Aufgaben:Exercise 3.7: Optimal Nyquist Equalization once again - Revision history

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	*However,  for $f = f_{\rm Nyq}$,  of the infinite sum, only the terms with  $\kappa = 0$  and  $\kappa = 1$  contribute relevantly to the result.		*However,  for $f = f_{\rm Nyq}$,  of the infinite sum, only the terms with  $\kappa = 0$  and  $\kappa = 1$  contribute relevantly to the result.
	*From this follows further with the result from subtask  '''(1)''':		*From this follows further with the result from subtask  '''(1)''':
	:$${\rm Max} \left [ H_{\rm TF}(f) \right ] \ = \ H_{\rm TF}(f = f_{\~~rmNyq~~})={1}/{2 \cdot \|H_{\rm SK}(f = f_{\~~rmNyq~~}) \|^2} = \ \frac{1}{2 \cdot (6.43 \cdot 10^{-5})^2}=\frac{10^{10}}{82.69} \hspace{0.15cm}\underline {\approx 1.21 \cdot 10^{8}}\hspace{0.05cm}.$$		:$${\rm Max} \left [ H_{\rm TF}(f) \right ] \ = \ H_{\rm TF}(f = f_{\rm Nyq})={1}/{2 \cdot \|H_{\rm SK}(f = f_{\rm Nyq}) \|^2} = \ \frac{1}{2 \cdot (6.43 \cdot 10^{-5})^2}=\frac{10^{10}}{82.69} \hspace{0.15cm}\underline {\approx 1.21 \cdot 10^{8}}\hspace{0.05cm}.$$


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	[[Category:Digital Signal Transmission: Exercises\|^3.5 Linear Nyquist Equalization^]]		[[Category:Digital Signal Transmission: Exercises\|^3.5 Linear Nyquist Equalization^]]
	[[de:Aufgaben:3.~~7_Nochmals_Optimale_Nyquistentzerrung~~]]		[[de:Aufgaben:Aufgabe 3.7: Nochmals Optimale Nyquistentzerrung]]

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	The normalized noise power at the decision is given by the following equations:		The normalized noise power at the decision is given by the following equations:
	:$$\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot		:$$\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot\int_{-\infty}^{+\infty} \|H_{\rm E}(f)\|^2 \,{\rm d} f\hspace{0.5cm} = \hspace{0.5cm}\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot\int_{-1/(2T)}^{+1/(2T)} H_{\rm TF}(f) \,{\rm d} f \hspace{0.5cm}= \hspace{0.5cm}T\cdot \int_{0}^{1/T} H_{\rm TF}(f) \,{\rm d} f\hspace{0.05cm}.$$
	\int_{-\infty}^{+\infty} \|H_{\rm E}(f)\|^2 \,{\rm d} f
	\hspace{0.5cm} = \hspace{0.5cm}
	\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot
	\int_{-1/(2T)}^{+1/(2T)} H_{\rm TF}(f) \,{\rm d} f \hspace{0.5cm}= \hspace{0.5cm}T
	\cdot \int_{0}^{1/T} H_{\rm TF}(f) \,{\rm d} f
	\hspace{0.05cm}.$$

	The validity of this equation follows from the periodicity of the transversal filter frequency response  $H_{\rm TF}(f)$.		The validity of this equation follows from the periodicity of the transversal filter frequency response  $H_{\rm TF}(f)$.

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	*However,  for $f = f_{\rm Nyq}$,  of the infinite sum, only the terms with  $\kappa = 0$  and  $\kappa = 1$  contribute relevantly to the result.		*However,  for $f = f_{\rm Nyq}$,  of the infinite sum, only the terms with  $\kappa = 0$  and  $\kappa = 1$  contribute relevantly to the result.
	*From this follows further with the result from subtask  '''(1)''':		*From this follows further with the result from subtask  '''(1)''':
	:$${\rm Max} \left [ H_{\rm TF}(f) \right ] \ = \ H_{\rm TF}(f = f_{\rm		:$${\rm Max} \left [ H_{\rm TF}(f) \right ] \ = \ H_{\rm TF}(f = f_{\rmNyq})={1}/{2 \cdot \|H_{\rm SK}(f = f_{\rmNyq}) \|^2} = \ \frac{1}{2 \cdot (6.43 \cdot 10^{-5})^2}=\frac{10^{10}}{82.69} \hspace{0.15cm}\underline {\approx 1.21 \cdot 10^{8}}\hspace{0.05cm}.$$
	~~Nyq~~})=
	{1}/{2 \cdot \|H_{\rm SK}(f = f_{\rm
	~~Nyq~~}) \|^2} = \ \frac{1}{2 \cdot (6.43 \cdot 10^{-5})^2}=
	\frac{10^{10}}{82.69} \hspace{0.15cm}\underline {\approx 1.21 \cdot 10^{8}}
	\hspace{0.05cm}.$$

