Digital Signal Transmission/Causes and Effects of Intersymbol Interference - Revision history

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	where the following parameters apply for these dimensions – referred to as  "normal coaxial cable":		where the following parameters apply for these dimensions – referred to as  "normal coaxial cable":
	:$$\alpha_0 = 0.00162 \hspace{0.15cm}\frac {\rm Np}{\rm km} \hspace{0.05cm},\hspace{0.2cm}		:$$\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},\hspace{0.2cm}\beta_1 = 21.78 \hspace{0.15cm}\frac {\rm rad}{{\rm km} \cdot {\rm MHz}} \hspace{0.05cm},\hspace{0.2cm}\beta_2 = 0.2722 \hspace{0.15cm}\frac {\rm rad}{{\rm km} \cdot \sqrt{\rm MHz}}\hspace{0.05cm}.$$
	\alpha_1 = 0.000435 \hspace{0.15cm}\frac {\rm Np}{{\rm km} \cdot {\rm MHz}} \hspace{0.05cm},
	\hspace{0.2cm}
	\alpha_2 = 0.2722 \hspace{0.15cm}\frac {\rm Np}{{\rm km} \cdot \sqrt{\rm MHz}} \hspace{0.05cm},\hspace{0.2cm}
	\beta_1 = 21.78 \hspace{0.15cm}\frac {\rm rad}{{\rm km} \cdot {\rm MHz}} \hspace{0.05cm},
	\hspace{0.2cm}
	\beta_2 = 0.2722 \hspace{0.15cm}\frac {\rm rad}{{\rm km} \cdot \sqrt{\rm MHz}}
	\hspace{0.05cm}.$$

	In the above equation the attenuation parameters  ("alpha")  are to be inserted in  "Np"  and the phase parameters  ("beta")  in  "rad".<br>		In the above equation the attenuation parameters  ("alpha")  are to be inserted in  "Np"  and the phase parameters  ("beta")  in  "rad".<br>

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	If we denote the samples of  $g_d(t)$  at multiples of the symbol duration  $T$  (spacing of the pulses)  as the  "basic detection pulse values",  it has been tacitly assumed so far:		If we denote the samples of  $g_d(t)$  at multiples of the symbol duration  $T$  (spacing of the pulses)  as the  "basic detection pulse values",  it has been tacitly assumed so far:
	[[File:EN_Dig_T_3_1_S1.png\|right\|frame\|Detection signals with and without intersymbol interferences\|class=fit]]		[[File:EN_Dig_T_3_1_S1.png\|right\|frame\|Detection signals with and without intersymbol interferences\|class=fit]]
	:$$g_\nu = g_d(\nu T) = \left\{ \begin{array}{c} g_0		:$$g_\nu = g_d(\nu T) = \left\{ \begin{array}{c} g_0\\ 0 \\ \end{array} \right.\quad\begin{array}{{1}c} {\rm{for}}\\ {\rm{for}} \\ \end{array}\begin{array}{{20}c}\nu = 0, \\ \nu \ne 0. \\\end{array}$$
	\\ 0 \\ \end{array} \right.\quad
	\begin{array}{*{1}c} {\rm{for}}\\ {\rm{for}} \\ \end{array}
	\begin{array}{*{20}c}\nu = 0, \\ \nu \ne 0. \\
	\end{array}$$

	As a consequence of this assumption it has resulted that in the binary case the signal component  (index "S")		As a consequence of this assumption it has resulted that in the binary case the signal component  (index "S")
	:$$d_{\rm S}(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g_d ( t - \nu \cdot T)		:$$d_{\rm S}(t) = \sum_{\nu = -\infty}^{+\infty} a_\nu \cdot g_d ( t - \nu \cdot T)\hspace{0.3cm}{\rm with}\hspace{0.3cm}a_\nu \in \{ -1, +1\}$$
	\hspace{0.3cm}{\rm with}\hspace{0.3cm}a_\nu \in \{ -1, +1\}$$

