Aufgaben:Exercise 4.1Z: Transmission Behavior of Short Cables - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:58:47Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:58, 16 March 2026
Line 78:		Line 78:
	:$$\alpha(f = 0) = [1\,{\rm Np}] \hspace{-0.03cm}\cdot \hspace{-0.03cm} \sqrt{{1}/{2}\hspace{-0.03cm}\cdot \hspace{-0.03cm} R\hspace{0.03cm}' \hspace{-0.03cm}\cdot \hspace{-0.03cm} G\hspace{0.03cm}'+ {1}/{2}\cdot R\hspace{0.03cm}' \hspace{-0.03cm}\cdot \hspace{-0.03cm}G\hspace{0.03cm}'} = [1\,{\rm Np}] \hspace{-0.03cm}\cdot \hspace{-0.03cm} \sqrt{ R\hspace{0.03cm}' \hspace{-0.03cm}\cdot \hspace{-0.03cm} G\hspace{0.03cm}'} = [1\,{\rm Np}] \hspace{-0.03cm}\cdot \hspace{-0.03cm} \sqrt{ 100\,{\rm \Omega/km} \hspace{-0.03cm}\cdot \hspace{-0.03cm} 10^{-6}\,{\rm (\Omega \hspace{-0.03cm}\cdot \hspace{-0.03cm} km})^{-1}}\hspace{0.15cm}\underline{= 0.01\,{\rm Np}/{ {\rm km}}}\hspace{0.05cm},$$		:$$\alpha(f = 0) = [1\,{\rm Np}] \hspace{-0.03cm}\cdot \hspace{-0.03cm} \sqrt{{1}/{2}\hspace{-0.03cm}\cdot \hspace{-0.03cm} R\hspace{0.03cm}' \hspace{-0.03cm}\cdot \hspace{-0.03cm} G\hspace{0.03cm}'+ {1}/{2}\cdot R\hspace{0.03cm}' \hspace{-0.03cm}\cdot \hspace{-0.03cm}G\hspace{0.03cm}'} = [1\,{\rm Np}] \hspace{-0.03cm}\cdot \hspace{-0.03cm} \sqrt{ R\hspace{0.03cm}' \hspace{-0.03cm}\cdot \hspace{-0.03cm} G\hspace{0.03cm}'} = [1\,{\rm Np}] \hspace{-0.03cm}\cdot \hspace{-0.03cm} \sqrt{ 100\,{\rm \Omega/km} \hspace{-0.03cm}\cdot \hspace{-0.03cm} 10^{-6}\,{\rm (\Omega \hspace{-0.03cm}\cdot \hspace{-0.03cm} km})^{-1}}\hspace{0.15cm}\underline{= 0.01\,{\rm Np}/{ {\rm km}}}\hspace{0.05cm},$$
	:$$\beta(f = 0) = [1\,{\rm rad}] \cdot \sqrt{-{1}/{2}\cdot R\hspace{0.03cm}' \cdot G\hspace{0.03cm}'+ {1}/{2}\cdot R\hspace{0.03cm}' \cdotG\hspace{0.03cm}'} \hspace{0.15cm}\underline{= 0 }\hspace{0.05cm},$$		:$$\beta(f = 0) = [1\,{\rm rad}] \cdot \sqrt{-{1}/{2}\cdot R\hspace{0.03cm}' \cdot G\hspace{0.03cm}'+ {1}/{2}\cdot R\hspace{0.03cm}' \cdotG\hspace{0.03cm}'} \hspace{0.15cm}\underline{= 0 }\hspace{0.05cm},$$
	:$$Z_{\rm W}(f = 0) = \sqrt{\frac {R\hspace{0.03cm}'}{G\hspace{0.03cm}'}} = \sqrt{\frac {100\,{\rm \Omega/km}}{{\rm 10^{-6}/(\Omega \cdot km})}}\hspace{0.15cm}\underline{= 10\, {\~~rmk~~ \Omega}}\hspace{0.05cm}.$$		:$$Z_{\rm W}(f = 0) = \sqrt{\frac {R\hspace{0.03cm}'}{G\hspace{0.03cm}'}} = \sqrt{\frac {100\,{\rm \Omega/km}}{{\rm 10^{-6}/(\Omega \cdot km})}}\hspace{0.15cm}\underline{= 10\, {\rm k \Omega}}\hspace{0.05cm}.$$

	The DC signal attenuation becomes relevant,		The DC signal attenuation becomes relevant,
Line 93:		Line 93:

	[[Category:Linear and Time-Invariant Systems: Exercises\|^4.1 Results of Line Transmission Theory^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^4.1 Results of Line Transmission Theory^]]
	[[de:~~Zusatzaufgaben~~:4.~~1_Übertragungsmaß~~]]		[[de:Aufgaben:Aufgabe 4.1Z: Übertragungsmaß]]

