Aufgaben:Exercise 3.4: Optimization of the Cutoff Frequency - Revision history

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	'''(2)'''  The optimum cutoff frequency is given when the quotient of the (half) eye opening and the noise rms value is maximum.		'''(2)'''  The optimum cutoff frequency is given when the quotient of the (half) eye opening and the noise rms value is maximum.
	*The optimum results for the binary system for  $f_{\rm G}/R_{\rm B} \underline {= 0.33}$:		*The optimum results for the binary system for  $f_{\rm G}/R_{\rm B} \underline {= 0.33}$:
	:$$\rho_{\rm U,\hspace{0.05cm} max} = \frac{0.184^2}{ 0.047^2} = 15.32\hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rmlg}\hspace{0.1cm}\rho_{\rm U,\hspace{0.05cm} max}\hspace{0.15cm}\underline { = 11.85\,{\~~rmdB~~}} \hspace{0.05cm}.$$		:$$\rho_{\rm U,\hspace{0.05cm} max} = \frac{0.184^2}{ 0.047^2} = 15.32\hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rmlg}\hspace{0.1cm}\rho_{\rm U,\hspace{0.05cm} max}\hspace{0.15cm}\underline { = 11.85\,{\rm dB}} \hspace{0.05cm}.$$

	*In contrast,  for the adjacent cutoff frequency values:		*In contrast,  for the adjacent cutoff frequency values:

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	* Taking into account the ideal channel equalizer,  the normalized noise power results in		* Taking into account the ideal channel equalizer,  the normalized noise power results in
	:$$\frac{ \sigma_d^2}{N_{\rm 0} \cdot R_{\rm B}} = \frac{ 1}{R_{\rm		:$$\frac{ \sigma_d^2}{N_{\rm 0} \cdot R_{\rm B}} = \frac{ 1}{R_{\rmB}} \cdot \int_{0}^{\infty}{\rm exp}\left [2 \cdot 9.2\cdot \sqrt{2 \cdot f/R_{\rm B}} - 2\pi \cdot \frac{(f/R_{\rmB})^2}{(2 f_{\rm G}/R_{\rm B})^2} \right ]{\rm d} \hspace{0.05cm} f\hspace{0.05cm}.$$
	B}} \cdot \int_{0}^{\infty}{\rm exp}\left [2 \cdot 9.2
	\cdot \sqrt{2 \cdot f/R_{\rm B}} - 2\pi \cdot \frac{(f/R_{\rm
	B})^2}{(2 f_{\rm G}/R_{\rm B})^2} \right ]{\rm d} \hspace{0.05cm} f
	\hspace{0.05cm}.$$
	* As will be derived in  [[Aufgaben:Exercise_3.4Z:_Eye_Opening_and_Level_Number\|Exercise 3.4Z]],  the following holds for the  (normalized)  half eye opening:		* As will be derived in  [[Aufgaben:Exercise_3.4Z:_Eye_Opening_and_Level_Number\|Exercise 3.4Z]],  the following holds for the  (normalized)  half eye opening:
	:$$\frac{\ddot{o}(T_{\rm D})}{ 2 \cdot s_0} = \frac{1}{ M-1}\cdot \big [1- 2 \cdot M \cdot {\rm Q} \left(		:$$\frac{\ddot{o}(T_{\rm D})}{ 2 \cdot s_0} = \frac{1}{ M-1}\cdot \big [1- 2 \cdot M \cdot {\rm Q} \left(\sqrt{2\pi} \cdot {\rm log_2}\hspace{0.1cm}(M) \cdot {f_{\rmG}}/{R_{\rm B}}\right)\big]\hspace{0.05cm}.$$
	\sqrt{2\pi} \cdot {\rm log_2}\hspace{0.1cm}(M) \cdot {f_{\rm
	G}}/{R_{\rm B}}
	\right)\big]
	\hspace{0.05cm}.$$
	* Thus,  for the worst case S/N ratio,  the last term can be interpreted as  "energy per bit related to noise power density"  for the NRZ rectangular pulse considered here:		* Thus,  for the worst case S/N ratio,  the last term can be interpreted as  "energy per bit related to noise power density"  for the NRZ rectangular pulse considered here:
	:$$\rho_{\rm U} = \left [\frac{\ddot{o}(T_{\rm D})}{ 2 \cdot s_0} \right ]^2 \cdot		:$$\rho_{\rm U} = \left [\frac{\ddot{o}(T_{\rm D})}{ 2 \cdot s_0} \right ]^2 \cdot\frac{N_{\rm 0} \cdot R_{\rm B}}{ \sigma_d^2} \cdot \frac{ s_0^2}{N_{\rm 0} \cdot R_{\rm B}}\hspace{0.05cm}.$$
	\frac{N_{\rm 0} \cdot R_{\rm B}}{ \sigma_d^2} \cdot \frac{ s_0^2}{N_{\rm 0} \cdot R_{\rm B}}
	\hspace{0.05cm}.$$