	of the detection signal at the instants  $\nu \cdot T$  can take only two different values,  namely  $\pm g_0$.<br>		of the detection signal at the instants  $\nu \cdot T$  can take only two different values,  namely  $\pm g_0$.<br>
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	The following applies to the  "frequency response"  and the   "impulse response"  of the channel  $(\rm exp[\hspace{0.05cm} . ]$  denotes the  "exponential function"$)$:		The following applies to the  "frequency response"  and the   "impulse response"  of the channel  $(\rm exp[\hspace{0.05cm} . ]$  denotes the  "exponential function"$)$:
	:$$H_{\rm K}(f) = {\rm exp} \left[ - a_{{\star} \hspace{0.01cm}({\rm Np})} \cdot \sqrt{\frac{f}{R_{\rm B}/2}}\hspace{0.1cm}\right] \cdot		:$$H_{\rm K}(f) = {\rm exp} \left[ - a_{{\star} \hspace{0.01cm}({\rm Np})} \cdot \sqrt{\frac{f}{R_{\rm B}/2}}\hspace{0.1cm}\right] \cdot{\rm exp} \left[ - {\rm j} \cdot a_{{\star} \hspace{0.01cm}({\rm Np})} \cdot \sqrt{\frac{f}{R_{\rm B}/2}}\hspace{0.1cm}\right]\hspace{0.05cm}, $$
	{\rm exp} \left[ - {\rm j} \cdot a_{{\star} \hspace{0.01cm}({\rm Np})} \cdot \sqrt{\frac{f}{R_{\rm B}/2}}\hspace{0.1cm}\right]
	\hspace{0.05cm}, $$
	[[File:EN_Dig_T_3_1_S5.png\|right\|frame\|Block diagram of a system with distorting channel\|class=fit]]		[[File:EN_Dig_T_3_1_S5.png\|right\|frame\|Block diagram of a system with distorting channel\|class=fit]]
	:$$h_{\rm K}(t) = \frac{ a_{{\star}\hspace{0.01cm}({\rm Np})}}{ \sqrt{2 \pi^2 \cdot R_{\rm B} \cdot t^3}}\hspace{0.1cm} \cdot		:$$h_{\rm K}(t) = \frac{ a_{{\star}\hspace{0.01cm}({\rm Np})}}{ \sqrt{2 \pi^2 \cdot R_{\rm B} \cdot t^3}}\hspace{0.1cm} \cdot{\rm exp} \left[ - \frac{a_{{\star} \hspace{0.01cm}({\rm Np})}^2}{2 \pi \cdot R_{\rm B} \cdot t}\hspace{0.1cm}\right]\hspace{0.05cm}.$$
	{\rm exp} \left[ - \frac{a_{{\star} \hspace{0.01cm}({\rm Np})}^2}{2 \pi \cdot R_{\rm B} \cdot t}\hspace{0.1cm}\right]
	\hspace{0.05cm}.$$

