Aufgaben:Exercise 3.7Z: Regenerator Field Length - Revision history

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	'''(3)'''  The sink SNR should be at least  $16.1 \ \rm dB$,  which means that it must be valid:		'''(3)'''  The sink SNR should be at least  $16.1 \ \rm dB$,  which means that it must be valid:
	:$$10 \cdot {\rm lg}\hspace{0.1cm}\rho = 10 \cdot {\rm lg}		:$$10 \cdot {\rm lg}\hspace{0.1cm}\rho = 10 \cdot {\rm lg}\hspace{0.1cm}\frac{s_0^2 }{N_0 \cdot R_{\rm B}} + 10 \cdot {\rmlg}\hspace{0.1cm}\eta \hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\eta \ >\ 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm min} - 10 \cdot {\rmlg}\hspace{0.1cm}\frac{s_0^2 }{N_0 \cdot R_{\rm B}} =\ 16.1- 100\hspace{0.15cm}\underline {= -83.9\,{\rm dB} = 10 \cdot {\rmlg}\hspace{0.1cm}\eta_{\hspace{0.05cm} \rm min}}\hspace{0.05cm}.$$
	\hspace{0.1cm}\frac{s_0^2 }{N_0 \cdot R_{\rm B}} + 10 \cdot {\rm
	lg}\hspace{0.1cm}\eta \hspace{0.3cm}
	\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\eta \ >
	\ 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm min} - 10 \cdot {\rm
	lg}
	\hspace{0.1cm}\frac{s_0^2 }{N_0 \cdot R_{\rm B}} =
	\ 16.1- 100\hspace{0.15cm}\underline {= -83.9\,{\rm dB} = 10 \cdot {\rm
	lg}\hspace{0.1cm}\eta_{\hspace{0.05cm} \rm min}}\hspace{0.05cm}.$$

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	* The logarithmized ratio of transmit energy  (per bit)  and AWGN noise power density is about  $100 \ \rm dB$,  for example:		* The logarithmized ratio of transmit energy  (per bit)  and AWGN noise power density is about  $100 \ \rm dB$,  for example:
	:$$s_0 = 3\,{\rm V},\hspace{0.2cm}R_{\rm B} = 1\,{\rmGbit/s},\hspace{0.2cm}N_{\rm 0} = 9 \cdot 10^{-19}\,{\rm V^2/Hz}$$		:$$s_0 = 3\,{\rm V},\hspace{0.2cm}R_{\rm B} = 1\,{\rmGbit/s},\hspace{0.2cm}N_{\rm 0} = 9 \cdot 10^{-19}\,{\rm V^2/Hz}$$
	:$$\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}		:$$\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\frac{s_0^2 }{N_0 \cdot R_{\rm B}}= 10 \cdot {\rmlg} \hspace{0.1cm} \frac{9\,{\rm V^2} } {9 \cdot 10^{-19}\,{\rm V^2/Hz}\cdot 10^{-9}\,{\rm 1/s}}= 100\,{\rmdB} \hspace{0.05cm}.$$
	\hspace{0.1cm}\frac{s_0^2 }{N_0 \cdot R_{\rm B}}= 10 \cdot {\rm
	lg} \hspace{0.1cm} \frac{9\,{\rm V^2} } {9 \cdot 10^{-19}\,{\rm V^2/Hz}
	\cdot 10^{-9}\,{\rm 1/s}}
	= 100\,{\rm
	dB} \hspace{0.05cm}.$$
	* A standard coaxial cable with dimensions  $2.6 \ \rm mm$  (inside)  and  $9.5 \ \rm mm$  (outside)  is to be used,  for which the following relationship is valid:		* A standard coaxial cable with dimensions  $2.6 \ \rm mm$  (inside)  and  $9.5 \ \rm mm$  (outside)  is to be used,  for which the following relationship is valid:
	:$$a_{\star} = \frac{2.36\,{\rm dB} } {{\rm km} \cdot \sqrt{{\rmMHz}}} \cdot l \cdot \sqrt{{R_{\rm B}}/{2}}\hspace{0.05cm}.$$		:$$a_{\star} = \frac{2.36\,{\rm dB} } {{\rm km} \cdot \sqrt{{\rmMHz}}} \cdot l \cdot \sqrt{{R_{\rm B}}/{2}}\hspace{0.05cm}.$$

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	[[File:EN_Dig_z_3_7.png\|right\|frame\|Results of a system simulation]]		[[File:EN_Dig_z_3_7.png\|right\|frame\|Results of a system simulation]]
	By simulation,  it was shown that there is approximately a linear relationship between the so-called  "system efficiency"   $\eta$  and the  "characteristic cable attenuation"   $a_$  of a coaxial cable – both plotted in  $\rm dB$  – if the characteristic cable attenuation is sufficiently large  $(a_ ≥ 40 \ \rm dB)$:		By simulation,  it was shown that there is approximately a linear relationship between the so-called  "system efficiency"   $\eta$  and the  "characteristic cable attenuation"   $a_$  of a coaxial cable – both plotted in  $\rm dB$  – if the characteristic cable attenuation is sufficiently large  $(a_ ≥ 40 \ \rm dB)$:
	:$$10 \cdot {\rm lg}\hspace{0.1cm}\eta \hspace{0.15cm} {\rm (in \hspace{0.15cm}dB)}= A + B \cdot a_{\star}		:$$10 \cdot {\rm lg}\hspace{0.1cm}\eta \hspace{0.15cm} {\rm (in \hspace{0.15cm}dB)}= A + B \cdot a_{\star}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	In the table,  the empirically found coefficients  $A$  and  $B$  are given for four exemplary system variants:		In the table,  the empirically found coefficients  $A$  and  $B$  are given for four exemplary system variants:
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	For the calculation of the  '''regenerator field length'''  (distance between two repeaters),  it should be noted:		For the calculation of the  '''regenerator field length'''  (distance between two repeaters),  it should be noted:
	* The worst-case error probability should not be larger than  $10^{-10}$, which results in the minimum sink SNR:		* The worst-case error probability should not be larger than  $10^{-10}$, which results in the minimum sink SNR:
	:$$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm min} \approx 16.1\,{\rm		:$$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm min} \approx 16.1\,{\rmdB} \hspace{0.05cm},$$
	dB} \hspace{0.05cm},$$
	* The logarithmized ratio of transmit energy  (per bit)  and AWGN noise power density is about  $100 \ \rm dB$,  for example:		* The logarithmized ratio of transmit energy  (per bit)  and AWGN noise power density is about  $100 \ \rm dB$,  for example:
	:$$s_0 = 3\,{\rm V},\hspace{0.2cm}R_{\rm B} = 1\,{\rm		:$$s_0 = 3\,{\rm V},\hspace{0.2cm}R_{\rm B} = 1\,{\rmGbit/s},\hspace{0.2cm}N_{\rm 0} = 9 \cdot 10^{-19}\,{\rm V^2/Hz}$$
	~~Gbit~~/s},\hspace{0.2cm}N_{\rm 0} = 9 \cdot 10^{-19}\,{\rm V^2/Hz}$$
	:$$\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}		:$$\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}
	\hspace{0.1cm}\frac{s_0^2 }{N_0 \cdot R_{\rm B}}= 10 \cdot {\rm		\hspace{0.1cm}\frac{s_0^2 }{N_0 \cdot R_{\rm B}}= 10 \cdot {\rm
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	dB} \hspace{0.05cm}.$$		dB} \hspace{0.05cm}.$$
	* A standard coaxial cable with dimensions  $2.6 \ \rm mm$  (inside)  and  $9.5 \ \rm mm$  (outside)  is to be used,  for which the following relationship is valid:		* A standard coaxial cable with dimensions  $2.6 \ \rm mm$  (inside)  and  $9.5 \ \rm mm$  (outside)  is to be used,  for which the following relationship is valid:
	:$$a_{\star} = \frac{2.36\,{\rm dB} } {{\rm km} \cdot \sqrt{{\rm		:$$a_{\star} = \frac{2.36\,{\rm dB} } {{\rm km} \cdot \sqrt{{\rmMHz}}} \cdot l \cdot \sqrt{{R_{\rm B}}/{2}}\hspace{0.05cm}.$$
	~~MHz~~}}} \cdot l \cdot \sqrt{{R_{\rm B}}/{2}}
	\hspace{0.05cm}.$$

