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

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	*The degradation due to the frequency response  $H_{\rm K}(f)$  is rather shown by a significant increase of the detection noise power, i.e. the variance of the noise component  $d_{\rm N}(t)$  – originating from the noise signal  $n(t)$  at the receiver input:		*The degradation due to the frequency response  $H_{\rm K}(f)$  is rather shown by a significant increase of the detection noise power, i.e. the variance of the noise component  $d_{\rm N}(t)$  – originating from the noise signal  $n(t)$  at the receiver input:
	:$$\sigma_d^2 = \frac{N_0}{2} \cdot \int_{-\infty}^{+\infty}		:$$\sigma_d^2 = \frac{N_0}{2} \cdot \int_{-\infty}^{+\infty}\|H_{\rm E}(f)\|^2 \,{\rm d} f = \frac{N_0}{2} \cdot\int_{-\infty}^{+\infty} \frac{1}{\|H_{\rm K}(f)\|^2}\cdot\|H_{\rm G}(f)\|^2 \,{\rm d} f \hspace{0.05cm}.$$
	\|H_{\rm E}(f)\|^2 \,{\rm d} f = \frac{N_0}{2} \cdot
	\int_{-\infty}^{+\infty} \frac{1}{\|H_{\rm K}(f)\|^2}\cdot
	\|H_{\rm G}(f)\|^2 \,{\rm d} f \hspace{0.05cm}.$$

	*The prerequisite for a finite noise power  $\sigma_d^2$  is that the Gaussian low-pass  $H_{\rm G}(f)$  attenuates the noise  $n(t)$  at  (very)  high frequencies more than it is raised by the ideal equalizer  $1/H_{\rm K}(f)$.  <br><br>		*The prerequisite for a finite noise power  $\sigma_d^2$  is that the Gaussian low-pass  $H_{\rm G}(f)$  attenuates the noise  $n(t)$  at  (very)  high frequencies more than it is raised by the ideal equalizer  $1/H_{\rm K}(f)$.  <br><br>
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	:$$H_{\rm K}(f) = 1 \ \ ⇒ \ \ 1/H_{\rm K}(f) = 1.$$		:$$H_{\rm K}(f) = 1 \ \ ⇒ \ \ 1/H_{\rm K}(f) = 1.$$
	It takes into account the AWGN noise,  but here it was assumed to be very small,  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 30 \ \rm dB$.  For this configuration,  the  worst-case  SNR it was determined by simulation:		It takes into account the AWGN noise,  but here it was assumed to be very small,  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 30 \ \rm dB$.  For this configuration,  the  worst-case  SNR it was determined by simulation:
	:$$10 \cdot {\rm		:$$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U}\approx 26.8\,{\rm dB}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U}<10^{-40}\hspace{0.05cm}.$$
	lg}\hspace{0.1cm}\rho_{\rm U}\approx 26.8\,{\rm dB}
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U}<
	10^{-40}\hspace{0.05cm}.$$

	⇒   The right  (also noisy)  diagram applies to a  [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference#Frequency_response_of_a_coaxial_cable\|coaxial cable]],  where the characteristic cable attenuation  $a_\star = 40 \ \rm dB$.  This results in significantly worse system performance for the same  $E_{\rm B}/N_0$:		⇒   The right  (also noisy)  diagram applies to a  [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference#Frequency_response_of_a_coaxial_cable\|coaxial cable]],  where the characteristic cable attenuation  $a_\star = 40 \ \rm dB$.  This results in significantly worse system performance for the same  $E_{\rm B}/N_0$:
	:$$10 \cdot {\rm		:$$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U}\approx -4.6\,{\rm dB}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U}\approx0.28\hspace{0.05cm}.$$
	lg}\hspace{0.1cm}\rho_{\rm U}\approx -4.6\,{\rm dB}
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U}\~~approx~~
	0.28\hspace{0.05cm}.$$

	This result can be interpreted as follows:		This result can be interpreted as follows:
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	*An acceptable worst-case error probability results only with smaller noise power density  $N_0$.  For example,  with  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 50 \ \rm dB$  $($instead of $30 \ \rm dB)$  the following result is obtained:		*An acceptable worst-case error probability results only with smaller noise power density  $N_0$.  For example,  with  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 50 \ \rm dB$  $($instead of $30 \ \rm dB)$  the following result is obtained:
	:$$10 \cdot {\rm		:$$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = -4.6 +20 \approx 15.4\,{\rm dB}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U}\approx 2 \cdot 10^{-9}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm S} \ge p_{\rm U}/4 \approx 0.5 \cdot 10^{-9}\hspace{0.05cm}.$$}}<br>
	lg}\hspace{0.1cm}\rho_{\rm U} = -4.6 +20 \approx 15.4\,{\rm dB}
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U}\approx 2 \cdot 10^{-9}
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm S} \ge p_{\rm U}/4 \approx 0.5 \cdot 10^{-9}\hspace{0.05cm}.$$}}<br>

	== Increase of the noise power by linear equalization==		== Increase of the noise power by linear equalization==
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	The eye diagrams on the last section impressively document the increase of the noise power  $\sigma_d^2$  with unchanged vertical eye opening,  if one compensates the channel frequency response  $H_{\rm K}(f)$  on the receiving side by its inverse.  This result shall now be interpreted in terms of the noise power density  ${\it \Phi}_{d{\rm N}}(f)$  after the receiver filter  (before the decision),  with the following settings:		The eye diagrams on the last section impressively document the increase of the noise power  $\sigma_d^2$  with unchanged vertical eye opening,  if one compensates the channel frequency response  $H_{\rm K}(f)$  on the receiving side by its inverse.  This result shall now be interpreted in terms of the noise power density  ${\it \Phi}_{d{\rm N}}(f)$  after the receiver filter  (before the decision),  with the following settings:
	*Let the channel be a  [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference#Frequency_response_of_a_coaxial_cable\|coaxial cable]]  with the magnitude frequency response		*Let the channel be a  [[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference#Frequency_response_of_a_coaxial_cable\|coaxial cable]]  with the magnitude frequency response
	:$$\|H_{\rm K}(f)\| = {\rm exp}\left [- a_{\star}\cdot \sqrt{2 f		:$$\|H_{\rm K}(f)\| = {\rm exp}\left [- a_{\star}\cdot \sqrt{2 fT}\hspace{0.05cm} \right ]\hspace{0.2cm}{\rm with}\hspace{0.2cm} a_{\star}= 1.7\,\,{\rm Np}\hspace{0.2cm} ({\rm corresponding \hspace{0.2cm} to} \hspace{0.2cm}15\,\,{\rm dB}) \hspace{0.05cm}.$$
	T}\hspace{0.05cm} \right ]\hspace{0.2cm}{\rm with}\hspace{0.2cm} a_{\star}
	= 1.7\,\,{\rm Np}\hspace{0.2cm} ({\rm corresponding \hspace{0.2cm} to} \hspace{0.2cm}
	15\,\,{\rm dB}) \hspace{0.05cm}.$$
	[[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,  for reasons of presentation,  the characteristic cable attenuation of  $a_\star = 15 \ \rm dB$   $($corresponding to  $1.7 \ \rm Np)$  is chosen to be significantly smaller than in the right eye diagram in  [[Digital_Signal_Transmission/Consideration_of_Channel_Distortion_and_Equalization#Ideal_channel_equalizer\|Example 1]]  in the last section  $($valid for  $a_\star = 40 \ \rm dB)$.\|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,  for reasons of presentation,  the characteristic cable attenuation of  $a_\star = 15 \ \rm dB$   $($corresponding to  $1.7 \ \rm Np)$  is chosen to be significantly smaller than in the right eye diagram in  [[Digital_Signal_Transmission/Consideration_of_Channel_Distortion_and_Equalization#Ideal_channel_equalizer\|Example 1]]  in the last section  $($valid for  $a_\star = 40 \ \rm dB)$.\|class=fit]]
	*The  [[Digital_Signal_Transmission/Consideration_of_Channel_Distortion_and_Equalization#Ideal_channel_equalizer\|ideal channel equalizer]]  $1/H_{\rm K}(f)$  compensates the channel frequency response completely.  No statement is made here about the realization of the attenuation and phase equalization.<br>		*The  [[Digital_Signal_Transmission/Consideration_of_Channel_Distortion_and_Equalization#Ideal_channel_equalizer\|ideal channel equalizer]]  $1/H_{\rm K}(f)$  compensates the channel frequency response completely.  No statement is made here about the realization of the attenuation and phase equalization.<br>
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	Thus,  the noise power-spectral density before the decision is:		Thus,  the noise power-spectral density before the decision is:
	:$${\it \Phi}_{d{\rm N}}(f) = \frac{N_0}{2} \cdot \frac{\|H_{\rm G		:$${\it \Phi}_{d{\rm N}}(f) = \frac{N_0}{2} \cdot \frac{\|H_{\rm G}(f)\|^2}{\|H_{\rm K}(f)\|^2} $$
	}(f)\|^2}{\|H_{\rm K}(f)\|^2} $$		:$$\Rightarrow \hspace{0.3cm} {\it \Phi}_{d{\rm N}}(f) = \frac{N_0}{2} \cdot {\rm exp}\left [2 \cdota_{\star}\cdot \sqrt{2 f T} - {\pi}/{2} \cdot \left ({f }/{f_{\rm G}}\right )^2 \right ] \hspace{0.05cm}.$$
	:$$\Rightarrow \hspace{0.3cm} {\it \Phi}_{d{\rm N}}(f) = \frac{N_0}{2} \cdot {\rm exp}\left [2 \~~cdot~~
	a_{\star}\cdot \sqrt{2 f T} - {\pi}/{2} \cdot \left ({f }/{f_{\rm G}}\right )^2 \right ] \hspace{0.05cm}.$$

	This curve is shown for two  (normalized)  cutoff frequencies		This curve is shown for two  (normalized)  cutoff frequencies
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	*A smaller cutoff frequency would result in a significantly smaller eye opening without equally reducing  $\sigma_d$.  For  $f_\text{G} \cdot T = 0.4$:		*A smaller cutoff frequency would result in a significantly smaller eye opening without equally reducing  $\sigma_d$.  For  $f_\text{G} \cdot T = 0.4$:
	:$$\ddot{o}/s_0 \approx 0.735,\hspace{0.2cm}\sigma_d/s_0 \approx 0.072$$		:$$\ddot{o}/s_0 \approx 0.735,\hspace{0.2cm}\sigma_d/s_0 \approx 0.072$$
	:$$\hspace{0.3cm}\Rightarrow		:$$\hspace{0.3cm}\Rightarrow\hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U}\approx14.1\,{\rm dB}\hspace{0.3cm}\Rightarrow\hspace{0.3cm} p_{\rm U}\approx 1.8 \cdot 10^{-7}\hspace{0.05cm}.$$
	\hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U}\~~approx~~
	14.1\,{\rm dB}
	\hspace{0.3cm}\Rightarrow
	\hspace{0.3cm} p_{\rm U}\approx 1.8 \cdot 10^{-7}\hspace{0.05cm}.$$
	*If the cutoff frequency  $f_\text{G}$  is too large, the noise is limited less effectively. <br>For example,  the values for the cutoff frequency are  $f_\text{G} \cdot T =0.8$:		*If the cutoff frequency  $f_\text{G}$  is too large, the noise is limited less effectively. <br>For example,  the values for the cutoff frequency are  $f_\text{G} \cdot T =0.8$:
	:$$\ddot{o}/s_0 \approx 1.819,\hspace{0.2cm}\sigma_d/s_0 \approx 0.178\hspace{0.3cm}\Rightarrow		:$$\ddot{o}/s_0 \approx 1.819,\hspace{0.2cm}\sigma_d/s_0 \approx 0.178\hspace{0.3cm}\Rightarrow\hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U}\approx14.2\,{\rm dB}\hspace{0.3cm}\Rightarrow\hspace{0.3cm} p_{\rm U}\approx 1.7 \cdot 10^{-7}\hspace{0.05cm}.$$
	\hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U}\~~approx~~
	14.2\,{\rm dB}
	\hspace{0.3cm}\Rightarrow
	\hspace{0.3cm} p_{\rm U}\approx 1.7 \cdot 10^{-7}\hspace{0.05cm}.$$
	*The optimal values  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{d} \approx 16.2 \ \rm dB$  and  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U} \approx \ \rm 15.9 dB$  are much more pronounced than for the ideal channel.		*The optimal values  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{d} \approx 16.2 \ \rm dB$  and  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U} \approx \ \rm 15.9 dB$  are much more pronounced than for the ideal channel.

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	Replacing  $\rho_d$  by  $\rho_{\rm U}$, i.e.,  $p_{\rm S}$  by  $p_{\rm U}$, the system efficiency can be represented as follows:		Replacing  $\rho_d$  by  $\rho_{\rm U}$, i.e.,  $p_{\rm S}$  by  $p_{\rm U}$, the system efficiency can be represented as follows:
	:$$\eta = \eta_{\rm A}=\frac{\rho_d}{\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm}\|\hspace{0.05cm}		:$$\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}.$$
	A}}}= \frac{\rho_d}{2 \cdot E_{\rm B}/N_0}\approx \frac{\rho_{\rm U}}{2 \cdot E_{\rm B}/N_0}.$$

	One can see from the arrangement of the blue circles:		One can see from the arrangement of the blue circles:

Maintenance script: Add German interlanguage link

2026-03-16T12:26:39Z

Add German interlanguage link

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

Guenter at 14:40, 28 June 2022

2022-06-28T14:40:35Z

← Older revision		Revision as of 16:40, 28 June 2022
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	*This is identical with the eye diagram shown in the section  [[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Definition_and_statements_of_the_eye_diagram\|"Definition and statements of the eye diagram"]]  in  $\text{Example 3}$  of the last chapter  (right diagram).<br>		*This is identical with the eye diagram shown in the section  [[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Definition_and_statements_of_the_eye_diagram\|"Definition and statements of the eye diagram"]]  in  $\text{Example 3}$  of the last chapter  (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  (noisy or noiseless)  with intersymbol interferences\|class=fit]]
	<br>		<br>
	⇒   The left  (noisy)  diagram is obtained for the ideal channel, i.e., for		⇒   The left  (noisy)  diagram is obtained for the ideal channel, i.e., for

Hwang at 13:05, 28 June 2022

2022-06-28T13:05:14Z

← Older revision		Revision as of 15:05, 28 June 2022
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	<br>		<br>
	The graph shows the signal-to-noise ratios as a function of the cutoff frequency  $f_{\rm G}$  of the overall Gaussian frequency response  $H_{\rm G}(f) = H_{\rm K}(f) \cdot H_{\rm E}(f)$.		The graph shows the signal-to-noise ratios as a function of the cutoff frequency  $f_{\rm G}$  of the overall Gaussian frequency response  $H_{\rm G}(f) = H_{\rm K}(f) \cdot H_{\rm E}(f)$.
	[[File:EN Dig T 3 3 S1b.png\|right\|frame\|Optimal ~~cut-off~~ frequency of the Gaussian low-pass with distorting channel  $(a_\star = 15 \ \rm dB).$<br>⇒   The circles show the dB values for  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_d$   ⇒   "mean detection SNR"   ⇒   measure of mean error probability  $p_{\rm S}$.<br>⇒   The squares show the dB values for  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U}$   ⇒   "worst-case SNR"   ⇒   measure of worst-case error probability  $p_{\rm U}$.\|class=fit]]		[[File:EN Dig T 3 3 S1b.png\|right\|frame\|Optimal cutoff frequency of the Gaussian low-pass with distorting channel  $(a_\star = 15 \ \rm dB).$<br>⇒   The circles show the dB values for  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_d$   ⇒   "mean detection SNR"   ⇒   measure of mean error probability  $p_{\rm S}$.<br>⇒   The squares show the dB values for  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U}$   ⇒   "worst-case SNR"   ⇒   measure of worst-case error probability  $p_{\rm U}$.\|class=fit]]

	This graphic is valid for		This graphic is valid for
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	When comparing the signal-to-noise ratios,  however,  it must be taken into account that  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 27 \ \rm dB$  is the basis here;  in contrast,  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 13 \ \rm dB$  was assumed in the  [[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Optimization_of_the_cutoff_frequency\|"corresponding graph"]]  for the ideal channel.<br>		When comparing the signal-to-noise ratios,  however,  it must be taken into account that  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 27 \ \rm dB$  is the basis here;  in contrast,  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 13 \ \rm dB$  was assumed in the  [[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Optimization_of_the_cutoff_frequency\|"corresponding graph"]]  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>		<br>
	We further consider		We further consider

Guenter at 14:13, 18 June 2022

2022-06-18T14:13:53Z

← Older revision		Revision as of 16:13, 18 June 2022
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	The blue circles  (left axis labels)  mark the optimal cutoff frequencies  $f_\text{G, opt}$  for the respective cable attenuation  $a_\star$.		The blue circles  (left axis labels)  mark the optimal cutoff frequencies  $f_\text{G, opt}$  for the respective cable attenuation  $a_\star$.

	In addition,  the graph shows with red squares the  [[Digital_Signal_Transmission/Optimization_of_Baseband_Transmission_Systems#System_optimization_with_peak_limitation\|"system efficiency"]]  (at peak limitation)   $\eta$,  which is the ratio of the SNR  $\rho_{d}$  achievable with the considered configuration to the maximum possible S/N ratio  $\rho_{d, \ {\rm max}}$.		In addition,  the graph shows with red squares the  [[Digital_Signal_Transmission/Optimization_of_Baseband_Transmission_Systems#System_optimization_with_peak_limitation\|"system efficiency"]]  (at peak limitation)   $\eta$,  which is the ratio of the SNR  $\rho_{d}$  achievable with the considered configuration to the maximum possible S/N ratio  $\rho_{d, \ {\rm max}}$.
	[[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>⇒   $10 · \lg \eta\hspace{0.05cm}(a_\star = 0 \ \rm dB) = -1.4 \ dB$;<br>⇒   $10 · \lg \eta\hspace{0.05cm}(a_\star = 80 \ \rm dB) = -78.2 \ dB.$\|class=fit\|center]]		[[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>⇒   $10 · \lg \eta\hspace{0.05cm}(a_\star = 0 \ \rm dB) = -1.4 \ dB$;<br>⇒   $10 · \lg \eta\hspace{0.05cm}(a_\star = 80 \ \rm dB) = -78.2 \ dB.$\|class=fit\|center]]



	Replacing  $\rho_d$  by  $\rho_{\rm U}$, i.e.,  $p_{\rm S}$  by  $p_{\rm U}$, the system efficiency can be represented as follows:		Replacing  $\rho_d$  by  $\rho_{\rm U}$, i.e.,  $p_{\rm S}$  by  $p_{\rm U}$, the system efficiency can be represented as follows:
	:$$\eta = \eta_{\rm A}=\frac{\rho_d}{\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm}\|\hspace{0.05cm}		:$$\eta = \eta_{\rm A}=\frac{\rho_d}{\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm}\|\hspace{0.05cm}

Guenter at 15:05, 16 June 2022

2022-06-16T15:05:34Z

← Older revision		Revision as of 17:05, 16 June 2022
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	= 1.7\,\,{\rm Np}\hspace{0.2cm} ({\rm corresponding \hspace{0.2cm} to} \hspace{0.2cm}		= 1.7\,\,{\rm Np}\hspace{0.2cm} ({\rm corresponding \hspace{0.2cm} to} \hspace{0.2cm}
	15\,\,{\rm dB}) \hspace{0.05cm}.$$		15\,\,{\rm dB}) \hspace{0.05cm}.$$
	[[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,  for reasons of presentation,  the characteristic cable attenuation of  $a_\star = 15 \ \rm dB$   $($corresponding to  $1.7 \ \rm Np)$  is chosen to be significantly smaller than in the right eye diagram in  [[Digital_Signal_Transmission/Consideration_of_Channel_Distortion_and_Equalization#Ideal_channel_equalizer\|Example 1]]  in the last section  $($valid for  $a_\star = 40 \ \rm dB)$.\|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,  for reasons of presentation,  the characteristic cable attenuation of  $a_\star = 15 \ \rm dB$   $($corresponding to  $1.7 \ \rm Np)$  is chosen to be significantly smaller than in the right eye diagram in  [[Digital_Signal_Transmission/Consideration_of_Channel_Distortion_and_Equalization#Ideal_channel_equalizer\|Example 1]]  in the last section  $($valid for  $a_\star = 40 \ \rm dB)$.\|class=fit]]
	*The  [[Digital_Signal_Transmission/Consideration_of_Channel_Distortion_and_Equalization#Ideal_channel_equalizer\|ideal channel equalizer]]  $1/H_{\rm K}(f)$  compensates the channel frequency response completely.  No statement is made here about the realization of the attenuation and phase equalization.<br>		*The  [[Digital_Signal_Transmission/Consideration_of_Channel_Distortion_and_Equalization#Ideal_channel_equalizer\|ideal channel equalizer]]  $1/H_{\rm K}(f)$  compensates the channel frequency response completely.  No statement is made here about the realization of the attenuation and phase equalization.<br>

Guenter at 12:49, 16 June 2022

2022-06-16T12:49:44Z

← Older revision		Revision as of 14:49, 16 June 2022
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	For a transmission system whose channel frequency response  $H_{\rm K}(f)$  causes severe distortion,  we assume the following block diagram  (upper graph)  and  equivalent circuit  (lower graph).  To these representations the following is to be noted:		For a transmission system whose channel frequency response  $H_{\rm K}(f)$  causes severe distortion,  we assume the following block diagram  (upper graph)  and  equivalent circuit  (lower graph).  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  $H_{\rm E}(f)$  is – at least mentally – composed of an  '''ideal equalizer'''  $1/H_{\rm K}(f)$  and a low-pass filter  $H_{\rm G}(f)$.  For the latter we use in this chapter exemplarily a  '''Gaussian low-pass'''  with the cutoff frequency  $f_{\rm G}$.<br>		*The receiver filter  $H_{\rm E}(f)$  is – at least mentally – composed of an  '''ideal equalizer'''  $1/H_{\rm K}(f)$  and a low-pass filter  $H_{\rm G}(f)$.  For the latter we use in this chapter exemplarily a  '''Gaussian low-pass'''  with the cutoff frequency  $f_{\rm G}$.<br>
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	*This is identical with the eye diagram shown in the section  [[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Definition_and_statements_of_the_eye_diagram\|"Definition and statements of the eye diagram"]]  in  $\text{Example 3}$  of the last chapter  (right diagram).<br>		*This is identical with the eye diagram shown in the section  [[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Definition_and_statements_of_the_eye_diagram\|"Definition and statements of the eye diagram"]]  in  $\text{Example 3}$  of the last chapter  (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>		<br>
	⇒   The left  (noisy)  diagram is obtained for the ideal channel, i.e., for		⇒   The left  (noisy)  diagram is obtained for the ideal channel, i.e., for
Line 137:		Line 137:
	== Optimal cut-off frequency as a function of cable attenuation==		== Optimal cut-off frequency as a function of cable attenuation==
	<br>		<br>
	[[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>$\hspace{0.8cm} 10 · \lg \eta\hspace{0.05cm}(a_\star = 0 \ \rm dB) = -1.4 \ dB;$   $\hspace{0.8cm} 10 · \lg \eta\hspace{0.05cm}(a_\star = 80 \ \rm dB) = -78.2 \ dB;$\|class=fit\|center]]
	We further consider		We further consider
	*a binary system with NRZ transmission pulses   ⇒   $E_{\rm B} = s_0^2 \cdot T$,<br>		*a binary system with NRZ transmission pulses   ⇒   $E_{\rm B} = s_0^2 \cdot T$,<br>
	*a coaxial cable $H_{\rm K}(f)$, characteristic attenuation  $a_\star$,<br>		*a coaxial cable $H_{\rm K}(f)$,  characteristic attenuation  $a_\star$,<br>
	*a Gaussian total frequency response  $H_{\rm G}(f) = H_{\rm K}(f) \cdot H_{\rm E}(f)$.		*a Gaussian total frequency response  $H_{\rm G}(f) = H_{\rm K}(f) \cdot H_{\rm E}(f)$.


	The blue circles (left axis labels) mark the optimal cutoff frequencies  $f_\text{G, opt}$  for the respective cable attenuation  $a_\star$.		The blue circles  (left axis labels)  mark the optimal cutoff frequencies  $f_\text{G, opt}$  for the respective cable attenuation  $a_\star$.

	In addition, the graph shows with red squares the  [[Digital_Signal_Transmission/Optimization_of_Baseband_Transmission_Systems#System_optimization_with_peak_limitation\|"system efficiency"]]  (at peak limitation)  $\eta$,  which is the ratio of the SNR  $\rho_{d}$  achievable with the considered configuration to the maximum possible S/N ratio  $\rho_{d, \ {\rm max}}$.

			In addition,  the graph shows with red squares the  [[Digital_Signal_Transmission/Optimization_of_Baseband_Transmission_Systems#System_optimization_with_peak_limitation\|"system efficiency"]]  (at peak limitation)   $\eta$,  which is the ratio of the SNR  $\rho_{d}$  achievable with the considered configuration to the maximum possible S/N ratio  $\rho_{d, \ {\rm max}}$.
			[[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>⇒   $10 · \lg \eta\hspace{0.05cm}(a_\star = 0 \ \rm dB) = -1.4 \ dB$;<br>⇒   $10 · \lg \eta\hspace{0.05cm}(a_\star = 80 \ \rm dB) = -78.2 \ dB.$\|class=fit\|center]]
	Replacing  $\rho_d$  by  $\rho_{\rm U}$, i.e.,  $p_{\rm S}$  by  $p_{\rm U}$, the system efficiency can be represented as follows:		Replacing  $\rho_d$  by  $\rho_{\rm U}$, i.e.,  $p_{\rm S}$  by  $p_{\rm U}$, the system efficiency can be represented as follows:
	:$$\eta = \eta_{\rm A}=\frac{\rho_d}{\rho_{d, \hspace{0.05cm}{\rm max \hspace{0.05cm}\|\hspace{0.05cm}		:$$\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}.$$		A}}}= \frac{\rho_d}{2 \cdot E_{\rm B}/N_0}\approx \frac{\rho_{\rm U}}{2 \cdot E_{\rm B}/N_0}.$$


	One can see from the arrangement of the blue circles:		One can see from the arrangement of the blue circles:
	*The optimal cutoff frequency  $f_\text{G, opt}$  depends significantly on the strength ~~of the distortions~~ of the coaxial cable, ~~more precisely:~~   exclusively on the characteristic cable attenuation  $a_\star$  at half the bit rate.		*The optimal cutoff frequency  $f_\text{G, opt}$  depends significantly on the strength of the coaxial cable distortions,    exclusively on the characteristic cable attenuation  $a_\star$  at half the bit rate.
	*The larger the cable attenuation  $a_\star$  and thus the noise influence, the lower the optimal cutoff frequency  $f_\text{G, opt}$.<br>
	*However, always  $f_\text{G, opt} > 0.27/T$. Otherwise, the eye would be closed, equivalent to the "worst–case" error probability  $p_{\rm U} = 0.5$.


			*The larger the cable attenuation  $a_\star$  and thus the noise influence,  the lower the optimal cutoff frequency  $f_\text{G, opt}$.<br>

	~~Let us now discuss the dependence of the system efficiency~~  $\~~eta~~$  ~~(red squares) on~~ the ~~characteristic cable attenuation~~  ~~$a_\star$. The right ordinate starts at~~ the ~~top at~~  ~~$0 \ \rm dB$~~&~~nbsp~~; ~~and extends downward to~~  $~~-100 \~~ \rm dB$.		*However,  always  $f_\text{G, opt} > 0.27/T$.  Otherwise,  the eye would be closed,  equivalent to the  "worst–case error probability"  $p_{\rm U} = 0.5$.

	~~As will now be illustrated by some numerical examples, the representation  $\eta = \eta\hspace{0.05cm}(a_\star)$ avoids some problems arising from the wide range of values of S/N ratios:~~
	* The ordinate value  $10 \cdot {\rm lg}\hspace{0.1cm} \eta\hspace{0.05cm}(a_\star = 0 \ \rm dB) = -1.4 \ \rm dB$  states that the best possible Gaussian low-pass with cutoff frequency  $f_\text{G} \cdot T = 0.8$  at ideal channel is  $1.4 \ \rm dB$  worse than the optimal (matched filter) receiver.<br>

			Let us now discuss the dependence of the  '''system efficiency'''  $\eta$  (red squares)  on the characteristic cable attenuation  $a_\star$.  The right ordinate starts at the top at  $0 \ \rm dB$  and extends downward to  $-100 \ \rm dB$.

	*If we assume ideal channel  ~~$(a_\star = 0 \ \rm dB)$  and~~  $10 \~~cdot {~~\~~rm lg}~~\hspace{0.~~1cm~~} ~~E_{~~\~~rm B}/N_0 = 10 \ \rm dB~~$,  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,  the representation  $\eta = \eta\hspace{0.05cm}(a_\star)$ 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  $10 \cdot {\rm lg}\hspace{0.1cm} \eta\hspace{0.05cm}(a_\star = 0 \ \rm dB) = -1.4 \ \rm dB$  states that the best possible Gaussian low-pass with cutoff frequency  $f_\text{G} \cdot T = 0.8$  at ideal channel is  $1.4 \ \rm dB$  worse than the optimal  (matched filter)  receiver.<br>

	*Accordingly, if this (worst-case) error probability  $p_{\rm U} = 7 \cdot 10^{-5}$   ⇒   $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U} = 11.6 \ \rm dB$  is not to be exceeded for the channel with characteristic cable attenuation  $a_\star = 80 \ \rm dB$,  then the following must apply to the  $E_{\rm B}/N_0$  ratio:		*If we assume an ideal channel  $(a_\star = 0 \ \rm dB)$  and  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 10 \ \rm dB$,  the above equation also states that this configuration will lead to the following  "worst-case error probability":
			:$$10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 10 \cdot {\rm lg}\hspace{0.1cm}{E_{\rm B}}/{N_0} + 10 \cdot {\rm lg}\hspace{0.1cm}(2) + 10 \cdot {\rm lg}\hspace{0.1cm}(\eta) \approx 10\,{\rm dB} \hspace{0.1cm}+\hspace{0.1cm}3\,{\rm dB} \hspace{0.1cm}-\hspace{0.1cm}1.4\, {\rm dB}= 11.6\,{\rm dB} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} p_{\rm U}\approx 7 \cdot 10^{-5}\hspace{0.05cm}.$$
			*Accordingly, if this worst-case error probability  $p_{\rm U} = 7 \cdot 10^{-5}$   ⇒   $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U} = 11.6 \ \rm dB$  is not to be exceeded for the channel with characteristic cable attenuation  $a_\star = 80 \ \rm dB$,  then the following must apply to the  $E_{\rm B}/N_0$  ratio:

	::<math>10 \cdot {\rm		::<math>10 \cdot {\rm
Line 179:		Line 176:
	10^{8}\hspace{0.05cm}.</math>		10^{8}\hspace{0.05cm}.</math>

	*To achieve this, however, the cutoff frequency of the Gaussian low-pass filter must be lowered to  $f_{\rm G}= 0.33/T$  according to the blue circles in the graph.<br>		*To achieve this,  however,  the cutoff frequency of the Gaussian low-pass filter must be lowered to  $f_{\rm G}= 0.33/T$  according to the blue circles in the graph.<br>

	== Exercises for the chapter==		== Exercises for the chapter==

Guenter at 07:51, 16 June 2022

2022-06-16T07:51:40Z

← Older revision		Revision as of 09:51, 16 June 2022
Line 114:		Line 114:

	One can see by comparison with the  [[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Optimization_of_the_cutoff_frequency\|corresponding plot]]  in the last chapter, which was valid for  $H_{\rm K}(f) = 1$,  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 13 \ \rm dB$:		One can see by comparison with the  [[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Optimization_of_the_cutoff_frequency\|corresponding plot]]  in the last chapter, which was valid for  $H_{\rm K}(f) = 1$,  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 13 \ \rm dB$:
	#Even with a highly distorting channel,  $\rho_{\rm U}$  is a suitable lower bound for  $\rho_d$   ⇒   $\rho_{d} \ge \rho_{\rm U}$. Correspondingly,  $p_{\rm U} \ge p_{\rm S} $  is also a reasonable upper bound for  $p_{\rm S}$.~~<br>~~<br>		*Even with a highly distorting channel,  $\rho_{\rm U}$  is a suitable lower bound for  $\rho_d$   ⇒   $\rho_{d} \ge \rho_{\rm U}$. Correspondingly,  $p_{\rm U} \ge p_{\rm S} $  is also a reasonable upper bound for  $p_{\rm S}$.<br>
	#For the considered cable attenuation  $a_\star = 15 \ \rm dB$,  the cutoff frequency  $f_\text{G} \cdot T \approx 0.55$  is optimal and  $\ddot{o}/s_0 \approx 1.327$  and  $\sigma_d/s_0 \approx 0.106$  hold.~~<br>~~<br>		*For the considered cable attenuation  $a_\star = 15 \ \rm dB$,  the cutoff frequency  $f_\text{G} \cdot T \approx 0.55$  is optimal and  $\ddot{o}/s_0 \approx 1.327$  and  $\sigma_d/s_0 \approx 0.106$  hold.<br>
	#This gives the  worst-case SNR  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U} \approx \ \rm 15.9 \ dB$  and the worst–case error probability  $p_{\rm U} \approx 2 \cdot 10^{-9}.$~~<br>~~<br>		*This gives the  worst-case SNR  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U} \approx \ \rm 15.9 \ dB$  and the worst–case error probability  $p_{\rm U} \approx 2 \cdot 10^{-9}.$<br>
	#A smaller cutoff frequency would result in a significantly smaller eye opening without equally reducing  $\sigma_d$.  For  $f_\text{G} \cdot T = 0.4$:~~<br><br>   ~~ $\ddot{o}/s_0 \approx 0.735,\hspace{0.2cm}\sigma_d/s_0 \approx 0.072$~~<br>   ~~$\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  $\sigma_d$.  For  $f_\text{G} \cdot T = 0.4$:
			:$$\ddot{o}/s_0 \approx 0.735,\hspace{0.2cm}\sigma_d/s_0 \approx 0.072$$
			:$$\hspace{0.3cm}\Rightarrow
			\hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U}\approx
	14.1\,{\rm dB}		14.1\,{\rm dB}
	\hspace{0.3cm}\Rightarrow		\hspace{0.3cm}\Rightarrow
	\hspace{0.3cm} p_{\rm U}\approx 1.8 \cdot 10^{-7}\hspace{0.05cm}.$~~<br><br>~~		\hspace{0.3cm} p_{\rm U}\approx 1.8 \cdot 10^{-7}\hspace{0.05cm}.$$
	#If the cutoff frequency  $f_\text{G}$  is too large, the noise is limited less effectively. <br>For example,  the values for the cutoff frequency are  $f_\text{G} \cdot T =0.8$:~~<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  $f_\text{G}$  is too large, the noise is limited less effectively. <br>For example,  the values for the cutoff frequency are  $f_\text{G} \cdot T =0.8$:
			:$$\ddot{o}/s_0 \approx 1.819,\hspace{0.2cm}\sigma_d/s_0 \approx 0.178\hspace{0.3cm}\Rightarrow
	\hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U}\approx		\hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U}\approx
	14.2\,{\rm dB}		14.2\,{\rm dB}
	\hspace{0.3cm}\Rightarrow		\hspace{0.3cm}\Rightarrow
	\hspace{0.3cm} p_{\rm U}\approx 1.7 \cdot 10^{-7}\hspace{0.05cm}.$~~<br>~~		\hspace{0.3cm} p_{\rm U}\approx 1.7 \cdot 10^{-7}\hspace{0.05cm}.$$
	#The optimal values of  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{d} \approx 16.2 \ \rm dB$  and  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U} \approx \ \rm 15.9 dB$  are much more pronounced than for ideal channel.		*The optimal values  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{d} \approx 16.2 \ \rm dB$  and  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U} \approx \ \rm 15.9 dB$  are much more pronounced than for the ideal channel.