	Here  $a_{{\star} \hspace{0.01cm}({\rm Np})}$  indicates the cable attenuation at half the bit rate.  We call this quantity the  '''characteristic cable attenuation'''  in Neper  $\rm (Np)$:		Here  $a_{{\star} \hspace{0.01cm}({\rm Np})}$  indicates the cable attenuation at half the bit rate.  We call this quantity the  '''characteristic cable attenuation'''  in Neper  $\rm (Np)$:
	:$$a_{{\star} \hspace{0.01cm}({\rm Np})} = a_{\rm K}(f = {R_{\rm B}}/{2})= 0.1151 \cdot a_{{\star} \hspace{0.01cm}({\rm dB})}		:$$a_{{\star} \hspace{0.01cm}({\rm Np})} = a_{\rm K}(f = {R_{\rm B}}/{2})= 0.1151 \cdot a_{{\star} \hspace{0.01cm}({\rm dB})}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	#The corresponding dB value   ⇒   $a_{{\star} \hspace{0.01cm}({\rm dB})}$  is larger by a factor of  $1/0.1151 = 8.686$.		#The corresponding dB value   ⇒   $a_{{\star} \hspace{0.01cm}({\rm dB})}$  is larger by a factor of  $1/0.1151 = 8.686$.
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	<br>		<br>
	A  '''coaxial cable'''  with core diameter 2.6 mm,  outer diameter 9.5 mm and length  $l$  has the following frequency response:		A  '''coaxial cable'''  with core diameter 2.6 mm,  outer diameter 9.5 mm and length  $l$  has the following frequency response:
	:$$H_{\rm K}(f) = {\rm e}^{-\left[ a_{\rm K}(f) + {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} b_{\rm K}(f)\right] } = {\rm e}^{- \alpha_0 \hspace{0.05cm} \cdot \hspace{0.05cm} l} \cdot		:$$H_{\rm K}(f) = {\rm e}^{-\left[ a_{\rm K}(f) + {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} b_{\rm K}(f)\right] } = {\rm e}^{- \alpha_0 \hspace{0.05cm} \cdot \hspace{0.05cm} l} \cdot{\rm e}^{- (\alpha_1 + {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_1) \hspace{0.05cm}\cdot f \hspace{0.05cm}\cdot \hspace{0.05cm}l} \cdot{\rm e}^{- (\alpha_2 + {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2) \hspace{0.05cm}\cdot \sqrt{f} \hspace{0.05cm}\cdot \hspace{0.05cm}l}\hspace{0.05cm},$$
	{\rm e}^{- (\alpha_1 + {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_1) \hspace{0.05cm}\cdot f \hspace{0.05cm}\cdot \hspace{0.05cm}l} \cdot
	{\rm e}^{- (\alpha_2 + {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2) \hspace{0.05cm}\cdot \sqrt{f} \hspace{0.05cm}\cdot \hspace{0.05cm}l}
	\hspace{0.05cm},$$

	where the following parameters apply for these dimensions – referred to as  "normal coaxial cable":		where the following parameters apply for these dimensions – referred to as  "normal coaxial cable":
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	[[File:EN_Dig_T_3_1_S3a.png\|right\|frame\|Attenuation curve of a coaxial cable and approximation  (skin effect only) \|class=fit]]<br>		[[File:EN_Dig_T_3_1_S3a.png\|right\|frame\|Attenuation curve of a coaxial cable and approximation  (skin effect only) \|class=fit]]<br>
	The graph shows the exact attenuation curve for a standard coaxial cable of one kilometer length and an approximation for frequencies up to  $f = 1000\ \rm MHz$:		The graph shows the exact attenuation curve for a standard coaxial cable of one kilometer length and an approximation for frequencies up to  $f = 1000\ \rm MHz$:
	:$$a_{\rm K}(f) = \alpha_0 \cdot l \hspace{0.05cm} + \hspace{0.05cm} \alpha_1 \cdot f \cdot l		:$$a_{\rm K}(f) = \alpha_0 \cdot l \hspace{0.05cm} + \hspace{0.05cm} \alpha_1 \cdot f \cdot l+ \hspace{0.05cm} \alpha_2 \cdot \sqrt{f} \cdot l \hspace{0.05cm},$$
	+ \hspace{0.05cm} \alpha_2 \cdot \sqrt{f} \cdot l \hspace{0.05cm},$$
	:$$a_{\rm K}(f) \approx \alpha_2 \cdot \sqrt{f} \cdot l.$$		:$$a_{\rm K}(f) \approx \alpha_2 \cdot \sqrt{f} \cdot l.$$