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	:$$\|H_{\rm S}(f)\|= {\rm sinc}(f T) \hspace{0.05cm},$$		:$$\|H_{\rm S}(f)\|= {\rm sinc}(f T) \hspace{0.05cm},$$
	* coaxial cable with characteristic cable attenuation  $a_* = 9.2 \ {\rm Np} \ (\approx 80 \ \rm dB)$:		* coaxial cable with characteristic cable attenuation  $a_* = 9.2 \ {\rm Np} \ (\approx 80 \ \rm dB)$:
	:$$\|H_{\rm K}(f)\|= {\rm e}^{ -9.2		:$$\|H_{\rm K}(f)\|= {\rm e}^{ -9.2\cdot \sqrt{2 \cdot \|f\| \cdot T} }\hspace{0.05cm},$$
	\cdot \sqrt{2 \cdot \|f\| \cdot T} }\hspace{0.05cm},$$
	* optimal Nyquist equalizer consisting of matched filter and transversal filter:		* optimal Nyquist equalizer consisting of matched filter and transversal filter:
	:$$H_{\rm E}(f) = H_{\rm MF}(f) \cdot H_{\rm TF}(f)$$		:$$H_{\rm E}(f) = H_{\rm MF}(f) \cdot H_{\rm TF}(f)$$
	:$$\hspace{0.8cm}{\rm where}\hspace{0.2cm}H_{\rm MF}(f) = H_{\rm S}^{\star}(f) \cdot H_{\rm K}^{\star}(f)\hspace{0.05cm},\hspace{0.2cm}		:$$\hspace{0.8cm}{\rm where}\hspace{0.2cm}H_{\rm MF}(f) = H_{\rm S}^{\star}(f) \cdot H_{\rm K}^{\star}(f)\hspace{0.05cm},\hspace{0.2cm}H_{\rm TF}(f) =\frac{1}{\sum\limits_{\kappa = -\infty}^{+\infty} \|H_{\rm SK}(f -{\kappa}/{T})\|^2}\hspace{0.05cm}.$$
	H_{\rm TF}(f) =
	\frac{1}{\sum\limits_{\kappa = -\infty}^{+\infty} \|H_{\rm SK}(f -
	{\kappa}/{T})
	\|^2}\hspace{0.05cm}.$$

	:Here,  $H_{\rm SK}(f) = H_{\rm S}(f) \cdot H_{\rm K}(f)$  denotes the product of transmitter and channel frequency response.		:Here,  $H_{\rm SK}(f) = H_{\rm S}(f) \cdot H_{\rm K}(f)$  denotes the product of transmitter and channel frequency response.
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	Because of Nyquist equalization,  the eye is maximally open.  For the error probability holds:		Because of Nyquist equalization,  the eye is maximally open.  For the error probability holds:
	:$$p_{\rm S} \left ( = p_{\rm U} \right ) = {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T}{N_0 \cdot \sigma_{d,\hspace{0.05cm} {\rm norm}}^2}} \right )		:$$p_{\rm S} \left ( = p_{\rm U} \right ) = {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T}{N_0 \cdot \sigma_{d,\hspace{0.05cm} {\rm norm}}^2}} \right )\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	The normalized noise power at the decision is given by the following equations:		The normalized noise power at the decision is given by the following equations:
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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  In general,  for all frequencies  $f \ge 0$:		'''(1)'''  In general,  for all frequencies  $f \ge 0$:
	:$$\|H_{\rm SK}(f)\|= {\rm sinc}(f T) \cdot {\rm e}^{ -9.2		:$$\|H_{\rm SK}(f)\|= {\rm sinc}(f T) \cdot {\rm e}^{ -9.2\cdot \sqrt{2 \cdot \|f\| \cdot T} }\hspace{0.05cm}.$$
	\cdot \sqrt{2 \cdot \|f\| \cdot T} }\hspace{0.05cm}.$$
	*This gives the special cases we are looking for:		*This gives the special cases we are looking for:
	:$$f= 0 \text{:} \ \hspace{0.1cm}\|H_{\rm SK}(f = 0)\|= {\rm sinc}(0) \cdot {\rm e}^0 \hspace{0.15cm}\underline {= 1}		:$$f= 0 \text{:} \ \hspace{0.1cm}\|H_{\rm SK}(f = 0)\|= {\rm sinc}(0) \cdot {\rm e}^0 \hspace{0.15cm}\underline {= 1}\hspace{0.05cm},$$
	\hspace{0.05cm},$$		:$$ f= f_{\rm Nyq}\text{:} \ \hspace{0.1cm}\|H_{\rm SK}(f = \frac{1}{2T})\|= {\rm sinc}({1}/{2}) \cdot {\rm e}^{-9.2}\hspace{0.15cm}\underline { \approx 6.43 \cdot 10^{-5}}\hspace{0.05cm},$$
	:$$ f= f_{\rm Nyq}\text{:} \ \hspace{0.1cm}\|H_{\rm SK}(f = \frac{1}{2T})\|= {\rm sinc}({1}/{2}) \cdot {\rm e}^{-9.2}		:$$ f= {1}/{T} \text{:}\ \hspace{0.1cm}\|H_{\rm SK}(f = \frac{1}{T})\|= {\rm sinc}({1}) \cdot {\rm e}^{...}\hspace{0.15cm}\underline { = 0} \hspace{0.05cm}.$$
	\hspace{0.15cm}\underline { \approx 6.43 \cdot 10^{-5}}
	\hspace{0.05cm},$$
	:$$ f= {1}/{T} \text{:}\ \hspace{0.1cm}\|H_{\rm SK}(f = \frac{1}{T})\|= {\rm sinc}({1}) \cdot {\rm e}^{...}
	\hspace{0.15cm}\underline { = 0} \hspace{0.05cm}.$$


