Digital Signal Transmission/Block Coding with 4B3T Codes - Revision history

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2026-03-16T13:46:46Z

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← Older revision		Revision as of 15:46, 16 March 2026
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	<br>		<br>
	In  '''block coding'''  each sequence of  $m_q$  binary source symbols  $(M_q = 2)$  is represented by a block of  $m_c$  encoder symbols with symbol set size  $M_c$.  In order to convert each source symbol sequence  $\langle q_\nu \rangle$  into another encoder symbol sequence  $\langle c_\nu \rangle$,  the following condition must be satisfied:		In  '''block coding'''  each sequence of  $m_q$  binary source symbols  $(M_q = 2)$  is represented by a block of  $m_c$  encoder symbols with symbol set size  $M_c$.  In order to convert each source symbol sequence  $\langle q_\nu \rangle$  into another encoder symbol sequence  $\langle c_\nu \rangle$,  the following condition must be satisfied:
	:$$M_c^{\hspace{0.1cm}m_c} \ge		:$$M_c^{\hspace{0.1cm}m_c} \geM_q^{\hspace{0.1cm}m_q}\hspace{0.05cm}.$$
	~~M_q~~^{\hspace{0.1cm}m_q}\hspace{0.05cm}.$$

	*For the  [[Digital_Signal_Transmission/Redundancy-Free_Coding\|"redundancy-free codes"]]  discussed in the last chapter,  the equal sign applies in this equation if  $M_q$  is a power of two.		*For the  [[Digital_Signal_Transmission/Redundancy-Free_Coding\|"redundancy-free codes"]]  discussed in the last chapter,  the equal sign applies in this equation if  $M_q$  is a power of two.
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	The best known block code for transmission coding is the   '''4B3T code'''   with the code parameters		The best known block code for transmission coding is the   '''4B3T code'''   with the code parameters
	:$$m_q = 4,\hspace{0.2cm}M_q = 2,\hspace{0.2cm}m_c =		:$$m_q = 4,\hspace{0.2cm}M_q = 2,\hspace{0.2cm}m_c =3,\hspace{0.2cm}M_c = 3\hspace{0.05cm},$$
	3,\hspace{0.2cm}M_c = 3\hspace{0.05cm},$$

	which was developed in the 1970s and is used,  for example,  in  [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_ISDN\|"ISDN"]]  ("Integrated Services Digital Networks").		which was developed in the 1970s and is used,  for example,  in  [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_ISDN\|"ISDN"]]  ("Integrated Services Digital Networks").
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	After the transmission of  $l$  coded blocks,  the  "running digital sum"  with ternary amplitude coefficients  $a_\nu \in \{ -1, \ 0, +1\}$:		After the transmission of  $l$  coded blocks,  the  "running digital sum"  with ternary amplitude coefficients  $a_\nu \in \{ -1, \ 0, +1\}$:
	[[File:EN_Dig_T_2_3_S2.png\|right\|frame\|Code tables for three 4B3T codes\|class=fit]]		[[File:EN_Dig_T_2_3_S2.png\|right\|frame\|Code tables for three 4B3T codes\|class=fit]]
	:$${\it \Sigma}_l = \sum_{\nu = 1}^{3 \hspace{0.02cm}\cdot		:$${\it \Sigma}_l = \sum_{\nu = 1}^{3 \hspace{0.02cm}\cdot\hspace{0.05cm} l}\hspace{0.02cm} a_\nu \hspace{0.05cm}.$$
	\hspace{0.05cm} l}\hspace{0.02cm} a_\nu \hspace{0.05cm}.$$