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	*With these simplifications,  the frequency response is thus determined by the skin effect alone.  Since the numerical values for  $\alpha_2$  $($in Np$)$  and  $\beta_2$  $($in rad$)$  are the same,  it follows:		*With these simplifications,  the frequency response is thus determined by the skin effect alone.  Since the numerical values for  $\alpha_2$  $($in Np$)$  and  $\beta_2$  $($in rad$)$  are the same,  it follows:
	:$$H_{\rm K}(f) ={\rm e}^{- (\alpha_2 + {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2) \hspace{0.05cm}\cdot \sqrt{f} \hspace{0.05cm}\cdot \hspace{0.05cm}l}		:$$H_{\rm K}(f) ={\rm e}^{- (\alpha_2 + {\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \beta_2) \hspace{0.05cm}\cdot \sqrt{f} \hspace{0.05cm}\cdot \hspace{0.05cm}l}\cdot {\rm e}^{- \alpha_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l \hspace{0.05cm}\cdot \sqrt{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}2f}} \hspace{0.05cm}.$$
	\cdot {\rm e}^{- \alpha_2 \hspace{0.05cm}\cdot \hspace{0.05cm}l \hspace{0.05cm}\cdot \sqrt{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}
	2f}} \hspace{0.05cm}.$$
	*Often in the literature – and also in this tutorial – the attenuation measure at half the bit rate is used,  which we call the  "characteristic cable attenuation"  $($in Neper$)$:		*Often in the literature – and also in this tutorial – the attenuation measure at half the bit rate is used,  which we call the  "characteristic cable attenuation"  $($in Neper$)$:
	:$$a_{\star} = a_{\rm K}(f ={R_{\rm B}}/{2})= a_{\rm K}(f = \frac{1}{2 \cdot T})\approx \frac{\alpha_2 \~~cdot~~		:$$a_{\star} = a_{\rm K}(f ={R_{\rm B}}/{2})= a_{\rm K}(f = \frac{1}{2 \cdot T})\approx \frac{\alpha_2 \cdotl }{ \sqrt {2\cdot T}}\hspace{0.05cm}.$$
	l }{ \sqrt {2\cdot T}}
	\hspace{0.05cm}.$$

	{{GraueBox\|TEXT=		{{GraueBox\|TEXT=
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	<br>		<br>
	We now consider the coaxial cable impulse response,  which is for a binary system  $(R_{\rm B} = 1/T)$  as follows:		We now consider the coaxial cable impulse response,  which is for a binary system  $(R_{\rm B} = 1/T)$  as follows:
	:$$h_{\rm K}(t) = \frac{ a_{\rm \star \hspace{0.01cm}(Np)}/T}{ \sqrt{2 \pi^2 \cdot (t/T)^3}}\hspace{0.1cm} \cdot		:$$h_{\rm K}(t) = \frac{ a_{\rm \star \hspace{0.01cm}(Np)}/T}{ \sqrt{2 \pi^2 \cdot (t/T)^3}}\hspace{0.1cm} \cdot{\rm exp} \left[ - \frac{a_{\rm \star \hspace{0.01cm}(Np)}^2}{2 \pi \cdot t/T}\hspace{0.1cm}\right]\hspace{0.05cm}.$$
	{\rm exp} \left[ - \frac{a_{\rm \star \hspace{0.01cm}(Np)}^2}{2 \pi \cdot t/T}\hspace{0.1cm}\right]
	\hspace{0.05cm}.$$

	This time course is shown here for characteristic cable attenuations  $(a_{\rm \star})$  between  $40 \ \rm dB$  and  $100 \ \rm dB$.  Note the conversion   $\rm 1 \ Np = 8.686 \ dB.$<br>		This time course is shown here for characteristic cable attenuations  $(a_{\rm \star})$  between  $40 \ \rm dB$  and  $100 \ \rm dB$.  Note the conversion   $\rm 1 \ Np = 8.686 \ dB.$<br>

Maintenance script: Add German interlanguage link

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	{{Display}}		{{Display}}
			[[de:Digitalsignalübertragung/Ursachen_und_Auswirkungen_von_Impulsinterferenzen]]

Hwang at 14:12, 23 January 2023

2023-01-23T14:12:02Z

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	#  the  »optimal Nyquist equalizer as an example of linear channel equalization,		#  the  »optimal Nyquist equalizer as an example of linear channel equalization,
	#  the  »decision feedback equalization«  $\rm (DFE)$ – an effective nonlinear decision realization,		#  the  »decision feedback equalization«  $\rm (DFE)$ – an effective nonlinear decision realization,
	#  the  »correlation receiver«  as an example of  »maximum likelihood or maximum a posteriori«  $\rm (MAP)$  decision strategy,		#  the  »correlation receiver«  as an example of  »maximum likelihood or maximum a-posteriori«  $\rm (MAP)$  decision strategy,
	#  the  »Viterbi receiver«,  a reduced-effort MAP decision algorithm.		#  the  »Viterbi receiver«,  a reduced-effort MAP decision algorithm.