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	*It follows with the given equation that		*It follows with the given equation that
	:$${\sum\limits_{\kappa = -\infty}^{+\infty} \|H_{\rm SK}(f -		:$${\sum\limits_{\kappa = -\infty}^{+\infty} \|H_{\rm SK}(f -\frac{\kappa}{T})\|^2}$$
	\frac{\kappa}{T})
	\|^2}$$

	:is minimal at the Nyquist frequency.		:is minimal at the Nyquist frequency.
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	'''(3)'''  Approximating the integral over  $H_{\rm TF}(f)$  by the triangular area plotted in the graph,  we obtain:		'''(3)'''  Approximating the integral over  $H_{\rm TF}(f)$  by the triangular area plotted in the graph,  we obtain:
	:$$\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot		:$$\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot\int_{0}^{1/T} H_{\rm TF}(f) \,{\rm d} f \approx T \cdot\frac{1}{2}\cdot 1.21 \cdot 10^{8}\cdot (0.64 -0.36)\hspace{0.15cm}\underline {\approx 1.7\cdot 10^{7}} \hspace{0.05cm}.$$
	\int_{0}^{1/T} H_{\rm TF}(f) \,{\rm d} f \approx T \cdot
	\frac{1}{2}\cdot 1.21 \cdot 10^{8}\cdot (0.64 -0.36)\hspace{0.15cm}\underline {\approx 1.7
	\cdot 10^{7}} \hspace{0.05cm}.$$


	'''(4)'''  According to the given equation,  we obtain for the (mean) symbol error probability:		'''(4)'''  According to the given equation,  we obtain for the (mean) symbol error probability:
	:$$p_{\rm S} = {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T}{N_0 \cdot \sigma_{d,\hspace{0.05cm} {\rm norm}}^2}} \right		:$$p_{\rm S} = {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T}{N_0 \cdot \sigma_{d,\hspace{0.05cm} {\rm norm}}^2}} \right) = {\rm Q}\left ( \sqrt{\frac{10^{8}}{1.7\cdot 10^{7}}} \right ) \approx {\rm Q}(2.42)\hspace{0.3cm}\Rightarrow\hspace{0.3cm} p_{\rm S} \hspace{0.15cm}\underline {\approx 0.8 \%} \hspace{0.05cm}.$$
	) = {\rm Q}\left ( \sqrt{\frac{10^{8}}{1.7
	\cdot 10^{7}}} \right ) \approx {\rm Q}(2.42)\hspace{0.3cm}
	\Rightarrow
	\hspace{0.3cm} p_{\rm S} \hspace{0.15cm}\underline {\approx 0.8 \%} \hspace{0.05cm}.$$

	:Since a binary Nyquist system is present,  the worst–case probability  $p_{\rm U}$  is just as large.		:Since a binary Nyquist system is present,  the worst–case probability  $p_{\rm U}$  is just as large.

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	[[Category:Digital Signal Transmission: Exercises\|^3.5 Linear Nyquist Equalization^]]		[[Category:Digital Signal Transmission: Exercises\|^3.5 Linear Nyquist Equalization^]]
			[[de:Aufgaben:3.7_Nochmals_Optimale_Nyquistentzerrung]]

Guenter at 09:38, 23 June 2022

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  In general, for all frequencies $f \ge 0$:		'''(1)'''  In general,  for all frequencies  $f \ge 0$:
	:$$\|H_{\rm SK}(f)\|= {\rm sinc}(f T) \cdot {\rm e}^{ -9.2		:$$\|H_{\rm SK}(f)\|= {\rm sinc}(f T) \cdot {\rm e}^{ -9.2
	\cdot \sqrt{2 \cdot \|f\| \cdot T} }\hspace{0.05cm}.$$		\cdot \sqrt{2 \cdot \|f\| \cdot T} }\hspace{0.05cm}.$$
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	'''(2)'''  The graph shows that $H_{\rm TF}(f)$ becomes maximal at $f = f_{\rm Nyq}$. It follows with the given equation that		'''(2)'''  The graph shows that  $H_{\rm TF}(f)$  becomes maximal at  $f = f_{\rm Nyq}$.

