Aufgaben:Exercise 3.11: Pre-Emphase and De-Emphase - Revision history

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	*Another gain results from the transition from   $3\ \rm kHz$  to  $1\ \rm kHz$:		*Another gain results from the transition from   $3\ \rm kHz$  to  $1\ \rm kHz$:
	:$$G_{\rm FM} (f_{\rm N} = 1\,{\rm kHz}) = 25.28\,{\rm dB} + 9.54\,{\~~rmdB~~}\hspace{0.15cm}\underline {= 34.82\,{\rm dB}}\hspace{0.05cm}.$$		:$$G_{\rm FM} (f_{\rm N} = 1\,{\rm kHz}) = 25.28\,{\rm dB} + 9.54\,{\rm dB}\hspace{0.15cm}\underline {= 34.82\,{\rm dB}}\hspace{0.05cm}.$$

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	:$$\rho_{{\rm AM} } = \frac{P_{\rm S}}{N_0 \cdot f_{\rm N} } = \xi\hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{{\rm AM} } = 10 \cdot {\rm lg} \hspace{0.15cm}\xi \hspace{0.05cm},$$		:$$\rho_{{\rm AM} } = \frac{P_{\rm S}}{N_0 \cdot f_{\rm N} } = \xi\hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{{\rm AM} } = 10 \cdot {\rm lg} \hspace{0.15cm}\xi \hspace{0.05cm},$$
	* Sink SNR and sink-to-noise ratio in frequency modulation $\rm (FM)$  without pre-emphasis/de-emphasis:		* Sink SNR and sink-to-noise ratio in frequency modulation $\rm (FM)$  without pre-emphasis/de-emphasis:
	:$$ \rho_{\rm FM} = {3}/{2 } \cdot \eta^2 \cdot \rho_{\rm AM } \hspace{0.3cm}\Rightarrow \hspace{0.3cm}		:$$ \rho_{\rm FM} = {3}/{2 } \cdot \eta^2 \cdot \rho_{\rm AM } \hspace{0.3cm}\Rightarrow \hspace{0.3cm}G_{\rm FM} = 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{\rm FM} -10 \cdot {\rm lg} \hspace{0.15cm}\rho_{\rm AM}= 10 \cdot {\rmlg} \hspace{0.15cm}{3}/{2 } \cdot \eta^2\hspace{0.05cm},$$
	G_{\rm FM} = 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{\rm FM} -
	10 \cdot {\rm lg} \hspace{0.15cm}\rho_{\rm AM}= 10 \cdot {\rm
	lg} \hspace{0.15cm}{3}/{2 } \cdot \eta^2
	\hspace{0.05cm},$$
	*Sink SNR and sink-to-noise ratio in frequency modulation $\rm (FM)$ using pre-emphasis/de-emphasis:		*Sink SNR and sink-to-noise ratio in frequency modulation $\rm (FM)$ using pre-emphasis/de-emphasis:
	:$$ \rho_{\rm DE} = \frac{(f_{\rm N}/f_{\rm G})^3}{3 \cdot (f_{\rm N}/f_{\rm G} - \arctan (f_{\rm N}/f_{\rm G}) }		:$$ \rho_{\rm DE} = \frac{(f_{\rm N}/f_{\rm G})^3}{3 \cdot (f_{\rm N}/f_{\rm G} - \arctan (f_{\rm N}/f_{\rm G}) }\hspace{0.3cm}\Rightarrow \hspace{0.3cm}G_{\rm DE} = 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{\rm DE} -10 \cdot {\rm lg} \hspace{0.15cm}\rho_{\rm FM}\hspace{0.05cm}$$
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}
	G_{\rm DE} = 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{\rm DE} -
	10 \cdot {\rm lg} \hspace{0.15cm}\rho_{\rm FM}\hspace{0.05cm}$$


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	'''(2)''' The frequency modulation is designed for the maximum frequency  $B_{\rm NF} = f_\text{N, max}= 9 \ \rm kHz$ .  Then the (maximum) frequency deviation should be  $Δf_{\rm A} = 45\ \rm kHz$ .		'''(2)''' The frequency modulation is designed for the maximum frequency  $B_{\rm NF} = f_\text{N, max}= 9 \ \rm kHz$ .  Then the (maximum) frequency deviation should be  $Δf_{\rm A} = 45\ \rm kHz$ .
	*From this it follows for the modulation index:		*From this it follows for the modulation index:
	:$$ \eta = \frac{\Delta f_{\rm A}}{f_{\rm N} } = 5 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}		:$$ \eta = \frac{\Delta f_{\rm A}}{f_{\rm N} } = 5 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}G_{\rm FM} (f_{\rm N} = 9\,{\rm kHz}) = 10 \cdot {\rm lg}\hspace{0.15cm}(1.5 \cdot 5^2) \hspace{0.15cm}\underline {\approx 15.74\,{\rm dB}}\hspace{0.05cm}.$$
	G_{\rm FM} (f_{\rm N} = 9\,{\rm kHz}) = 10 \cdot {\rm lg}
	\hspace{0.15cm}(1.5 \cdot 5^2) \hspace{0.15cm}\underline {\approx 15.74\,{\rm dB}}
	\hspace{0.05cm}.$$