Hwang at 14:11, 23 January 2023

2023-01-23T14:11:26Z

← Older revision		Revision as of 16:11, 23 January 2023
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	In the main chapter 4:   "Properties of electrical lines" of the book  [[Linear_and_Time_Invariant_Systems\|"Linear and Time Invariant Systems"]]  it is shown that these equations reproduce the conditions with good approximation for wireline communication systems via coaxial cable.  For a two-wire line,  the deviation between this very simple  (analytically manageable)  formula and the actual conditions is somewhat larger.<br>		In the main chapter 4:   "Properties of electrical lines" of the book  [[Linear_and_Time_Invariant_Systems\|"Linear and Time Invariant Systems"]]  it is shown that these equations reproduce the conditions with good approximation for wireline communication systems via coaxial cable.  For a two-wire line,  the deviation between this very simple  (analytically manageable)  formula and the actual conditions is somewhat larger.<br>

	A short summary of these derivations follows on the next two ~~pages~~.		A short summary of these derivations follows in the next two sections.
	*For simplicity,  we will use a redundancy-free binary system.		*For simplicity,  we will use a redundancy-free binary system.
	*Thus,  the bit rate  $R_{\rm B}$  is equal to the reciprocal of the symbol duration  $T$.<br>		*Thus,  the bit rate  $R_{\rm B}$  is equal to the reciprocal of the symbol duration  $T$.<br>

Guenter at 12:50, 31 August 2022

2022-08-31T12:50:45Z

← Older revision		Revision as of 14:50, 31 August 2022
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	In detail,  this chapter deals with:		In detail,  this chapter deals with:
	*the  »causes and effects«  of intersymbol interference,		#  the  »causes and effects«  of intersymbol interference,
	*the  »eye diagram«  as a suitable tool for the description of intersymbol interferences,		#  the  »eye diagram«  as a suitable tool for the description of intersymbol interferences,
	*the  »error probability calculation«  considering channel distortions,		#  the  »error probability calculation«  considering channel distortions,
	*the  »influence of intersymbol interference in multilevel and/or coded transmission«,		#  the  »influence of intersymbol interference in multilevel and/or coded transmission«,
	*the  »optimal Nyquist equalizer as an example of linear channel equalization,		#  the  »optimal Nyquist equalizer as an example of linear channel equalization,
	*the  »decision feedback equalization«  $\rm (DFE)$ – an effective nonlinear decision realization,		#  the  »decision feedback equalization«  $\rm (DFE)$ – an effective nonlinear decision realization,
	*the  »correlation receiver«  as an example of  »maximum likelihood or maximum a posteriori«  $\rm (MAP)$  decision strategy,		#  the  »correlation receiver«  as an example of  »maximum likelihood or maximum a posteriori«  $\rm (MAP)$  decision strategy,
	*the  »Viterbi receiver«,  a reduced-effort MAP decision algorithm.		#  the  »Viterbi receiver«,  a reduced-effort MAP decision algorithm.