	The selection of the table for encoding the  $(l + 1)$–th block is done depending on the current   ${\it \Sigma}_l$  value.		The selection of the table for encoding the  $(l + 1)$–th block is done depending on the current   ${\it \Sigma}_l$  value.
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	For the eye diagram and the following calculations, a signal-to-noise ratio of  $10 \cdot \lg \hspace{0.05cm}(s_0^2 \cdot T/N_0) = 13 \ \rm dB$  is assumed in each case. This gives:		For the eye diagram and the following calculations, a signal-to-noise ratio of  $10 \cdot \lg \hspace{0.05cm}(s_0^2 \cdot T/N_0) = 13 \ \rm dB$  is assumed in each case. This gives:
	:$$10 \cdot {\rm lg } \hspace{0.1cm}{s_0^2 \cdot T}/{N_0} = 13 \, {\rm dB } \hspace{0.3cm}		:$$10 \cdot {\rm lg } \hspace{0.1cm}{s_0^2 \cdot T}/{N_0} = 13 \, {\rm dB } \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{s_0^2 \cdot T}/{N_0} = 10^{1.3} \approx 20 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\sigma_1^2 = 0.56 \cdot {s_0^2}/{20} \approx 0.028 \cdot s_0^2 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{ \sigma_1}/{s_0}\approx 0.167 \hspace{0.05cm}.$$
	\Rightarrow \hspace{0.3cm}{s_0^2 \cdot T}/{N_0} = 10^{1.3} \approx 20 \hspace{0.3cm}
	\Rightarrow \hspace{0.3cm}\sigma_1^2 = 0.56 \cdot {s_0^2}/{20} \approx 0.028 \cdot s_0^2 \hspace{0.3cm}
	\Rightarrow \hspace{0.3cm}{ \sigma_1}/{s_0}\approx 0.167 \hspace{0.05cm}.$$

	Row  $\rm B$  shows the associated symbol error probability		Row  $\rm B$  shows the associated symbol error probability
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	The two other eye diagrams can be interpreted as follows:		The two other eye diagrams can be interpreted as follows:
	*For the redundancy-free ternary system,  the eye opening is only half as large as in the binary case,  and the noise power  $\sigma_2^2$  is smaller than  $\sigma_1^2$ by a factor of  $\log_2 \hspace{0.05cm}(3)$.  The factor  $4/3$  in front of the Q–function takes into account that the ternary  "0"  can be falsified in both directions.  This results in the following numerical values:		*For the redundancy-free ternary system,  the eye opening is only half as large as in the binary case,  and the noise power  $\sigma_2^2$  is smaller than  $\sigma_1^2$ by a factor of  $\log_2 \hspace{0.05cm}(3)$.  The factor  $4/3$  in front of the Q–function takes into account that the ternary  "0"  can be falsified in both directions.  This results in the following numerical values:
	:$$\frac{ \sigma_2}{s_0}\hspace{-0.05cm} =\hspace{-0.05cm} \frac{ \sigma_1/s_0}{\sqrt{{\rm log_2} (3)}}\hspace{-0.05cm} =\hspace{-0.05cm}\frac{ 0.167}{1.259}		:$$\frac{ \sigma_2}{s_0}\hspace{-0.05cm} =\hspace{-0.05cm} \frac{ \sigma_1/s_0}{\sqrt{{\rm log_2} (3)}}\hspace{-0.05cm} =\hspace{-0.05cm}\frac{ 0.167}{1.259}\approx 0.133 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm S}\hspace{-0.05cm}=\hspace{-0.05cm} {4}/{3} \cdot {\rm Q}\left (\frac{ 0.5}{0.133}\right)\approx{4}/{3} \cdot {\rm Q}(3.76) \hspace{-0.05cm}= \hspace{-0.05cm}1.1 \cdot 10^{-4} .$$
	\approx 0.133 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm S}\hspace{-0.05cm}=\hspace{-0.05cm} {4}/{3} \cdot {\rm Q}\left (\frac{ 0.5}{0.133}\right)\approx
	{4}/{3} \cdot {\rm Q}(3.76) \hspace{-0.05cm}= \hspace{-0.05cm}1.1 \cdot 10^{-4} .$$
	*The 4B3T code yields even slightly less favorable results, since here the noise power  $(\sigma_3^2)$  is reduced less than in the redundancy-free ternary code  $(\sigma_2^2)$ for the same eye opening:		*The 4B3T code yields even slightly less favorable results, since here the noise power  $(\sigma_3^2)$  is reduced less than in the redundancy-free ternary code  $(\sigma_2^2)$ for the same eye opening:
	:$$\frac{ \sigma_3}{s_0} = \frac{ \sigma_1/s_0}{\sqrt{4/3}} =\frac{ 0.167}{1.155}		:$$\frac{ \sigma_3}{s_0} = \frac{ \sigma_1/s_0}{\sqrt{4/3}} =\frac{ 0.167}{1.155}\approx 0.145 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm S}= {4}/{3} \cdot {\rm Q}\left (\frac{ 0.5}{0.145}\right)\approx{4}/{3} \cdot {\rm Q}(3.45) = 3.7 \cdot 10^{-4} \hspace{0.05cm}.$$
	\approx 0.145 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm S}= {4}/{3} \cdot {\rm Q}\left (\frac{ 0.5}{0.145}\right)\approx
	{4}/{3} \cdot {\rm Q}(3.45) = 3.7 \cdot 10^{-4} \hspace{0.05cm}.$$

Maintenance script: Add German interlanguage link

2026-03-16T12:26:41Z

Add German interlanguage link

← Older revision		Revision as of 14:26, 16 March 2026
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	{{Display}}		{{Display}}
			[[de:Digitalsignalübertragung/Blockweise_Codierung_mit_4B3T-Codes]]

Guenter at 15:19, 24 August 2022

2022-08-24T15:19:23Z

← Older revision		Revision as of 17:19, 24 August 2022
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	The statements of this graph can be summarized as follows:		The statements of this graph can be summarized as follows:

	[[File:~~P_ID1336__Dig_T_2_3_S3b_v1~~.png\|right\|frame\|Power-spectral density (of amplitude coefficients) of 4B3T compared to redundancy-free and AMI coding ~~'''Korrektur'''~~\|class=fit]]		[[File:EN_Dig_T_2_4_S3b_v23.png\|right\|frame\|Power-spectral density (of amplitude coefficients) of 4B3T compared to redundancy-free and AMI coding\|class=fit]]

	*The graph shows the power-spectral density  ${\it \Phi}_a(f)$  of the amplitude coefficients  $a_\nu$  of the 4B3T code   ⇒   red curve.		*The graph shows the power-spectral density  ${\it \Phi}_a(f)$  of the amplitude coefficients  $a_\nu$  of the 4B3T code   ⇒   red curve.

Guenter at 15:22, 3 June 2022

2022-06-03T15:22:06Z

Hwang at 08:37, 24 May 2022

2022-05-24T08:37:42Z

← Older revision		Revision as of 10:37, 24 May 2022
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	== General description of block codes ==		== General description of block codes ==
	<br>		<br>
	In  '''block coding'''  each sequence of  $m_q$  binary source symbols  $(M_q = 2)$  is represented by a block of  $m_c$  coded symbols with symbol set size  $M_c$.  In order to convert each source symbol sequence  $\langle q_\nu \rangle$  into another encoder symbol sequence  $\langle c_\nu \rangle$,  the following condition must be satisfied:		In  '''block coding'''  each sequence of  $m_q$  binary source symbols  $(M_q = 2)$  is represented by a block of  $m_c$  coded '''KORREKTUR: encoder''' symbols with symbol set size  $M_c$.  In order to convert each source symbol sequence  $\langle q_\nu \rangle$  into another encoder symbol sequence  $\langle c_\nu \rangle$,  the following condition must be satisfied:
	:$$M_c^{\hspace{0.1cm}m_c} \ge		:$$M_c^{\hspace{0.1cm}m_c} \ge
	M_q^{\hspace{0.1cm}m_q}\hspace{0.05cm}.$$		M_q^{\hspace{0.1cm}m_q}\hspace{0.05cm}.$$