	*Using the message frequency  $ f_{\rm N} = 3 \ \rm kHz$  results in a modulation index larger by a factor of   $3$  and thus a signal-to-noise ratio larger by a factor of  $10 · \lg \ 9 = 9.54 \ \rm dB$ :		*Using the message frequency  $ f_{\rm N} = 3 \ \rm kHz$  results in a modulation index larger by a factor of   $3$  and thus a signal-to-noise ratio larger by a factor of  $10 · \lg \ 9 = 9.54 \ \rm dB$ :
	:$$G_{\rm FM} (f_{\rm N} = 3\,{\rm kHz}) = 10 \cdot {\rm lg}		:$$G_{\rm FM} (f_{\rm N} = 3\,{\rm kHz}) = 10 \cdot {\rm lg}\hspace{0.15cm}(1.5 \cdot 15^2) \hspace{0.15cm}\underline {\approx 25.28\,{\rm dB}}\hspace{0.05cm}.$$
	\hspace{0.15cm}(1.5 \cdot 15^2) \hspace{0.15cm}\underline {\approx 25.28\,{\rm dB}}
	\hspace{0.05cm}.$$

	*Another gain results from the transition from   $3\ \rm kHz$  to  $1\ \rm kHz$:		*Another gain results from the transition from   $3\ \rm kHz$  to  $1\ \rm kHz$:
	:$$G_{\rm FM} (f_{\rm N} = 1\,{\rm kHz}) = 25.28\,{\rm dB} + 9.54\,{\rm		:$$G_{\rm FM} (f_{\rm N} = 1\,{\rm kHz}) = 25.28\,{\rm dB} + 9.54\,{\rmdB}\hspace{0.15cm}\underline {= 34.82\,{\rm dB}}\hspace{0.05cm}.$$
	dB}\hspace{0.15cm}\underline {= 34.82\,{\rm dB}}
	\hspace{0.05cm}.$$


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	'''(4)'''  Using the given formula, the following "gains due to pre-emphasis/de-emphasis" are obtained:		'''(4)'''  Using the given formula, the following "gains due to pre-emphasis/de-emphasis" are obtained:
	:$$G_{\rm DE} (f_{\rm N} = 9\,{\rm kHz}) = 10 \cdot {\rm		:$$G_{\rm DE} (f_{\rm N} = 9\,{\rm kHz}) = 10 \cdot {\rmlg}\hspace{0.15cm} \frac{(f_{\rm N}/f_{\rm G})^3}{3 \cdot (f_{\rmN}/f_{\rm G} - \arctan (f_{\rm N}/f_{\rm G}) }= 10 \cdot {\rm lg}\hspace{0.15cm} \frac{3^3}{3 \cdot (3 -1.249) }\hspace{0.15cm}\underline {\approx 7.1\,{\rm dB}}\hspace{0.05cm},$$
	lg}\hspace{0.15cm} \frac{(f_{\rm N}/f_{\rm G})^3}{3 \cdot (f_{\rm
	N}/f_{\rm G} - \arctan (f_{\rm N}/f_{\rm G}) }= 10 \cdot {\rm lg}\hspace{0.15cm} \frac{3^3}{3 \cdot (3 -
	1.249) }\hspace{0.15cm}\underline {\approx 7.1\,{\rm dB}}\hspace{0.05cm},$$
	:$$ G_{\rm DE} (f_{\rm N} = 3\,{\rm kHz}) = 10 \cdot {\rm lg}\hspace{0.15cm} \frac{1^3}{3 \cdot (1 - \pi/4) }\hspace{0.15cm}\underline {\approx 1.9\,{\rm dB}}\hspace{0.05cm},$$		:$$ G_{\rm DE} (f_{\rm N} = 3\,{\rm kHz}) = 10 \cdot {\rm lg}\hspace{0.15cm} \frac{1^3}{3 \cdot (1 - \pi/4) }\hspace{0.15cm}\underline {\approx 1.9\,{\rm dB}}\hspace{0.05cm},$$
	:$$G_{\rm DE} (f_{\rm N} = 1\,{\rm kHz}) = 10 \cdot {\rm lg}\hspace{0.15cm} \frac{(1/3)^3}{3 \cdot (1/3 - 0.322) }\hspace{0.15cm}\underline {\approx 0.28\,{\rm dB}}\hspace{0.05cm}.$$		:$$G_{\rm DE} (f_{\rm N} = 1\,{\rm kHz}) = 10 \cdot {\rm lg}\hspace{0.15cm} \frac{(1/3)^3}{3 \cdot (1/3 - 0.322) }\hspace{0.15cm}\underline {\approx 0.28\,{\rm dB}}\hspace{0.05cm}.$$

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	[[Category:Modulation Methods: Exercises\|^3.3 Noise Influence with PM and FM^]]		[[Category:Modulation Methods: Exercises\|^3.3 Noise Influence with PM and FM^]]
			[[de:Aufgaben:3.10_Preemphase–Deemphase]]

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	}}		}}

	[[File:P_ID1116__Mod_A_3_10.png\|right\|frame\|~~Realisierung einer Preemphase~~]]		[[File:P_ID1116__Mod_A_3_10.png\|right\|frame\|Realization of a pre-emphase]]
	In voice and audio signal transmission, the signal frequency band is pre-distorted before the FM modulator via an RC high-pass filter according to the diagram. This measure is called "pre-emphasis"   $\rm (PE)$.		In voice and audio signal transmission, the signal frequency band is pre-distorted before the FM modulator via an RC high-pass filter according to the diagram. This measure is called "pre-emphasis"   $\rm (PE)$.

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Richtig sind~~ <u>~~der erste und der letzte Lösungsvorschlag~~</u>:		'''(1)'''  The <u>first and last answer</u> are correct:
	*~~Der Betragsfrequenzgang des Deemphase–Netzwerks ist wie folgt festgelegt~~:		*The magnitude frequency response of the de-emphasis network is defined as follows:
	:$$ \|H_{\rm DE} (f)\| = \frac{1}{\|H_{\rm PE} (f)\|}= \frac{1}{\alpha}\cdot \frac{1}{\sqrt{1 + (f/f_{\rm G})^2}} \hspace{0.05cm}.$$		:$$ \|H_{\rm DE} (f)\| = \frac{1}{\|H_{\rm PE} (f)\|}= \frac{1}{\alpha}\cdot \frac{1}{\sqrt{1 + (f/f_{\rm G})^2}} \hspace{0.05cm}.$$
	*~~Der Frequenzgang eines einfachen RC–Tiefpasses – auch bekannt als Tiefpass erster Ordnung – lautet~~:		*The frequency response of a simple RC low-pass filter - also known as a first-order low-pass filter - is:
	:$$ H_{\rm RC-TP} (f) = \frac{1}{{1 + {\rm j}\cdot f/f_{\rm G}}} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \|H_{\rm RC-TP} (f)\| = \frac{1}{\sqrt{1 + (f/f_{\rm G})^2}}\hspace{0.05cm}.$$		:$$ H_{\rm RC-TP} (f) = \frac{1}{{1 + {\rm j}\cdot f/f_{\rm G}}} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} \|H_{\rm RC-TP} (f)\| = \frac{1}{\sqrt{1 + (f/f_{\rm G})^2}}\hspace{0.05cm}.$$



	'''(2)'''  ~~Die Frequenzmodulation ist auf die maximale Frequenz~~  $B_{\rm NF} = f_\text{N, max}= 9 \ \rm kHz$  ~~ausgelegt~~.  ~~Dann soll der~~ (~~maximale~~) ~~Frequenzhub~~  $Δf_{\rm A} = 45\ \rm kHz$  ~~betragen~~.		'''(2)''' The frequency modulation is designed for the maximum frequency  $B_{\rm NF} = f_\text{N, max}= 9 \ \rm kHz$ .  Then the (maximum) frequency deviation should be  $Δf_{\rm A} = 45\ \rm kHz$ .
	*~~Daraus folgt für den Modulationsindex~~:		*From this it follows for the modulation index:
	:$$ \eta = \frac{\Delta f_{\rm A}}{f_{\rm N} } = 5 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}		:$$ \eta = \frac{\Delta f_{\rm A}}{f_{\rm N} } = 5 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
	G_{\rm FM} (f_{\rm N} = 9\,{\rm kHz}) = 10 \cdot {\rm lg}		G_{\rm FM} (f_{\rm N} = 9\,{\rm kHz}) = 10 \cdot {\rm lg}
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*~~Mit der Nachrichtenfrequenz~~  $ f_{\rm N} = 3 \ \rm kHz$  ~~ergibt sich ein um den Faktor~~  $3$  ~~größerer Modulationsindex und damit ein um den Faktor~~  $10 · \lg \ 9 = 9.54 \ \rm dB$  ~~größerer Störabstand~~:		*Using the message frequency  $ f_{\rm N} = 3 \ \rm kHz$  results in a modulation index larger by a factor of   $3$  and thus a signal-to-noise ratio larger by a factor of  $10 · \lg \ 9 = 9.54 \ \rm dB$ :
	:$$G_{\rm FM} (f_{\rm N} = 3\,{\rm kHz}) = 10 \cdot {\rm lg}		:$$G_{\rm FM} (f_{\rm N} = 3\,{\rm kHz}) = 10 \cdot {\rm lg}
	\hspace{0.15cm}(1.5 \cdot 15^2) \hspace{0.15cm}\underline {\approx 25.28\,{\rm dB}}		\hspace{0.15cm}(1.5 \cdot 15^2) \hspace{0.15cm}\underline {\approx 25.28\,{\rm dB}}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*~~Ein weiterer Zugewinn ergibt sich durch den Übergang von~~  $3\ \rm kHz$  ~~auf~~  $1\ \rm kHz$:		*Another gain results from the transition from   $3\ \rm kHz$  to  $1\ \rm kHz$:
	:$$G_{\rm FM} (f_{\rm N} = 1\,{\rm kHz}) = 25.28\,{\rm dB} + 9.54\,{\rm		:$$G_{\rm FM} (f_{\rm N} = 1\,{\rm kHz}) = 25.28\,{\rm dB} + 9.54\,{\rm
	dB}\hspace{0.15cm}\underline {= 34.82\,{\rm dB}}		dB}\hspace{0.15cm}\underline {= 34.82\,{\rm dB}}
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	'''(3)'''  ~~Der maximale Frequenzhub ergibt sich für~~ $f_{\rm N} = B_{\rm NF}$.		'''(3)'''  The maximum frequency deviation is obtained for $f_{\rm N} = B_{\rm NF}$.
	*~~Daraus folgt mit~~ $f_{\rm G} = 3 \ \rm kHz$ ~~und~~ $B_{\rm NF} = 9 \ \rm kHz$:		*It follows, with $f_{\rm G} = 3 \ \rm kHz$ and $B_{\rm NF} = 9 \ \rm kHz$:
	:$$\Delta f_{\rm A} (B_{\rm NF}) = \Delta f_{\rm A, \hspace{0.08cm}min} \cdot \sqrt{{1 + \left(\frac{B_{\rm NF}}{f_{\rm G}}\right)^2}} = \sqrt {10} \cdot \Delta f_{\rm A, \hspace{0.08cm}min}= \Delta f_{\rm A, \hspace{0.08cm}max} = 45\,{\rm kHz}$$		:$$\Delta f_{\rm A} (B_{\rm NF}) = \Delta f_{\rm A, \hspace{0.08cm}min} \cdot \sqrt{{1 + \left(\frac{B_{\rm NF}}{f_{\rm G}}\right)^2}} = \sqrt {10} \cdot \Delta f_{\rm A, \hspace{0.08cm}min}= \Delta f_{\rm A, \hspace{0.08cm}max} = 45\,{\rm kHz}$$
	:$$\Rightarrow \hspace{0.3cm} \Delta f_{\rm A, \hspace{0.08cm}min} = \frac{45\,{\rm kHz}}{\sqrt {10}}\hspace{0.15cm}\underline {\approx 14.23\,{\rm kHz}}\hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} \Delta f_{\rm A, \hspace{0.08cm}min} = \frac{45\,{\rm kHz}}{\sqrt {10}}\hspace{0.15cm}\underline {\approx 14.23\,{\rm kHz}}\hspace{0.05cm}.$$


	'''(4)'''  ~~Mit der angegebenen Formel erhält man folgende~~ "~~Gewinne durch Premphase~~/~~Deemphase~~":		'''(4)'''  Using the given formula, the following "gains due to pre-emphasis/de-emphasis" are obtained:
	:$$G_{\rm DE} (f_{\rm N} = 9\,{\rm kHz}) = 10 \cdot {\rm		:$$G_{\rm DE} (f_{\rm N} = 9\,{\rm kHz}) = 10 \cdot {\rm
	lg}\hspace{0.15cm} \frac{(f_{\rm N}/f_{\rm G})^3}{3 \cdot (f_{\rm		lg}\hspace{0.15cm} \frac{(f_{\rm N}/f_{\rm G})^3}{3 \cdot (f_{\rm

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	<quiz display=simple>		<quiz display=simple>
	{~~Geben Sie eine mögliche Realisierung des Deemphase–Netzwerks~~  $H_{\rm DE}(f)$  an.~~  Welche der folgenden Aussagen sind richtig~~?		{Give a possible realization of the de-emphasis network  $H_{\rm DE}(f)$ . Which of the following statements are correct?
	\|type="[]"}		\|type="[]"}
	+ $H_{\rm DE}(f)$  ~~ist ein Tiefpass erster Ordnung~~.		+ $H_{\rm DE}(f)$  is a first-order low-pass filter.
	- $H_{\rm DE}(f)$  ~~ist ein Hochpass erster Ordnung~~.		- $H_{\rm DE}(f)$  is a first-order high-pass filter.
	- $H_{\rm DE}(f)$  ~~ist ein Bandpass~~.		- $H_{\rm DE}(f)$  is a bandpass.
	+ ~~Zusätzlich muss der Faktor~~  $α$  ~~korrigiert werden~~.		+ In addition, the factor  $α$  must be corrected.


	{~~Wie groß ist der Störabstandsgewinn~~  $G_{\rm FM}$  ~~der herkömmlichen~~ FM ~~gegenüber~~ AM ~~bei den genannten Nachrichtenfrequenzen~~  $ f_{\rm N}$?		{What is the signal-to-noise ratio advantage  $G_{\rm FM}$  of conventional FM over AM at the given message frequencies  $ f_{\rm N}$?
	\|type="{}"}		\|type="{}"}
	$ f_{\rm N} = \text{9 kHz:} \hspace{0.2cm} G_{\rm FM} \ = \ $ { 15.74 3% } $\ \rm dB$		$ f_{\rm N} = \text{9 kHz:} \hspace{0.2cm} G_{\rm FM} \ = \ $ { 15.74 3% } $\ \rm dB$
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	$ f_{\rm N} = \text{1 kHz:} \hspace{0.2cm} G_{\rm FM} \ = \ $ { 34.82 3% } $\ \rm dB$		$ f_{\rm N} = \text{1 kHz:} \hspace{0.2cm} G_{\rm FM} \ = \ $ { 34.82 3% } $\ \rm dB$

	{~~Wie groß ist~~  $Δf_\text{A, min}$   ~~mit~~   $Δf_\text{A, max} = 45 \ \rm kHz$   ~~und~~   $B_{\rm NF} = f_\text{N, max}= 9 \ \rm kHz$   ~~zu wählen~~?		{What  $Δf_\text{A, min}$   should we choose when   $Δf_\text{A, max} = 45 \ \rm kHz$   and   $B_{\rm NF} = f_\text{N, max}= 9 \ \rm kHz$  ?
	\|type="{}"}		\|type="{}"}
	$Δf_\text{A, min} \ = \ $ { 14.23 3% } $\ \rm kHz$		$Δf_\text{A, min} \ = \ $ { 14.23 3% } $\ \rm kHz$

	{~~Welcher zusätzliche Gewinn ist durch Preemphase~~/~~Deemphase zu erzielen~~?		{What is the additional efficiency gain to be obtained by pre-emphasis/de-emphasis??
	\|type="{}"}		\|type="{}"}
	$ f_{\rm N} = \text{9 kHz:} \hspace{0.2cm} G_{\rm DE} \ = \ $ { 7.1 3% } $\ \rm dB$		$ f_{\rm N} = \text{9 kHz:} \hspace{0.2cm} G_{\rm DE} \ = \ $ { 7.1 3% } $\ \rm dB$

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	[[File:P_ID1116__Mod_A_3_10.png\|right\|frame\|Realisierung einer Preemphase]]		[[File:P_ID1116__Mod_A_3_10.png\|right\|frame\|Realisierung einer Preemphase]]
	~~Bei der Sprach– und Tonsignalübertragung wird das Signalfrequenzband vor dem FM–Modulator über ein RC–Hochpassglied gemäß der Skizze vorverzerrt~~. ~~Man bezeichnet diese Maßnahme als~~ "~~Preemphase~~"  $\rm (PE)$.		In voice and audio signal transmission, the signal frequency band is pre-distorted before the FM modulator via an RC high-pass filter according to the diagram. This measure is called "pre-emphasis"   $\rm (PE)$.

	~~Der Amplitudengang des Preemphase–Netzwerks lautet~~		The amplitude response of the preemphasis network, together with
	*~~mit den beiden Grenzfrequenzen~~  $f_{\rm G1} = (2π · R_1 · C)^{–1}$  ~~und~~  $f_{\rm G2} = f_{\rm G1}/α_0$, ~~sowie~~		*the two cutoff frequencies  $f_{\rm G1} = (2π · R_1 · C)^{–1}$  and  $f_{\rm G2} = f_{\rm G1}/α_0$, as well as
	*~~dem Gleichsignalübertragungsfaktor~~  $α_0 = R_2/(R_1 + R_2)$:		*the DC signal transmission factor  $α_0 = R_2/(R_1 + R_2)$
			is given by:
	:$$ \|H_{\rm PE} (f)\| = \alpha_0 \cdot \sqrt{\frac{1 + (f/f_{\rm G1})^2}{1 + (f/f_{\rm G2})^2}} \hspace{0.05cm}.$$		:$$ \|H_{\rm PE} (f)\| = \alpha_0 \cdot \sqrt{\frac{1 + (f/f_{\rm G1})^2}{1 + (f/f_{\rm G2})^2}} \hspace{0.05cm}.$$

	~~Für die Praxis kann man davon ausgehen~~, ~~dass die maximale Nachrichtenfrequenz~~  $f_{\rm N}$  ~~sehr viel kleiner als~~  $f_{\rm G2}$  ~~ist~~.		For practical purposes, we can assume that the maximum message frequency  $f_{\rm N}$  is much smaller than  $f_{\rm G2}$ .

	~~Berücksichtigt man weiter, dass der Gleichsignalübertragungsfaktor~~  $α_0$  ~~durch eine Verstärkung um &~~nbsp;$α$  ~~verändert werden kann~~, ~~so ist im Weiteren von folgendem Preemphase–Frequenzgang auszugehen~~  $(f_{\rm G} = f_{\rm G1} = 3 \ \rm kHz)$:		If we further consider that the DC signal transmission factor $α_0$  can be changed by an amplification of nbsp;$α$ , we can further assume the following pre-emphasis frequency response
			where  $(f_{\rm G} = f_{\rm G1} = 3 \ \rm kHz)$:
	:$$\|H_{\rm PE} (f)\| \approx \alpha \cdot \sqrt{{1 + \left({f}/{f_{\rm G}}\right)^2}} \hspace{0.05cm}.$$		:$$\|H_{\rm PE} (f)\| \approx \alpha \cdot \sqrt{{1 + \left({f}/{f_{\rm G}}\right)^2}} \hspace{0.05cm}.$$
	~~Mit diesem Netzwerk lautet der Frequenzhub~~  $Δf_{\rm A}$  ~~in Abhängigkeit der Nachrichtenfrequenz~~ $f_{\rm N}$:		In this network, the frequency deviation is  $Δf_{\rm A}$  as a function of the message frequency $f_{\rm N}$:
	:$$ \Delta f_{\rm A} (f_{\rm N}) = \Delta f_{\rm A, \hspace{0.08cm}min} \cdot \sqrt{{1 + \left({f_{\rm N}}/{f_{\rm G}}\right)^2}} \hspace{0.05cm}.$$		:$$ \Delta f_{\rm A} (f_{\rm N}) = \Delta f_{\rm A, \hspace{0.08cm}min} \cdot \sqrt{{1 + \left({f_{\rm N}}/{f_{\rm G}}\right)^2}} \hspace{0.05cm}.$$
	*~~Hierbei ist~~  $Δf_\text{A, min}$  ~~der Frequenzhub für sehr kleine Frequenzen~~  $(f_{\rm N} → 0)$.		*Here  $Δf_\text{A, min}$  is the frequency deviation for very small frequencies  $(f_{\rm N} → 0)$.
	*~~Dieser Parameter ist~~ so ~~zu wählen, dass der maximale Frequenzhub~~  $Δf_\text{A, max}$  ~~nicht größer wird als~~  $45 \ \rm kHz$.		*This parameter should be chosen so that the maximum frequency deviation  $Δf_\text{A, max}$  does not exceed  $45 \ \rm kHz$.


	~~Um das Nutzsignal nicht zu verfälschen~~, ~~muss diese Vorverzerrung durch ein ~~ "~~Deemphase~~"~~–Netzwerk beim Empfänger wieder ausgeglichen werden~~.~~  Ziel und Zweck von Preemphase~~/~~Deemphase ist es allein, die Abhängigkeit des Signal–zu–Rausch–Leistungsverhältnisses von der Signalfrequenz zu vermindern~~.		In order not to distort the useful signal, this pre-emphasis must be rebalanced by a "de-emphasis" network at the receiver. The goal and purpose of preemphasis/deemphasis is solely to reduce the dependence of the signal-to-noise power ratio on the signal frequency.

	In ~~dieser Aufgabe werden folgende Größen verwendet~~:		In this task, the following quantities are used:
	* ~~Sinken–SNR bei Zweiseitenband~~-~~Amplitudenmodulation~~  $\rm (~~ZSB–AM~~)$:		* Sink SNR in double-sideband amplitude modulation (DSB-AM)  $\rm (DSB–AM)$:
	:$$\rho_{{\rm AM} } = \frac{P_{\rm S}}{N_0 \cdot f_{\rm N} } = \xi\hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{{\rm AM} } = 10 \cdot {\rm lg} \hspace{0.15cm}\xi \hspace{0.05cm},$$		:$$\rho_{{\rm AM} } = \frac{P_{\rm S}}{N_0 \cdot f_{\rm N} } = \xi\hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{{\rm AM} } = 10 \cdot {\rm lg} \hspace{0.15cm}\xi \hspace{0.05cm},$$
	* ~~Sinken–SNR und Störabstandsgewinn bei Frequenzmodulation ~~ $\rm (FM)$  ~~ohne Preemphase~~/~~Deemphase~~:		* Sink SNR and sink-to-noise ratio in frequency modulation $\rm (FM)$  without pre-emphasis/de-emphasis:
	:$$ \rho_{\rm FM} = {3}/{2 } \cdot \eta^2 \cdot \rho_{\rm AM } \hspace{0.3cm}\Rightarrow \hspace{0.3cm}		:$$ \rho_{\rm FM} = {3}/{2 } \cdot \eta^2 \cdot \rho_{\rm AM } \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
	G_{\rm FM} = 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{\rm FM} -		G_{\rm FM} = 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{\rm FM} -
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	lg} \hspace{0.15cm}{3}/{2 } \cdot \eta^2		lg} \hspace{0.15cm}{3}/{2 } \cdot \eta^2
	\hspace{0.05cm},$$		\hspace{0.05cm},$$
	*~~Sinken–SNR und Störabstandsgewinn bei Frequenzmodulation ~~ $\rm (FM)$  ~~durch Preemphase~~/~~Deemphase~~:		*Sink SNR and sink-to-noise ratio in frequency modulation $\rm (FM)$ using pre-emphasis/de-emphasis:
	:$$ \rho_{\rm DE} = \frac{(f_{\rm N}/f_{\rm G})^3}{3 \cdot (f_{\rm N}/f_{\rm G} - \arctan (f_{\rm N}/f_{\rm G}) }		:$$ \rho_{\rm DE} = \frac{(f_{\rm N}/f_{\rm G})^3}{3 \cdot (f_{\rm N}/f_{\rm G} - \arctan (f_{\rm N}/f_{\rm G}) }
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}		\hspace{0.3cm}\Rightarrow \hspace{0.3cm}
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	''~~Hinweise~~:''		''Hints:''
	*~~Die Aufgabe gehört zum Kapitel~~  [[Modulation_Methods/~~Rauscheinfluss_bei_Winkelmodulation~~\|~~Rauscheinfluss bei Winkelmodulation~~]].		*This exercise belongs to the chapter  [[Modulation_Methods/Influence_of_Noise_on_Systems_with_Angle_Modulation\|Influence of Noise on Systems with Angle Modulation]].
	*~~Bezug genommen wirdinsbesondere auf den Abschnitt~~  [[Modulation_Methods/~~Rauscheinfluss_bei_Winkelmodulation~~#~~Preemphase_und_Deemphase~~\|~~Preemphase und Deemphase~~]].		*Particular reference is made to the section  [[Modulation_Methods/Influence_of_Noise_on_Systems_with_Angle_Modulation#Pre-emphasis_and_de-emphasis\|Pre-emphasis and de-emphasis]].
	*~~Gehen Sie in der gesamten Aufgabe von einem Nachrichtensignal aus~~, ~~das Frequenzen bis einschließlich~~  $B_{\rm NF}= 9 \ \rm kHz$  ~~beinhaltet~~.		*Throughout the task, assume a message signal containing frequencies up to and including  $B_{\rm NF}= 9 \ \rm kHz$ .




	===~~Fragebogen~~===		===Questions===

	<quiz display=simple>		<quiz display=simple>
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	</quiz>		</quiz>

	===~~Musterlösung~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Richtig sind <u>der erste und der letzte Lösungsvorschlag</u>:		'''(1)'''  Richtig sind <u>der erste und der letzte Lösungsvorschlag</u>:

Guenter at 13:39, 16 March 2022

2022-03-16T13:39:45Z

← Older revision		Revision as of 15:39, 16 March 2022
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	[[Category:Modulation Methods: Exercises\|^3.3 ~~Rauscheinfluss bei~~ PM ~~und~~ FM^]]		[[Category:Modulation Methods: Exercises\|^3.3 Noise Influence with PM and FM^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 3.11: Preemphase und Deemphase to Aufgaben:Exercise 3.11: Pre-Emphase and De-Emphase

2022-03-16T13:37:32Z

Guenter moved page Aufgaben:Aufgabe 3.11: Preemphase und Deemphase to Aufgaben:Exercise 3.11: Pre-Emphase and De-Emphase

← Older revision	Revision as of 15:37, 16 March 2022
(No difference)

← Older revision		Revision as of 17:57, 16 March 2026
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	:$$\rho_{{\rm AM} } = \frac{P_{\rm S}}{N_0 \cdot f_{\rm N} } = \xi\hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{{\rm AM} } = 10 \cdot {\rm lg} \hspace{0.15cm}\xi \hspace{0.05cm},$$		:$$\rho_{{\rm AM} } = \frac{P_{\rm S}}{N_0 \cdot f_{\rm N} } = \xi\hspace{0.3cm}\Rightarrow \hspace{0.3cm} 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{{\rm AM} } = 10 \cdot {\rm lg} \hspace{0.15cm}\xi \hspace{0.05cm},$$
	* Sink SNR and sink-to-noise ratio in frequency modulation $\rm (FM)$  without pre-emphasis/de-emphasis:		* Sink SNR and sink-to-noise ratio in frequency modulation $\rm (FM)$  without pre-emphasis/de-emphasis:
	:$$ \rho_{\rm FM} = {3}/{2 } \cdot \eta^2 \cdot \rho_{\rm AM } \hspace{0.3cm}\Rightarrow \hspace{0.3cm}G_{\rm FM} = 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{\rm FM} -10 \cdot {\rm lg} \hspace{0.15cm}\rho_{\rm AM}= 10 \cdot {\~~rmlg~~} \hspace{0.15cm}{3}/{2 } \cdot \eta^2\hspace{0.05cm},$$		:$$ \rho_{\rm FM} = {3}/{2 } \cdot \eta^2 \cdot \rho_{\rm AM } \hspace{0.3cm}\Rightarrow \hspace{0.3cm}G_{\rm FM} = 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{\rm FM} -10 \cdot {\rm lg} \hspace{0.15cm}\rho_{\rm AM}= 10 \cdot {\rm lg} \hspace{0.15cm}{3}/{2 } \cdot \eta^2\hspace{0.05cm},$$
	*Sink SNR and sink-to-noise ratio in frequency modulation $\rm (FM)$ using pre-emphasis/de-emphasis:		*Sink SNR and sink-to-noise ratio in frequency modulation $\rm (FM)$ using pre-emphasis/de-emphasis:
	:$$ \rho_{\rm DE} = \frac{(f_{\rm N}/f_{\rm G})^3}{3 \cdot (f_{\rm N}/f_{\rm G} - \arctan (f_{\rm N}/f_{\rm G}) }\hspace{0.3cm}\Rightarrow \hspace{0.3cm}G_{\rm DE} = 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{\rm DE} -10 \cdot {\rm lg} \hspace{0.15cm}\rho_{\rm FM}\hspace{0.05cm}$$		:$$ \rho_{\rm DE} = \frac{(f_{\rm N}/f_{\rm G})^3}{3 \cdot (f_{\rm N}/f_{\rm G} - \arctan (f_{\rm N}/f_{\rm G}) }\hspace{0.3cm}\Rightarrow \hspace{0.3cm}G_{\rm DE} = 10 \cdot {\rm lg} \hspace{0.15cm}\rho_{\rm DE} -10 \cdot {\rm lg} \hspace{0.15cm}\rho_{\rm FM}\hspace{0.05cm}$$
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	'''(4)'''  Using the given formula, the following "gains due to pre-emphasis/de-emphasis" are obtained:		'''(4)'''  Using the given formula, the following "gains due to pre-emphasis/de-emphasis" are obtained:
	:$$G_{\rm DE} (f_{\rm N} = 9\,{\rm kHz}) = 10 \cdot {\~~rmlg~~}\hspace{0.15cm} \frac{(f_{\rm N}/f_{\rm G})^3}{3 \cdot (f_{\~~rmN~~}/f_{\rm G} - \arctan (f_{\rm N}/f_{\rm G}) }= 10 \cdot {\rm lg}\hspace{0.15cm} \frac{3^3}{3 \cdot (3 -1.249) }\hspace{0.15cm}\underline {\approx 7.1\,{\rm dB}}\hspace{0.05cm},$$		:$$G_{\rm DE} (f_{\rm N} = 9\,{\rm kHz}) = 10 \cdot {\rm lg}\hspace{0.15cm} \frac{(f_{\rm N}/f_{\rm G})^3}{3 \cdot (f_{\rm N}/f_{\rm G} - \arctan (f_{\rm N}/f_{\rm G}) }= 10 \cdot {\rm lg}\hspace{0.15cm} \frac{3^3}{3 \cdot (3 -1.249) }\hspace{0.15cm}\underline {\approx 7.1\,{\rm dB}}\hspace{0.05cm},$$
	:$$ G_{\rm DE} (f_{\rm N} = 3\,{\rm kHz}) = 10 \cdot {\rm lg}\hspace{0.15cm} \frac{1^3}{3 \cdot (1 - \pi/4) }\hspace{0.15cm}\underline {\approx 1.9\,{\rm dB}}\hspace{0.05cm},$$		:$$ G_{\rm DE} (f_{\rm N} = 3\,{\rm kHz}) = 10 \cdot {\rm lg}\hspace{0.15cm} \frac{1^3}{3 \cdot (1 - \pi/4) }\hspace{0.15cm}\underline {\approx 1.9\,{\rm dB}}\hspace{0.05cm},$$
	:$$G_{\rm DE} (f_{\rm N} = 1\,{\rm kHz}) = 10 \cdot {\rm lg}\hspace{0.15cm} \frac{(1/3)^3}{3 \cdot (1/3 - 0.322) }\hspace{0.15cm}\underline {\approx 0.28\,{\rm dB}}\hspace{0.05cm}.$$		:$$G_{\rm DE} (f_{\rm N} = 1\,{\rm kHz}) = 10 \cdot {\rm lg}\hspace{0.15cm} \frac{(1/3)^3}{3 \cdot (1/3 - 0.322) }\hspace{0.15cm}\underline {\approx 0.28\,{\rm dB}}\hspace{0.05cm}.$$
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	[[Category:Modulation Methods: Exercises\|^3.3 Noise Influence with PM and FM^]]		[[Category:Modulation Methods: Exercises\|^3.3 Noise Influence with PM and FM^]]
	[[de:Aufgaben:3.~~10_Preemphase–Deemphase~~]]		[[de:Aufgaben:Aufgabe 3.11: Pre-emphase und De-emphase]]