Guenter at 11:40, 8 June 2022

2022-06-08T11:40:43Z

← Older revision		Revision as of 13:40, 8 June 2022
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	This time course is shown here for characteristic cable attenuations  $(a_{\rm \star})$  between  $40 \ \rm dB$  and  $100 \ \rm dB$.  Note the conversion   $\rm 1 \ Np = 8.686 \ dB.$<br>		This time course is shown here for characteristic cable attenuations  $(a_{\rm \star})$  between  $40 \ \rm dB$  and  $100 \ \rm dB$.  Note the conversion   $\rm 1 \ Np = 8.686 \ dB.$<br>
	[[File:EN_Dig_T_3_1_S3b.png\|right\|frame\|Impulse and rectangular response of the coaxial cable  (skin effect only);   <u>Note:</u> <br>Here,  the basic receiver pulse  $g_r(t)$  is identical ~~withe~~ the  "rectangular response"]]		[[File:EN_Dig_T_3_1_S3b.png\|right\|frame\|Impulse and rectangular pulse response of the coaxial cable  (skin effect only);   <u>Note:</u> <br>Here,  the basic receiver pulse  $g_r(t)$  is identical with the  "rectangular pulse response"]]

	One can see from this time domain plot:		One can see from this time domain plot:

Guenter at 11:20, 8 June 2022

2022-06-08T11:20:52Z

← Older revision		Revision as of 13:20, 8 June 2022
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	*the  »optimal Nyquist equalizer as an example of linear channel equalization,		*the  »optimal Nyquist equalizer as an example of linear channel equalization,
	*the  »decision feedback equalization«  $\rm (DFE)$ – an effective nonlinear decision realization,		*the  »decision feedback equalization«  $\rm (DFE)$ – an effective nonlinear decision realization,
	*the  »correlation receiver«  as an example of  »maximum likelihood or maximum a posteriori«  $\rm (MAP)$  decision ~~making~~,		*the  »correlation receiver«  as an example of  »maximum likelihood or maximum a posteriori«  $\rm (MAP)$  decision strategy,
	*the  »Viterbi receiver«,  a reduced-effort MAP decision algorithm.		*the  »Viterbi receiver«,  a reduced-effort MAP decision algorithm.

Guenter at 11:19, 8 June 2022

2022-06-08T11:19:06Z

← Older revision		Revision as of 13:19, 8 June 2022
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	This time course is shown here for characteristic cable attenuations  $(a_{\rm \star})$  between  $40 \ \rm dB$  and  $100 \ \rm dB$.  Note the conversion   $\rm 1 \ Np = 8.686 \ dB.$<br>		This time course is shown here for characteristic cable attenuations  $(a_{\rm \star})$  between  $40 \ \rm dB$  and  $100 \ \rm dB$.  Note the conversion   $\rm 1 \ Np = 8.686 \ dB.$<br>
	[[File:EN_Dig_T_3_1_S3b.png\|right\|frame\|Impulse response of the coaxial cable  (skin effect only) ~~'''KORREKTUR~~: ~~Rechteckantwort =~~ rectangular response~~?'''~~]]		[[File:EN_Dig_T_3_1_S3b.png\|right\|frame\|Impulse and rectangular response of the coaxial cable  (skin effect only);   <u>Note:</u> <br>Here,  the basic receiver pulse  $g_r(t)$  is identical withe the  "rectangular response"]]

	One can see from this time domain plot:		One can see from this time domain plot:

Hwang at 07:35, 6 June 2022

2022-06-06T07:35:12Z

← Older revision		Revision as of 09:35, 6 June 2022
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	This time course is shown here for characteristic cable attenuations  $(a_{\rm \star})$  between  $40 \ \rm dB$  and  $100 \ \rm dB$.  Note the conversion   $\rm 1 \ Np = 8.686 \ dB.$<br>		This time course is shown here for characteristic cable attenuations  $(a_{\rm \star})$  between  $40 \ \rm dB$  and  $100 \ \rm dB$.  Note the conversion   $\rm 1 \ Np = 8.686 \ dB.$<br>
	[[File:EN_Dig_T_3_1_S3b.png\|right\|frame\|Impulse response of the coaxial cable  (skin effect only) '''KORREKTUR: Rechteckantwort = ~~rechtangular~~ response?''']]		[[File:EN_Dig_T_3_1_S3b.png\|right\|frame\|Impulse response of the coaxial cable  (skin effect only) '''KORREKTUR: Rechteckantwort = rectangular response?''']]

	One can see from this time domain plot:		One can see from this time domain plot: