Digital Signal Transmission/Consideration of Channel Distortion and Equalization - Revision history

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@@ Line 37: / Line 37: @@
 *This is identical with the eye diagram shown in the section &nbsp;[[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Definition_and_statements_of_the_eye_diagram|"Definition and statements of the eye diagram"]]&nbsp; in&nbsp; $\text{Example 3}$&nbsp; of the last chapter&nbsp; (right diagram).<br>
-[[File:P ID1397 Dig T 3 3 S1b version1.png|right|frame|Binary eye diagrams with intersymbol interferences '''Korrektur'''|class=fit]]
+[[File:EN_Dig_T_3_3_S1b_neu.png|right|frame|Binary eye diagrams&nbsp; (noisy or noiseless)&nbsp; with intersymbol interferences|class=fit]]
 <br>
 &rArr; &nbsp; The left&nbsp; (noisy)&nbsp; diagram is obtained for the ideal channel, i.e., for&nbsp;

@@ Line 105: / Line 105: @@
 <br>
 The graph shows the signal-to-noise ratios as a function of the cutoff frequency &nbsp;$f_{\rm G}$&nbsp; of the overall Gaussian frequency response &nbsp;$H_{\rm G}(f) = H_{\rm K}(f) \cdot H_{\rm E}(f)$.
-[[File:EN Dig T 3 3 S1b.png|right|frame|Optimal cut-off frequency of the Gaussian low-pass with distorting channel &nbsp;$(a_\star = 15 \ \rm dB).$<br>&rArr; &nbsp; The circles show the dB values for &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm} \rho_d$ &nbsp; &#8658; &nbsp; "mean detection SNR" &nbsp; &#8658; &nbsp; measure of mean error probability &nbsp;$p_{\rm S}$.<br>&rArr; &nbsp; The squares show the dB values for &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U}$ &nbsp; &#8658; &nbsp; "worst-case SNR" &nbsp; &#8658; &nbsp; measure of worst-case error probability &nbsp;$p_{\rm U}$.|class=fit]]
+[[File:EN Dig T 3 3 S1b.png|right|frame|Optimal cutoff frequency of the Gaussian low-pass with distorting channel &nbsp;$(a_\star = 15 \ \rm dB).$<br>&rArr; &nbsp; The circles show the dB values for &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm} \rho_d$ &nbsp; &#8658; &nbsp; "mean detection SNR" &nbsp; &#8658; &nbsp; measure of mean error probability &nbsp;$p_{\rm S}$.<br>&rArr; &nbsp; The squares show the dB values for &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U}$ &nbsp; &#8658; &nbsp; "worst-case SNR" &nbsp; &#8658; &nbsp; measure of worst-case error probability &nbsp;$p_{\rm U}$.|class=fit]]
 This graphic is valid for
@@ Line 135: / Line 135: @@
 When comparing the signal-to-noise ratios,&nbsp; however,&nbsp; it must be taken into account that &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 27 \ \rm dB$&nbsp; is the basis here;&nbsp; in contrast, &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 13 \ \rm dB$&nbsp; was assumed in the &nbsp;[[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Optimization_of_the_cutoff_frequency|"corresponding graph"]]&nbsp; for the ideal channel.<br>
-== Optimal cut-off frequency as a function of cable attenuation==
+== Optimal cutoff frequency as a function of cable attenuation==
 <br>
 We further consider

@@ Line 145: / Line 145: @@
 The blue circles&nbsp; (left axis labels)&nbsp; mark the optimal cutoff frequencies &nbsp;$f_\text{G, opt}$&nbsp; for the respective cable attenuation &nbsp;$a_\star$.
-In addition,&nbsp; the graph shows with red squares the &nbsp;[[Digital_Signal_Transmission/Optimization_of_Baseband_Transmission_Systems#System_optimization_with_peak_limitation|"system efficiency"]]&nbsp; (at peak limitation) &nbsp; $\eta$,&nbsp; which is the ratio of the SNR &nbsp;$\rho_{d}$&nbsp; achievable with the considered configuration to the maximum possible S/N ratio &nbsp;$\rho_{d, \ {\rm max}}$.&nbsp;
+In addition,&nbsp; the graph shows with red squares the &nbsp;[[Digital_Signal_Transmission/Optimization_of_Baseband_Transmission_Systems#System_optimization_with_peak_limitation|"system efficiency"]]&nbsp; (at peak limitation) &nbsp;  $\eta$,&nbsp; which is the ratio of the SNR &nbsp;$\rho_{d}$&nbsp; achievable with the considered configuration to the maximum possible S/N ratio &nbsp;$\rho_{d, \ {\rm max}}$.&nbsp;
 [[File:Dig_T_3_3_S3b_version2.png|right|frame|Optimal cutoff frequency and system efficiency as a function of characteristic cable attenuation. In particular:     <br>&rArr; &nbsp; $10 &middot; \lg \eta\hspace{0.05cm}(a_\star = 0 \ \rm dB) =  -1.4 \ dB$;<br>&rArr; &nbsp; $10 &middot; \lg \eta\hspace{0.05cm}(a_\star = 80 \ \rm dB) =  -78.2 \ dB.$|class=fit|center]]
 Replacing &nbsp;$\rho_d$&nbsp; by &nbsp;$\rho_{\rm U}$, i.e., &nbsp;$p_{\rm S}$&nbsp; by &nbsp;$p_{\rm U}$, the system efficiency can be represented as follows:
 :$$\eta = \eta_{\rm A}=\frac{\rho_d}{\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm}|\hspace{0.05cm}

@@ Line 74: / Line 74: @@
 = 1.7\,\,{\rm Np}\hspace{0.2cm} ({\rm corresponding \hspace{0.2cm} to} \hspace{0.2cm}
 \,\,{\rm dB}) \hspace{0.05cm}.$$
-[[File:P ID1399 Dig T 3 3 S2 version1.png|right|frame|Noise power-spectral density before the decision due to distorting channel;<br>note that here,&nbsp; for reasons of presentation,&nbsp; the characteristic cable attenuation of &nbsp;$a_\star = 15 \ \rm dB$&nbsp; &nbsp;$($corresponding to &nbsp;$1.7 \ \rm Np)$&nbsp; is chosen to be significantly smaller than in the right eye diagram in &nbsp;[[Digital_Signal_Transmission/Consideration_of_Channel_Distortion_and_Equalization#Ideal_channel_equalizer|Example 1]]&nbsp; in the last section &nbsp;$($valid for &nbsp;$a_\star = 40 \ \rm dB)$.|class=fit]]
+[[File:P ID1399 Dig T 3 3 S2 version1.png|right|frame|Noise power-spectral density before the decision due to distorting channel.<br>Note that here,&nbsp; for reasons of presentation,&nbsp; the characteristic cable attenuation of &nbsp;$a_\star = 15 \ \rm dB$&nbsp; &nbsp;$($corresponding to &nbsp;$1.7 \ \rm Np)$&nbsp; is chosen to be significantly smaller than in the right eye diagram in &nbsp;[[Digital_Signal_Transmission/Consideration_of_Channel_Distortion_and_Equalization#Ideal_channel_equalizer|Example 1]]&nbsp; in the last section &nbsp;$($valid for &nbsp;$a_\star = 40 \ \rm dB)$.|class=fit]]
 *The &nbsp;[[Digital_Signal_Transmission/Consideration_of_Channel_Distortion_and_Equalization#Ideal_channel_equalizer|ideal channel equalizer]]&nbsp; $1/H_{\rm K}(f)$&nbsp; compensates the channel frequency response completely.&nbsp; No statement is made here about the realization of the attenuation and phase equalization.<br>

@@ Line 10: / Line 10: @@
 For a transmission system whose channel frequency response &nbsp;$H_{\rm K}(f)$&nbsp; causes severe distortion,&nbsp; we assume the following block diagram&nbsp; (upper graph)&nbsp; and&nbsp; equivalent circuit&nbsp; (lower graph).&nbsp; To these representations the following is to be noted:
-[[File:EN_Dig_T_3_3_S1_neu.png|right|frame|Block diagram and equivalent circuit diagram for consideration of a channel frequency response|class=fit]]
+[[File:EN_Dig_T_3_3_S1_neu.png|right|frame|Block diagram & equivalent circuit diagram for consideration of a channel frequency response|class=fit]]
 *The receiver filter &nbsp;$H_{\rm E}(f)$&nbsp; is &ndash; at least mentally &ndash; composed of an &nbsp;'''ideal equalizer'''&nbsp; $1/H_{\rm K}(f)$&nbsp; and a low-pass filter &nbsp;$H_{\rm G}(f)$.&nbsp; For the latter we use in this chapter exemplarily a&nbsp; '''Gaussian low-pass'''&nbsp; with the cutoff frequency &nbsp;$f_{\rm G}$.<br>
@@ Line 37: / Line 37: @@
 *This is identical with the eye diagram shown in the section &nbsp;[[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Definition_and_statements_of_the_eye_diagram|"Definition and statements of the eye diagram"]]&nbsp; in&nbsp; $\text{Example 3}$&nbsp; of the last chapter&nbsp; (right diagram).<br>
-[[File:P ID1397 Dig T 3 3 S1b version1.png|right|frame|Binary eye diagrams with intersymbol interferences|class=fit]]
+[[File:P ID1397 Dig T 3 3 S1b version1.png|right|frame|Binary eye diagrams with intersymbol interferences '''Korrektur'''|class=fit]]
 <br>
 &rArr; &nbsp; The left&nbsp; (noisy)&nbsp; diagram is obtained for the ideal channel, i.e., for&nbsp;
@@ Line 137: / Line 137: @@
 == Optimal cut-off frequency as a function of cable attenuation==
 <br>
 We further consider
 *a binary system with NRZ transmission pulses &nbsp; &rArr; &nbsp; $E_{\rm B} = s_0^2 \cdot T$,<br>
-*a coaxial cable $H_{\rm K}(f)$, characteristic attenuation &nbsp;$a_\star$,<br>
+*a coaxial cable $H_{\rm K}(f)$,&nbsp; characteristic attenuation &nbsp;$a_\star$,<br>
 *a Gaussian total frequency response &nbsp;$H_{\rm G}(f) = H_{\rm K}(f) \cdot H_{\rm E}(f)$.
-The blue circles (left axis labels) mark the optimal cutoff frequencies &nbsp;$f_\text{G, opt}$&nbsp; for the respective cable attenuation &nbsp;$a_\star$.
+The blue circles&nbsp; (left axis labels)&nbsp; mark the optimal cutoff frequencies &nbsp;$f_\text{G, opt}$&nbsp; for the respective cable attenuation &nbsp;$a_\star$.
 Replacing &nbsp;$\rho_d$&nbsp; by &nbsp;$\rho_{\rm U}$, i.e., &nbsp;$p_{\rm S}$&nbsp; by &nbsp;$p_{\rm U}$, the system efficiency can be represented as follows:
 :$$\eta = \eta_{\rm A}=\frac{\rho_d}{\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm}|\hspace{0.05cm}
   A}}}=  \frac{\rho_d}{2 \cdot E_{\rm B}/N_0}\approx \frac{\rho_{\rm U}}{2 \cdot E_{\rm B}/N_0}.$$
 One can see from the arrangement of the blue circles:
-*The optimal cutoff frequency &nbsp;$f_\text{G, opt}$&nbsp; depends significantly on the strength of the distortions of the coaxial cable, more precisely: &nbsp; exclusively on the characteristic cable attenuation &nbsp;$a_\star$&nbsp; at half the bit rate.
+*The optimal cutoff frequency &nbsp;$f_\text{G, opt}$&nbsp; depends significantly on the strength of the coaxial cable distortions,&nbsp; &nbsp; exclusively on the characteristic cable attenuation &nbsp;$a_\star$&nbsp; at half the bit rate.
-Let us now discuss the dependence of the system efficiency &nbsp;$\eta$&nbsp; (red squares) on the characteristic cable attenuation &nbsp;$a_\star$. The right ordinate starts at the top at &nbsp;$0 \ \rm dB$&nbsp; and extends downward to &nbsp;$-100 \ \rm dB$.
+*However,&nbsp; always &nbsp;$f_\text{G, opt} > 0.27/T$.&nbsp; Otherwise,&nbsp; the eye would be closed,&nbsp; equivalent to the&nbsp; "worst&ndash;case error probability" &nbsp;$p_{\rm U} = 0.5$.
-*If we assume ideal channel &nbsp;$(a_\star = 0 \ \rm dB)$&nbsp; and &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 10 \ \rm dB$,&nbsp; the above equation also states that this configuration will lead to the following (worst-case) error probability:
+As will now be illustrated by some numerical examples,&nbsp; the representation &nbsp;$\eta = \eta\hspace{0.05cm}(a_\star)$&nbsp;avoids some problems arising from the wide value range  of S/N ratios:
-:$$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U}  =  10 \cdot {\rm lg}\hspace{0.1cm}{E_{\rm B}}/{N_0} + 10 \cdot {\rm lg}\hspace{0.1cm}(2) + 10 \cdot {\rm lg}\hspace{0.1cm}(\eta) \approx  \approx  10\,{\rm dB} \hspace{0.1cm}+\hspace{0.1cm}3\,{\rm dB} \hspace{0.1cm}-\hspace{0.1cm}1.4\, {\rm dB}= 11.6\,{\rm dB} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U}\approx 7 \cdot 10^{-5}\hspace{0.05cm}.$$
+* The ordinate value &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm}  \eta\hspace{0.05cm}(a_\star = 0 \ \rm dB) = -1.4 \ \rm dB$&nbsp; states that the best possible Gaussian low-pass with cutoff frequency &nbsp;$f_\text{G} \cdot T = 0.8$&nbsp; at ideal channel is &nbsp;$1.4 \ \rm dB$&nbsp; worse than the optimal&nbsp; (matched filter)&nbsp; receiver.<br>
-*Accordingly, if this (worst-case) error probability &nbsp;$p_{\rm U} =  7 \cdot 10^{-5}$  &nbsp; &#8658; &nbsp; $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U} = 11.6  \ \rm dB$&nbsp; is not to be exceeded for the channel with characteristic cable attenuation &nbsp;$a_\star = 80 \ \rm dB$,&nbsp; then the following must apply to the &nbsp;$E_{\rm B}/N_0$&nbsp; ratio:
+*If we assume an ideal channel &nbsp;$(a_\star = 0 \ \rm dB)$&nbsp; and &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 10 \ \rm dB$,&nbsp; the above equation also states that this configuration will lead to the following&nbsp; "worst-case error probability":
 ::<math>10 \cdot {\rm
@@ Line 179: / Line 176: @@
 ^{8}\hspace{0.05cm}.</math>
-*To achieve this, however, the cutoff frequency of the Gaussian low-pass filter must be lowered to &nbsp;$f_{\rm G}= 0.33/T$&nbsp; according to the blue circles in the graph.<br>
+*To achieve this,&nbsp; however,&nbsp; the cutoff frequency of the Gaussian low-pass filter must be lowered to &nbsp;$f_{\rm G}= 0.33/T$&nbsp; according to the blue circles in the graph.<br>
 == Exercises for the chapter==

@@ Line 114: / Line 114: @@
 One can see by comparison with the &nbsp;[[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Optimization_of_the_cutoff_frequency|corresponding plot]]&nbsp; in the last chapter, which was valid for &nbsp;$H_{\rm K}(f) = 1$,&nbsp; $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 13 \ \rm dB$:&nbsp;
-#Even with a highly distorting channel, &nbsp;$\rho_{\rm U}$&nbsp; is a suitable lower bound for &nbsp;$\rho_d$ &nbsp; &rArr; &nbsp; $\rho_{d} \ge \rho_{\rm U}$. Correspondingly, &nbsp;$p_{\rm U} \ge p_{\rm S} $&nbsp; is also a reasonable upper bound for &nbsp;$p_{\rm S}$.<br><br>
+*Even with a highly distorting channel, &nbsp;$\rho_{\rm U}$&nbsp; is a suitable lower bound for &nbsp;$\rho_d$ &nbsp; &rArr; &nbsp; $\rho_{d} \ge \rho_{\rm U}$. Correspondingly, &nbsp;$p_{\rm U} \ge p_{\rm S} $&nbsp; is also a reasonable upper bound for &nbsp;$p_{\rm S}$.<br>
-#For the considered cable attenuation &nbsp;$a_\star = 15 \ \rm dB$,&nbsp; the cutoff frequency &nbsp;$f_\text{G} \cdot T \approx 0.55$&nbsp; is optimal and &nbsp;$\ddot{o}/s_0 \approx 1.327$&nbsp; and &nbsp;$\sigma_d/s_0 \approx 0.106$&nbsp; hold.<br><br>
+*For the considered cable attenuation &nbsp;$a_\star = 15 \ \rm dB$,&nbsp; the cutoff frequency &nbsp;$f_\text{G} \cdot T \approx 0.55$&nbsp; is optimal and &nbsp;$\ddot{o}/s_0 \approx 1.327$&nbsp; and &nbsp;$\sigma_d/s_0 \approx 0.106$&nbsp; hold.<br>
-#This gives the&nbsp;  worst-case SNR  &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U} \approx \ \rm 15.9 \ dB$&nbsp; and the  worst&ndash;case error probability &nbsp;$p_{\rm U} \approx  2 \cdot  10^{-9}.$<br><br>
+*This gives the&nbsp;  worst-case SNR  &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U} \approx \ \rm 15.9 \ dB$&nbsp; and the  worst&ndash;case error probability &nbsp;$p_{\rm U} \approx  2 \cdot  10^{-9}.$<br>
-#A smaller cutoff frequency would result in a significantly smaller eye opening without equally reducing &nbsp;$\sigma_d$.&nbsp; For &nbsp;$f_\text{G} \cdot T = 0.4$:<br><br>&nbsp; &nbsp; $\ddot{o}/s_0 \approx 0.735,\hspace{0.2cm}\sigma_d/s_0 \approx 0.072$<br>&nbsp; &nbsp;$\Rightarrow \hspace{0.3cm}  10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U}\approx
+*A smaller cutoff frequency would result in a significantly smaller eye opening without equally reducing &nbsp;$\sigma_d$.&nbsp; For &nbsp;$f_\text{G} \cdot T = 0.4$:
 .1\,{\rm dB}
   \hspace{0.3cm}\Rightarrow
-\hspace{0.3cm} p_{\rm U}\approx 1.8 \cdot 10^{-7}\hspace{0.05cm}.$<br><br>
+\hspace{0.3cm} p_{\rm U}\approx 1.8 \cdot 10^{-7}\hspace{0.05cm}.$$
-#If the cutoff frequency &nbsp;$f_\text{G}$&nbsp; is too large, the noise is limited less effectively. <br>For example,&nbsp; the values for the cutoff frequency are &nbsp;$f_\text{G} \cdot T =0.8$:<br><br>$\ddot{o}/s_0 \approx 1.819,\hspace{0.2cm}\sigma_d/s_0 \approx 0.178\hspace{0.3cm}\Rightarrow
+*If the cutoff frequency &nbsp;$f_\text{G}$&nbsp; is too large, the noise is limited less effectively. <br>For example,&nbsp; the values for the cutoff frequency are &nbsp;$f_\text{G} \cdot T =0.8$:
 \hspace{0.3cm}  10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U}\approx
 .2\,{\rm dB}
   \hspace{0.3cm}\Rightarrow
-\hspace{0.3cm} p_{\rm U}\approx 1.7 \cdot 10^{-7}\hspace{0.05cm}.$<br>
+\hspace{0.3cm} p_{\rm U}\approx 1.7 \cdot 10^{-7}\hspace{0.05cm}.$$
-#The optimal values of &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm} \rho_{d} \approx 16.2 \ \rm dB$&nbsp; and &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U} \approx \ \rm 15.9 dB$&nbsp; are much more pronounced than for ideal channel.
+*The optimal values &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm} \rho_{d} \approx 16.2 \ \rm dB$&nbsp; and &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U} \approx \ \rm 15.9 dB$&nbsp; are much more pronounced than for the ideal channel.
-When comparing the signal-to-noise ratios, however, it must be taken into account that &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 27 \ \rm dB$&nbsp; is the basis here; in contrast, &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 13 \ \rm dB$&nbsp; was assumed in the &nbsp;[[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Optimization_of_the_cutoff_frequency|"corresponding graph"]]&nbsp; for the ideal channel.<br>
+When comparing the signal-to-noise ratios,&nbsp; however,&nbsp; it must be taken into account that &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 27 \ \rm dB$&nbsp; is the basis here;&nbsp; in contrast, &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 13 \ \rm dB$&nbsp; was assumed in the &nbsp;[[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Optimization_of_the_cutoff_frequency|"corresponding graph"]]&nbsp; for the ideal channel.<br>
 == Optimal cut-off frequency as a function of cable attenuation==

@@ Line 35: / Line 35: @@
 *The middle graph shows the eye diagram of the signal component &nbsp;$d_{\rm S}(t)$&nbsp;of the detection signal  for this case &ndash; i.e. without taking the noise into account.
-*This is identical with the eye diagram shown in the chapter &nbsp;[[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Definition_and_statements_of_the_eye_diagram|"Definition and statements of the eye diagram"]]&nbsp; in $\text{example 3}$ (right diagram).<br>
+*This is identical with the eye diagram shown in the section &nbsp;[[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Definition_and_statements_of_the_eye_diagram|"Definition and statements of the eye diagram"]]&nbsp; in&nbsp; $\text{Example 3}$&nbsp; of the last chapter&nbsp; (right diagram).<br>
 [[File:P ID1397 Dig T 3 3 S1b version1.png|right|frame|Binary eye diagrams with intersymbol interferences|class=fit]]
 <br>
-The left eye diagram is obtained for ideal channel, i.e., for&nbsp; $H_{\rm K}(f) = 1$ &nbsp; &rArr; &nbsp; $1/H_{\rm K}(f) = 1$. It takes into account the AWGN noise, but here it was assumed to be very small, &nbsp;$10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 30 \ \rm dB$.&nbsp; For this configuration, it was determined by simulation:
+&rArr; &nbsp; The left&nbsp; (noisy)&nbsp; diagram is obtained for the ideal channel, i.e., for&nbsp;
 :$$10 \cdot {\rm
 lg}\hspace{0.1cm}\rho_{\rm U}\approx 26.8\,{\rm dB}
@@ Line 45: / Line 47: @@
 ^{-40}\hspace{0.05cm}.$$
-In contrast, the right diagram applies to a &nbsp;[[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference#Frequency_response_of_a_coaxial_cable|"coaxial cable"]], where the characteristic cable attenuation &nbsp;$a_\star = 40 \ \rm dB$.&nbsp; This results in significantly worse system sizes for the same &nbsp;$E_{\rm B}/N_0$:&nbsp;
+&rArr; &nbsp;  The right&nbsp; (also noisy)&nbsp; diagram applies to a &nbsp;[[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference#Frequency_response_of_a_coaxial_cable|"coaxial cable"]], where the characteristic cable attenuation &nbsp;$a_\star = 40 \ \rm dB$.&nbsp; This results in significantly worse system sizes for the same &nbsp;$E_{\rm B}/N_0$:&nbsp;
 :$$10 \cdot {\rm
 lg}\hspace{0.1cm}\rho_{\rm U}\approx -4.6\,{\rm dB}

@@ Line 8: / Line 8: @@
 == Ideal channel equalizer==
 <br>
-For a transmission system whose channel frequency response &nbsp;$H_{\rm K}(f)$&nbsp; causes severe distortion, we assume the following block diagram (upper graph) and equivalent circuit (lower graph).<br>
+For a transmission system whose channel frequency response &nbsp;$H_{\rm K}(f)$&nbsp; causes severe distortion,&nbsp; we assume the following block diagram&nbsp; (upper graph)&nbsp; and&nbsp; equivalent circuit&nbsp; (lower graph).&nbsp; To these representations the following is to be noted:
-[[File:EN_Dig_T_3_3_S1_neu.png|center|frame|Block diagram and equivalent circuit diagram for consideration of a channel frequency response|class=fit]]
+[[File:EN_Dig_T_3_3_S1_neu.png|right|frame|Block diagram and equivalent circuit diagram for consideration of a channel frequency response|class=fit]]
-To these representations the following is to be noted:
+*The receiver filter &nbsp;$H_{\rm E}(f)$&nbsp; is &ndash; at least mentally &ndash; composed of an &nbsp;'''ideal equalizer'''&nbsp; $1/H_{\rm K}(f)$&nbsp; and a low-pass filter &nbsp;$H_{\rm G}(f)$.&nbsp; For the latter we use in this chapter exemplarily a&nbsp; '''Gaussian low-pass'''&nbsp; with the cutoff frequency &nbsp;$f_{\rm G}$.<br>
-*If we now move the ideal equalizer &ndash; again purely mentally &ndash; to the left side of the noise addition point, nothing changes with respect to the S/N ratio at the sink and with respect to the error probability compared to the block diagram drawn above.<br>
+*If we now move the ideal equalizer &ndash; again purely mentally &ndash; to the left side of the noise addition point,&nbsp; nothing changes with respect to the S/N ratio before the decision device and with respect to the error probability compared to the block diagram drawn above.<br>
-*From the equivalent circuit below it can be seen that the channel frequency response &nbsp;$H_{\rm K}(f)$&nbsp; does not change anything with respect to the signal component of the detection signal &nbsp;$d_{\rm S}(t)$&nbsp; &ndash; originating from the transmitted signal &nbsp;$s(t)$&nbsp; &ndash; if it is fully compensated with &nbsp;$1/H_{\rm K}(f)$.&nbsp; Thus, the signal component has exactly the same shape as calculated in the chapter &nbsp;[[Digital_Signal_Transmission/Fehlerwahrscheinlichkeit_unter_Berücksichtigung_von_Impulsinterferenzen|"Error Probability with Intersymbol Interference"]].&nbsp; <br>
+*From the equivalent diagram below it can be seen that the channel frequency response &nbsp;$H_{\rm K}(f)$&nbsp; does not change anything with respect to the signal component &nbsp;$d_{\rm S}(t)$&nbsp; of the detection signal   if it is fully compensated with &nbsp;$1/H_{\rm K}(f)$.&nbsp;
-*The degradation due to the channel frequency response &nbsp;$H_{\rm K}(f)$&nbsp; is rather shown by a significant increase of the detection noise power, i.e. the variance of the signal &nbsp;$d_{\rm N}(t)$&nbsp; &ndash; originating from the noise signal &nbsp;$n(t)$:
+*Thus,&nbsp; the signal component &nbsp;$d_{\rm S}(t)$&nbsp; &ndash; originating from the transmitted signal &nbsp;$s(t)$&nbsp; &ndash; has exactly the same shape as calculated in the chapter &nbsp;[[Digital_Signal_Transmission/Fehlerwahrscheinlichkeit_unter_Berücksichtigung_von_Impulsinterferenzen|"Error Probability with Intersymbol Interference"]].&nbsp; <br>
 :$$\sigma_d^2 = \frac{N_0}{2} \cdot \int_{-\infty}^{+\infty}
 |H_{\rm E}(f)|^2 \,{\rm d} f = \frac{N_0}{2} \cdot
@@ Line 25: / Line 26: @@
 |H_{\rm G}(f)|^2 \,{\rm d} f \hspace{0.05cm}.$$
-*The prerequisite for a finite noise power &nbsp;$\sigma_d^2$&nbsp; is that the low-pass &nbsp;$H_{\rm G}(f)$&nbsp; attenuates the noise &nbsp;$n(t)$&nbsp; at (very) high frequencies more than it is raised by the ideal equalizer &nbsp;$1/H_{\rm K}(f)$.&nbsp; <br><br>
+*The prerequisite for a finite noise power &nbsp;$\sigma_d^2$&nbsp; is that the Gaussian low-pass &nbsp;$H_{\rm G}(f)$&nbsp; attenuates the noise &nbsp;$n(t)$&nbsp; at&nbsp; (very)&nbsp; high frequencies more than it is raised by the ideal equalizer &nbsp;$1/H_{\rm K}(f)$.&nbsp; <br><br>
-<i>Note</i>: &nbsp; The channel frequency response &nbsp;$H_{\rm K}(f)$&nbsp; can be equalized according to magnitude and phase, but only in a limited frequency range given by &nbsp;$H_{\rm G}(f)$.&nbsp; However, a complete phase equalization is only possible at the expense of a (frequency-independent) running time, which will not be considered further in the following.
+<u>Note</u>: &nbsp; The channel frequency response &nbsp;$H_{\rm K}(f)$&nbsp; must be equalized for magnitude and phase,&nbsp; but only in a limited frequency range specified by &nbsp;$H_{\rm G}(f)$&nbsp;.&nbsp; However,&nbsp; a complete phase equalization is only possible at the expense of a&nbsp; (frequency-independent)&nbsp; delay time,&nbsp; which will not be considered further in the following.
 {{GraueBox|TEXT=
-$\text{Example 1:}$&nbsp; We consider again a binary system with NRZ rectangular pulses and Gaussian receiver filter &nbsp;$H_{\rm E}(f) = H_{\rm G}(f)$&nbsp; with (normalized) cutoff frequency &nbsp;$f_\text{G, opt} \cdot T = 0.4$.
+$\text{Example 1:}$&nbsp; We consider again a binary system with NRZ rectangular pulses and a Gaussian receiver filter &nbsp;$H_{\rm E}(f) = H_{\rm G}(f)$&nbsp; with&nbsp; (normalized)&nbsp; cutoff frequency &nbsp;$f_\text{G, opt} \cdot T = 0.4$.
 *This is identical with the eye diagram shown in the chapter &nbsp;[[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Definition_and_statements_of_the_eye_diagram|"Definition and statements of the eye diagram"]]&nbsp; in $\text{example 3}$ (right diagram).<br>