Hwang at 08:36, 24 May 2022

2022-05-24T08:36:11Z

← Older revision		Revision as of 10:36, 24 May 2022
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	Such a 4B3T code has the following properties:		Such a 4B3T code has the following properties:
	*Because of  $m_q \cdot T_{\rm B} = m_c \cdot T$,  the symbol duration  $T$  of the ternary encoder signal is larger than the bit duration  $T_{\rm B}$  of the binary source signal by a factor of  $4/3$.  This results in the favorable property that the bandwidth requirement is a quarter less than for redundancy-free binary transmission.		*Because of  $m_q \cdot T_{\rm B} = m_c \cdot T$,  the symbol duration  $T$  of the ternary encoder '''KORREKTUR: encoded''' signal is larger than the bit duration  $T_{\rm B}$  of the binary source signal by a factor of  $4/3$.  This results in the favorable property that the bandwidth requirement is a quarter less than for redundancy-free binary transmission.

	*The relative redundancy can be calculated with the above equation and results in  $r_c \approx 16\%$.  This redundancy is used in the 4B3T code to achieve DC freedom.		*The relative redundancy can be calculated with the above equation and results in  $r_c \approx 16\%$.  This redundancy is used in the 4B3T code to achieve DC freedom.

Hwang at 08:34, 24 May 2022

2022-05-24T08:34:56Z

← Older revision		Revision as of 10:34, 24 May 2022
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	The two other eye diagrams can be interpreted as follows:		The two other eye diagrams can be interpreted as follows:
	*For the redundancy-free ternary system,  the eye opening is only half as large as in the binary case,  and the noise power  $\sigma_2^2$  is smaller than  $\sigma_1^2$ by a factor of  $\log_2 \hspace{0.05cm}(3)$.  The factor  $4/3$  in front of the Q–function takes into account that the ternary  "0"  can be distorted in both directions.  This results in the following numerical values:		*For the redundancy-free ternary system,  the eye opening is only half as large as in the binary case,  and the noise power  $\sigma_2^2$  is smaller than  $\sigma_1^2$ by a factor of  $\log_2 \hspace{0.05cm}(3)$.  The factor  $4/3$  in front of the Q–function takes into account that the ternary  "0"  can be distorted '''KORREKTUR: falsified''' in both directions.  This results in the following numerical values:
	:$$\frac{ \sigma_2}{s_0}\hspace{-0.05cm} =\hspace{-0.05cm} \frac{ \sigma_1/s_0}{\sqrt{{\rm log_2} (3)}}\hspace{-0.05cm} =\hspace{-0.05cm}\frac{ 0.167}{1.259}		:$$\frac{ \sigma_2}{s_0}\hspace{-0.05cm} =\hspace{-0.05cm} \frac{ \sigma_1/s_0}{\sqrt{{\rm log_2} (3)}}\hspace{-0.05cm} =\hspace{-0.05cm}\frac{ 0.167}{1.259}
	\approx 0.133 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm S}\hspace{-0.05cm}=\hspace{-0.05cm} {4}/{3} \cdot {\rm Q}\left (\frac{ 0.5}{0.133}\right)\approx		\approx 0.133 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm S}\hspace{-0.05cm}=\hspace{-0.05cm} {4}/{3} \cdot {\rm Q}\left (\frac{ 0.5}{0.133}\right)\approx

Hwang at 08:31, 24 May 2022

2022-05-24T08:31:09Z

← Older revision		Revision as of 10:31, 24 May 2022
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	*The zero point at  $f = 0$  has the advantage that the 4B3T signal can also be transmitted without major losses via a so-called  "telephone channel",  which is not suitable for a DC signal due to transformers.<br>		*The zero point at  $f = 0$  has the advantage that the 4B3T signal can also be transmitted without major losses via a so-called  "telephone channel",  which is not suitable for a DC signal due to transformers.<br>