	When comparing the signal-to-noise ratios, however, it must be taken into account that  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 27 \ \rm dB$  is the basis here; in contrast,  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 13 \ \rm dB$  was assumed in the  [[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Optimization_of_the_cutoff_frequency\|"corresponding graph"]]  for the ideal channel.<br>		When comparing the signal-to-noise ratios,  however,  it must be taken into account that  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 27 \ \rm dB$  is the basis here;  in contrast,  $10 \cdot {\rm lg}\hspace{0.1cm} E_{\rm B}/N_0 = 13 \ \rm dB$  was assumed in the  [[Digital_Signal_Transmission/Error_Probability_with_Intersymbol_Interference#Optimization_of_the_cutoff_frequency\|"corresponding graph"]]  for the ideal channel.<br>

	== Optimal cut-off frequency as a function of cable attenuation==		== Optimal cut-off frequency as a function of cable attenuation==

Guenter at 16:59, 15 June 2022

2022-06-15T16:59:30Z

← Older revision		Revision as of 18:59, 15 June 2022
Line 117:		Line 117:
	#For the considered cable attenuation  $a_\star = 15 \ \rm dB$,  the cutoff frequency  $f_\text{G} \cdot T \approx 0.55$  is optimal and  $\ddot{o}/s_0 \approx 1.327$  and  $\sigma_d/s_0 \approx 0.106$  hold.<br><br>		#For the considered cable attenuation  $a_\star = 15 \ \rm dB$,  the cutoff frequency  $f_\text{G} \cdot T \approx 0.55$  is optimal and  $\ddot{o}/s_0 \approx 1.327$  and  $\sigma_d/s_0 \approx 0.106$  hold.<br><br>
	#This gives the  worst-case SNR  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U} \approx \ \rm 15.9 \ dB$  and the worst–case error probability  $p_{\rm U} \approx 2 \cdot 10^{-9}.$<br><br>		#This gives the  worst-case SNR  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U} \approx \ \rm 15.9 \ dB$  and the worst–case error probability  $p_{\rm U} \approx 2 \cdot 10^{-9}.$<br><br>
	#A smaller cutoff frequency would result in a significantly smaller eye opening without equally reducing  $\sigma_d$.  For  $f_\text{G} \cdot T = 0.4$:<br>    $\ddot{o}/s_0 \approx 0.735,\hspace{0.2cm}\sigma_d/s_0 \approx 0.072$<br>   $\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  $\sigma_d$.  For  $f_\text{G} \cdot T = 0.4$:<br><br>    $\ddot{o}/s_0 \approx 0.735,\hspace{0.2cm}\sigma_d/s_0 \approx 0.072$<br>   $\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U}\approx
	14.1\,{\rm dB}		14.1\,{\rm dB}
	\hspace{0.3cm}\Rightarrow		\hspace{0.3cm}\Rightarrow
	\hspace{0.3cm} p_{\rm U}\approx 1.8 \cdot 10^{-7}\hspace{0.05cm}.$		\hspace{0.3cm} p_{\rm U}\approx 1.8 \cdot 10^{-7}\hspace{0.05cm}.$<br><br>
	#If the cutoff frequency  $f_\text{G}$  is too large, the noise is limited less effectively. For example, the values for the cutoff frequency are  $f_\text{G} \cdot T =0.8$:		#If the cutoff frequency  $f_\text{G}$  is too large, the noise is limited less effectively. <br>For example,  the values for the cutoff frequency are  $f_\text{G} \cdot T =0.8$:<br><br>$\ddot{o}/s_0 \approx 1.819,\hspace{0.2cm}\sigma_d/s_0 \approx 0.178\hspace{0.3cm}\Rightarrow
	:$$\ddot{o}/s_0 \approx 1.819,\hspace{0.2cm}\sigma_d/s_0 \approx 0.178\hspace{0.3cm}\Rightarrow
	\hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U}\approx		\hspace{0.3cm} 10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U}\approx
	14.2\,{\rm dB}		14.2\,{\rm dB}
	\hspace{0.3cm}\Rightarrow		\hspace{0.3cm}\Rightarrow
	\hspace{0.3cm} p_{\rm U}\approx 1.7 \cdot 10^{-7}\hspace{0.05cm}.$$		\hspace{0.3cm} p_{\rm U}\approx 1.7 \cdot 10^{-7}\hspace{0.05cm}.$<br>
			#The optimal values of  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{d} \approx 16.2 \ \rm dB$  and  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U} \approx \ \rm 15.9 dB$  are much more pronounced than for ideal channel.
	*The optimal values of  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{d} \approx 16.2 \ \rm dB$  and  $10 \cdot {\rm lg}\hspace{0.1cm} \rho_{\rm U} \approx \ \rm 15.9 dB$  are much more pronounced than for ideal channel.

Guenter at 16:52, 15 June 2022

2022-06-15T16:52:57Z

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