]"}
	+ One can also do without the phase term  $\beta_1$		+ One can also do without the phase term  $\beta_1$.
	- One can also do without the phase term  $\beta_2$		- One can also do without the phase term  $\beta_2$.
	- $a_* ≈ 40\,{\rm dB}$  is valid for a system with  $R_{\rm B} = 70\,{\rm Mbit/s}$  and  $l = 2\,{\rm km}$.		- $a_* ≈ 40\,{\rm dB}$  is valid for a system with  $R_{\rm B} = 70\,{\rm Mbit/s}$  and  $l = 2\,{\rm km}$.
	+ $a_* ≈ 40\,{\rm dB}$  is valid for a system with  $R_{\rm B} = 140\,{\rm Mbit/s}$  and  $l = 2\,{\rm km}$.		+ $a_* ≈ 40\,{\rm dB}$  is valid for a system with  $R_{\rm B} = 140\,{\rm Mbit/s}$  and  $l = 2\,{\rm km}$.

Hwang at 12:45, 17 May 2022

2022-05-17T12:45:54Z

← Older revision		Revision as of 14:45, 17 May 2022
Line 19:		Line 19:
	\alpha_2 = 0.2722 \hspace{0.15cm}\frac{\rm Np}{\rm km\cdot\sqrt{\rm MHz}} \hspace{0.05cm}.$$		\alpha_2 = 0.2722 \hspace{0.15cm}\frac{\rm Np}{\rm km\cdot\sqrt{\rm MHz}} \hspace{0.05cm}.$$

	Often, to describe a  [[Linear_and_Time_Invariant_Systems/Systembeschreibung_im_Frequenzbereich\|linear time-invariant system]]  (LZI) in terms of system theory, one uses		Often, to describe a  [[Linear_and_Time_Invariant_Systems/Systembeschreibung_im_Frequenzbereich\|"linear time-invariant system"]]  (LZI) in terms of system theory, one uses
	* the attenuation function $($in  $\rm Np$  or  $\rm dB)$:		* the attenuation function $($in  $\rm Np$  or  $\rm dB)$:
	:$$a_{\rm K}(f) = - {\rm ln} \hspace{0.10cm}\|H_{\rm K}(f)\|= - 20 \cdot {\rm lg} \hspace{0.10cm}\|H_{\rm K}(f)\|		:$$a_{\rm K}(f) = - {\rm ln} \hspace{0.10cm}\|H_{\rm K}(f)\|= - 20 \cdot {\rm lg} \hspace{0.10cm}\|H_{\rm K}(f)\|

Hwang at 15:53, 16 May 2022

2022-05-16T15:53:49Z

← Older revision		Revision as of 17:53, 16 May 2022
Line 47:		Line 47:

	''Notes:''		''Notes:''
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference\|Causes and Effects of Intersymbol Interference]].		*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference\|"Causes and Effects of Intersymbol Interference"]].
	*Reference is made in particular to the section  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces\|Signals, Basis Functions and Vector Spaces]].		*Reference is made in particular to the section  [[Digital_Signal_Transmission/Signals,_Basis_Functions_and_Vector_Spaces\|"Signals, Basis Functions and Vector Spaces"]].