	*The zero point at  $f = 1/T$  has the disadvantage that this makes clock recovery at the receiver more difficult.  Outside of these zeros,  the 4B3T codes have a flatter  ${\it \Phi}_a(f)$  than the  [[Digital_Signal_Transmission/~~Symbol~~-~~Wise_Coding_with_Pseudo_Ternary_Codes~~#Properties_of_the_AMI_code\|"AMI code"]]  discussed in the next chapter  (blue curve), which is advantageous.<br>		*The zero point at  $f = 1/T$  has the disadvantage that this makes clock recovery at the receiver more difficult.  Outside of these zeros,  the 4B3T codes have a flatter  ${\it \Phi}_a(f)$  than the  [[Digital_Signal_Transmission/Symbolwise_Coding_with_Pseudo-Ternary_Codes#Properties_of_the_AMI_code\|"AMI code"]]  discussed in the next chapter  (blue curve), which is advantageous.<br>

	*The reason for the flatter PSD curve at medium frequencies as well as the steeper drop towards the zeros is that for the 4B3T codes up to five  $+1$  coefficients  (resp.  $-1$ coefficients)  can follow each other.  With the AMI code,  these symbols occur only in isolation.<br>		*The reason for the flatter PSD curve at medium frequencies as well as the steeper drop towards the zeros is that for the 4B3T codes up to five  $+1$  coefficients  (resp.  $-1$ coefficients)  can follow each other.  With the AMI code,  these symbols occur only in isolation.<br>

Hwang at 08:27, 24 May 2022

2022-05-24T08:27:06Z

Guenter at 14:13, 18 May 2022

2022-05-18T14:13:29Z

← Older revision		Revision as of 16:13, 18 May 2022
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	You can ~~seee~~ from the graph:		You can see from the graph:
	#The two code tables of the Jessop–Waters code are selected in such a way that the running digital sum  ${\it \Sigma}_l$  always lies between  $0$  and  $5$.<br><br>		#The two code tables of the Jessop–Waters code are selected in such a way that the running digital sum  ${\it \Sigma}_l$  always lies between  $0$  and  $5$.<br><br>
	#For the other two codes  (MS43,  FoMoT),  the restriction of the running digital sum to the range  $0 \le {\it \Sigma}_l \le 3$  is achieved by three resp. four alternative tables.		#For the other two codes  (MS43,  FoMoT),  the restriction of the running digital sum to the range  $0 \le {\it \Sigma}_l \le 3$  is achieved by three resp. four alternative tables.
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	The ~~following~~ graph shows the  [[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Definition_and_statements_of_the_eye_diagram\|eye diagrams]]  (with noise) of the three systems to be compared.  It also contains  $($in row  $\rm A)$  the equations for calculating the symbol error probability.  Approximately  2000  eye lines are drawn for each diagram.		The graph shows the  [[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Definition_and_statements_of_the_eye_diagram\|eye diagrams]]  (with noise) of the three systems to be compared.  It also contains  $($in row  $\rm A)$  the equations for calculating the symbol error probability.  Approximately  2000  eye lines are drawn for each diagram.

	The first two rows of the table describe the system comparison at peak limitation.  For the binary system  (first column),  the noise power  (taking into account the  $12\%$ increase)  is given by		The first two rows of the table describe the system comparison at peak limitation.  For the binary system  (first column),  the noise power  (taking into account the  $12\%$ increase)  is given by