Maintenance script: Fix 1 split formula blocks

2026-03-16T15:26:31Z

Fix 1 split formula blocks

@@ Line 87: / Line 87: @@
 '''(2)'''&nbsp; With&nbsp; $f = 10^{5} \ \rm  Hz$&nbsp; and the specified values,&nbsp; the following holds:
-:$$f \cdot  2\pi  L'  =  10^5\,\frac{1}{ {\rm s}} \cdot 2 \cdot
+:$$f \cdot  2\pi  L'  =  10^5\,\frac{1}{ {\rm s}} \cdot 2 \cdot10^{-3}\,\frac{\rm \Omega \cdot s}{ {\rm km}}= 200 \,\frac{\rm\Omega}{ {\rm km}} \hspace{0.05cm},\hspace{1.05cm}f \cdot  2\pi  C'  =  10^5\,\frac{1}{ {\rm s}} \cdot 2 \cdot10^{-7}\,\frac{\rm  s}{ {\rm \Omega \cdot km}}= 0.02\,\frac{\rm 1 }{ {\rm \Omega \cdot km}} \hspace{0.05cm}.$$This results in the following for the attenuation function in "Np/km"::$$\alpha(f = 100\,{\rm kHz})=  \sqrt{ {1}/{2}\cdot \left (100 \cdot 10^{-6} - 200 \cdot 0.02 \right)+{1}/{2} \cdot  \sqrt{(100^2 + 200^2) \cdot (10^{-12} + 0.02^2)}} $$:$$ \Rightarrow \; \;  \alpha(f = 100\,{\rm kHz}) \approx   \sqrt{{1}/{2}\cdot \left (10^{-4} - 4 \right)+{1}/{2}\cdot \sqrt{5 \cdot 10^{4} \cdot 4 \cdot 10^{-4}}} \approx \sqrt {-2 + \frac{\sqrt{20}}{2}} \hspace{0.15cm}\underline{\approx 0.486 \ {\rm Np/km}} \hspace{0.05cm}.$$'''(3)'''&nbsp; The limit for&nbsp; $f  &#8594; \infty$&nbsp; results if one neglects the second terms in the numerator&nbsp; $R\hspace{0.03cm}'$&nbsp; and in the denominator&nbsp; $G\hspace{0.08cm}'$&nbsp;::$$\lim_{f \rightarrow \infty} \hspace{0.1cm} Z_{\rm W}(f)= \lim_{\omega \rightarrow \infty} \hspace{0.1cm}  \sqrt{\frac {R\hspace{0.03cm}' + {\rm j}   \cdot \omega L'}{G' + {\rm j}   \cdot \omega  C\hspace{0.03cm}'}}=\sqrt{\frac {2 \pi L\hspace{0.03cm}' }{2 \pi C\hspace{0.03cm}'}}=\sqrt{\frac {2 \cdot 10^{-3}\,{\rm \Omega \cdot s} }{2 \cdot 10^{-73}\,{\rm s/\Omega} }} \hspace{0.15cm}\underline{= 100\,{\rm \Omega }}\hspace{0.05cm}.$$*The approximation for the attenuation function is more difficult to derive.&nbsp; Starting from:$$\alpha(\omega)  =  \sqrt{ {1}/{2}\cdot \left (R\hspace{0.03cm}' G\hspace{0.03cm}' - \omega^2 \cdot L\hspace{0.03cm}'  C\hspace{0.03cm}'\right)+{1}/{2}\sqrt{(R\hspace{0.03cm}'\hspace{0.03cm}^2 + \omega^2 \cdot L\hspace{0.03cm}'\hspace{0.03cm}^2) \cdot (G\hspace{0.03cm}'\hspace{0.03cm}^2 + \omega^2 \cdot C\hspace{0.03cm}\hspace{0.03cm}'^2)}}$$:then also the following applies::$$2 \cdot \alpha^2(\omega)    =   R\hspace{0.03cm}' G\hspace{0.03cm}' + \omega^2 \cdot L'C'\cdot\left [-1 +\sqrt{(1 + \frac{R\hspace{0.03cm}'^2}{ \omega^2 \cdot L\hspace{0.03cm}'^2}) \cdot (1 + \frac{G\hspace{0.03cm}'^2}{ \omega^2 \cdot C\hspace{0.03cm}'^2})} \hspace{0.1cm}\right]$$:$$\Rightarrow \; \; 2 \cdot \alpha^2(\omega)      \approx   R\hspace{0.03cm}' G\hspace{0.03cm}' + \omega^2 \cdot L'C\hspace{0.03cm}'\cdot\left [-1 +\sqrt{1 + \frac{R'^2}{ \omega^2 \cdot L'^2}+ \frac{G\hspace{0.03cm}'^2}{ \omega^2 \cdot C\hspace{0.03cm}'^2}} \hspace{0.1cm}\right].$$*Using the approximation&nbsp; $\sqrt{1 + x}\approx 1+x/2$&nbsp; valid for small&nbsp; $x$,&nbsp; one arrives at the intermediate result for (infinitely) large frequencies::$$2 \cdot \alpha^2(\omega \rightarrow \infty)    =   R\hspace{0.03cm}' G\hspace{0.05cm}' + \omega^2 \cdot L'C\hspace{0.05cm}'\cdot\left [ -1 +1 + {1}/{2} \cdot  \left ( \frac{R\hspace{0.03cm}'\hspace{0.03cm}^2}{ \omega^2 \cdot L\hspace{0.03cm}'\hspace{0.03cm}^2}+ \frac{G\hspace{0.03cm}'\hspace{0.03cm}^2}{ \omega^2 \cdot C\hspace{0.03cm}'\hspace{0.03cm}^2}\right) \hspace{0.1cm}\right]  $$:$$\Rightarrow \hspace{0.3cm}  2  \cdot \alpha^2(\omega \rightarrow \infty) =  \frac{2 \cdot  R\hspace{0.03cm}'  G\hspace{0.03cm}'  C\hspace{0.03cm}'  L'+ R\hspace{0.03cm}'\hspace{0.03cm}^2  