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	'''(1)'''  Normalizing the cutoff frequency  $f_{\rm G}$  to the bit rate  $R_{\rm B}$  $($and not to the symbol rate $1/T)$,  the given noise rms values hold independently of the level number.  Thus one obtains:		'''(1)'''  Normalizing the cutoff frequency  $f_{\rm G}$  to the bit rate  $R_{\rm B}$  $($and not to the symbol rate $1/T)$,  the given noise rms values hold independently of the level number.  Thus one obtains:
	:$$M = 2, \hspace{0.1cm}f_{\rm G}/R_{\rm B} = 0.33\text{:} \hspace{0.2cm}		:$$M = 2, \hspace{0.1cm}f_{\rm G}/R_{\rm B} = 0.33\text{:} \hspace{0.2cm}\sigma_d/s_0 \ \hspace{0.15cm}\underline { = 0.047}\hspace{0.05cm},$$
	\sigma_d/s_0 \ \hspace{0.15cm}\underline { = 0.047}		:$$M = 4, \hspace{0.1cm}f_{\rm G}/R_{\rm B} = 0.28\text{:} \hspace{0.2cm}\sigma_d/s_0 \ \hspace{0.15cm}\underline { = 0.021}\hspace{0.05cm}.$$
	\hspace{0.05cm},$$
	:$$M = 4, \hspace{0.1cm}f_{\rm G}/R_{\rm B} = 0.28\text{:} \hspace{0.2cm}
	\sigma_d/s_0 \ \hspace{0.15cm}\underline { = 0.021}
	\hspace{0.05cm}.$$

	'''(2)'''  The optimum cutoff frequency is given when the quotient of the (half) eye opening and the noise rms value is maximum.		'''(2)'''  The optimum cutoff frequency is given when the quotient of the (half) eye opening and the noise rms value is maximum.
	*The optimum results for the binary system for  $f_{\rm G}/R_{\rm B} \underline {= 0.33}$:		*The optimum results for the binary system for  $f_{\rm G}/R_{\rm B} \underline {= 0.33}$:
	:$$\rho_{\rm U,\hspace{0.05cm} max} = \frac{0.184^2}{ 0.047^2} = 15.32		:$$\rho_{\rm U,\hspace{0.05cm} max} = \frac{0.184^2}{ 0.047^2} = 15.32\hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rmlg}\hspace{0.1cm}\rho_{\rm U,\hspace{0.05cm} max}\hspace{0.15cm}\underline { = 11.85\,{\rmdB}} \hspace{0.05cm}.$$
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}
	10 \cdot {\rm
	lg}\hspace{0.1cm}\rho_{\rm U,\hspace{0.05cm} max}\hspace{0.15cm}\underline { = 11.85\,{\rm
	dB}} \hspace{0.05cm}.$$

	*In contrast,  for the adjacent cutoff frequency values:		*In contrast,  for the adjacent cutoff frequency values:
	:$$f_{\rm G}/R_{\rm B} = 0.32\text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.155^2}{ 0.040^2} = 15.02		:$$f_{\rm G}/R_{\rm B} = 0.32\text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.155^2}{ 0.040^2} = 15.02\hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rmlg}\hspace{0.1cm}\rho_{\rm U} = 11.76\,{\rm dB} \hspace{0.05cm},$$
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}		:$$f_{\rm G}/R_{\rm B} = 0.34 \text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.212^2}{ 0.055^2} =14.86\hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rmlg}\hspace{0.1cm}\rho_{\rm U} = 11.72\,{\rm dB} \hspace{0.05cm}.$$
	10 \cdot {\rm
	lg}\hspace{0.1cm}\rho_{\rm U} = 11.76\,{\rm dB} \hspace{0.05cm},$$
	:$$f_{\rm G}/R_{\rm B} = 0.34 \text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.212^2}{ 0.055^2} =
	14.86
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}
	10 \cdot {\rm
	lg}\hspace{0.1cm}\rho_{\rm U} = 11.72\,{\rm dB} \hspace{0.05cm}.$$
	*From this,  one can see the optimum,  albeit a flat one.		*From this,  one can see the optimum,  albeit a flat one.