	:Here,  $a_*$  denotes the characteristic attenuation at half the bit rate – in the example at  $500 \ \rm MHz$  – and  $l$  denotes the cable length.		:Here,  $a_*$  denotes the characteristic attenuation at half the bit rate – in the example at  $500 \ \rm MHz$  – and  $l$  denotes the cable length.
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	'''(1)'''  Calculating the system efficiency under the assumption  $a_* = 40 \ \rm dB$,  we obtain for the four system variants:		'''(1)'''  Calculating the system efficiency under the assumption  $a_* = 40 \ \rm dB$,  we obtain for the four system variants:
	:$$({\rm GLP},\hspace{0.1cm}M=2) \text{:}\hspace{0.3cm} 10 \cdot {\rm		:$$({\rm GLP},\hspace{0.1cm}M=2) \text{:}\hspace{0.3cm} 10 \cdot {\rmlg}\hspace{0.1cm}\eta= +9.4\,{\rm dB} -1.10 \cdot 40\,{\rm dB} = -34.6\,{\rm dB}\hspace{0.05cm},$$
	lg}\hspace{0.1cm}\eta		:$$({\rm GLP},\hspace{0.1cm}M=8) \text{:}\hspace{0.3cm}10 \cdot {\rmlg}\hspace{0.1cm}\eta= -1.3\,{\rm dB} -0.91 \cdot 40\,{\rm dB} = -37.7\,{\rm dB}\hspace{0.05cm},$$
	= +9.4\,{\rm dB} -1.10 \cdot 40\,{\rm dB} = -34.6\,{\rm dB}\hspace{0.05cm},$$		:$$({\rm ONE},\hspace{0.1cm}M=2) \text{:}\hspace{0.3cm}10 \cdot {\rmlg}\hspace{0.1cm}\eta= +4.5\,{\rm dB} -0.96 \cdot 40 \,{\rm dB}= -33.9\,{\rm dB}\hspace{0.05cm},$$
	:$$({\rm GLP},\hspace{0.1cm}M=8) \text{:}\hspace{0.3cm}10 \cdot {\rm		:$$({\rm ONE},\hspace{0.1cm}M=8) \text{:}\hspace{0.3cm} 10 \cdot {\rmlg}\hspace{0.1cm}\eta = -9.3\,{\rm dB} -0.54 \cdot 40\,{\rm dB} = -30.9\,{\rmdB}\hspace{0.05cm}.$$
	lg}\hspace{0.1cm}\eta
	= -1.3\,{\rm dB} -0.91 \cdot 40\,{\rm dB} = -37.7\,{\rm dB}\hspace{0.05cm},$$
	:$$({\rm ONE},\hspace{0.1cm}M=2) \text{:}\hspace{0.3cm}10 \cdot {\rm
	lg}\hspace{0.1cm}\eta
	= +4.5\,{\rm dB} -0.96 \cdot 40 \,{\rm dB}= -33.9\,{\rm dB}\hspace{0.05cm},$$
	:$$({\rm ONE},\hspace{0.1cm}M=8) \text{:}\hspace{0.3cm} 10 \cdot {\rm
	lg}\hspace{0.1cm}\eta = -9.3\,{\rm dB} -0.54 \cdot 40\,{\rm dB} = -30.9\,{\rm
	dB}\hspace{0.05cm}.$$

	From this it follows:		From this it follows:
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	'''(2)'''  As a determination equation,  we use here:		'''(2)'''  As a determination equation,  we use here:
	:$$-1.3\,{\rm dB} -0.91 \cdot a_{\star} = +4.5 \,{\rm dB}-0.96 \cdot a_{\star}\hspace{0.3cm}		:$$-1.3\,{\rm dB} -0.91 \cdot a_{\star} = +4.5 \,{\rm dB}-0.96 \cdot a_{\star}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} 0.05 \cdot a_{\star} = 5.8\,{\rm dB}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}a_{\star,\hspace{0.05cm}{\rm limit}} \hspace{0.15cm}\underline {= 116\,{\rm dB}}\hspace{0.05cm}.$$
	\Rightarrow \hspace{0.3cm} 0.05 \cdot a_{\star} = 5.8\,{\rm dB}
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}a_{\star,\hspace{0.05cm}{\rm limit}} \hspace{0.15cm}\underline {= 116\,{\rm dB}}\hspace{0.05cm}.$$