			*It follows with the given equation that
	:$${\sum\limits_{\kappa = -\infty}^{+\infty} \|H_{\rm SK}(f -		:$${\sum\limits_{\kappa = -\infty}^{+\infty} \|H_{\rm SK}(f -
	\frac{\kappa}{T})		\frac{\kappa}{T})
	\|^2}$$		\|^2}$$

	is minimal at the Nyquist frequency. However, for $f = f_{\rm Nyq}$, of the infinite sum, only the terms with $\kappa = 0$ and $\kappa = 1$ contribute relevantly to the result.		:is minimal at the Nyquist frequency.
	*From this follows further with the result from '''(1)''':		*However,  for $f = f_{\rm Nyq}$,  of the infinite sum, only the terms with  $\kappa = 0$  and  $\kappa = 1$  contribute relevantly to the result.
			*From this follows further with the result from subtask  '''(1)''':
	:$${\rm Max} \left [ H_{\rm TF}(f) \right ] \ = \ H_{\rm TF}(f = f_{\rm		:$${\rm Max} \left [ H_{\rm TF}(f) \right ] \ = \ H_{\rm TF}(f = f_{\rm
	Nyq})=		Nyq})=
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	'''(3)'''  Approximating the integral over $H_{\rm TF}(f)$ by the triangular area plotted in the graph, we obtain:		'''(3)'''  Approximating the integral over  $H_{\rm TF}(f)$  by the triangular area plotted in the graph,  we obtain:
	:$$\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot		:$$\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot
	\int_{0}^{1/T} H_{\rm TF}(f) \,{\rm d} f \approx T \cdot		\int_{0}^{1/T} H_{\rm TF}(f) \,{\rm d} f \approx T \cdot
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	'''(4)'''  According to the given equation, we obtain for the (mean) symbol error probability:		'''(4)'''  According to the given equation,  we obtain for the (mean) symbol error probability:
	:$$p_{\rm S} = {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T}{N_0 \cdot \sigma_{d,\hspace{0.05cm} {\rm norm}}^2}} \right		:$$p_{\rm S} = {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T}{N_0 \cdot \sigma_{d,\hspace{0.05cm} {\rm norm}}^2}} \right
	) = {\rm Q}\left ( \sqrt{\frac{10^{8}}{1.7		) = {\rm Q}\left ( \sqrt{\frac{10^{8}}{1.7
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	\hspace{0.3cm} p_{\rm S} \hspace{0.15cm}\underline {\approx 0.8 \%} \hspace{0.05cm}.$$		\hspace{0.3cm} p_{\rm S} \hspace{0.15cm}\underline {\approx 0.8 \%} \hspace{0.05cm}.$$

	:Since a Nyquist system is present, the worst–case probability $p_{\rm U}$ is just as large.		:Since a binary Nyquist system is present,  the worst–case probability  $p_{\rm U}$  is just as large.
	{{ML-Fuß}}		{{ML-Fuß}}


	[[Category:Digital Signal Transmission: Exercises\|^3.5 Linear Nyquist Equalization^]]		[[Category:Digital Signal Transmission: Exercises\|^3.5 Linear Nyquist Equalization^]]

Guenter at 09:12, 23 June 2022

2022-06-23T09:12:16Z

Hwang at 08:15, 26 May 2022

2022-05-26T08:15:56Z

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	''~~Hinweise~~:''		''Notes:''
	*~~Die Aufgabe gehört zum Kapitel~~  [[Digital_Signal_Transmission/Linear_Nyquist_Equalization\|"Linear Nyquist Equalization"]].		*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Linear_Nyquist_Equalization\|"Linear Nyquist Equalization"]].

	* ~~Zur Bestimmung der Fehlerwahrscheinlichkeit können Sie das interaktive Berechnungsmodul~~  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen\|~~Komplementäre Gaußsche Fehlerfunktionen~~]]  ~~benutzen.~~		* To determine the error probability you can use the interactive calculation module  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen\|"Complementary Gaussian Error Functions"]].



	===~~Fragebogen~~===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{~~Berechnen Sie den Betrag des Sender–Kanal–Frequenzgangs für die Frequenzen~~  $f = 0$,  $f = 1/(2T)= f_{\rm Nyq}$  ~~und~~  $f = 1/T = 2 \cdot f_{\rm Nyq}$.		{Calculate the magnitude of the transmitter channel frequency response for the frequencies  $f = 0$,  $f = 1/(2T)= f_{\rm Nyq}$  and  $f = 1/T = 2 \cdot f_{\rm Nyq}$.
	\|type="{}"}		\|type="{}"}
	$\|H_{\rm SK} (f = 0)\| \hspace{0.8cm} = \ $ { 1 3% }		$\|H_{\rm SK} (f = 0)\| \hspace{0.8cm} = \ $ { 1 3% }
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	$\|H_{\rm SK} (f = 1/T)\| \hspace{0.25cm} = \ $ { 0. }		$\|H_{\rm SK} (f = 1/T)\| \hspace{0.25cm} = \ $ { 0. }

	{~~Berechnen Sie den Maximalwert von~~  $H_{\rm TF}(f)$  ~~bei der Frequenz~~  $f = f_{\rm Nyq}$.		{Calculate the maximum value of  $H_{\rm TF}(f)$  at frequency  $f = f_{\rm Nyq}$.
	\|type="{}"}		\|type="{}"}
	$\|H_{\rm TF} (f = f_{\rm Nyq})\| \hspace{0.2cm} = \ $ { 1.21 3% } $\ \cdot 10^8$		$\|H_{\rm TF} (f = f_{\rm Nyq})\| \hspace{0.2cm} = \ $ { 1.21 3% } $\ \cdot 10^8$

	{~~Berechnen Sie die normierte Störleistung entsprechend der Dreiecknäherung~~.		{Calculate the normalized noise power according to the triangular approximation.
	\|type="{}"}		\|type="{}"}
	$\sigma_{d, \ \rm norm}^2 \hspace{0.2cm} = \ $ { 1.7 3% } $\ \cdot 10^7$		$\sigma_{d, \ \rm norm}^2 \hspace{0.2cm} = \ $ { 1.7 3% } $\ \cdot 10^7$

	{~~Welche Symbolfehlerwahrscheinlichkeit ergibt sich mit~~  $s_0^2 \cdot T/N_0 = 10^8$?		{What is the symbol error probability with  $s_0^2 \cdot T/N_0 = 10^8$?
	\|type="{}"}		\|type="{}"}
	$p_{\rm S} \hspace{0.2cm} = \ $ { 0.8 3% } $\ \%$		$p_{\rm S} \hspace{0.2cm} = \ $ { 0.8 3% } $\ \%$
	</quiz>		</quiz>

	===~~Musterlösung~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Allgemein gilt für alle Frequenzen~~ $f \ge 0$:		'''(1)'''  In general, for all frequencies $f \ge 0$:
	:$$\|H_{\rm SK}(f)\|= {\rm sinc}(f T) \cdot {\rm e}^{ -9.2		:$$\|H_{\rm SK}(f)\|= {\rm sinc}(f T) \cdot {\rm e}^{ -9.2
	\cdot \sqrt{2 \cdot \|f\| \cdot T} }\hspace{0.05cm}.$$		\cdot \sqrt{2 \cdot \|f\| \cdot T} }\hspace{0.05cm}.$$
	*~~Daraus ergeben sich die gesuchten Sonderfälle~~:		*This gives the special cases we are looking for:
	:$$f= 0 \text{:} \ \hspace{0.1cm}\|H_{\rm SK}(f = 0)\|= {\rm sinc}(0) \cdot {\rm e}^0 \hspace{0.15cm}\underline {= 1}		:$$f= 0 \text{:} \ \hspace{0.1cm}\|H_{\rm SK}(f = 0)\|= {\rm sinc}(0) \cdot {\rm e}^0 \hspace{0.15cm}\underline {= 1}
	\hspace{0.05cm},$$		\hspace{0.05cm},$$
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	'''(2)'''  ~~Die Grafik zeigt, dass~~ $H_{\rm TF}(f)$ ~~bei~~ $f = f_{\rm Nyq}$ ~~maximal wird~~. ~~Daraus folgt mit der angegebenen Gleichung, dass~~		'''(2)'''  The graph shows that $H_{\rm TF}(f)$ becomes maximal at $f = f_{\rm Nyq}$. It follows with the given equation that
	:$${\sum\limits_{\kappa = -\infty}^{+\infty} \|H_{\rm SK}(f -		:$${\sum\limits_{\kappa = -\infty}^{+\infty} \|H_{\rm SK}(f -
	\frac{\kappa}{T})		\frac{\kappa}{T})
	\|^2}$$		\|^2}$$

	~~bei der Nyquistfrequenz~~ minimal ~~ist~~. ~~Für~~ $f = f_{\rm Nyq}$ ~~tragen von der unendlichen Summe allerdings nur die Terme mit~~ $\kappa = 0$ ~~und~~ $\kappa = 1$ ~~relevant zum Ergebnis bei~~.		is minimal at the Nyquist frequency. However, for $f = f_{\rm Nyq}$, of the infinite sum, only the terms with $\kappa = 0$ and $\kappa = 1$ contribute relevantly to the result.
	*~~Daraus folgt weiter mit dem Ergebnis aus~~ '''(1)''':		*From this follows further with the result from '''(1)''':
	:$${\rm Max} \left [ H_{\rm TF}(f) \right ] \ = \ H_{\rm TF}(f = f_{\rm		:$${\rm Max} \left [ H_{\rm TF}(f) \right ] \ = \ H_{\rm TF}(f = f_{\rm
	Nyq})=		Nyq})=
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	'''(3)'''  ~~Nähert man das Integral über~~ $H_{\rm TF}(f)$ ~~durch die~~ in ~~der Grafik eingezeichnete Dreieckfläche an~~, ~~so erhält man~~:		'''(3)'''  Approximating the integral over $H_{\rm TF}(f)$ by the triangular area plotted in the graph, we obtain:
	:$$\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot		:$$\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot
	\int_{0}^{1/T} H_{\rm TF}(f) \,{\rm d} f \approx T \cdot		\int_{0}^{1/T} H_{\rm TF}(f) \,{\rm d} f \approx T \cdot
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	'''(4)'''  ~~Gemäß der gegebenen Gleichung erhält man für die~~ (~~mittlere~~) ~~Symbolfehlerwahrscheinlichkeit~~:		'''(4)'''  According to the given equation, we obtain for the (mean) symbol error probability:
	:$$p_{\rm S} = {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T}{N_0 \cdot \sigma_{d,\hspace{0.05cm} {\rm norm}}^2}} \right		:$$p_{\rm S} = {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T}{N_0 \cdot \sigma_{d,\hspace{0.05cm} {\rm norm}}^2}} \right
	) = {\rm Q}\left ( \sqrt{\frac{10^{8}}{1.7		) = {\rm Q}\left ( \sqrt{\frac{10^{8}}{1.7
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	\hspace{0.3cm} p_{\rm S} \hspace{0.15cm}\underline {\approx 0.8 \%} \hspace{0.05cm}.$$		\hspace{0.3cm} p_{\rm S} \hspace{0.15cm}\underline {\approx 0.8 \%} \hspace{0.05cm}.$$

	:~~Da ein Nyquistsystem vorliegt~~, ~~ist die ungünstigste (~~worst–case~~) Fehlerwahrscheinlichkeit~~ $p_{\rm U}$ ~~genau so groß~~.		:Since a Nyquist system is present, the worst–case probability $p_{\rm U}$ is just as large.
	{{ML-Fuß}}		{{ML-Fuß}}


	[[Category:Digital Signal Transmission: Exercises\|^3.5 Linear Nyquist Equalization^]]		[[Category:Digital Signal Transmission: Exercises\|^3.5 Linear Nyquist Equalization^]]

Hwang at 08:08, 26 May 2022

2022-05-26T08:08:31Z

← Older revision		Revision as of 10:08, 26 May 2022
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	{{quiz-Header\|Buchseite=~~Digitalsignalübertragung~~/~~Lineare Nyquistentzerrung~~}}		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Linear_Nyquist_Equalization}}

	[[File:P_ID1435__Dig_A_3_7.png\|right\|frame\|~~Transversalfilterfrequenzgang~~]]		[[File:P_ID1435__Dig_A_3_7.png\|right\|frame\|Transversal filter frequency response]]
	~~Wir gehen bei dieser Aufgabe von folgenden Voraussetzungen aus~~:		We assume the following for this exercise:
	* ~~binäre bipolare~~ NRZ~~–Rechteckimpulse~~		* binary bipolar NRZ rectangular pulses
	:$$\|H_{\rm S}(f)\|= {\rm si}(~~\pi~~ f T) \hspace{0.05cm},$$		:$$\|H_{\rm S}(f)\|= {\rm sinc}(f T) \hspace{0.05cm},$$
	* ~~Koaxialkabel mit der Kabeldämpfung~~  $a_* = 9.2 \ {\rm Np} \ (\approx 80 \ \rm dB)$:		* coaxial cable with cable attenuation  $a_* = 9.2 \ {\rm Np} \ (\approx 80 \ \rm dB)$:
	:$$\|H_{\rm K}(f)\|= {\rm e}^{ -9.2		:$$\|H_{\rm K}(f)\|= {\rm e}^{ -9.2
	\cdot \sqrt{2 \cdot \|f\| \cdot T} }\hspace{0.05cm},$$		\cdot \sqrt{2 \cdot \|f\| \cdot T} }\hspace{0.05cm},$$
	* ~~optimaler Nyquistentzerrer, bestehend aus Matched–Filter und Transversalfilter~~:		* optimal Nyquist equalizer consisting of matched filter and transversal filter:
	:$$H_{\rm E}(f) = H_{\rm MF}(f) \cdot H_{\rm TF}(f)$$		:$$H_{\rm E}(f) = H_{\rm MF}(f) \cdot H_{\rm TF}(f)$$
	:$$\hspace{0.8cm}{\rm ~~mit~~}\hspace{0.2cm}H_{\rm MF}(f) = H_{\rm S}^{\star}(f) \cdot H_{\rm K}^{\star}(f)\hspace{0.05cm},\hspace{0.2cm}		:$$\hspace{0.8cm}{\rm where}\hspace{0.2cm}H_{\rm MF}(f) = H_{\rm S}^{\star}(f) \cdot H_{\rm K}^{\star}(f)\hspace{0.05cm},\hspace{0.2cm}
	H_{\rm TF}(f) =		H_{\rm TF}(f) =
	\frac{1}{\sum\limits_{\kappa = -\infty}^{+\infty} \|H_{\rm SK}(f -		\frac{1}{\sum\limits_{\kappa = -\infty}^{+\infty} \|H_{\rm SK}(f -
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	\|^2}\hspace{0.05cm}.$$		\|^2}\hspace{0.05cm}.$$

	:~~Hierbei bezeichnet~~  $H_{\rm SK}(f) = H_{\rm S}(f) \cdot H_{\rm K}(f)$  ~~das Produkt von Sender– und Kanalfrequenzgang~~.		:Here,  $H_{\rm SK}(f) = H_{\rm S}(f) \cdot H_{\rm K}(f)$  denotes the product of transmitter and channel frequency response.