C\hspace{0.03cm}'\hspace{0.03cm}^2+G\hspace{0.03cm}'\hspace{0.03cm}^2  L\hspace{0.03cm}'\hspace{0.03cm}^2}{2 \cdot C\hspace{0.03cm}'  L\hspace{0.03cm}'}=  \frac{(R\hspace{0.03cm}'  C\hspace{0.03cm}' + G\hspace{0.03cm}'  L\hspace{0.03cm}')^2}{2 \cdot C\hspace{0.03cm}'  L\hspace{0.03cm}' }$$:$$\Rightarrow \hspace{0.3cm} \alpha(\omega \rightarrow \infty)   ={1}/{2}\cdot \frac{R\hspace{0.03cm}' C\hspace{0.03cm}' + G\hspace{0.03cm}'  L\hspace{0.03cm}'}{\sqrt{ C\hspace{0.03cm}'  L\hspace{0.03cm}' }}={1}/{2}\cdot \left [R\hspace{0.03cm}' \cdot \sqrt{\frac{C\hspace{0.03cm}'}{L\hspace{0.03cm}'}}+G\hspace{0.03cm}' \cdot \sqrt{\frac{L\hspace{0.03cm}'}{C\hspace{0.03cm}'}}\right]\hspace{0.05cm}.$$*With the numerical values inserted,&nbsp; we get:$$\alpha(f \rightarrow \infty)   =  \alpha(\omega \rightarrow \infty)=  {0.5\,{\rm Np/km}}\cdot \left [100 \cdot \sqrt{\frac{2 \cdot 10^{-7}}{2 \cdot10^{-3}}}+10^{-6} \cdot\sqrt{\frac{2 \cdot10^{-3}}{2 \cdot10^{-7}}}\right]\hspace{0.15cm}\underline{\approx 0.5 \, {\rm Np}/{\rm km}}\hspace{0.05cm}.$$'''(4)'''&nbsp; For small frequencies, &nbsp;$\omega L\hspace{0.03cm}' \ll R\hspace{0.03cm}'$&nbsp; and &nbsp;$ \omega C\hspace{0.03cm}' \gg G\hspace{0.03cm}'$ apply.*Neglecting the&nbsp; $\omega^2$&ndash;part, one obtains::$$\alpha(f)    =  \sqrt{\frac {1}{2}\cdot \left (R\hspace{0.03cm}' G\hspace{0.03cm}' - \omega^2 \cdot L\hspace{0.03cm}'  C\hspace{0.03cm}'\right)+\frac {1}{2}\sqrt{(R\hspace{0.03cm}'\hspace{0.03cm}^2 + \omega^2 \cdot L\hspace{0.03cm}'\hspace{0.03cm}^2) \cdot (G\hspace{0.03cm}'\hspace{0.03cm}^2+ \omega^2 \cdot C\hspace{0.03cm}'\hspace{0.03cm}^2)}}\hspace{0.1cm}\bigg |_{\hspace{0.05cm}\omega \hspace{0.05cm}= \hspace{0.05cm}2\pif}$$:$$ \Rightarrow \hspace{0.3cm} \alpha(f)     \approx  \sqrt{\frac {R\hspace{0.03cm}' G\hspace{0.03cm}'}{2}+\frac {R\hspace{0.03cm}' \cdot \omega C\hspace{0.03cm}'}{2}}\hspace{0.1cm}\bigg |_{\hspace{0.03cm}\omega \hspace{0.05cm}= \hspace{0.05cm}2\pif} \approx \sqrt{{1}/{2} \cdot f \cdot R\hspace{0.03cm}' \cdot 2 \pi C\hspace{0.03cm}'}\hspace{0.05cm}.$$*Here it is considered that the first part can be neglected according to subtask&nbsp; '''(1)'''&nbsp; except for the frequency&nbsp; $f = 0$&nbsp;.*For the frequency&nbsp; $f = 1 \ \rm kHz$&nbsp; we get the approximation:$$\alpha(f = 1\,{\rm kHz})   = \sqrt{{1}/{2} \cdot 10^{3}\,{\rm Hz} \cdot 100\,\frac{\rm \Omega }{ {\rm km}} \cdot 2 \cdot 10^{-7}\,\frac{\rm s }{ {\rm \Omega \cdot km}}}\hspace{0.15cm}\underline{= 0.1\,{\rm Np }/{ {\rm km}}}\hspace{0.05cm}.$$*For frequency&nbsp; $f = 4 \ \rm kHz$&nbsp; the attenuation function per unit length is twice as large::$$\alpha(f = 4\,{\rm kHz})  \hspace{0.15cm}\underline{ = 0.2\,{\rm Np }/{ {\rm km}}}\hspace{0.05cm}.$$'''(5)'''&nbsp; The wave impedance at low frequencies is approximated by::$$Z_{\rm W}(f)  =  \sqrt{\frac {R\hspace{0.03cm}' + {\rm j}   \cdot f \cdot 2 \pi  L\hspace{0.03cm}'}{G\hspace{0.03cm}' + {\rm j}    \cdot f \cdot 2 \pi  C\hspace{0.03cm}'}}\approx \sqrt\frac{1 }{  {\rm j}} \cdot \sqrt{\frac {R\hspace{0.03cm}' }{  f \cdot 2 \piC\hspace{0.03cm}'}}= (1 - {\rm j})\cdot \sqrt{\frac {R\hspace{0.03cm}' }{  2 \cdot f \cdot 2 \piC\hspace{0.03cm}'}}\hspace{0.05cm}.$$*With the specified line fittings we obtain::$${\rm Re}\{Z_{\rm W}(f= 1\,{\rm kHz})\}  =   \sqrt{\frac {100\,{\rm \Omega/km }}{  2 \cdot 10^{3}\,{\rm Hz} \cdot 2 \cdot 10^{-7}\,{\rm s/(\Omega \cdot km) }}} \hspace{0.15cm}\underline{= 500\,{\rm\Omega}}\hspace{0.05cm},$$:$$ {\rm Im}\{Z_{\rm W}(f= 1\,{\rm kHz})\} =  -{\rm Re}\{Z_{\rm W}(f= 1\,{\rm kHz})\}\hspace{0.15cm}\underline{= -500\,{\rm\Omega}}\hspace{0.05cm}.$$
 {{ML-Fuß}}