	That is:		That is:
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	'''(4)'''  For the system considered here:   $10 \cdot {\rm lg}\hspace{0.1cm}\eta = -9.3\,{\rm dB} -0.54 \cdot a_{\star}.$		'''(4)'''  For the system considered here:   $10 \cdot {\rm lg}\hspace{0.1cm}\eta = -9.3\,{\rm dB} -0.54 \cdot a_{\star}.$
	*Thus,  from the system efficiency condition   ⇒   $10 \cdot {\rm lg} \, \eta > \hspace{0.1cm}–83.9 \ \rm dB $,  the characteristic cable attenuation condition is:		*Thus,  from the system efficiency condition   ⇒   $10 \cdot {\rm lg} \, \eta > \hspace{0.1cm}–83.9 \ \rm dB $,  the characteristic cable attenuation condition is:
	:$$a_{\star} < \frac{-83.9\,{\rm dB} + 9.3\,{\rm dB}} {-0.54} \~~approx~~		:$$a_{\star} < \frac{-83.9\,{\rm dB} + 9.3\,{\rm dB}} {-0.54} \approx138.1\,{\rm dB} \hspace{0.05cm}.$$
	~~138~~.1\,{\rm dB} \hspace{0.05cm}.$$

	*With the given equation		*With the given equation
	:$$a_{\star} = \frac{2.36\,{\rm dB} } {{\rm km} \cdot \sqrt{{\rm		:$$a_{\star} = \frac{2.36\,{\rm dB} } {{\rm km} \cdot \sqrt{{\rmMHz}}} \cdot l \cdot \sqrt{{R_{\rm B}}/{2}}\hspace{0.05cm}.$$
	~~MHz~~}}} \cdot l \cdot \sqrt{{R_{\rm B}}/{2}}
	\hspace{0.05cm}.$$

	:thus the maximum cable length  ("regenerator field length")  can be specified:		:thus the maximum cable length  ("regenerator field length")  can be specified:
	:$$l_{\rm max} = \frac{138.1\,{\rm dB} } {2.36\,{\rm dB}/{\rm		:$$l_{\rm max} = \frac{138.1\,{\rm dB} } {2.36\,{\rm dB}/{\rmkm} \cdot \sqrt{\rm MHz})\cdot \sqrt{500\,{\rm MHz}}} \hspace{0.15cm}\underline {\approx 2.62\,{\rm km}} \hspace{0.05cm}.$$
	km} \cdot \sqrt{\rm MHz})\cdot \sqrt{500\,{\rm MHz}}} \hspace{0.15cm}\underline {\approx 2.62\,
	{\rm km}} \hspace{0.05cm}.$$


	'''(5)'''  Following the same procedure,  but in a more compact notation,  results in a smaller  "regenerator field length"  for this  "worse"  system:		'''(5)'''  Following the same procedure,  but in a more compact notation,  results in a smaller  "regenerator field length"  for this  "worse"  system:
	:$$l_{\rm max} = \frac{-(83.9\,{\rm dB}+A)/B } {2.36\,{\rm dB}/{\rm		:$$l_{\rm max} = \frac{-(83.9\,{\rm dB}+A)/B } {2.36\,{\rm dB}/{\rmkm} \cdot \sqrt{500}} = \frac{+(83.9+9.4)/1.10 } {2.36\cdot\sqrt{500}}\hspace{0.1cm}{\rm km}\hspace{0.15cm}\underline {\approx 1.61\, {\rm km}}\hspace{0.05cm}.$$
	km} \cdot \sqrt{500}} = \frac{+(83.9+9.4)/1.10 } {2.36\cdot
	\sqrt{500}}\hspace{0.1cm}{\rm km}\hspace{0.15cm}\underline {\approx 1.61\, {\rm km}}
	\hspace{0.05cm}.$$
	{{ML-Fuß}}		{{ML-Fuß}}

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	[[Category:Digital Signal Transmission: Exercises\|^3.5 Linear Nyquist Equalization^]]		[[Category:Digital Signal Transmission: Exercises\|^3.5 Linear Nyquist Equalization^]]
			[[de:3.7Z_Regeneratorfeldlänge]]

Guenter at 14:51, 28 June 2022

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	{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Linear_Nyquist_Equalization}}		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Linear_Nyquist_Equalization}}

	[[File:~~P_ID1438__Dig_Z_3_7~~.png\|right\|frame\|Results of a system simulation ~~'''Korrektur''': GLP~~]]		[[File:EN_Dig_z_3_7.png\|right\|frame\|Results of a system simulation]]
	By simulation,  it was shown that there is approximately a linear relationship between the so-called  "system efficiency"   $\eta$  and the  "characteristic cable attenuation"   $a_$  of a coaxial cable – both plotted in  $\rm dB$  – if the characteristic cable attenuation is sufficiently large  $(a_ ≥ 40 \ \rm dB)$:		By simulation,  it was shown that there is approximately a linear relationship between the so-called  "system efficiency"   $\eta$  and the  "characteristic cable attenuation"   $a_$  of a coaxial cable – both plotted in  $\rm dB$  – if the characteristic cable attenuation is sufficiently large  $(a_ ≥ 40 \ \rm dB)$:
	:$$10 \cdot {\rm lg}\hspace{0.1cm}\eta \hspace{0.15cm} {\rm (in \hspace{0.15cm}dB)}= A + B \cdot a_{\star}		:$$10 \cdot {\rm lg}\hspace{0.1cm}\eta \hspace{0.15cm} {\rm (in \hspace{0.15cm}dB)}= A + B \cdot a_{\star}