← Older revision		Revision as of 17:22, 3 June 2022
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	== General description of block codes ==		== General description of block codes ==
	<br>		<br>
	In  '''block coding'''  each sequence of  $m_q$  binary source symbols  $(M_q = 2)$  is represented by a block of  $m_c$  ~~coded '''KORREKTUR:~~ encoder~~'''~~ symbols with symbol set size  $M_c$.  In order to convert each source symbol sequence  $\langle q_\nu \rangle$  into another encoder symbol sequence  $\langle c_\nu \rangle$,  the following condition must be satisfied:		In  '''block coding'''  each sequence of  $m_q$  binary source symbols  $(M_q = 2)$  is represented by a block of  $m_c$  encoder symbols with symbol set size  $M_c$.  In order to convert each source symbol sequence  $\langle q_\nu \rangle$  into another encoder symbol sequence  $\langle c_\nu \rangle$,  the following condition must be satisfied:
	:$$M_c^{\hspace{0.1cm}m_c} \ge		:$$M_c^{\hspace{0.1cm}m_c} \ge
	M_q^{\hspace{0.1cm}m_q}\hspace{0.05cm}.$$		M_q^{\hspace{0.1cm}m_q}\hspace{0.05cm}.$$
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	Such a 4B3T code has the following properties:		Such a 4B3T code has the following properties:
	*Because of  $m_q \cdot T_{\rm B} = m_c \cdot T$,  the symbol duration  $T$  of the ternary ~~encoder '''KORREKTUR:~~ encoded~~'''~~ signal is larger than the bit duration  $T_{\rm B}$  of the binary source signal by a factor of  $4/3$.  This results in the favorable property that the bandwidth requirement is a quarter less than for redundancy-free binary transmission.		*Because of  $m_q \cdot T_{\rm B} = m_c \cdot T$,  the symbol duration  $T$  of the ternary encoded signal is larger than the bit duration  $T_{\rm B}$  of the binary source signal by a factor of  $4/3$.  This results in the favorable property that the bandwidth requirement is a quarter less than for redundancy-free binary transmission.

	*The relative redundancy can be calculated with the above equation and results in  $r_c \approx 16\%$.  This redundancy is used in the 4B3T code to achieve DC freedom.		*The relative redundancy can be calculated with the above equation and results in  $r_c \approx 16\%$.  This redundancy is used in the 4B3T code to achieve DC freedom.
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	The two other eye diagrams can be interpreted as follows:		The two other eye diagrams can be interpreted as follows:
	*For the redundancy-free ternary system,  the eye opening is only half as large as in the binary case,  and the noise power  $\sigma_2^2$  is smaller than  $\sigma_1^2$ by a factor of  $\log_2 \hspace{0.05cm}(3)$.  The factor  $4/3$  in front of the Q–function takes into account that the ternary  "0"  can be ~~distorted '''KORREKTUR:~~ falsified~~'''~~ in both directions.  This results in the following numerical values:		*For the redundancy-free ternary system,  the eye opening is only half as large as in the binary case,  and the noise power  $\sigma_2^2$  is smaller than  $\sigma_1^2$ by a factor of  $\log_2 \hspace{0.05cm}(3)$.  The factor  $4/3$  in front of the Q–function takes into account that the ternary  "0"  can be falsified in both directions.  This results in the following numerical values:
	:$$\frac{ \sigma_2}{s_0}\hspace{-0.05cm} =\hspace{-0.05cm} \frac{ \sigma_1/s_0}{\sqrt{{\rm log_2} (3)}}\hspace{-0.05cm} =\hspace{-0.05cm}\frac{ 0.167}{1.259}		:$$\frac{ \sigma_2}{s_0}\hspace{-0.05cm} =\hspace{-0.05cm} \frac{ \sigma_1/s_0}{\sqrt{{\rm log_2} (3)}}\hspace{-0.05cm} =\hspace{-0.05cm}\frac{ 0.167}{1.259}
	\approx 0.133 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm S}\hspace{-0.05cm}=\hspace{-0.05cm} {4}/{3} \cdot {\rm Q}\left (\frac{ 0.5}{0.133}\right)\approx		\approx 0.133 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}p_{\rm S}\hspace{-0.05cm}=\hspace{-0.05cm} {4}/{3} \cdot {\rm Q}\left (\frac{ 0.5}{0.133}\right)\approx

← Older revision		Revision as of 10:27, 24 May 2022
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	M_q^{\hspace{0.1cm}m_q}\hspace{0.05cm}.$$		M_q^{\hspace{0.1cm}m_q}\hspace{0.05cm}.$$

	*For the  [[Digital_Signal_Transmission/Redundancy-Free_Coding\|redundancy-free codes]]  discussed in the last chapter,  the equal sign applies in this equation if  $M_q$  is a power of two.		*For the  [[Digital_Signal_Transmission/Redundancy-Free_Coding\|"redundancy-free codes"]]  discussed in the last chapter,  the equal sign applies in this equation if  $M_q$  is a power of two.