Maintenance script: Join split $$ formula lines for correct rendering

2026-03-16T13:48:31Z

Join split $$ formula lines for correct rendering

Show changes

Maintenance script: Add German interlanguage link

2026-03-16T12:29:22Z

Add German interlanguage link

← Older revision		Revision as of 14:29, 16 March 2026
Line 196:		Line 196:

	[[Category:Linear and Time-Invariant Systems: Exercises\|^4.1 Results of Line Transmission Theory^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^4.1 Results of Line Transmission Theory^]]
			[[de:Zusatzaufgaben:4.1_Übertragungsmaß]]

Guenter at 15:14, 9 November 2021

2021-11-09T15:14:33Z

← Older revision		Revision as of 17:14, 9 November 2021
Line 3:		Line 3:
	}}		}}

	[[File:~~LZI_Z_4_1~~.png\|right\|frame\|Short line section]]		[[File:EN_LZI_Z_4_1.png\|right\|frame\|Short line section]]
	We assume a homogeneous and reflection-free terminated line of length  $l$  so that the following applies to the spectral function at the output:		We assume a homogeneous and reflection-free terminated line of length  $l$  so that the following applies to the spectral function at the output:
	:$$U_2(f) = U_1(f) \cdot {\rm e}^{-\hspace{0.02cm}\gamma(f) \hspace{0.05cm} \cdot \hspace{0.05cm}l} \hspace{0.05cm}.$$		:$$U_2(f) = U_1(f) \cdot {\rm e}^{-\hspace{0.02cm}\gamma(f) \hspace{0.05cm} \cdot \hspace{0.05cm}l} \hspace{0.05cm}.$$

Guenter at 16:35, 6 November 2021

2021-11-06T16:35:28Z

Guenter at 16:23, 6 November 2021

2021-11-06T16:23:01Z

← Older revision		Revision as of 18:23, 6 November 2021
Line 6:		Line 6:
	We assume a homogeneous and reflection-free terminated line of length  $l$  so that the following applies to the spectral function at the output:		We assume a homogeneous and reflection-free terminated line of length  $l$  so that the following applies to the spectral function at the output:
	:$$U_2(f) = U_1(f) \cdot {\rm e}^{-\hspace{0.02cm}\gamma(f) \hspace{0.05cm} \cdot \hspace{0.05cm}l} \hspace{0.05cm}.$$		:$$U_2(f) = U_1(f) \cdot {\rm e}^{-\hspace{0.02cm}\gamma(f) \hspace{0.05cm} \cdot \hspace{0.05cm}l} \hspace{0.05cm}.$$
	Here  $\gamma(f)$  describes the complex propagation function of an extremely short line of infinitesimal length  $dx$, which can be represented with the parameters  $R\hspace{0.05cm}'$,  $L\hspace{0.05cm}'$,  $G\hspace{0.08cm}'$ and  $C\hspace{0.08cm}'$ (see diagram) as follows:		Here  $\gamma(f)$  describes the  '''complex propagation function'''  of an extremely short line of infinitesimal length  $dx$,  which can be represented with the parameters  $R\hspace{0.05cm}'$,  $L\hspace{0.05cm}'$,  $G\hspace{0.08cm}'$ and  $C\hspace{0.08cm}'$ (see diagram) as follows:
	:$$\gamma(f) = \sqrt{(R\hspace{0.05cm}' + {\rm j} \cdot 2\pi f \cdot L\hspace{0.05cm}') \cdot (G\hspace{0.08cm}' + {\rm j} \cdot 2\pi f \cdot C\hspace{0.08cm}')} =		:$$\gamma(f) = \sqrt{(R\hspace{0.05cm}' + {\rm j} \cdot 2\pi f \cdot L\hspace{0.05cm}') \cdot (G\hspace{0.08cm}' + {\rm j} \cdot 2\pi f \cdot C\hspace{0.08cm}')} =
	\alpha (f) + {\rm j} \cdot \beta (f)\hspace{0.05cm}.$$		\alpha (f) + {\rm j} \cdot \beta (f)\hspace{0.05cm}.$$
	The real part of  $\gamma(f)$  results in		The real part of  $\gamma(f)$  results in
	*the attenuation function ~~per unit length~~  $\alpha(f)$ ~~, and~~		*The real part of  $\gamma(f)$  results in the attenuation function $\alpha(f)$  (per unit length).
	*~~the~~ imaginary part of the phase function  $\beta(f)$.		*The imaginary part of  $\gamma(f)$  results in the phase function  $\beta(f)$ (per unit length).


Line 22:		Line 22:
	{1}/{2}\cdot \sqrt{(R\hspace{0.05cm}'\hspace{0.05cm}^2 + \omega^2 \cdot L\hspace{0.05cm}'\hspace{0.05cm}^2) \cdot (G\hspace{0.08cm}'\hspace{0.05cm}^2 + \omega^2 \cdot C\hspace{0.08cm}'\hspace{0.05cm}^2)}}		{1}/{2}\cdot \sqrt{(R\hspace{0.05cm}'\hspace{0.05cm}^2 + \omega^2 \cdot L\hspace{0.05cm}'\hspace{0.05cm}^2) \cdot (G\hspace{0.08cm}'\hspace{0.05cm}^2 + \omega^2 \cdot C\hspace{0.08cm}'\hspace{0.05cm}^2)}}
	\hspace{0.1cm}\bigg \|_{\hspace{0.05cm}\omega \hspace{0.05cm}= \hspace{0.05cm}2\pi f}.$$		\hspace{0.1cm}\bigg \|_{\hspace{0.05cm}\omega \hspace{0.05cm}= \hspace{0.05cm}2\pi f}.$$
	For the attenuation function  $a(f)$  the pseudo unit "Neper"  (Np) has to be added additionally and for the phase function  $b(f)$  "Radian" (rad).   Since the primary line parameters are each related to the line length,  $\alpha(f)$  and  $\beta(f)$  have the units "Np/km" and "rad/km", respectively.

	Another important descriptive quantity besides  $\gamma(f)$  is the wave impedance  $Z_{\rm W}(f)$, which gives the relationship between voltage and current of the two running waves at each location. It holds:		*For the attenuation function  $a(f)$  the pseudo unit  "Neper"  (Np)  has to be added additionally and for the phase function  $b(f)$  "Radian"  (rad).
			*Since the primary line parameters are each related to the line length,  $\alpha(f)$  and  $\beta(f)$  have the units  "Np/km"  and  "rad/km",  respectively.


			Another important descriptive quantity besides  $\gamma(f)$  is the  '''wave impedance'''  $Z_{\rm W}(f)$,  which gives the relationship between voltage and current of the two running waves at each location.  It holds:
	:$$Z_{\rm W}(f) = \sqrt{\frac {R\hspace{0.05cm}' + {\rm j} \cdot \omega L\hspace{0.05cm}'}{G\hspace{0.08cm}' + {\rm j} \cdot \omega C\hspace{0.08cm}'}}		:$$Z_{\rm W}(f) = \sqrt{\frac {R\hspace{0.05cm}' + {\rm j} \cdot \omega L\hspace{0.05cm}'}{G\hspace{0.08cm}' + {\rm j} \cdot \omega C\hspace{0.08cm}'}}
	\hspace{0.1cm}\bigg \|_{\hspace{0.05cm} \omega \hspace{0.05cm}= \hspace{0.05cm}2\pi f}.$$		\hspace{0.1cm}\bigg \|_{\hspace{0.05cm} \omega \hspace{0.05cm}= \hspace{0.05cm}2\pi f}.$$
Line 35:		Line 38:
	*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Some_Results_from_Line_Transmission_Theory\|Some Results from Line Transmission Theory]].		*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Some_Results_from_Line_Transmission_Theory\|Some Results from Line Transmission Theory]].

	*Use the following ~~numerical~~ values for the numerical calculations:		*Use the following values for the numerical calculations:
	:$$R\hspace{0.05cm}' = 100\,\,{\rm \Omega}/{ {\rm km} }\hspace{0.05cm},\hspace{0.3cm}		:$$R\hspace{0.05cm}' = 100\,\,{\rm \Omega}/{ {\rm km} }\hspace{0.05cm},\hspace{0.3cm}
	G\hspace{0.08cm}' = 1\,\,{\rm µ S}/{ {\rm km}}\hspace{0.05cm},\hspace{0.3cm}		G\hspace{0.08cm}' = 1\,\,{\rm µ S}/{ {\rm km}}\hspace{0.05cm},\hspace{0.3cm}
Line 48:		Line 51:

	<quiz display=simple>		<quiz display=simple>
	{Specify  $\alpha(f)$,  $\beta(f)$ and  $Z_{\rm W}(f)$  for frequency  $f = 0$ (direct current).		{Specify  $\alpha(f)$,  $\beta(f)$ and  $Z_{\rm W}(f)$  for frequency  $f = 0$  ("direct current").
	\|type="{}"}		\|type="{}"}
	$\alpha(f =0) \ =$ { 0.01 3% } $\ \rm Np/km$		$\alpha(f =0) \ =$ { 0.01 3% } $\ \rm Np/km$
Line 55:		Line 58:


	{Calculate the attenuation function ~~per unit length~~  $\alpha(f)$  for  $f = 100\ \rm kHz$.		{Calculate the attenuation function  $\alpha(f)$  (per unit length)  for  $f = 100\ \rm kHz$.
	\|type="{}"}		\|type="{}"}
	$\alpha(f = 100\ \rm kHz) \ = \ $ { 0.486 3% } $\ \rm Np/km$		$\alpha(f = 100\ \rm kHz) \ = \ $ { 0.486 3% } $\ \rm Np/km$


	{Give the approximations of  $Z_{\rm W}(f)$  and  $\alpha(f)$  valid for  $f → \infty$ .		{Give the approximations of  $Z_{\rm W}(f)$  and  $\alpha(f)$,  valid for  $f → \infty$ .
	\|type="{}"}		\|type="{}"}
	$ Z_{\rm W}(f → \infty) \ = \ $ { 100 3% } $\ \rm \Omega$		$ Z_{\rm W}(f → \infty) \ = \ $ { 100 3% } $\ \rm \Omega$
Line 66:		Line 69:


	{Use  $\omega L\hspace{0.03cm}' \ll R\hspace{0.05cm}'$  and  $\omega C\hspace{0.08cm}' \gg G\hspace{0.08cm}'$  to derive an  $\alpha(f)$&~~ndash~~; approximation for (not too) small frequencies. <br>What is the attenuation function for  $ f = 1 \ \rm kHz$  and  $ f = 4 \ \rm kHz$?		{Use  $\omega L\hspace{0.03cm}' \ll R\hspace{0.05cm}'$  and  $\omega C\hspace{0.08cm}' \gg G\hspace{0.08cm}'$  to derive an  $\alpha(f)$  approximation for  (not too)  small frequencies. <br>What is the attenuation function per unit length for  $ f = 1 \ \rm kHz$  and  $ f = 4 \ \rm kHz$?
	\|type="{}"}		\|type="{}"}
	$\alpha(f = 1\ \rm kHz) \ = \ $ { 0.1 3% } $\ \rm Np/km$		$\alpha(f = 1\ \rm kHz) \ = \ $ { 0.1 3% } $\ \rm Np/km$
Line 72:		Line 75:


	{For the same frequency range, give a suitable approximation for the wave impedance  $Z_{\rm W}(f)$ . <br>What value results for  $ f = 1 \ \rm kHz$?		{For the same frequency range,  give a suitable approximation for the wave impedance  $Z_{\rm W}(f)$ . <br>What value results for  $ f = 1 \ \rm kHz$?
	\|type="{}"}		\|type="{}"}
	${\rm Re}\{Z_{\rm W}(f = 1\ \rm kHz)\} \ = \ $ { 500 3% } $\ \rm \Omega$		${\rm Re}\{Z_{\rm W}(f = 1\ \rm kHz)\} \ = \ $ { 500 3% } $\ \rm \Omega$

Hwang at 14:26, 5 November 2021

2021-11-05T14:26:45Z

Show changes

Guenter at 15:00, 15 October 2021

2021-10-15T15:00:38Z

← Older revision		Revision as of 17:00, 15 October 2021
Line 194:		Line 194:


	[[Category:Linear and Time-Invariant Systems: Exercises\|^4.1 ~~Einige Ergebnisse der Leitungstheorie~~^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^4.1 Results of Line Transmission Theory^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 4.1Z: Übertragungsmaß to Aufgaben:Exercise 4.1Z: Transmission Behavior of Short Cables

2021-10-15T14:48:44Z

Guenter moved page Aufgaben:Aufgabe 4.1Z: Übertragungsmaß to Aufgaben:Exercise 4.1Z: Transmission Behavior of Short Cables

← Older revision	Revision as of 16:48, 15 October 2021
(No difference)

← Older revision		Revision as of 18:35, 6 November 2021
Line 35:		Line 35:


	''Notes:''		Notes:
	*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Some_Results_from_Line_Transmission_Theory\|Some Results from Line Transmission Theory]].		*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Some_Results_from_Line_Transmission_Theory\|Some Results from Line Transmission Theory]].

Line 97:		Line 97:

	The DC signal attenuation becomes relevant,		The DC signal attenuation becomes relevant,
	*if the useful signal is to be transmitted in the baseband and has a DC component, or		*if the useful signal is to be transmitted in the baseband and has a DC component,  or
	*if the network termination at the participant must be supplied with power from the local exchange (remote power supply).		*if the network termination at the participant must be supplied with power from the local exchange  ("remote power supply").



	'''(2)'''  With  $f = 10^{5} \ \rm Hz$  and the specified values, the following holds:		'''(2)'''  With  $f = 10^{5} \ \rm Hz$  and the specified values,  the following holds:
	:$$f \cdot 2\pi L' = 10^5\,\frac{1}{ {\rm s}} \cdot 2 \cdot		:$$f \cdot 2\pi L' = 10^5\,\frac{1}{ {\rm s}} \cdot 2 \cdot
	10^{-3}\,\frac{\rm \Omega \cdot s}{ {\rm km}}= 200 \,\frac{\rm		10^{-3}\,\frac{\rm \Omega \cdot s}{ {\rm km}}= 200 \,\frac{\rm
Line 119:		Line 119:


			'''(3)'''  The limit for  $f → \infty$  results if one neglects the second terms in the numerator  $R\hspace{0.03cm}'$  and in the denominator  $G\hspace{0.08cm}'$ :
	'''(3)'''  The limit for  $f → \infty$  results if one neglects the second ~~term~~ in the numerator  $R\hspace{0.03cm}'$  and in the denominator  $G\hspace{0.08cm}'$ :
	:$$\lim_{f \rightarrow \infty} \hspace{0.1cm} Z_{\rm W}(f)		:$$\lim_{f \rightarrow \infty} \hspace{0.1cm} Z_{\rm W}(f)
	= \lim_{\omega \rightarrow \infty} \hspace{0.1cm} \sqrt{\frac {R\hspace{0.03cm}' + {\rm j} \cdot \omega L'}{G' + {\rm j} \cdot \omega C\hspace{0.03cm}'}}		= \lim_{\omega \rightarrow \infty} \hspace{0.1cm} \sqrt{\frac {R\hspace{0.03cm}' + {\rm j} \cdot \omega L'}{G' + {\rm j} \cdot \omega C\hspace{0.03cm}'}}
	=\sqrt{\frac {2 \pi L\hspace{0.03cm}' }{2 \pi C\hspace{0.03cm}'}}=\sqrt{\frac {2 \cdot 10^{-3}\,{\rm \Omega \cdot s} }		=\sqrt{\frac {2 \pi L\hspace{0.03cm}' }{2 \pi C\hspace{0.03cm}'}}=\sqrt{\frac {2 \cdot 10^{-3}\,{\rm \Omega \cdot s} }
	{2 \cdot 10^{-73}\,{\rm s/\Omega} }} \hspace{0.15cm}\underline{= 100\,{\rm \Omega }}\hspace{0.05cm}.$$		{2 \cdot 10^{-73}\,{\rm s/\Omega} }} \hspace{0.15cm}\underline{= 100\,{\rm \Omega }}\hspace{0.05cm}.$$
	*The approximation for the attenuation function is more difficult to derive. Starting from		*The approximation for the attenuation function is more difficult to derive.  Starting from
	:$$\alpha(\omega) = \sqrt{ {1}/{2}\cdot \left (R\hspace{0.03cm}' G\hspace{0.03cm}' - \omega^2 \cdot L\hspace{0.03cm}' C\hspace{0.03cm}'\right)+		:$$\alpha(\omega) = \sqrt{ {1}/{2}\cdot \left (R\hspace{0.03cm}' G\hspace{0.03cm}' - \omega^2 \cdot L\hspace{0.03cm}' C\hspace{0.03cm}'\right)+
	{1}/{2}\sqrt{(R\hspace{0.03cm}'\hspace{0.03cm}^2 + \omega^2 \cdot L\hspace{0.03cm}'\hspace{0.03cm}^2) \cdot (G\hspace{0.03cm}'\hspace{0.03cm}^2 + \omega^2 \cdot C\hspace{0.03cm}\hspace{0.03cm}'^2)}}$$		{1}/{2}\sqrt{(R\hspace{0.03cm}'\hspace{0.03cm}^2 + \omega^2 \cdot L\hspace{0.03cm}'\hspace{0.03cm}^2) \cdot (G\hspace{0.03cm}'\hspace{0.03cm}^2 + \omega^2 \cdot C\hspace{0.03cm}\hspace{0.03cm}'^2)}}$$
Line 137:		Line 136:
	\left [-1 +\sqrt{1 + \frac{R'^2}{ \omega^2 \cdot L'^2}+ \frac{G\hspace{0.03cm}'^2}{ \omega^2 \cdot C\hspace{0.03cm}'^2}} \hspace{0.1cm}		\left [-1 +\sqrt{1 + \frac{R'^2}{ \omega^2 \cdot L'^2}+ \frac{G\hspace{0.03cm}'^2}{ \omega^2 \cdot C\hspace{0.03cm}'^2}} \hspace{0.1cm}
	\right].$$		\right].$$
	*Using the approximation $\sqrt{1 + x}\approx 1+x/2$  valid for small  $x$ , one arrives at the intermediate result for (infinitely) large frequencies:		*Using the approximation  $\sqrt{1 + x}\approx 1+x/2$  valid for small  $x$,  one arrives at the intermediate result for (infinitely) large frequencies:
	:$$2 \cdot \alpha^2(\omega \rightarrow \infty) = R\hspace{0.03cm}' G\hspace{0.05cm}' + \omega^2 \cdot L'		:$$2 \cdot \alpha^2(\omega \rightarrow \infty) = R\hspace{0.03cm}' G\hspace{0.05cm}' + \omega^2 \cdot L'
	C\hspace{0.05cm}'\cdot		C\hspace{0.05cm}'\cdot
Line 149:		Line 148:
	{1}/{2}\cdot \frac{R\hspace{0.03cm}' C\hspace{0.03cm}' + G\hspace{0.03cm}' L\hspace{0.03cm}'}{\sqrt{ C\hspace{0.03cm}' L\hspace{0.03cm}' }}=		{1}/{2}\cdot \frac{R\hspace{0.03cm}' C\hspace{0.03cm}' + G\hspace{0.03cm}' L\hspace{0.03cm}'}{\sqrt{ C\hspace{0.03cm}' L\hspace{0.03cm}' }}=
	{1}/{2}\cdot \left [R\hspace{0.03cm}' \cdot \sqrt{\frac{C\hspace{0.03cm}'}{L\hspace{0.03cm}'}}+G\hspace{0.03cm}' \cdot \sqrt{\frac{L\hspace{0.03cm}'}{C\hspace{0.03cm}'}}\right]\hspace{0.05cm}.$$		{1}/{2}\cdot \left [R\hspace{0.03cm}' \cdot \sqrt{\frac{C\hspace{0.03cm}'}{L\hspace{0.03cm}'}}+G\hspace{0.03cm}' \cdot \sqrt{\frac{L\hspace{0.03cm}'}{C\hspace{0.03cm}'}}\right]\hspace{0.05cm}.$$
	*With the numerical values inserted, we get		*With the numerical values inserted,  we get
	:$$\alpha(f \rightarrow \infty) = \alpha(\omega \rightarrow \infty)		:$$\alpha(f \rightarrow \infty) = \alpha(\omega \rightarrow \infty)
	= {0.5\,{\rm Np/km}}\cdot \left [100 \cdot \sqrt{\frac{2 \cdot 10^{-7}}{2 \cdot10^{-3}}}+10^{-6} \cdot		= {0.5\,{\rm Np/km}}\cdot \left [100 \cdot \sqrt{\frac{2 \cdot 10^{-7}}{2 \cdot10^{-3}}}+10^{-6} \cdot
	\sqrt{\frac{2 \cdot10^{-3}}{2 \cdot10^{-7}}}\right]		\sqrt{\frac{2 \cdot10^{-3}}{2 \cdot10^{-7}}}\right]
	\hspace{0.15cm}\underline{\approx 0.5 \, {\rm Np}/{\rm km}}\hspace{0.05cm}.$$		\hspace{0.15cm}\underline{\approx 0.5 \, {\rm Np}/{\rm km}}\hspace{0.05cm}.$$