Guenter at 12:13, 23 June 2022

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Calculating the system efficiency under the assumption $a_* = 40 \ \rm dB$, we obtain for the four system variants:		'''(1)'''  Calculating the system efficiency under the assumption  $a_* = 40 \ \rm dB$,  we obtain for the four system variants:
	:$$({\rm ~~GTP~~},\hspace{0.1cm}M=2) \text{:}\hspace{0.3cm} 10 \cdot {\rm		:$$({\rm GLP},\hspace{0.1cm}M=2) \text{:}\hspace{0.3cm} 10 \cdot {\rm
	lg}\hspace{0.1cm}\eta		lg}\hspace{0.1cm}\eta
	= +9.4\,{\rm dB} -1.10 \cdot 40\,{\rm dB} = -34.6\,{\rm dB}\hspace{0.05cm},$$		= +9.4\,{\rm dB} -1.10 \cdot 40\,{\rm dB} = -34.6\,{\rm dB}\hspace{0.05cm},$$
	:$$({\rm ~~GTP~~},\hspace{0.1cm}M=8) \text{:}\hspace{0.3cm}10 \cdot {\rm		:$$({\rm GLP},\hspace{0.1cm}M=8) \text{:}\hspace{0.3cm}10 \cdot {\rm
	lg}\hspace{0.1cm}\eta		lg}\hspace{0.1cm}\eta
	= -1.3\,{\rm dB} -0.91 \cdot 40\,{\rm dB} = -37.7\,{\rm dB}\hspace{0.05cm},$$		= -1.3\,{\rm dB} -0.91 \cdot 40\,{\rm dB} = -37.7\,{\rm dB}\hspace{0.05cm},$$
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	From this it follows:		From this it follows:
	*The <u>first statement</u> is true because the system $({\rm ONE},\hspace{0.1cm} M = 8)$ is already best at $40 \ \rm dB$ cable attenuation and has the most favorable $\rm B$ coefficient.		*The  <u>first statement</u>  is true because the system  $({\rm ONE},\hspace{0.1cm} M = 8)$  is already best at  $40 \ \rm dB$  cable attenuation and has the most favorable  $\rm B$  coefficient.
	*In contrast, the second statement is not true because, for example, at $40 \ \rm dB$ cable attenuation, the octal $\rm GTP$ system is worse than the binary one.

			*In contrast,  the second statement is false because,  for example,  at $40 \ \rm dB$ cable attenuation,  the octal $\rm GLP$ system is worse than the binary one.

	'''(2)'''  As a determination equation, we use here:
			'''(2)'''  As a determination equation,  we use here:
	:$$-1.3\,{\rm dB} -0.91 \cdot a_{\star} = +4.5 \,{\rm dB}-0.96 \cdot a_{\star}\hspace{0.3cm}		:$$-1.3\,{\rm dB} -0.91 \cdot a_{\star} = +4.5 \,{\rm dB}-0.96 \cdot a_{\star}\hspace{0.3cm}
	\Rightarrow \hspace{0.3cm} 0.05 \cdot a_{\star} = 5.8\,{\rm dB}		\Rightarrow \hspace{0.3cm} 0.05 \cdot a_{\star} = 5.8\,{\rm dB}
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	That is:		That is:
	Up to the characteristic cable attenuation $a_ = 116 \ \rm dB$ (note: this is an unrealistically large value for currently realized systems), the binary Nyquist system is superior to the system $({\rm ~~GTP~~},\hspace{0.1cm} M = 8)$.		Up to the characteristic cable attenuation  $a_ = 116 \ \rm dB$  $($note:  this is an unrealistically large value for currently realized systems$)$,  the binary Nyquist system is superior to the system $({\rm GLP},\hspace{0.1cm} M = 8)$.
	Only for larger values than $a_{\rm , \ limit} = 116 \ \rm dB$ does the advantage of the latter $(M = 8$ and thus significantly lower symbol rate$)$ outweigh the disadvantage $($octal decision and thus greater weight of intersymbol interference$)$.
			Only for larger values than  $a_{\rm , \ limit} = 116 \ \rm dB$  does the advantage of the latter  $(M = 8$  and thus significantly lower symbol rate$)$ outweigh the disadvantage  $($octal decision and thus greater weight of intersymbol interference$)$.



	'''(3)'''  The sink SNR should be at least $16.1 \ \rm dB$, which means that it must be valid:		'''(3)'''  The sink SNR should be at least  $16.1 \ \rm dB$,  which means that it must be valid:
	:$$10 \cdot {\rm lg}\hspace{0.1cm}\rho = 10 \cdot {\rm lg}		:$$10 \cdot {\rm lg}\hspace{0.1cm}\rho = 10 \cdot {\rm lg}
	\hspace{0.1cm}\frac{s_0^2 }{N_0 \cdot R_{\rm B}} + 10 \cdot {\rm		\hspace{0.1cm}\frac{s_0^2 }{N_0 \cdot R_{\rm B}} + 10 \cdot {\rm
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	\ 16.1- 100\hspace{0.15cm}\underline {= -83.9\,{\rm dB} = 10 \cdot {\rm		\ 16.1- 100\hspace{0.15cm}\underline {= -83.9\,{\rm dB} = 10 \cdot {\rm
	lg}\hspace{0.1cm}\eta_{\hspace{0.05cm} \rm min}}\hspace{0.05cm}.$$		lg}\hspace{0.1cm}\eta_{\hspace{0.05cm} \rm min}}\hspace{0.05cm}.$$



	'''(4)'''  For the system considered here:   $10 \cdot {\rm lg}\hspace{0.1cm}\eta = -9.3\,{\rm dB} -0.54 \cdot a_{\star}.$		'''(4)'''  For the system considered here:   $10 \cdot {\rm lg}\hspace{0.1cm}\eta = -9.3\,{\rm dB} -0.54 \cdot a_{\star}.$
	*Thus, from the system efficiency condition   ⇒   $10 \cdot {\rm lg} \, \eta > \hspace{0.1cm}–83.9 \ \rm dB $, the characteristic cable attenuation condition is:		*Thus,  from the system efficiency condition   ⇒   $10 \cdot {\rm lg} \, \eta > \hspace{0.1cm}–83.9 \ \rm dB $,  the characteristic cable attenuation condition is:
	:$$a_{\star} < \frac{-83.9\,{\rm dB} + 9.3\,{\rm dB}} {-0.54} \approx		:$$a_{\star} < \frac{-83.9\,{\rm dB} + 9.3\,{\rm dB}} {-0.54} \approx
	138.1\,{\rm dB} \hspace{0.05cm}.$$		138.1\,{\rm dB} \hspace{0.05cm}.$$
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	:thus the maximum cable length (regenerator field length) can be specified:		:thus the maximum cable length  ("regenerator field length")  can be specified:
	:$$l_{\rm max} = \frac{138.1\,{\rm dB} } {2.36\,{\rm dB}/{\rm		:$$l_{\rm max} = \frac{138.1\,{\rm dB} } {2.36\,{\rm dB}/{\rm
	km} \cdot \sqrt{\rm MHz})\cdot \sqrt{500\,{\rm MHz}}} \hspace{0.15cm}\underline {\approx 2.62\,		km} \cdot \sqrt{\rm MHz})\cdot \sqrt{500\,{\rm MHz}}} \hspace{0.15cm}\underline {\approx 2.62\,
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	'''(5)'''  Following the same procedure, but in a more compact notation, results in a smaller regenerator field length for this "worse" system:		'''(5)'''  Following the same procedure,  but in a more compact notation,  results in a smaller  "regenerator field length"  for this  "worse"  system:
	:$$l_{\rm max} = \frac{-(83.9\,{\rm dB}+A)/B } {2.36\,{\rm dB}/{\rm		:$$l_{\rm max} = \frac{-(83.9\,{\rm dB}+A)/B } {2.36\,{\rm dB}/{\rm
	km} \cdot \sqrt{500}} = \frac{+(83.9+9.4)/1.10 } {2.36\cdot		km} \cdot \sqrt{500}} = \frac{+(83.9+9.4)/1.10 } {2.36\cdot

Guenter at 12:01, 23 June 2022

2022-06-23T12:01:23Z

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	{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Linear_Nyquist_Equalization}}		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Linear_Nyquist_Equalization}}

	[[File:P_ID1438__Dig_Z_3_7.png\|right\|frame\|Results of a system simulation]]		[[File:P_ID1438__Dig_Z_3_7.png\|right\|frame\|Results of a system simulation '''Korrektur''': GLP]]
	By simulation, it was shown that there is approximately a linear relationship between the so-called system efficiency  $\eta$  and the characteristic cable attenuation  $a_$  of a coaxial cable – both plotted in  $\rm dB$  – if the characteristic cable attenuation is sufficiently large  $(a_ ≥ 40 \ \rm dB)$:		By simulation,  it was shown that there is approximately a linear relationship between the so-called  "system efficiency"   $\eta$  and the  "characteristic cable attenuation"   $a_$  of a coaxial cable – both plotted in  $\rm dB$  – if the characteristic cable attenuation is sufficiently large  $(a_ ≥ 40 \ \rm dB)$:
	:$$10 \cdot {\rm lg}\hspace{0.1cm}\eta \hspace{0.15cm} {\rm (in \hspace{0.15cm}dB)}= A + B \cdot a_{\star}		:$$10 \cdot {\rm lg}\hspace{0.1cm}\eta \hspace{0.15cm} {\rm (in \hspace{0.15cm}dB)}= A + B \cdot a_{\star}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	In the table, the empirically found coefficients  $A$  and  $B$  are given for four exemplary system variants:		In the table,  the empirically found coefficients  $A$  and  $B$  are given for four exemplary system variants:
	* for the intersymbol interference binary system  $(M = 2)$  with Gaussian low-pass filter  $\rm (~~GTP~~)$, see chapter  <br>[[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference\|"Error Probability with Intersymbol Interference"]],		* for the intersymbol interference binary system  $(M = 2)$  with Gaussian low-pass filter  $\rm (GLP)$,  see chapter  <br>[[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference\|"Error Probability with Intersymbol Interference"]],
	* for the intersymbol interference octal system  $(M = 8)$  with Gaussian low-pass filter  $\rm (~~GTP~~)$, see chapter  <br>[[Digital_Signal_Transmission/Intersymbol_Interference_for_Multi-Level_Transmission\|"Intersymbol Interference for Multi-Level Transmission"]],
	*for optimal ~~intersymbol interference~~ free systems  $\rm (ONE)$, see chapter  [[Digital_Signal_Transmission/Linear_Nyquist_Equalization\|"Linear Nyquist Equalization"]];  $M = 2$  and  $M = 8$.		* for the intersymbol interference octal system  $(M = 8)$  with Gaussian low-pass filter  $\rm (GLP)$,  see chapter  <br>[[Digital_Signal_Transmission/Intersymbol_Interference_for_Multi-Level_Transmission\|"Intersymbol Interference for Multi-Level Transmission"]],

			*for optimal ISI–free systems  $\rm (ONE)$,  see chapter  [[Digital_Signal_Transmission/Linear_Nyquist_Equalization\|"Linear Nyquist Equalization"]];  $M = 2$  and  $M = 8$.


	The larger the system efficiency  $\eta$,  the better a system is for a given value  $a_*$  (and thus a fixed cable length).		The larger the system efficiency  $\eta$,  the better a system is for a given value  $a_*$  (and thus a fixed cable length).

			For the calculation of the  '''regenerator field length'''  (distance between two repeaters),  it should be noted:
	For the calculation of the regenerator field length (distance between two repeaters), it should be noted ~~that~~		* The worst-case error probability should not be larger than  $10^{-10}$, which results in the minimum sink SNR:
	* ~~the~~ worst-case error probability should not be larger than  $10^{-10}$, which results in the minimum sink~~-noise distance~~:
	:$$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm min} \approx 16.1\,{\rm		:$$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm min} \approx 16.1\,{\rm
	dB} \hspace{0.05cm},$$		dB} \hspace{0.05cm},$$
	* ~~the~~ logarithmized ratio of transmit energy (per bit) and AWGN noise power density is about  $100 \ \rm dB$,  for example:		* The logarithmized ratio of transmit energy  (per bit)  and AWGN noise power density is about  $100 \ \rm dB$,  for example:
	:$$s_0 = 3\,{\rm V},\hspace{0.2cm}R_{\rm B} = 1\,{\rm		:$$s_0 = 3\,{\rm V},\hspace{0.2cm}R_{\rm B} = 1\,{\rm
	Gbit/s},\hspace{0.2cm}N_{\rm 0} = 9 \cdot 10^{-19}\,{\rm V^2/Hz}$$		Gbit/s},\hspace{0.2cm}N_{\rm 0} = 9 \cdot 10^{-19}\,{\rm V^2/Hz}$$
Line 27:		Line 28:
	\cdot 10^{-9}\,{\rm 1/s}}		\cdot 10^{-9}\,{\rm 1/s}}
	= 100\,{\rm		= 100\,{\rm
	dB} \hspace{0.05cm},$$		dB} \hspace{0.05cm}.$$
	* a standard coaxial cable with dimensions  $2.6 \ \rm mm$  (inside) and  $9.5 \ \rm mm$  (outside) is to be used, for which the following relationship is valid:		* A standard coaxial cable with dimensions  $2.6 \ \rm mm$  (inside)  and  $9.5 \ \rm mm$  (outside)  is to be used,  for which the following relationship is valid:
	:$$a_{\star} = \frac{2.36\,{\rm dB} } {{\rm km} \cdot \sqrt{{\rm		:$$a_{\star} = \frac{2.36\,{\rm dB} } {{\rm km} \cdot \sqrt{{\rm
	MHz}}} \cdot l \cdot \sqrt{{R_{\rm B}}/{2}}		MHz}}} \cdot l \cdot \sqrt{{R_{\rm B}}/{2}}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	:Here,  $a_*$  denotes the characteristic attenuation at half the bit rate – at  $500 \ \rm MHz$  ~~in the example~~ – and  $l$  denotes the cable length.		:Here,  $a_*$  denotes the characteristic attenuation at half the bit rate – in the example at  $500 \ \rm MHz$  – and  $l$  denotes the cable length.