	*Using the greater sign results in a redundant digital signal,  and the  "relative encoder redundancy"  can be calculated as follows:		*Using the greater sign results in a redundant digital signal,  and the  "relative encoder redundancy"  can be calculated as follows:
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	3,\hspace{0.2cm}M_c = 3\hspace{0.05cm},$$		3,\hspace{0.2cm}M_c = 3\hspace{0.05cm},$$

	which was developed in the 1970s and is used,  for example,  in  [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_ISDN\| ISDN]]  ("Integrated Services Digital Networks").		which was developed in the 1970s and is used,  for example,  in  [[Examples_of_Communication_Systems/Allgemeine_Beschreibung_von_ISDN\|"ISDN"]]  ("Integrated Services Digital Networks").

	Such a 4B3T code has the following properties:		Such a 4B3T code has the following properties:
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	*The zero point at  $f = 0$  has the advantage that the 4B3T signal can also be transmitted without major losses via a so-called  "telephone channel",  which is not suitable for a DC signal due to transformers.<br>		*The zero point at  $f = 0$  has the advantage that the 4B3T signal can also be transmitted without major losses via a so-called  "telephone channel",  which is not suitable for a DC signal due to transformers.<br>

	*The zero point at  $f = 1/T$  has the disadvantage that this makes clock recovery at the receiver more difficult.  Outside of these zeros,  the 4B3T codes have a flatter  ${\it \Phi}_a(f)$  than the  [[Digital_Signal_Transmission/Symbol-Wise_Coding_with_Pseudo_Ternary_Codes#Properties_of_the_AMI_code\|AMI code]]  discussed in the next chapter  (blue curve), which is advantageous.<br>		*The zero point at  $f = 1/T$  has the disadvantage that this makes clock recovery at the receiver more difficult.  Outside of these zeros,  the 4B3T codes have a flatter  ${\it \Phi}_a(f)$  than the  [[Digital_Signal_Transmission/Symbol-Wise_Coding_with_Pseudo_Ternary_Codes#Properties_of_the_AMI_code\|"AMI code"]]  discussed in the next chapter  (blue curve), which is advantageous.<br>

	*The reason for the flatter PSD curve at medium frequencies as well as the steeper drop towards the zeros is that for the 4B3T codes up to five  $+1$  coefficients  (resp.  $-1$ coefficients)  can follow each other.  With the AMI code,  these symbols occur only in isolation.<br>		*The reason for the flatter PSD curve at medium frequencies as well as the steeper drop towards the zeros is that for the 4B3T codes up to five  $+1$  coefficients  (resp.  $-1$ coefficients)  can follow each other.  With the AMI code,  these symbols occur only in isolation.<br>
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	The graph shows the  [[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Definition_and_statements_of_the_eye_diagram\|eye diagrams]]  (with noise) of the three systems to be compared.  It also contains  $($in row  $\rm A)$  the equations for calculating the symbol error probability.  Approximately  2000  eye lines are drawn for each diagram.		The graph shows the  [[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Definition_and_statements_of_the_eye_diagram\|"eye diagrams"]]  (with noise) of the three systems to be compared.  It also contains  $($in row  $\rm A)$  the equations for calculating the symbol error probability.  Approximately  2000  eye lines are drawn for each diagram.

	The first two rows of the table describe the system comparison at peak limitation.  For the binary system  (first column),  the noise power  (taking into account the  $12\%$ increase)  is given by		The first two rows of the table describe the system comparison at peak limitation.  For the binary system  (first column),  the noise power  (taking into account the  $12\%$ increase)  is given by