Aufgaben:Exercise 4.6: k-parameters and alpha-parameters - Revision history

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 *From&nbsp; $C_1 \cdot \alpha_2 + C_2  = D_1 \cdot \alpha_2 + D_2$&nbsp; we obtain a linear equation for&nbsp; $\alpha_2$.&nbsp; With the result from&nbsp; '''(2)'''&nbsp; we can write:
-:$$\alpha_2   =  \frac{D_2 - C_2}{C_1 - D_1} = \frac{- \frac{2.5 \cdot k_2 }{k_3
+:$$\alpha_2   =  \frac{D_2 - C_2}{C_1 - D_1} = \frac{- \frac{2.5 \cdot k_2 }{k_3+1.5} \cdot \frac{B^{k_3-1}}{f_0^{k_3}} + \frac{3 k_2 }{k_3 +2}\cdot \frac{B^{k_3-1}}{f_0^{k_3}}}{{6}/{5}\cdot B^{-0.5} -{5}/{4}\cdot B^{-0.5}} =  \frac{- {2.5 \cdot k_2}\cdot(k_3 +2)  + {3 k_2 }\cdot (k_3 +1.5) }{({6}/{5} -{5}/{4})(k_3 +1.5)(k_3 +2)} \cdot\frac{B^{k_3-0.5}}{f_0^{k_3}}$$:$$  \Rightarrow \hspace{0.3cm}\alpha_2    =  10 \cdot (B/f_0)^{k_3-0.5}\cdot \frac{1-k_3}{(k_3 + 1.5)(k_3 +2)}\cdot \frac {k_2}{\sqrt{f_0}}\hspace{0.05cm} .$$*For the parameter &nbsp; $\alpha_1$&nbsp; then holds::$$\alpha_1   =  - C_1 \cdot \alpha_2 - C_2 =-\frac{6}{5}\cdot B^{-0.5} \cdot 10 \cdot (B/f_0)^{k_3-0.5}\cdot \frac{1-k_3}{(k_3 + 1.5)(k_3 +2)}\cdot \frac {k_2}{\sqrt{f_0}} +\frac{3 k_2 }{k_3 +2}\cdot \frac{B^{k_3-1}}{f_0^{k_3}}$$:$$ \Rightarrow \hspace{0.3cm}\alpha_1  =   (B/f_0)^{k_3 -1}\cdot\frac{-12 \cdot (1-k_3) + 3 \cdot (k_3 + 1.5)}{(k_3 + 1.5)(k_3 +2)} \cdot \frac {k_2}{f_0} \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\alpha_1  =15 \cdot (B/f_0)^{k_3-1}\cdot \frac{k_3 -0.5}{(k_3 + 1.5)(k_3 +2)}\cdot \frac {k_2}{f_0}\hspace{0.05cm} .$$*Regardless of the bandwidth,&nbsp; we obtain for&nbsp; $k_3 = 1$::$$\alpha_1    =   (B/f_0)^{k_3-1}\cdot \frac{15 \cdot (k_3 -0.5)}{(k_3 + 1.5)(k_3 +2)}\cdot \frac {k_2}{f_0} = \frac{15 \cdot 0.5}{2.5 \cdot 3}\cdot \frac {k_2}{f_0}\hspace{0.15cm}\underline{ = {k_2}/{f_0}}\hspace{0.05cm},$$:$$ \alpha_2   =   (B/f_0)^{k_3-0.5}\cdot \frac{10 \cdot (1-k_3)}{(k_3 + 1.5)(k_3 +2)}\cdot \frac {k_2}{\sqrt{f_0}}\hspace{0.15cm}\underline{= 0} \hspace{0.05cm}.$$*In contrast,&nbsp; for&nbsp; $k_3 = 0.5$::$$\alpha_1    =   (B/f_0)^{k_3-1}\cdot \frac{15 \cdot (k_3 -0.5)}{(k_3 + 1.5)(k_3 +2)}\cdot \frac {k_2}{f_0} \hspace{0.15cm}\underline{= 0}\hspace{0.05cm},$$:$$ \alpha_2   =  (B/f_0)^{k_3-0.5}\cdot \frac{10 \cdot (1-k_3)}{(k_3 + 1.5)(k_3 +2)}\cdot \frac {k_2}{\sqrt{f_0}}=  \frac{10 \cdot 0.5}{2 \cdot 2.5}\cdot \frac {k_2}{\sqrt{f_0}} = \hspace{0.15cm}\underline{ {k_2}/{\sqrt{f_0}}} \hspace{0.05cm}.$$'''(4)'''&nbsp; For the two coefficients, with&nbsp; $k_2 = 10.8 \ \rm dB/km$,&nbsp; $k_3 = 0.6 \ \rm dB/km$&nbsp; and&nbsp; $B/f_0 = 30$::$$\alpha_1    =   (B/f_0)^{k_3-1}\cdot \frac{15 \cdot (k_3 -0.5)}{(k_3 + 1.5)(k_3 +2)}\cdot \frac {k_2}{f_0}  =   30^{-0.4}\cdot \frac{15 \cdot 0.1}{2.1 \cdot 2.6}\cdot\frac {10.8 \, {\rm dB/km} }{1 \, {\rm MHz}}\hspace{0.15cm}\underline{ \approx 0.761\,{{\rm dB} }/{({\rm km \cdot MHz})}}\hspace{0.05cm},$$:$$ \alpha_2   =  (B/f_0)^{k_3-0.5}\cdot \frac{10 \cdot (1-k_3)}{(k_3 + 1.5)(k_3 +2)}\cdot \frac {k_2}{\sqrt{f_0}}=  \frac {k_2}{\sqrt{f_0}}=   30^{0.1}\cdot \frac{10 \cdot 0.4}{2.1 \cdot 2.6}\cdot \frac{10.8 \, {\rm dB/km} }{1 \, {\rm MHz^{0.5}}}\hspace{0.15cm}\underline{ \approx 11.1\,{{\rm dB} }/{({\rm km \cdot \sqrt{MHz}}})}\hspace{0.05cm}.$$'''(5)'''&nbsp; According to the given equation&nbsp; $\alpha_{\rm II}(f)$&nbsp; thus also holds::$$\alpha_{\rm II}(f = 30 \, {\rm MHz})    =  \alpha_0 + \alpha_1 \cdot f +  \alpha_2 \cdot \sqrt {f}=  \big [ \hspace{0.05cm} 4.4 + 0.761 \cdot 30 +  11.1 \cdot \sqrt {30}\hspace{0.05cm}\big ]\frac{\rm dB}{\rm km }\hspace{0.15cm}\underline{\approx 88.1\, {\rm dB}/{\rm km }}\hspace{0.05cm}.$$
 {{ML-Fuß}}