	'''(3)'''  For $M = 4$, the following results are obtained:		'''(3)'''  For $M = 4$, the following results are obtained:
	:$$f_{\rm G}/R_{\rm B} = 0.27 \text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.097^2}{ 0.017^2} =		:$$f_{\rm G}/R_{\rm B} = 0.27 \text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.097^2}{ 0.017^2} =32.56 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rmlg}\hspace{0.1cm}\rho_{\rm U} = 15.13\,{\rm dB} \hspace{0.05cm},$$
	32.56 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}		:$$f_{\rm G}/R_{\rm B} = 0.28 \text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.121^2}{ 0.021^2} =33.20 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rmlg}\hspace{0.1cm}\rho_{\rm U} \hspace{0.15cm}\underline {= 15.21\,{\rm dB}} \hspace{0.05cm},$$
	10 \cdot {\rm		:$$f_{\rm G}/R_{\rm B} = 0.29 \text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.139^2}{ 0.025^2} =30.91 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rmlg}\hspace{0.1cm}\rho_{\rm U} = 14.90\,{\rm dB} \hspace{0.05cm}.$$
	lg}\hspace{0.1cm}\rho_{\rm U} = 15.13\,{\rm dB} \hspace{0.05cm},$$
	:$$f_{\rm G}/R_{\rm B} = 0.28 \text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.121^2}{ 0.021^2} =
	33.20 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
	10 \cdot {\rm
	lg}\hspace{0.1cm}\rho_{\rm U} \hspace{0.15cm}\underline {= 15.21\,{\rm dB}} \hspace{0.05cm},$$
	:$$f_{\rm G}/R_{\rm B} = 0.29 \text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.139^2}{ 0.025^2} =
	30.91 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
	10 \cdot {\rm
	lg}\hspace{0.1cm}\rho_{\rm U} = 14.90\,{\rm dB} \hspace{0.05cm}.$$

	*Thus,  the optimum cutoff frequency is  $f_{\rm G}/R_{\rm B} \underline {= 0.28}$  for the quaternary system.		*Thus,  the optimum cutoff frequency is  $f_{\rm G}/R_{\rm B} \underline {= 0.28}$  for the quaternary system.
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	In contrast,  no statement is possible for the characteristic cable attenuation  $a_ = 40 \, {\rm dB}$  based on the available numerical material.  A system simulation provided the following results for this  $($for $E_{\rm B}/N_0 = 50 \, {\rm dB})$:		In contrast,  no statement is possible for the characteristic cable attenuation  $a_ = 40 \, {\rm dB}$  based on the available numerical material.  A system simulation provided the following results for this  $($for $E_{\rm B}/N_0 = 50 \, {\rm dB})$:
	:$$M =2\text{:} \hspace{0.2cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 15.43\,{\rm dB} \hspace{0.2cm}{\rm with}\hspace{0.2cm}		:$$M =2\text{:} \hspace{0.2cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 15.43\,{\rm dB} \hspace{0.2cm}{\rm with}\hspace{0.2cm}f_{\rm G}/R_{\rm B} \approx 0.4 \hspace{0.05cm},$$
	f_{\rm G}/R_{\rm B} \approx 0.4 \hspace{0.05cm},$$		:$$M =4\text{:} \hspace{0.2cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 14.65\,{\rm dB} \hspace{0.2cm}{\rm with}\hspace{0.2cm}f_{\rm G}/R_{\rm B} \approx 0.32 \hspace{0.05cm}.$$
	:$$M =4\text{:} \hspace{0.2cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 14.65\,{\rm dB} \hspace{0.2cm}{\rm with}\hspace{0.2cm}
	f_{\rm G}/R_{\rm B} \approx 0.32 \hspace{0.05cm}.$$
	{{ML-Fuß}}		{{ML-Fuß}}

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	[[Category:Digital Signal Transmission: Exercises\|^3.4 Eye Opening with Multilevel Systems^]]		[[Category:Digital Signal Transmission: Exercises\|^3.4 Eye Opening with Multilevel Systems^]]
			[[de:Aufgaben:3.4_Grenzfrequenzoptimierung]]

Hwang at 09:08, 27 June 2022

2022-06-27T09:08:43Z

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	In contrast,  no statement is possible for the characteristic cable attenuation  $a_ = 40 \, {\rm dB}$  based on the available numerical material.  A system simulation provided the following results for this  $($for $E_{\rm B}/N_0 = 50 \, {\rm dB})$:		In contrast,  no statement is possible for the characteristic cable attenuation  $a_ = 40 \, {\rm dB}$  based on the available numerical material.  A system simulation provided the following results for this  $($for $E_{\rm B}/N_0 = 50 \, {\rm dB})$:
	:$$M =2\text{:} \hspace{0.2cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 15.43\,{\rm dB} \hspace{0.2cm}{\rm ~~mit~~}\hspace{0.2cm}		:$$M =2\text{:} \hspace{0.2cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 15.43\,{\rm dB} \hspace{0.2cm}{\rm with}\hspace{0.2cm}
	f_{\rm G}/R_{\rm B} \approx 0.4 \hspace{0.05cm},$$		f_{\rm G}/R_{\rm B} \approx 0.4 \hspace{0.05cm},$$
	:$$M =4\text{:} \hspace{0.2cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 14.65\,{\rm dB} \hspace{0.2cm}{\rm ~~mit~~}\hspace{0.2cm}		:$$M =4\text{:} \hspace{0.2cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 14.65\,{\rm dB} \hspace{0.2cm}{\rm with}\hspace{0.2cm}
	f_{\rm G}/R_{\rm B} \approx 0.32 \hspace{0.05cm}.$$		f_{\rm G}/R_{\rm B} \approx 0.32 \hspace{0.05cm}.$$
	{{ML-Fuß}}		{{ML-Fuß}}

Hwang at 09:05, 27 June 2022

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	B})^2}{(2 f_{\rm G}/R_{\rm B})^2} \right ]{\rm d} \hspace{0.05cm} f		B})^2}{(2 f_{\rm G}/R_{\rm B})^2} \right ]{\rm d} \hspace{0.05cm} f
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
	* As will be derived in  [[Aufgaben:Exercise_3.4Z:~~_Eye_Opening_and_Number_of_Steps~~\|Exercise 3.4Z]],  the following holds for the  (normalized)  half eye opening:		* As will be derived in  [[Aufgaben:Exercise_3.4Z:_Eye_Opening_and_Level_Number\|Exercise 3.4Z]],  the following holds for the  (normalized)  half eye opening:
	:$$\frac{\ddot{o}(T_{\rm D})}{ 2 \cdot s_0} = \frac{1}{ M-1}\cdot \big [1- 2 \cdot M \cdot {\rm Q} \left(		:$$\frac{\ddot{o}(T_{\rm D})}{ 2 \cdot s_0} = \frac{1}{ M-1}\cdot \big [1- 2 \cdot M \cdot {\rm Q} \left(
	\sqrt{2\pi} \cdot {\rm log_2}\hspace{0.1cm}(M) \cdot {f_{\rm		\sqrt{2\pi} \cdot {\rm log_2}\hspace{0.1cm}(M) \cdot {f_{\rm

Guenter at 13:43, 20 June 2022

2022-06-20T13:43:57Z

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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	'''(2)'''  The optimum cutoff frequency is given when the quotient of (half) eye opening and noise rms value is maximum.		'''(2)'''  The optimum cutoff frequency is given when the quotient of the (half) eye opening and the noise rms value is maximum.
	*The optimum results for the binary system for $f_{\rm G}/R_{\rm B} \underline {= 0.33}$:		*The optimum results for the binary system for  $f_{\rm G}/R_{\rm B} \underline {= 0.33}$:
	:$$\rho_{\rm U,\hspace{0.05cm} max} = \frac{0.184^2}{ 0.047^2} = 15.32		:$$\rho_{\rm U,\hspace{0.05cm} max} = \frac{0.184^2}{ 0.047^2} = 15.32
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}		\hspace{0.3cm}\Rightarrow \hspace{0.3cm}
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	dB}} \hspace{0.05cm}.$$		dB}} \hspace{0.05cm}.$$

	*In contrast, for the adjacent cutoff frequency values:		*In contrast,  for the adjacent cutoff frequency values:
	:$$f_{\rm G}/R_{\rm B} = 0.32\text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.155^2}{ 0.040^2} = 15.02		:$$f_{\rm G}/R_{\rm B} = 0.32\text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.155^2}{ 0.040^2} = 15.02
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}		\hspace{0.3cm}\Rightarrow \hspace{0.3cm}
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	10 \cdot {\rm		10 \cdot {\rm
	lg}\hspace{0.1cm}\rho_{\rm U} = 11.72\,{\rm dB} \hspace{0.05cm}.$$		lg}\hspace{0.1cm}\rho_{\rm U} = 11.72\,{\rm dB} \hspace{0.05cm}.$$
	*From this, one can see the optimum, albeit a flat one.		*From this,  one can see the optimum,  albeit a flat one.


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	lg}\hspace{0.1cm}\rho_{\rm U} = 14.90\,{\rm dB} \hspace{0.05cm}.$$		lg}\hspace{0.1cm}\rho_{\rm U} = 14.90\,{\rm dB} \hspace{0.05cm}.$$

	*Thus, the optimum cutoff frequency is $f_{\rm G}/R_{\rm B} \underline {= 0.28}$ for the quaternary system.		*Thus,  the optimum cutoff frequency is  $f_{\rm G}/R_{\rm B} \underline {= 0.28}$  for the quaternary system.
	*The signal-to-noise ratio is then more than $3 \, {\rm dB}$ larger than for the binary system with optimized cutoff frequency.		*The signal-to-noise ratio is then more than  $3 \, {\rm dB}$  larger than for the binary system with optimized cutoff frequency.


	'''(4)'''  <u>Statements 1, 3 and 4</u> are correct:		'''(4)'''  <u>Statements 1, 3 and 4</u>  are correct:
	*The correctness of the first statement is confirmed by the table. For $f_{\rm G}/R_{\rm B} ≥ 0.35$, the binary system has a larger eye opening than the quaternary system. Moreover, by normalizing all frequencies to the bit rate, the noise rms value is independent of the level number $M$, so that the optimization can be restricted to the eye opening.		*The correctness of the first statement is confirmed by the table.  For  $f_{\rm G}/R_{\rm B} ≥ 0.35$,  the binary system has a larger eye opening than the quaternary system.  Moreover,  by normalizing all frequencies to the bit rate,  the noise rms value is independent of the level number  $M$,  so that the optimization can be restricted to the eye opening.

	*For $f_{\rm G}/R_{\rm B} < 0.35$, on the other hand, the quaternary system is better, so also for $f_{\rm G}/R_{\rm B} = 0.33$. Although this cutoff frequency is optimal for $M = 2$ and not for $M = 4$, with $f_{\rm G}/R_{\rm B} = 0.33$ the quaternary system is better than the binary system by about $0.85 \, {\rm dB}$.		*For  $f_{\rm G}/R_{\rm B} < 0.35$,  on the other hand,  the quaternary system is better,  so also for  $f_{\rm G}/R_{\rm B} = 0.33$.  Although this cutoff frequency is optimal for  $M = 2$  and not for  $M = 4$,  with $f_{\rm G}/R_{\rm B} = 0.33$  the quaternary system is better than the binary system by about  $0.85 \, {\rm dB}$.

	*The third statement is true. Due to the lower (more precisely: half) symbol rate, for $M = 4$ the eye is still open even with $f_{\rm G}/R_{\rm B} = 0.23$, while for a binary system there is already an (almost) closed eye for $f_{\rm G}/R_{\rm B} = 0.27$.		*The third statement is true.  Due to the lower (more precisely:  half) symbol rate,  for  $M = 4$  the eye is still open even with  $f_{\rm G}/R_{\rm B} = 0.23$,  while for a binary system there is already an (almost) closed eye for  $f_{\rm G}/R_{\rm B} = 0.27$.

	With larger characteristic cable attenuation, the tendency goes to smaller and smaller cutoff frequency in order to keep the increase of the noise as small as possible. But if already at $a_ = 80 \, {\rm dB}$ the quaternary system (optimized with respect to cutoff frequency) is better, the same is true for $100 \, {\rm dB}$. The gain is larger than $15.21 ~~\, – \,~~ 11.85 \approx 3.4 \, {\rm dB}$. These values were obtained in questions (2) and (3).		With larger characteristic cable attenuation,  the tendency goes to smaller and smaller cutoff frequency in order to keep the increase of the noise as small as possible.  But if already at  $a_ = 80 \, {\rm dB}$  the quaternary system  (optimized with respect to cutoff frequency)  is better,  the same is true for  $100 \, {\rm dB}$.  The gain is larger  than $15.21 - 11.85 \approx 3.4 \, {\rm dB}$.  These values were obtained in questions  '''(2)'''  and  '''(3)'''.

	In contrast, no statement is possible for the characteristic cable attenuation $a_ = 40 \, {\rm dB}$ based on the available numerical material. A system simulation provided the following results for this $($for $E_{\rm B}/N_0 = 50 \, {\rm dB})$:		In contrast,  no statement is possible for the characteristic cable attenuation  $a_ = 40 \, {\rm dB}$  based on the available numerical material.  A system simulation provided the following results for this  $($for $E_{\rm B}/N_0 = 50 \, {\rm dB})$:
	:$$M =2\text{:} \hspace{0.2cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 15.43\,{\rm dB} \hspace{0.2cm}{\rm mit}\hspace{0.2cm}		:$$M =2\text{:} \hspace{0.2cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 15.43\,{\rm dB} \hspace{0.2cm}{\rm mit}\hspace{0.2cm}
	f_{\rm G}/R_{\rm B} \approx 0.4 \hspace{0.05cm},$$		f_{\rm G}/R_{\rm B} \approx 0.4 \hspace{0.05cm},$$

Guenter at 13:00, 20 June 2022

2022-06-20T13:00:14Z

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	:$$\rho_{\rm U} = \frac{\big[\ddot{o}(T_{\rm D})/2 \big]^2}{ \sigma_d^2} \hspace{0.05cm}.$$		:$$\rho_{\rm U} = \frac{\big[\ddot{o}(T_{\rm D})/2 \big]^2}{ \sigma_d^2} \hspace{0.05cm}.$$

	$\ddot{o}(T_{\rm D})$  is the vertical eye opening and  $\sigma_d^2$  indicates the detection noise power.  The same boundary conditions apply to both system configurations  (similar to  [[Exercise_3.4Z:_Eye_Opening_and_Level_Number\|Exercise 3.4Z]]):		$\ddot{o}(T_{\rm D})$  is the vertical eye opening and  $\sigma_d^2$  indicates the detection noise power.  The same boundary conditions apply to both system configurations  (similar to  [[Aufgaben:Exercise_3.4Z:_Eye_Opening_and_Level_Number\|Exercise 3.4Z]]):
	* The rectangular basic transmission pulse  $g_s(t)$  in NRZ format has the height  $s_0 = 1 \, {\rm V}$.		* The rectangular basic transmission pulse  $g_s(t)$  in NRZ format has the height  $s_0 = 1 \, {\rm V}$.

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	In contrast to  [[Exercise_3.4Z:_Eye_Opening_and_Level_Number\|Exercise 3.4Z]]  $($fixed cutoff frequency  $f_{\rm G} = 30 \, {\rm MHz}),$  the cutoff frequency of the low-pass filter is variable here.  This is to be determined in such a way that the worst-case S/N ratio  $\rho_{\rm U}$  is maximized and thus the worst-case error probability  $p_{\rm U}$  is minimized.		In contrast to  [[Aufgaben:Exercise_3.4Z:_Eye_Opening_and_Level_Number\|Exercise 3.4Z]]  $($fixed cutoff frequency  $f_{\rm G} = 30 \, {\rm MHz}),$  the cutoff frequency of the low-pass filter is variable here.  This is to be determined in such a way that the worst-case S/N ratio  $\rho_{\rm U}$  is maximized and thus the worst-case error probability  $p_{\rm U}$  is minimized.

	The table shows		The table shows

Guenter at 12:59, 20 June 2022

2022-06-20T12:59:00Z

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	:$$\rho_{\rm U} = \frac{\big[\ddot{o}(T_{\rm D})/2 \big]^2}{ \sigma_d^2} \hspace{0.05cm}.$$		:$$\rho_{\rm U} = \frac{\big[\ddot{o}(T_{\rm D})/2 \big]^2}{ \sigma_d^2} \hspace{0.05cm}.$$

	$\ddot{o}(T_{\rm D})$  is the vertical eye opening and  $\sigma_d^2$  indicates the detection noise power.  The same boundary conditions apply to both system configurations  (similar to  [[~~Aufgaben:~~Exercise_3.4Z:~~_Eye_Opening_and_Number_of_Steps~~\|Exercise 3.4Z]]):		$\ddot{o}(T_{\rm D})$  is the vertical eye opening and  $\sigma_d^2$  indicates the detection noise power.  The same boundary conditions apply to both system configurations  (similar to  [[Exercise_3.4Z:_Eye_Opening_and_Level_Number\|Exercise 3.4Z]]):
	* The rectangular basic transmission pulse  $g_s(t)$  in NRZ format has the height  $s_0 = 1 \, {\rm V}$.		* The rectangular basic transmission pulse  $g_s(t)$  in NRZ format has the height  $s_0 = 1 \, {\rm V}$.

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	In contrast to  [[~~Aufgaben:~~Exercise_3.4Z:~~_Eye_Opening_and_Number_of_Steps~~\|Exercise 3.4Z]]  $($fixed cutoff frequency  $f_{\rm G} = 30 \, {\rm MHz}),$  the cutoff frequency of the low-pass filter is variable here.  This is to be determined in such a way that the worst-case S/N ratio  $\rho_{\rm U}$  is maximized and thus the worst-case error probability  $p_{\rm U}$  is minimized.		In contrast to  [[Exercise_3.4Z:_Eye_Opening_and_Level_Number\|Exercise 3.4Z]]  $($fixed cutoff frequency  $f_{\rm G} = 30 \, {\rm MHz}),$  the cutoff frequency of the low-pass filter is variable here.  This is to be determined in such a way that the worst-case S/N ratio  $\rho_{\rm U}$  is maximized and thus the worst-case error probability  $p_{\rm U}$  is minimized.

	The table shows		The table shows

Guenter at 12:57, 20 June 2022

2022-06-20T12:57:18Z

← Older revision		Revision as of 14:57, 20 June 2022
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Normalizing the cutoff frequency $f_{\rm G}$ to the bit rate $R_{\rm B}$ (and not to the symbol rate $1/T$), the given noise rms values hold independently of the level number. Thus one obtains:		'''(1)'''  Normalizing the cutoff frequency  $f_{\rm G}$  to the bit rate  $R_{\rm B}$  $($and not to the symbol rate $1/T)$,  the given noise rms values hold independently of the level number.  Thus one obtains:
	:$$M = 2, \hspace{0.1cm}f_{\rm G}/R_{\rm B} = 0.33\text{:} \hspace{0.2cm}		:$$M = 2, \hspace{0.1cm}f_{\rm G}/R_{\rm B} = 0.33\text{:} \hspace{0.2cm}
	\sigma_d/s_0 \ \hspace{0.15cm}\underline { = 0.047}		\sigma_d/s_0 \ \hspace{0.15cm}\underline { = 0.047}

← Older revision		Revision as of 17:58, 16 March 2026
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	* Taking into account the ideal channel equalizer,  the normalized noise power results in		* Taking into account the ideal channel equalizer,  the normalized noise power results in
	:$$\frac{ \sigma_d^2}{N_{\rm 0} \cdot R_{\rm B}} = \frac{ 1}{R_{\~~rmB~~}} \cdot \int_{0}^{\infty}{\rm exp}\left [2 \cdot 9.2\cdot \sqrt{2 \cdot f/R_{\rm B}} - 2\pi \cdot \frac{(f/R_{\~~rmB~~})^2}{(2 f_{\rm G}/R_{\rm B})^2} \right ]{\rm d} \hspace{0.05cm} f\hspace{0.05cm}.$$		:$$\frac{ \sigma_d^2}{N_{\rm 0} \cdot R_{\rm B}} = \frac{ 1}{R_{\rm B}} \cdot \int_{0}^{\infty}{\rm exp}\left [2 \cdot 9.2\cdot \sqrt{2 \cdot f/R_{\rm B}} - 2\pi \cdot \frac{(f/R_{\rm B})^2}{(2 f_{\rm G}/R_{\rm B})^2} \right ]{\rm d} \hspace{0.05cm} f\hspace{0.05cm}.$$
	* As will be derived in  [[Aufgaben:Exercise_3.4Z:_Eye_Opening_and_Level_Number\|Exercise 3.4Z]],  the following holds for the  (normalized)  half eye opening:		* As will be derived in  [[Aufgaben:Exercise_3.4Z:_Eye_Opening_and_Level_Number\|Exercise 3.4Z]],  the following holds for the  (normalized)  half eye opening:
	:$$\frac{\ddot{o}(T_{\rm D})}{ 2 \cdot s_0} = \frac{1}{ M-1}\cdot \big [1- 2 \cdot M \cdot {\rm Q} \left(\sqrt{2\pi} \cdot {\rm log_2}\hspace{0.1cm}(M) \cdot {f_{\~~rmG~~}}/{R_{\rm B}}\right)\big]\hspace{0.05cm}.$$		:$$\frac{\ddot{o}(T_{\rm D})}{ 2 \cdot s_0} = \frac{1}{ M-1}\cdot \big [1- 2 \cdot M \cdot {\rm Q} \left(\sqrt{2\pi} \cdot {\rm log_2}\hspace{0.1cm}(M) \cdot {f_{\rm G}}/{R_{\rm B}}\right)\big]\hspace{0.05cm}.$$
	* Thus,  for the worst case S/N ratio,  the last term can be interpreted as  "energy per bit related to noise power density"  for the NRZ rectangular pulse considered here:		* Thus,  for the worst case S/N ratio,  the last term can be interpreted as  "energy per bit related to noise power density"  for the NRZ rectangular pulse considered here:
	:$$\rho_{\rm U} = \left [\frac{\ddot{o}(T_{\rm D})}{ 2 \cdot s_0} \right ]^2 \cdot\frac{N_{\rm 0} \cdot R_{\rm B}}{ \sigma_d^2} \cdot \frac{ s_0^2}{N_{\rm 0} \cdot R_{\rm B}}\hspace{0.05cm}.$$		:$$\rho_{\rm U} = \left [\frac{\ddot{o}(T_{\rm D})}{ 2 \cdot s_0} \right ]^2 \cdot\frac{N_{\rm 0} \cdot R_{\rm B}}{ \sigma_d^2} \cdot \frac{ s_0^2}{N_{\rm 0} \cdot R_{\rm B}}\hspace{0.05cm}.$$
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	'''(2)'''  The optimum cutoff frequency is given when the quotient of the (half) eye opening and the noise rms value is maximum.		'''(2)'''  The optimum cutoff frequency is given when the quotient of the (half) eye opening and the noise rms value is maximum.
	*The optimum results for the binary system for  $f_{\rm G}/R_{\rm B} \underline {= 0.33}$:		*The optimum results for the binary system for  $f_{\rm G}/R_{\rm B} \underline {= 0.33}$:
	:$$\rho_{\rm U,\hspace{0.05cm} max} = \frac{0.184^2}{ 0.047^2} = 15.32\hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\~~rmlg~~}\hspace{0.1cm}\rho_{\rm U,\hspace{0.05cm} max}\hspace{0.15cm}\underline { = 11.85\,{\rm dB}} \hspace{0.05cm}.$$		:$$\rho_{\rm U,\hspace{0.05cm} max} = \frac{0.184^2}{ 0.047^2} = 15.32\hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U,\hspace{0.05cm} max}\hspace{0.15cm}\underline { = 11.85\,{\rm dB}} \hspace{0.05cm}.$$

	*In contrast,  for the adjacent cutoff frequency values:		*In contrast,  for the adjacent cutoff frequency values:
	:$$f_{\rm G}/R_{\rm B} = 0.32\text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.155^2}{ 0.040^2} = 15.02\hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\~~rmlg~~}\hspace{0.1cm}\rho_{\rm U} = 11.76\,{\rm dB} \hspace{0.05cm},$$		:$$f_{\rm G}/R_{\rm B} = 0.32\text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.155^2}{ 0.040^2} = 15.02\hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 11.76\,{\rm dB} \hspace{0.05cm},$$
	:$$f_{\rm G}/R_{\rm B} = 0.34 \text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.212^2}{ 0.055^2} =14.86\hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\~~rmlg~~}\hspace{0.1cm}\rho_{\rm U} = 11.72\,{\rm dB} \hspace{0.05cm}.$$		:$$f_{\rm G}/R_{\rm B} = 0.34 \text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.212^2}{ 0.055^2} =14.86\hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 11.72\,{\rm dB} \hspace{0.05cm}.$$
	*From this,  one can see the optimum,  albeit a flat one.		*From this,  one can see the optimum,  albeit a flat one.


	'''(3)'''  For $M = 4$, the following results are obtained:		'''(3)'''  For $M = 4$, the following results are obtained:
	:$$f_{\rm G}/R_{\rm B} = 0.27 \text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.097^2}{ 0.017^2} =32.56 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\~~rmlg~~}\hspace{0.1cm}\rho_{\rm U} = 15.13\,{\rm dB} \hspace{0.05cm},$$		:$$f_{\rm G}/R_{\rm B} = 0.27 \text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.097^2}{ 0.017^2} =32.56 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 15.13\,{\rm dB} \hspace{0.05cm},$$
	:$$f_{\rm G}/R_{\rm B} = 0.28 \text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.121^2}{ 0.021^2} =33.20 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\~~rmlg~~}\hspace{0.1cm}\rho_{\rm U} \hspace{0.15cm}\underline {= 15.21\,{\rm dB}} \hspace{0.05cm},$$		:$$f_{\rm G}/R_{\rm B} = 0.28 \text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.121^2}{ 0.021^2} =33.20 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} \hspace{0.15cm}\underline {= 15.21\,{\rm dB}} \hspace{0.05cm},$$
	:$$f_{\rm G}/R_{\rm B} = 0.29 \text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.139^2}{ 0.025^2} =30.91 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\~~rmlg~~}\hspace{0.1cm}\rho_{\rm U} = 14.90\,{\rm dB} \hspace{0.05cm}.$$		:$$f_{\rm G}/R_{\rm B} = 0.29 \text{:} \hspace{0.2cm}\rho_{\rm U} = \frac{0.139^2}{ 0.025^2} =30.91 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}10 \cdot {\rm lg}\hspace{0.1cm}\rho_{\rm U} = 14.90\,{\rm dB} \hspace{0.05cm}.$$

	*Thus,  the optimum cutoff frequency is  $f_{\rm G}/R_{\rm B} \underline {= 0.28}$  for the quaternary system.		*Thus,  the optimum cutoff frequency is  $f_{\rm G}/R_{\rm B} \underline {= 0.28}$  for the quaternary system.
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	[[Category:Digital Signal Transmission: Exercises\|^3.4 Eye Opening with Multilevel Systems^]]		[[Category:Digital Signal Transmission: Exercises\|^3.4 Eye Opening with Multilevel Systems^]]
	[[de:Aufgaben:3.~~4_Grenzfrequenzoptimierung~~]]		[[de:Aufgaben:Aufgabe 3.4: Grenzfrequenzoptimierung]]