	''Note:''		Note:  The exercise belongs to the chapter  [[Digital_Signal_Transmission/Linear_Nyquist_Equalization\|"Linear Nyquist Equalization"]].
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Linear_Nyquist_Equalization\|"Linear Nyquist Equalization"]].


Line 47:		Line 47:
	\|type="[]"}		\|type="[]"}
	+ The system  $({\rm ONE}, \ M = 8)$  is best for any  $a_*$.		+ The system  $({\rm ONE}, \ M = 8)$  is best for any  $a_*$.
	- The system  $({\rm ~~GTP~~}, \ M = 2)$  is worst for  $a_* ≥ 40 \ \rm dB$.		- The system  $({\rm GLP}, \ M = 2)$  is worst for  $a_* ≥ 40 \ \rm dB$.

	{Starting from which cable attenuation is  $({\rm ~~GTP~~}, \ M = 8)$  better than  $({\rm ONE}, \ M = 2)$?		{Starting from which cable attenuation is the system  $({\rm GLP}, \ M = 8)$  better than the system  $({\rm ONE}, \ M = 2)$?
	\|type="{}"}		\|type="{}"}
	$a_{\rm *, \ limit}\ = \ $ { 116 3% } $\ \rm dB$		$a_{\rm *, \ limit}\ = \ $ { 116 3% } $\ \rm dB$
Line 57:		Line 57:
	$10 \cdot {\rm lg} \ \eta_{\hspace{0.05cm}\rm min} \ = \ $ { -86.417--81.383 } $\ \rm dB$		$10 \cdot {\rm lg} \ \eta_{\hspace{0.05cm}\rm min} \ = \ $ { -86.417--81.383 } $\ \rm dB$

	{What is the maximum length of the coaxial cable for  $({\rm ONE}, \ M = 8)$?		{What is the maximum length of the coaxial cable for the system  $({\rm ONE}, \ M = 8)$?
	\|type="{}"}		\|type="{}"}
	$l_{\hspace{0.05cm}\rm max}\ = \ $ { 2.62 3% } $\ \rm km$		$l_{\hspace{0.05cm}\rm max}\ = \ $ { 2.62 3% } $\ \rm km$

	{What is the maximum length of the coaxial cable at  $({\rm GTP}, \ M = 2)$?		{What is the maximum length of the coaxial cable for the system  $({\rm GTP}, \ M = 2)$?
	\|type="{}"}		\|type="{}"}
	$l_{\hspace{0.05cm}\rm max}\ = \ $ { 1.61 3% } $\ \rm km$		$l_{\hspace{0.05cm}\rm max}\ = \ $ { 1.61 3% } $\ \rm km$

Hwang at 08:49, 26 May 2022

2022-05-26T08:49:43Z

Show changes

← Older revision		Revision as of 17:57, 16 March 2026
Line 19:		Line 19:
	:$$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm min} \approx 16.1\,{\rm dB} \hspace{0.05cm},$$		:$$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm min} \approx 16.1\,{\rm dB} \hspace{0.05cm},$$
	* The logarithmized ratio of transmit energy  (per bit)  and AWGN noise power density is about  $100 \ \rm dB$,  for example:		* The logarithmized ratio of transmit energy  (per bit)  and AWGN noise power density is about  $100 \ \rm dB$,  for example:
	:$$s_0 = 3\,{\rm V},\hspace{0.2cm}R_{\rm B} = 1\,{\~~rmGbit~~/s},\hspace{0.2cm}N_{\rm 0} = 9 \cdot 10^{-19}\,{\rm V^2/Hz}$$		:$$s_0 = 3\,{\rm V},\hspace{0.2cm}R_{\rm B} = 1\,{\rm Gbit/s},\hspace{0.2cm}N_{\rm 0} = 9 \cdot 10^{-19}\,{\rm V^2/Hz}$$
	:$$\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\frac{s_0^2 }{N_0 \cdot R_{\rm B}}= 10 \cdot {\~~rmlg~~} \hspace{0.1cm} \frac{9\,{\rm V^2} } {9 \cdot 10^{-19}\,{\rm V^2/Hz}\cdot 10^{-9}\,{\rm 1/s}}= 100\,{\rm dB} \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\frac{s_0^2 }{N_0 \cdot R_{\rm B}}= 10 \cdot {\rm lg} \hspace{0.1cm} \frac{9\,{\rm V^2} } {9 \cdot 10^{-19}\,{\rm V^2/Hz}\cdot 10^{-9}\,{\rm 1/s}}= 100\,{\rm dB} \hspace{0.05cm}.$$
	* A standard coaxial cable with dimensions  $2.6 \ \rm mm$  (inside)  and  $9.5 \ \rm mm$  (outside)  is to be used,  for which the following relationship is valid:		* A standard coaxial cable with dimensions  $2.6 \ \rm mm$  (inside)  and  $9.5 \ \rm mm$  (outside)  is to be used,  for which the following relationship is valid:
	:$$a_{\star} = \frac{2.36\,{\rm dB} } {{\rm km} \cdot \sqrt{{\rm MHz}}} \cdot l \cdot \sqrt{{R_{\rm B}}/{2}}\hspace{0.05cm}.$$		:$$a_{\star} = \frac{2.36\,{\rm dB} } {{\rm km} \cdot \sqrt{{\rm MHz}}} \cdot l \cdot \sqrt{{R_{\rm B}}/{2}}\hspace{0.05cm}.$$
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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Calculating the system efficiency under the assumption  $a_* = 40 \ \rm dB$,  we obtain for the four system variants:		'''(1)'''  Calculating the system efficiency under the assumption  $a_* = 40 \ \rm dB$,  we obtain for the four system variants:
	:$$({\rm GLP},\hspace{0.1cm}M=2) \text{:}\hspace{0.3cm} 10 \cdot {\~~rmlg~~}\hspace{0.1cm}\eta= +9.4\,{\rm dB} -1.10 \cdot 40\,{\rm dB} = -34.6\,{\rm dB}\hspace{0.05cm},$$		:$$({\rm GLP},\hspace{0.1cm}M=2) \text{:}\hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\eta= +9.4\,{\rm dB} -1.10 \cdot 40\,{\rm dB} = -34.6\,{\rm dB}\hspace{0.05cm},$$
	:$$({\rm GLP},\hspace{0.1cm}M=8) \text{:}\hspace{0.3cm}10 \cdot {\~~rmlg~~}\hspace{0.1cm}\eta= -1.3\,{\rm dB} -0.91 \cdot 40\,{\rm dB} = -37.7\,{\rm dB}\hspace{0.05cm},$$		:$$({\rm GLP},\hspace{0.1cm}M=8) \text{:}\hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\eta= -1.3\,{\rm dB} -0.91 \cdot 40\,{\rm dB} = -37.7\,{\rm dB}\hspace{0.05cm},$$
	:$$({\rm ONE},\hspace{0.1cm}M=2) \text{:}\hspace{0.3cm}10 \cdot {\~~rmlg~~}\hspace{0.1cm}\eta= +4.5\,{\rm dB} -0.96 \cdot 40 \,{\rm dB}= -33.9\,{\rm dB}\hspace{0.05cm},$$		:$$({\rm ONE},\hspace{0.1cm}M=2) \text{:}\hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\eta= +4.5\,{\rm dB} -0.96 \cdot 40 \,{\rm dB}= -33.9\,{\rm dB}\hspace{0.05cm},$$
	:$$({\rm ONE},\hspace{0.1cm}M=8) \text{:}\hspace{0.3cm} 10 \cdot {\~~rmlg~~}\hspace{0.1cm}\eta = -9.3\,{\rm dB} -0.54 \cdot 40\,{\rm dB} = -30.9\,{\rm dB}\hspace{0.05cm}.$$		:$$({\rm ONE},\hspace{0.1cm}M=8) \text{:}\hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\eta = -9.3\,{\rm dB} -0.54 \cdot 40\,{\rm dB} = -30.9\,{\rm dB}\hspace{0.05cm}.$$

	From this it follows:		From this it follows:
Line 81:		Line 81:

	'''(3)'''  The sink SNR should be at least  $16.1 \ \rm dB$,  which means that it must be valid:		'''(3)'''  The sink SNR should be at least  $16.1 \ \rm dB$,  which means that it must be valid:
	:$$10 \cdot {\rm lg}\hspace{0.1cm}\rho = 10 \cdot {\rm lg}\hspace{0.1cm}\frac{s_0^2 }{N_0 \cdot R_{\rm B}} + 10 \cdot {\~~rmlg~~}\hspace{0.1cm}\eta \hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\eta \ >\ 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm min} - 10 \cdot {\~~rmlg~~}\hspace{0.1cm}\frac{s_0^2 }{N_0 \cdot R_{\rm B}} =\ 16.1- 100\hspace{0.15cm}\underline {= -83.9\,{\rm dB} = 10 \cdot {\~~rmlg~~}\hspace{0.1cm}\eta_{\hspace{0.05cm} \rm min}}\hspace{0.05cm}.$$		:$$10 \cdot {\rm lg}\hspace{0.1cm}\rho = 10 \cdot {\rm lg}\hspace{0.1cm}\frac{s_0^2 }{N_0 \cdot R_{\rm B}} + 10 \cdot {\rm lg}\hspace{0.1cm}\eta \hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\eta \ >\ 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm min} - 10 \cdot {\rm lg}\hspace{0.1cm}\frac{s_0^2 }{N_0 \cdot R_{\rm B}} =\ 16.1- 100\hspace{0.15cm}\underline {= -83.9\,{\rm dB} = 10 \cdot {\rm lg}\hspace{0.1cm}\eta_{\hspace{0.05cm} \rm min}}\hspace{0.05cm}.$$


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	[[Category:Digital Signal Transmission: Exercises\|^3.5 Linear Nyquist Equalization^]]		[[Category:Digital Signal Transmission: Exercises\|^3.5 Linear Nyquist Equalization^]]
	[[de:3.~~7Z_Regeneratorfeldlänge~~]]		[[de:Aufgaben:Aufgabe 3.7Z: Regeneratorfeldlänge]]

← Older revision		Revision as of 16:03, 16 March 2026
Line 17:		Line 17:
	For the calculation of the  '''regenerator field length'''  (distance between two repeaters),  it should be noted:		For the calculation of the  '''regenerator field length'''  (distance between two repeaters),  it should be noted:
	* The worst-case error probability should not be larger than  $10^{-10}$, which results in the minimum sink SNR:		* The worst-case error probability should not be larger than  $10^{-10}$, which results in the minimum sink SNR:
	:$$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm min} \approx 16.1\,{\~~rmdB~~} \hspace{0.05cm},$$		:$$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm min} \approx 16.1\,{\rm dB} \hspace{0.05cm},$$
	* The logarithmized ratio of transmit energy  (per bit)  and AWGN noise power density is about  $100 \ \rm dB$,  for example:		* The logarithmized ratio of transmit energy  (per bit)  and AWGN noise power density is about  $100 \ \rm dB$,  for example:
	:$$s_0 = 3\,{\rm V},\hspace{0.2cm}R_{\rm B} = 1\,{\rmGbit/s},\hspace{0.2cm}N_{\rm 0} = 9 \cdot 10^{-19}\,{\rm V^2/Hz}$$		:$$s_0 = 3\,{\rm V},\hspace{0.2cm}R_{\rm B} = 1\,{\rmGbit/s},\hspace{0.2cm}N_{\rm 0} = 9 \cdot 10^{-19}\,{\rm V^2/Hz}$$
	:$$\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\frac{s_0^2 }{N_0 \cdot R_{\rm B}}= 10 \cdot {\rmlg} \hspace{0.1cm} \frac{9\,{\rm V^2} } {9 \cdot 10^{-19}\,{\rm V^2/Hz}\cdot 10^{-9}\,{\rm 1/s}}= 100\,{\~~rmdB~~} \hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\frac{s_0^2 }{N_0 \cdot R_{\rm B}}= 10 \cdot {\rmlg} \hspace{0.1cm} \frac{9\,{\rm V^2} } {9 \cdot 10^{-19}\,{\rm V^2/Hz}\cdot 10^{-9}\,{\rm 1/s}}= 100\,{\rm dB} \hspace{0.05cm}.$$
	* A standard coaxial cable with dimensions  $2.6 \ \rm mm$  (inside)  and  $9.5 \ \rm mm$  (outside)  is to be used,  for which the following relationship is valid:		* A standard coaxial cable with dimensions  $2.6 \ \rm mm$  (inside)  and  $9.5 \ \rm mm$  (outside)  is to be used,  for which the following relationship is valid:
	:$$a_{\star} = \frac{2.36\,{\rm dB} } {{\rm km} \cdot \sqrt{{\~~rmMHz~~}}} \cdot l \cdot \sqrt{{R_{\rm B}}/{2}}\hspace{0.05cm}.$$		:$$a_{\star} = \frac{2.36\,{\rm dB} } {{\rm km} \cdot \sqrt{{\rm MHz}}} \cdot l \cdot \sqrt{{R_{\rm B}}/{2}}\hspace{0.05cm}.$$

	:Here,  $a_*$  denotes the characteristic attenuation at half the bit rate – in the example at  $500 \ \rm MHz$  – and  $l$  denotes the cable length.		:Here,  $a_*$  denotes the characteristic attenuation at half the bit rate – in the example at  $500 \ \rm MHz$  – and  $l$  denotes the cable length.
Line 62:		Line 62:
	:$$({\rm GLP},\hspace{0.1cm}M=8) \text{:}\hspace{0.3cm}10 \cdot {\rmlg}\hspace{0.1cm}\eta= -1.3\,{\rm dB} -0.91 \cdot 40\,{\rm dB} = -37.7\,{\rm dB}\hspace{0.05cm},$$		:$$({\rm GLP},\hspace{0.1cm}M=8) \text{:}\hspace{0.3cm}10 \cdot {\rmlg}\hspace{0.1cm}\eta= -1.3\,{\rm dB} -0.91 \cdot 40\,{\rm dB} = -37.7\,{\rm dB}\hspace{0.05cm},$$
	:$$({\rm ONE},\hspace{0.1cm}M=2) \text{:}\hspace{0.3cm}10 \cdot {\rmlg}\hspace{0.1cm}\eta= +4.5\,{\rm dB} -0.96 \cdot 40 \,{\rm dB}= -33.9\,{\rm dB}\hspace{0.05cm},$$		:$$({\rm ONE},\hspace{0.1cm}M=2) \text{:}\hspace{0.3cm}10 \cdot {\rmlg}\hspace{0.1cm}\eta= +4.5\,{\rm dB} -0.96 \cdot 40 \,{\rm dB}= -33.9\,{\rm dB}\hspace{0.05cm},$$
	:$$({\rm ONE},\hspace{0.1cm}M=8) \text{:}\hspace{0.3cm} 10 \cdot {\rmlg}\hspace{0.1cm}\eta = -9.3\,{\rm dB} -0.54 \cdot 40\,{\rm dB} = -30.9\,{\~~rmdB~~}\hspace{0.05cm}.$$		:$$({\rm ONE},\hspace{0.1cm}M=8) \text{:}\hspace{0.3cm} 10 \cdot {\rmlg}\hspace{0.1cm}\eta = -9.3\,{\rm dB} -0.54 \cdot 40\,{\rm dB} = -30.9\,{\rm dB}\hspace{0.05cm}.$$

	From this it follows:		From this it follows:
Line 97:		Line 97:

	*With the given equation		*With the given equation
	:$$a_{\star} = \frac{2.36\,{\rm dB} } {{\rm km} \cdot \sqrt{{\~~rmMHz~~}}} \cdot l \cdot \sqrt{{R_{\rm B}}/{2}}\hspace{0.05cm}.$$		:$$a_{\star} = \frac{2.36\,{\rm dB} } {{\rm km} \cdot \sqrt{{\rm MHz}}} \cdot l \cdot \sqrt{{R_{\rm B}}/{2}}\hspace{0.05cm}.$$

	:thus the maximum cable length  ("regenerator field length")  can be specified:		:thus the maximum cable length  ("regenerator field length")  can be specified:
	:$$l_{\rm max} = \frac{138.1\,{\rm dB} } {2.36\,{\rm dB}/{\~~rmkm~~} \cdot \sqrt{\rm MHz})\cdot \sqrt{500\,{\rm MHz}}} \hspace{0.15cm}\underline {\approx 2.62\,{\rm km}} \hspace{0.05cm}.$$		:$$l_{\rm max} = \frac{138.1\,{\rm dB} } {2.36\,{\rm dB}/{\rm km} \cdot \sqrt{\rm MHz})\cdot \sqrt{500\,{\rm MHz}}} \hspace{0.15cm}\underline {\approx 2.62\,{\rm km}} \hspace{0.05cm}.$$


	'''(5)'''  Following the same procedure,  but in a more compact notation,  results in a smaller  "regenerator field length"  for this  "worse"  system:		'''(5)'''  Following the same procedure,  but in a more compact notation,  results in a smaller  "regenerator field length"  for this  "worse"  system:
	:$$l_{\rm max} = \frac{-(83.9\,{\rm dB}+A)/B } {2.36\,{\rm dB}/{\~~rmkm~~} \cdot \sqrt{500}} = \frac{+(83.9+9.4)/1.10 } {2.36\cdot\sqrt{500}}\hspace{0.1cm}{\rm km}\hspace{0.15cm}\underline {\approx 1.61\, {\rm km}}\hspace{0.05cm}.$$		:$$l_{\rm max} = \frac{-(83.9\,{\rm dB}+A)/B } {2.36\,{\rm dB}/{\rm km} \cdot \sqrt{500}} = \frac{+(83.9+9.4)/1.10 } {2.36\cdot\sqrt{500}}\hspace{0.1cm}{\rm km}\hspace{0.15cm}\underline {\approx 1.61\, {\rm km}}\hspace{0.05cm}.$$
	{{ML-Fuß}}		{{ML-Fuß}}