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 * The difference &nbsp;$\varepsilon^2(f)$&nbsp; and the mean square error &nbsp;${\rm E}\big[\varepsilon^2(f)\big]$&nbsp; are obtained as follows:
 :$$\varepsilon^2(f) =  \big [ \alpha_1 \cdot f +  \alpha_2 \cdot \sqrt {f} - k_2  \cdot (f/f_0)^{k_3}\big ]^2=\alpha_1^2 \hspace{0.05cm}\cdot\hspace{0.05cm}  f^2  + 2  \alpha_1 \alpha_2 \hspace{0.05cm}\cdot\hspace{0.05cm}  f^{1.5} +\alpha_1^2 \hspace{0.05cm}\cdot\hspace{0.05cm}  f + k_2^2\hspace{0.05cm}\cdot\hspace{0.05cm} \frac{f^{2k_3}}{f_0^{2k_3}} - 2 k_2 \alpha_1 \hspace{0.05cm}\cdot\hspace{0.05cm}\frac{f^{k_3+1}} {f_0^{k_3}}-{2 k_2 \alpha_2} \hspace{0.05cm}\cdot\hspace{0.05cm}  \frac{f^{k_3+0.5}}{f_0^{k_3}}$$
-:$$\Rightarrow
+:$$\Rightarrow\hspace{0.3cm}{\rm E}\big[\varepsilon^2(f)\big]   =   \alpha_1^2\hspace{0.05cm}\cdot\hspace{0.05cm}\frac{B^3}{3} + \frac{4}{5} \hspace{0.05cm}\cdot\hspace{0.05cm}\alpha_1 \alpha_2  \hspace{0.05cm}\cdot\hspace{0.05cm}B^{2.5} +\alpha_1^2 \hspace{0.05cm}\cdot\hspace{0.05cm} \frac{B^2}{2} + \frac{k_2^2}{2k_3 +1} \hspace{0.05cm}\cdot\hspace{0.05cm}\frac{B^{2k_3+1}}{f_0^{2k_3}} - \hspace{0.15cm}\frac{2 k_2 \alpha_1}{k_3 + 2}\hspace{0.05cm}\cdot\hspace{0.05cm}$$:This equation contains the cable parameters &nbsp;$\alpha_1$, &nbsp;$\alpha_2$, &nbsp;$k_2$&nbsp; and &nbsp;$k_3$&nbsp; to be calculated as well as the bandwidth &nbsp;$B$,&nbsp; within which the approximation should be valid.* By setting the derivatives of &nbsp;${\rm E}\big[\varepsilon^2(f)\big]$&nbsp; to &nbsp;$\alpha_1$&nbsp; and &nbsp;$\alpha_2$&nbsp; to zero, two equations are obtained for the best possible coefficients &nbsp;$\alpha_1$&nbsp; and &nbsp;$\alpha_2$ that minimize the mean square error. These can be represented in the following form::$$\frac{{\rm d}\,{\rm E}\big[\varepsilon^2(f)\big]}{{\rm d}\,{\alpha_1}}  = 0 \hspace{0.2cm}\Rightarrow\hspace{0.2cm} \alpha_1 + C_1 \cdot \alpha_2 + C_2  = 0 \hspace{0.05cm},$$:$$\frac{{\rm d}\,{\rm E}\big[\varepsilon^2(f)\big]}{{\rm d}\,{\alpha_2}}  =0 \hspace{0.2cm}\Rightarrow\hspace{0.2cm} \alpha_1 + D_1 \cdot \alpha_2 + D_2 = 0 \hspace{0.05cm}. $$* From the equation &nbsp;$C_1 \cdot \alpha_2 + C_2 = D_1 \cdot \alpha_2 + D_2$,&nbsp; the coefficient &nbsp;$\alpha_2$&nbsp; can be calculated and then the coefficient &nbsp;$\alpha_1$ can be calculated from each of the two equations above.The graph shows the attenuation function per unit length for a copper twin wire with&nbsp; $\text{0.5 mm}$&nbsp; diameter, whose&nbsp; $k$&ndash;parameters are::$$k_1  = 4.4\, {\rm dB}/{\rm km} \hspace{0.05cm}, \hspace{0.2cm}k_2  = 10.8\, {\rm dB}/{\rm km}\hspace{0.05cm}, \hspace{0.2cm}k_3  = 0.60\hspace{0.05cm}\hspace{0.05cm}.$$
 *The red curve shows the function &nbsp;$\alpha(f)$&nbsp; calculated with this parameters.&nbsp; For &nbsp;$f = 30 \ \rm MHz$&nbsp; the attenuation function per unit length is &nbsp;$\alpha(f)= 87.5 \ \rm dB/km$.

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← Older revision		Revision as of 16:16, 16 March 2026
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	:$$\frac{{\rm d}\,{\rm E}[\varepsilon^2(f)]}{{\rm d}\,{\alpha_1}} =\frac{2}{3}\cdot B^3 \cdot \alpha_1 + \frac{4}{5}\cdot B^{2.5} \cdot \alpha_2- \frac{2 k_2 }{k_3+ 2} \cdot \frac{B^{k_3+2}}{f_0^{k_3}}= 0\hspace{0.05cm} .$$		:$$\frac{{\rm d}\,{\rm E}[\varepsilon^2(f)]}{{\rm d}\,{\alpha_1}} =\frac{2}{3}\cdot B^3 \cdot \alpha_1 + \frac{4}{5}\cdot B^{2.5} \cdot \alpha_2- \frac{2 k_2 }{k_3+ 2} \cdot \frac{B^{k_3+2}}{f_0^{k_3}}= 0\hspace{0.05cm} .$$
	*By setting it to zero and dividing by  $2B^2/3$,  we obtain:		*By setting it to zero and dividing by  $2B^2/3$,  we obtain:
	:$$\alpha_1 + \frac{6}{5}\cdot B^{-0.5} \cdot \alpha_2		:$$\alpha_1 + \frac{6}{5}\cdot B^{-0.5} \cdot \alpha_2- \frac{3 k_2 }{k_3+2} \cdot \frac{B^{k_3-1}}{f_0^{k_3}}= 0\hspace{0.3cm}\Rightarrow \hspace{0.3cm} C_1 = \frac{6}{5}\cdot B^{-0.5} \hspace{0.05cm} ,\hspace{0.5cm} C_2 =- \frac{3 k_2 }{k_3+2} \cdot \frac{B^{k_3-1}}{f_0^{k_3}}\hspace{0.05cm} .$$
	- \frac{3 k_2 }{k_3
	+2} \cdot \frac{B^{k_3-1}}{f_0^{k_3}}= 0
	\hspace{0.3cm}
	\Rightarrow \hspace{0.3cm} C_1 = \frac{6}{5}\cdot B^{-0.5} \hspace{0.05cm} ,
	\hspace{0.5cm} C_2 =
	- \frac{3 k_2 }{k_3
	+2} \cdot \frac{B^{k_3-1}}{f_0^{k_3}}
	\hspace{0.05cm} .$$


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	*Using the same procedure as in subtask  '''(1)''',  we obtain:		*Using the same procedure as in subtask  '''(1)''',  we obtain:
	:$$\frac{{\rm d}\,{\rm E}[\varepsilon^2(f)]}{{\rm d}\,{\alpha_2}} =\frac{4}{5}\cdot B^{2.5} \cdot \alpha_1 + B^{2} \cdot \alpha_2- \frac{2 k_2 }{k_3+ 1.5} \cdot \frac{B^{k_3+1.5}}{f_0^{k_3}}= 0$$		:$$\frac{{\rm d}\,{\rm E}[\varepsilon^2(f)]}{{\rm d}\,{\alpha_2}} =\frac{4}{5}\cdot B^{2.5} \cdot \alpha_1 + B^{2} \cdot \alpha_2- \frac{2 k_2 }{k_3+ 1.5} \cdot \frac{B^{k_3+1.5}}{f_0^{k_3}}= 0$$
	:$$\Rightarrow \hspace{0.3cm} \alpha_1 + \frac{5}{4}\cdot B^{-0.5} \cdot \alpha_2		:$$\Rightarrow \hspace{0.3cm} \alpha_1 + \frac{5}{4}\cdot B^{-0.5} \cdot \alpha_2- \frac{2.5 \cdot k_2 }{k_3+1.5} \cdot \frac{B^{k_3-1}}{f_0^{k_3}}= 0\hspace{0.3cm}\Rightarrow \hspace{0.3cm}D_1 = \frac{5}{4}\cdot B^{-0.5} \hspace{0.05cm} ,\hspace{0.3cm}D_2 =- \frac{2.5 \cdot k_2 }{k_3+1.5} \cdot \frac{B^{k_3-1}}{f_0^{k_3}}\hspace{0.05cm} .$$
	- \frac{2.5 \cdot k_2 }{k_3
	+1.5} \cdot \frac{B^{k_3-1}}{f_0^{k_3}}= 0
	\hspace{0.3cm}
	\Rightarrow \hspace{0.3cm}D_1 = \frac{5}{4}\cdot B^{-0.5} \hspace{0.05cm} ,
	\hspace{0.3cm}D_2 =
	- \frac{2.5 \cdot k_2 }{k_3
	+1.5} \cdot \frac{B^{k_3-1}}{f_0^{k_3}}
	\hspace{0.05cm} .$$

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← Older revision		Revision as of 14:33, 16 March 2026
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	[[Category:Linear and Time-Invariant Systems: Exercises\|^4.3 Balanced Copper Twisted Pair^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^4.3 Balanced Copper Twisted Pair^]]
			[[de:Aufgaben:4.6_$k$-_und_$\\alpha$-Parameter]]

Guenter at 16:11, 23 November 2021

2021-11-23T16:11:41Z

← Older revision		Revision as of 18:11, 23 November 2021
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>Solutions 1 and 6</u> are correct:		'''(1)'''  <u>Solutions 1 and 6</u>  are correct:
	*The derivative of the given expected value with respect to  $\alpha_1$  gives:		*The derivative of the given expected value with respect to  $\alpha_1$  gives:
	:$$\frac{{\rm d}\,{\rm E}[\varepsilon^2(f)]}{{\rm d}\,{\alpha_1}} =		:$$\frac{{\rm d}\,{\rm E}[\varepsilon^2(f)]}{{\rm d}\,{\alpha_1}} =
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	'''(2)'''  <u>Solutions 2 and 5</u> are correct:		'''(2)'''  <u>Solutions 2 and 5</u>  are correct:
	*Using the same procedure as in subtask  '''(1)''',  we obtain:		*Using the same procedure as in subtask  '''(1)''',  we obtain:
	:$$\frac{{\rm d}\,{\rm E}[\varepsilon^2(f)]}{{\rm d}\,{\alpha_2}} =		:$$\frac{{\rm d}\,{\rm E}[\varepsilon^2(f)]}{{\rm d}\,{\alpha_2}} =
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	'''(3)'''  From  $C_1 \cdot \alpha_2 + C_2 = D_1 \cdot \alpha_2 + D_2$  we obtain a linear equation for  $\alpha_2$. With the result from  '''(2)'''  we can write:		'''(3)'''  <u>Both solutions</u>  are correct:

			*From  $C_1 \cdot \alpha_2 + C_2 = D_1 \cdot \alpha_2 + D_2$  we obtain a linear equation for  $\alpha_2$.  With the result from  '''(2)'''  we can write:
	:$$\alpha_2 = \frac{D_2 - C_2}{C_1 - D_1} = \frac{- \frac{2.5 \cdot k_2 }{k_3		:$$\alpha_2 = \frac{D_2 - C_2}{C_1 - D_1} = \frac{- \frac{2.5 \cdot k_2 }{k_3
	+1.5} \cdot \frac{B^{k_3-1}}{f_0^{k_3}} + \frac{3 k_2 }{k_3 +2}		+1.5} \cdot \frac{B^{k_3-1}}{f_0^{k_3}} + \frac{3 k_2 }{k_3 +2}
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	-1}\cdot \frac{k_3 -0.5}{(k_3 + 1.5)(k_3 +		-1}\cdot \frac{k_3 -0.5}{(k_3 + 1.5)(k_3 +
	2)}\cdot \frac {k_2}{f_0}\hspace{0.05cm} .$$		2)}\cdot \frac {k_2}{f_0}\hspace{0.05cm} .$$
	~~<u>Both solutions</u> are correct.~~
	*Regardless of the bandwidth, we obtain for  $k_3 = 1$:		*Regardless of the bandwidth,  we obtain for  $k_3 = 1$:
	:$$\alpha_1 = (B/f_0)^{k_3		:$$\alpha_1 = (B/f_0)^{k_3
	-1}\cdot \frac{15 \cdot (k_3 -0.5)}{(k_3 + 1.5)(k_3 +		-1}\cdot \frac{15 \cdot (k_3 -0.5)}{(k_3 + 1.5)(k_3 +
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	2)}\cdot \frac {k_2}{\sqrt{f_0}}\hspace{0.15cm}\underline{= 0} \hspace{0.05cm}		2)}\cdot \frac {k_2}{\sqrt{f_0}}\hspace{0.15cm}\underline{= 0} \hspace{0.05cm}
	.$$		.$$
	*In contrast, for  $k_3 = 0.5$:		*In contrast,  for  $k_3 = 0.5$:
	:$$\alpha_1 = (B/f_0)^{k_3		:$$\alpha_1 = (B/f_0)^{k_3
	-1}\cdot \frac{15 \cdot (k_3 -0.5)}{(k_3 + 1.5)(k_3 +		-1}\cdot \frac{15 \cdot (k_3 -0.5)}{(k_3 + 1.5)(k_3 +

Guenter at 16:03, 23 November 2021

2021-11-23T16:03:15Z

← Older revision		Revision as of 18:03, 23 November 2021
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	}}		}}

	[[File:EN_LZI_A_4_6.png\|right\|frame\|Attenuation function per unit length  ~~$\text{~~(0.5 mm~~}$  twin wire$~~)~~$ with  $k$– and  $\alpha$-parameters~~]]		[[File:EN_LZI_A_4_6.png\|right\|frame\|Attenuation function per unit length, <br>valid for  "copper twin wire"  (0.5 mm)]]
	For symmetrical copper twisted pairs, the following empirical formula can be found in  [PW95],  which is valid for the frequency range  $0 \le f \le 30 \ \rm MHz$:		For symmetrical copper twisted pairs,  the following empirical formula can be found in  [PW95],  which is valid for the frequency range  $0 \le f \le 30 \ \rm MHz$:
	:$$\alpha_{\rm I} (f) = k_1 + k_2 \cdot (f/f_0)^{k_3} , \hspace{0.15cm}		:$$\alpha_{\rm I} (f) = k_1 + k_2 \cdot (f/f_0)^{k_3} , \hspace{0.15cm}
	f_0 = 1\,{\rm MHz} .$$		f_0 = 1\,{\rm MHz} .$$
	In contrast, the attenuation function per unit length of a coaxial cable is usually given in the following form:		In contrast,  the attenuation function per unit length of a coaxial cable is usually given in the following form:
	:$$\alpha_{\rm II}(f) = \alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot \sqrt {f}\hspace{0.05cm}.$$		:$$\alpha_{\rm II}(f) = \alpha_0 + \alpha_1 \cdot f + \alpha_2 \cdot \sqrt {f}\hspace{0.05cm}.$$
	Especially for the calculation of impulse response and rectangular response it is advantageous also for the copper twisted pairs to choose the second representation form with the cable parameters  $\alpha_0$,  $\alpha_1$  and  $\alpha_2$  instead of the representation with  $k_1$,  $k_2$  and  $k_3$.		Especially for the calculation of impulse response and rectangular response it is advantageous also for the copper twisted pairs to choose the second representation form with the cable parameters  $\alpha_0$,  $\alpha_1$  and  $\alpha_2$  instead of the representation with  $k_1$,  $k_2$  and  $k_3$.

	For the conversion, one proceeds as follows:		For the conversion,  one proceeds as follows:
	* From above equations, it is obvious that the coefficient characterizing the DC signal attenuation is  $\alpha_0 = k_1$~~ ~~.		* From above equations,  it is obvious that the coefficient characterizing the DC signal attenuation is  $\alpha_0 = k_1$.
	* To determine  $\alpha_1$  and  $\alpha_2$,  it is assumed that the mean square error should be minimum in the range of a given bandwidth  $B$~~ ~~:		* To determine  $\alpha_1$  and  $\alpha_2$,  it is assumed that the mean square error should be minimum in the range of a given bandwidth  $B$:
	:$${\rm E}\big[\varepsilon^2(f)\big] = \int_{0}^{		:$${\rm E}\big[\varepsilon^2(f)\big] = \int_{0}^{
	B} \left [ \alpha_{\rm II} (f) - \alpha_{\rm I} (f)\right ]^2		B} \left [ \alpha_{\rm II} (f) - \alpha_{\rm I} (f)\right ]^2
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*The red curve shows the function  $\alpha(f)$  calculated with ~~the~~ parameters. For  $f = 30 \ \rm MHz$  the attenuation function per unit length is  $\alpha(f)= 87.5 \ \rm dB/km$.		*The red curve shows the function  $\alpha(f)$  calculated with this parameters.  For  $f = 30 \ \rm MHz$  the attenuation function per unit length is  $\alpha(f)= 87.5 \ \rm dB/km$.
	*The blue curve gives the approximation with the  $\alpha$–coefficients.  This is almost indistinguishable from the red curve within the drawing accuracy.		*The blue curve gives the approximation with the  $\alpha$–coefficients.  This is almost indistinguishable from the red curve within the drawing accuracy.

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			Notes:


	''Notes:''
	*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Eigenschaften_von_Kupfer–Doppeladern\|Properties of Balanced Copper Pairs]].		*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Eigenschaften_von_Kupfer–Doppeladern\|Properties of Balanced Copper Pairs]].

	*You can use the  (German language)  interactive SWF applet  [[Applets:Dämpfung_von_Kupferkabeln\|"Dämpfung von Kupferkabeln"]]  ⇒   "Attenuation of copper cables" .		*You can use the  (German language)  interactive SWF applet  [[Applets:Dämpfung_von_Kupferkabeln\|"Dämpfung von Kupferkabeln"]]  ⇒   "Attenuation of copper cables" .
	*[PW95]  denotes the following literature reference:   Pollakowski, P.; Wellhausen, H.-W.: ''Eigenschaften symmetrischer Ortsanschlusskabel im Frequenzbereich bis 30 MHz.'' Deutsche Telekom AG, Forschungs- und Technologiezentrum Darmstadt, 1995.		*[PW95]  denotes the following literature reference:   Pollakowski, P.; Wellhausen, H.-W.:  Eigenschaften symmetrischer Ortsanschlusskabel im Frequenzbereich bis 30 MHz.  Deutsche Telekom AG, Forschungs- und Technologiezentrum Darmstadt, 1995.


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	<quiz display=simple>		<quiz display=simple>
	{Calculate the parameters  $C_1$  and  $C_2$  of the equation  $\alpha_1 + C_1 \cdot \alpha_2 + C_2 = 0$ resulting from the derivative  ${\rm dE\big[\text{...}\big]/d}\alpha_1$.  <br>Which results are correct?		{Calculate the parameters  $C_1$  and  $C_2$  of the equation  $\alpha_1 + C_1 \cdot \alpha_2 + C_2 = 0$  resulting from the derivative  ${\rm dE\big[\text{...}\big]/d}\alpha_1$.  <br>Which results are correct?
	\|type="[]"}		\|type="[]"}
	+ $C_1 = 6/5 \cdot B^{-0.5}$,		+ $C_1 = 6/5 \cdot B^{-0.5}$,
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	{Calculate the parameters  $D_1$  and  $D_2$  of the equation  $ \alpha_1 + D_1 \cdot \alpha_2 + D_2 = 0$ resulting from the derivative  ${\rm dE\big[\text{...}\big]/d}\alpha_2$.  <br> Which results are correct?		{Calculate the parameters  $D_1$  and  $D_2$  of the equation  $ \alpha_1 + D_1 \cdot \alpha_2 + D_2 = 0$  resulting from the derivative  ${\rm dE\big[\text{...}\big]/d}\alpha_2$.  <br> Which results are correct?
	\|type="[]"}		\|type="[]"}
	- $D_1 = 6/5 \cdot B^{-0.5}$,		- $D_1 = 6/5 \cdot B^{-0.5}$,
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	{Using the  $\alpha$–parameters, calculate the attenuation function per unit length for the frequency  $f = 30\ \rm MHz$.		{Using the  $\alpha$–parameters,  calculate the attenuation function per unit length for the frequency  $f = 30\ \rm MHz$.
	\|type="{}"}		\|type="{}"}
	$\alpha_{\rm II}(f = 30\ \rm MHz) \ = \ $ { 88.1 3% } $\ \rm dB/km$		$\alpha_{\rm II}(f = 30\ \rm MHz) \ = \ $ { 88.1 3% } $\ \rm dB/km$

Hwang at 08:56, 22 November 2021

2021-11-22T08:56:16Z

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Guenter: Guenter moved page Aufgaben:Exercise 4.6: k-Parameters and Alpha-Parameters to Aufgaben:Exercise 4.6: k-parameters and alpha-parameters

2021-11-20T13:59:27Z

Guenter moved page Aufgaben:Exercise 4.6: k-Parameters and Alpha-Parameters to Aufgaben:Exercise 4.6: k-parameters and alpha-parameters

← Older revision	Revision as of 15:59, 20 November 2021
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Guenter: Guenter moved page Aufgaben:Exercise 4.6: K-Parameters and Alpha-Parameters to Aufgaben:Exercise 4.6: k-Parameters and Alpha-Parameters

2021-11-19T09:18:33Z

Guenter moved page Aufgaben:Exercise 4.6: K-Parameters and Alpha-Parameters to Aufgaben:Exercise 4.6: k-Parameters and Alpha-Parameters

← Older revision	Revision as of 11:18, 19 November 2021
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