	~~Wegen der Nyquistentzerrung ist das Auge maximal geöffnet~~. ~~Für die Fehlerwahrscheinlichkeit gilt~~:		Because of Nyquist equalization, the eye is maximally open. For the error probability holds:
	:$$p_{\rm S} \left ( = p_{\rm U} \right ) = {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T}{N_0 \cdot \sigma_{d,\hspace{0.05cm} {\rm norm}}^2}} \right )		:$$p_{\rm S} \left ( = p_{\rm U} \right ) = {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T}{N_0 \cdot \sigma_{d,\hspace{0.05cm} {\rm norm}}^2}} \right )
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	~~Die normierte Störleistung am Entscheider ist durch folgende Gleichungen gegeben~~:		The normalized interference power at the decision is given by the following equations:
	:$$\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot		:$$\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot
	\int_{-\infty}^{+\infty} \|H_{\rm E}(f)\|^2 \,{\rm d} f		\int_{-\infty}^{+\infty} \|H_{\rm E}(f)\|^2 \,{\rm d} f
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	~~Die Gültigkeit dieser Gleichung ergibt sich aus der Periodizität des Transversalfilterfrequenzgangs~~  $H_{\rm TF}(f)$.		The validity of this equation follows from the periodicity of the transversal filter frequency response  $H_{\rm TF}(f)$.
	*In ~~der Grafik erkennt man die normierte Störleistung als die rot hinterlegte Fläche~~.		*In the graph, the normalized noise power can be seen as the area highlighted in red.
	*~~Näherungsweise kann die normierte Störleistung durch die~~ in ~~der Grafik blau eingezeichnete Dreieckfläche berechnet werden~~.		*As an approximation, the normalized noise power can be calculated by the triangular area shown in blue in the graph.


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	''Hinweise:''		''Hinweise:''
	*Die Aufgabe gehört zum Kapitel  [[~~Digitalsignal%C3%BCbertragung~~/~~Lineare_Nyquistentzerrung~~\|~~Linare Nyquistentzerrung~~]].		*Die Aufgabe gehört zum Kapitel  [[Digital_Signal_Transmission/Linear_Nyquist_Equalization\|"Linear Nyquist Equalization"]].

	* Zur Bestimmung der Fehlerwahrscheinlichkeit können Sie das interaktive Berechnungsmodul  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen\|Komplementäre Gaußsche Fehlerfunktionen]]  benutzen.		* Zur Bestimmung der Fehlerwahrscheinlichkeit können Sie das interaktive Berechnungsmodul  [[Applets:Komplementäre_Gaußsche_Fehlerfunktionen\|Komplementäre Gaußsche Fehlerfunktionen]]  benutzen.
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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Allgemein gilt für alle Frequenzen $f \ge 0$:		'''(1)'''  Allgemein gilt für alle Frequenzen $f \ge 0$:
	:$$\|H_{\rm SK}(f)\|= {\rm si}(~~\pi~~ f T) \cdot {\rm e}^{ -9.2		:$$\|H_{\rm SK}(f)\|= {\rm sinc}(f T) \cdot {\rm e}^{ -9.2
	\cdot \sqrt{2 \cdot \|f\| \cdot T} }\hspace{0.05cm}.$$		\cdot \sqrt{2 \cdot \|f\| \cdot T} }\hspace{0.05cm}.$$
	*Daraus ergeben sich die gesuchten Sonderfälle:		*Daraus ergeben sich die gesuchten Sonderfälle:
	:$$f= 0 \text{:} \ \hspace{0.1cm}\|H_{\rm SK}(f = 0)\|= {\rm si}(0) \cdot {\rm e}^0 \hspace{0.15cm}\underline {= 1}		:$$f= 0 \text{:} \ \hspace{0.1cm}\|H_{\rm SK}(f = 0)\|= {\rm sinc}(0) \cdot {\rm e}^0 \hspace{0.15cm}\underline {= 1}
	\hspace{0.05cm},$$		\hspace{0.05cm},$$
	:$$ f= f_{\rm Nyq}\text{:} \ \hspace{0.1cm}\|H_{\rm SK}(f = \frac{1}{2T})\|= {\rm si}({~~\pi~~}/{2}) \cdot {\rm e}^{-9.2}		:$$ f= f_{\rm Nyq}\text{:} \ \hspace{0.1cm}\|H_{\rm SK}(f = \frac{1}{2T})\|= {\rm sinc}({1}/{2}) \cdot {\rm e}^{-9.2}
	\hspace{0.15cm}\underline { \approx 6.43 \cdot 10^{-5}}		\hspace{0.15cm}\underline { \approx 6.43 \cdot 10^{-5}}
	\hspace{0.05cm},$$		\hspace{0.05cm},$$
	:$$ f= {1}/{T} \text{:}\ \hspace{0.1cm}\|H_{\rm SK}(f = \frac{1}{T})\|= {\rm si}({~~\pi~~}) \cdot {\rm e}^{...}		:$$ f= {1}/{T} \text{:}\ \hspace{0.1cm}\|H_{\rm SK}(f = \frac{1}{T})\|= {\rm sinc}({1}) \cdot {\rm e}^{...}
	\hspace{0.15cm}\underline { = 0} \hspace{0.05cm}.$$		\hspace{0.15cm}\underline { = 0} \hspace{0.05cm}.$$

Guenter at 15:15, 7 May 2022

2022-05-07T15:15:36Z

← Older revision		Revision as of 17:15, 7 May 2022
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	[[Category:Digital Signal Transmission: Exercises\|^3.5 ~~Lineare Nyquistentzerrung~~^]]		[[Category:Digital Signal Transmission: Exercises\|^3.5 Linear Nyquist Equalization^]]

← Older revision		Revision as of 11:12, 23 June 2022
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	{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Linear_Nyquist_Equalization}}		{{quiz-Header\|Buchseite=Digital_Signal_Transmission/Linear_Nyquist_Equalization}}

	[[File:P_ID1435__Dig_A_3_7.png\|right\|frame\|Transversal filter frequency response]]		[[File:P_ID1435__Dig_A_3_7.png\|right\|frame\|Transversal filter <br>frequency response]]
	We assume the following for this exercise:		We assume the following for this exercise:
	* binary bipolar NRZ rectangular pulses		* binary bipolar NRZ rectangular pulses
	:$$\|H_{\rm S}(f)\|= {\rm sinc}(f T) \hspace{0.05cm},$$		:$$\|H_{\rm S}(f)\|= {\rm sinc}(f T) \hspace{0.05cm},$$
	* coaxial cable with cable attenuation  $a_* = 9.2 \ {\rm Np} \ (\approx 80 \ \rm dB)$:		* coaxial cable with characteristic cable attenuation  $a_* = 9.2 \ {\rm Np} \ (\approx 80 \ \rm dB)$:
	:$$\|H_{\rm K}(f)\|= {\rm e}^{ -9.2		:$$\|H_{\rm K}(f)\|= {\rm e}^{ -9.2
	\cdot \sqrt{2 \cdot \|f\| \cdot T} }\hspace{0.05cm},$$		\cdot \sqrt{2 \cdot \|f\| \cdot T} }\hspace{0.05cm},$$
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	Because of Nyquist equalization, the eye is maximally open. For the error probability holds:		Because of Nyquist equalization,  the eye is maximally open.  For the error probability holds:
	:$$p_{\rm S} \left ( = p_{\rm U} \right ) = {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T}{N_0 \cdot \sigma_{d,\hspace{0.05cm} {\rm norm}}^2}} \right )		:$$p_{\rm S} \left ( = p_{\rm U} \right ) = {\rm Q}\left ( \sqrt{\frac{s_0^2 \cdot T}{N_0 \cdot \sigma_{d,\hspace{0.05cm} {\rm norm}}^2}} \right )
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	The normalized ~~interference~~ power at the decision is given by the following equations:		The normalized noise power at the decision is given by the following equations:
	:$$\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot		:$$\sigma_{d,\hspace{0.05cm} {\rm norm}}^2 = T \cdot
	\int_{-\infty}^{+\infty} \|H_{\rm E}(f)\|^2 \,{\rm d} f		\int_{-\infty}^{+\infty} \|H_{\rm E}(f)\|^2 \,{\rm d} f
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	The validity of this equation follows from the periodicity of the transversal filter frequency response  $H_{\rm TF}(f)$.		The validity of this equation follows from the periodicity of the transversal filter frequency response  $H_{\rm TF}(f)$.
	*In the graph, the normalized noise power can be seen as the area highlighted in red.		*In the graph,  the normalized noise power can be seen as the area highlighted in red.
	*As an approximation, the normalized noise power can be calculated by the triangular area shown in blue in the graph.

			*As an approximation,  the normalized noise power can be calculated by the triangular area shown in blue in the graph.



			Notes:
	''Notes:''
	*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Linear_Nyquist_Equalization\|"Linear Nyquist Equalization"]].		*The exercise belongs to the chapter  [[Digital_Signal_Transmission/Linear_Nyquist_Equalization\|"Linear Nyquist Equalization"]].