Aufgaben:Exercise 4.13: FSK Demodulation - Revision history

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	[[File:EN_Mod_A_4_12_v2.png\|right\|frame\|Signal waveforms of FSK demodulation]]		[[File:EN_Mod_A_4_12_v2.png\|right\|frame\|Signal waveforms of FSK demodulation]]
	The  [[Modulation_Methods/~~Nonlinear_Digital_Modulation~~#Coherent_demodulation_of_FSK\| block diagram]]  of the coherent FSK demodulator has already been given in the theory section, but in this exercise we will assume the lower system variant. Noise components are not considered here.		The  [[Modulation_Methods/Non-Linear_Digital_Modulation#Coherent_demodulation_of_FSK\| block diagram]]  of the coherent FSK demodulator has already been given in the theory section, but in this exercise we will assume the lower system variant. Noise components are not considered here.

	The figure on the right shows the signal waveforms at different points of the block diagram, where three symbols are drawn in each case, separated by dashed lines:		The figure on the right shows the signal waveforms at different points of the block diagram, where three symbols are drawn in each case, separated by dashed lines:
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	''Hints:''		''Hints:''
	*This exercise belongs to the chapter  [[Modulation_Methods/Nonlinear_Digital_Modulation\|Nonlinear Digital Modulation]].		*This exercise belongs to the chapter  [[Modulation_Methods/Nonlinear_Digital_Modulation\|Nonlinear Digital Modulation]].
	*Particular reference is made to the page  [[Modulation_Methods/~~Nonlinear_Digital_Modulation~~#Coherent_demodulation_of_FSK\|Coherent demodulation of FSK]].		*Particular reference is made to the page  [[Modulation_Methods/Non-Linear_Digital_Modulation#Coherent_demodulation_of_FSK\|Coherent demodulation of FSK]].
	* A technical hint: By enlarging the image, one can recognize the signal curves a little better.		* A technical hint: By enlarging the image, one can recognize the signal curves a little better.
	Please accept an apology on behalf of the author: blue on a black background is not great.		Please accept an apology on behalf of the author: blue on a black background is not great.
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	[[Category:Modulation Methods: Exercises\|^4.4 Non-linear Digital Modulation^]]		[[Category:Modulation Methods: Exercises\|^4.4 Non-linear Digital Modulation^]]
	[[de:Aufgaben:4.~~12_FSK–Demodulation~~]]		[[de:Aufgaben:Aufgabe 4.13: FSK–Demodulation]]

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	'''(3)'''  In this range, where the first term represents the received signal $r(t)$ and the second term represents the oscillation added in the modulator:		'''(3)'''  In this range, where the first term represents the received signal $r(t)$ and the second term represents the oscillation added in the modulator:
	:$$b_{\rm +1}(t) = s_0 \cdot \sin (2 \pi \cdot 5/T \cdot t )\cdot 2 \cdot \sin (2 \pi \cdot 5/T \cdot t )=		:$$b_{\rm +1}(t) = s_0 \cdot \sin (2 \pi \cdot 5/T \cdot t )\cdot 2 \cdot \sin (2 \pi \cdot 5/T \cdot t )=s_0 \cdot \big [ 1 - \cos (2 \pi \cdot 10/T \cdot t )\big ] \hspace{0.05cm}.$$
	s_0 \cdot \big [ 1 - \cos (2 \pi \cdot 10/T \cdot t )\big ] \hspace{0.05cm}.$$
	*Therefore, <u>the last two answers</u> are correct.		*Therefore, <u>the last two answers</u> are correct.

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	[[Category:Modulation Methods: Exercises\|^4.4 Non-linear Digital Modulation^]]		[[Category:Modulation Methods: Exercises\|^4.4 Non-linear Digital Modulation^]]
			[[de:Aufgaben:4.12_FSK–Demodulation]]

Guenter at 10:27, 5 April 2023

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	}}		}}

	[[File:~~P_ID1739__Mod_A_4_12~~.png\|right\|frame\|Signal waveforms of FSK demodulation ~~<b>KORREKTUR</b>~~]]		[[File:EN_Mod_A_4_12_v2.png\|right\|frame\|Signal waveforms of FSK demodulation]]
	The  [[Modulation_Methods/Nonlinear_Digital_Modulation#Coherent_demodulation_of_FSK\| block diagram]]  of the coherent FSK demodulator has already been given in the theory section, but in this exercise we will assume the lower system variant. Noise components are not considered here.		The  [[Modulation_Methods/Nonlinear_Digital_Modulation#Coherent_demodulation_of_FSK\| block diagram]]  of the coherent FSK demodulator has already been given in the theory section, but in this exercise we will assume the lower system variant. Noise components are not considered here.

Reed at 10:57, 21 March 2022

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	(Integration over multiples of the period of a sinusoidal functions).		(Integration over multiples of the period of a sinusoidal functions).
	*The first integral is equal to  $s_0 · T$.  The decision value in the first time interval follows from this:   $E_1\hspace{0.15cm}\underline { = +1}$.  Similarly,  $E_3\hspace{0.15cm}\underline { = +1}$.		*The first integral is equal to  $s_0 · T$.  The decision value in the first time interval follows from this:   $E_1\hspace{0.15cm}\underline { = +1}$.  Similarly,  $E_3\hspace{0.15cm}\underline { = +1}$.
	*On the other hand, when calculating $E_2$, the first integral is zero and the second has the value   $s_0 · T$. . Thus, we obtain the value  $E_2\hspace{0.15cm}\underline { = -1}$.		*On the other hand, when calculating $E_2$, the first integral is zero and the second has the value   $s_0 · T$.  Thus, we obtain the value  $E_2\hspace{0.15cm}\underline { = -1}$.

Reed: /* Solution */

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Solution

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Im mittleren Bereich ist eine niedrigere Frequenz als~~ in ~~den beiden äußeren Bereichen zu erkennen~~:		'''(1)'''  A lower frequency can be seen in the middle region than in the two outer regions:
	:$$a_1 \hspace{0.15cm}\underline {= +1},\hspace{0.2cm}a_2 \hspace{0.15cm}\underline {= -1},\hspace{0.2cm}a_3\hspace{0.15cm}\underline { = +1} \hspace{0.05cm}.$$		:$$a_1 \hspace{0.15cm}\underline {= +1},\hspace{0.2cm}a_2 \hspace{0.15cm}\underline {= -1},\hspace{0.2cm}a_3\hspace{0.15cm}\underline { = +1} \hspace{0.05cm}.$$



	'''(2)'''  ~~Aus der Grafik erkennt man im ersten und im letzten Zeitintervall fünf Schwingungen und im zweiten Intervall drei Schwingungen~~:		'''(2)''' From the graph, we can see five oscillations in the first and last time intervals and three in the second interval:
	:$$f_{\rm +1} \hspace{0.15cm}\underline {= 5 \cdot 1/T},\hspace{0.2cm}f_{\rm -1}\hspace{0.15cm}\underline { = 3 \cdot 1/T} \hspace{0.05cm}.$$		:$$f_{\rm +1} \hspace{0.15cm}\underline {= 5 \cdot 1/T},\hspace{0.2cm}f_{\rm -1}\hspace{0.15cm}\underline { = 3 \cdot 1/T} \hspace{0.05cm}.$$
	*~~Die Trägerfrequenz ist somit~~ $f_{\rm T} = 4/T$ ~~und der Frequenzhub~~ $Δf_{\rm A} = 1/T$.		*Thus, the carrier frequency is $f_{\rm T} = 4/T$ and the frequency deviation is $Δf_{\rm A} = 1/T$.



	'''(3)'''  In ~~diesem Bereich gilt~~, ~~wobei der erste Term das Empfangssignal~~ $r(t)$ ~~beschreibt und der zweite Term die im Modulator zugesetzte Schwingung~~:		'''(3)'''  In this range, where the first term represents the received signal $r(t)$ and the second term represents the oscillation added in the modulator:
	:$$b_{\rm +1}(t) = s_0 \cdot \sin (2 \pi \cdot 5/T \cdot t )\cdot 2 \cdot \sin (2 \pi \cdot 5/T \cdot t )=		:$$b_{\rm +1}(t) = s_0 \cdot \sin (2 \pi \cdot 5/T \cdot t )\cdot 2 \cdot \sin (2 \pi \cdot 5/T \cdot t )=
	s_0 \cdot \big [ 1 - \cos (2 \pi \cdot 10/T \cdot t )\big ] \hspace{0.05cm}.$$		s_0 \cdot \big [ 1 - \cos (2 \pi \cdot 10/T \cdot t )\big ] \hspace{0.05cm}.$$
	*~~Richtig sind demnach~~ <u>~~die beiden letzten Lösungsvorschläge~~</u>.		*Therefore, <u>the last two answers</u> are correct.



	'''(4)'''  ~~Für das untere Signal~~  $b_{-1}(t)$  ~~gilt im gleichen Zeitintervall entsprechend~~:		'''(4)'''  Analogously, the lower signal  $b_{-1}(t)$  in the same time interval, is characterized by:
	:$$b_{\rm -1}(t) = s_0 \cdot \sin (2 \pi \cdot 5/T \cdot t )\cdot 2 \cdot \sin (2 \pi \cdot 3/T \cdot t )= s_0 \cdot \big [ \cos (2 \pi \cdot 2/T \cdot t ) - \cos (2 \pi \cdot 8/T \cdot t )\big ] \hspace{0.05cm}$$		:$$b_{\rm -1}(t) = s_0 \cdot \sin (2 \pi \cdot 5/T \cdot t )\cdot 2 \cdot \sin (2 \pi \cdot 3/T \cdot t )= s_0 \cdot \big [ \cos (2 \pi \cdot 2/T \cdot t ) - \cos (2 \pi \cdot 8/T \cdot t )\big ] \hspace{0.05cm}$$
	*~~Richtig sind hier~~ <u>~~die Lösungsvorschläge~~ 1 ~~und~~ 4</u>.		*<u>Answers 1 and 4</u> are correct.



	'''(5)'''  ~~Für den ersten Entscheidungswert gilt mit~~  $b(t) = b_{+1}(t) - b_{–1}(t)$:		'''(5)'''  For the first decision value with  $b(t) = b_{+1}(t) - b_{–1}(t)$, it holds that:
	:$$E_{1} = \frac{1}{s_0 \cdot T} \cdot \int_{T/2 }^{3T/2} b_{\rm +1} (t )\hspace{0.1cm} {\rm d}t - \frac{1}{s_0 \cdot T} \cdot \int_{T/2 }^{3T/2} b_{\rm -1} (t )\hspace{0.1cm} {\rm d}t\hspace{0.05cm}.$$		:$$E_{1} = \frac{1}{s_0 \cdot T} \cdot \int_{T/2 }^{3T/2} b_{\rm +1} (t )\hspace{0.1cm} {\rm d}t - \frac{1}{s_0 \cdot T} \cdot \int_{T/2 }^{3T/2} b_{\rm -1} (t )\hspace{0.1cm} {\rm d}t\hspace{0.05cm}.$$
	*~~Aus dem Ergebnis der Teilaufgabe~~  '''(4)'''  ~~erkennt man~~, ~~dass das zweite Integral Null ergibt~~ <br>(Integration ~~über Vielfache der Periodendauer von Sinusfunktionen~~).		*From the result of question   '''(4)''' , we can see that the second integral yields zero <br>
	*~~Das erste Integral ist gleich~~  $s_0 · T$.  ~~Daraus folgt für den Entscheidungswert im ersten Zeitintervall~~:   $E_1\hspace{0.15cm}\underline { = +1}$.  ~~Ebenso ist~~  $E_3\hspace{0.15cm}\underline { = +1}$.		(Integration over multiples of the period of a sinusoidal functions).
	*~~Dagegen ist bei der Berechnung von~~ $E_2$ ~~das erste Integral Null und das zweite hat den Wert~~  $s_0 · T$.  ~~Somit erhält man hierfür den Wert~~  $E_2\hspace{0.15cm}\underline { = -1}$.		*The first integral is equal to  $s_0 · T$.  The decision value in the first time interval follows from this:   $E_1\hspace{0.15cm}\underline { = +1}$.  Similarly,  $E_3\hspace{0.15cm}\underline { = +1}$.
			*On the other hand, when calculating $E_2$, the first integral is zero and the second has the value   $s_0 · T$. . Thus, we obtain the value  $E_2\hspace{0.15cm}\underline { = -1}$.

Reed at 10:45, 21 March 2022

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	</quiz>		</quiz>

	===~~Musterlösung~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Im mittleren Bereich ist eine niedrigere Frequenz als in den beiden äußeren Bereichen zu erkennen:		'''(1)'''  Im mittleren Bereich ist eine niedrigere Frequenz als in den beiden äußeren Bereichen zu erkennen:

Reed: /* Questions */

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Questions

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	<quiz display=simple>		<quiz display=simple>
	{~~Wie lauten die Amplitudenkoeffizienten~~?		{What are the amplitude coefficients?
	\|type="{}"}		\|type="{}"}
	$a_1 \ = \ $ { 1 }		$a_1 \ = \ $ { 1 }
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	$a_3 \ = \ $ { 1 }		$a_3 \ = \ $ { 1 }

	{~~Geben Sie die beiden Frequenzen~~  $f_{+1}$  ~~und~~  $f_{–1}$  an.		{Give the two frequencies  $f_{+1}$  and  $f_{–1}$ .
	\|type="{}"}		\|type="{}"}
	$ f_{+1} \ = \ $ { 5 3% } $\ \cdot 1/T$		$ f_{+1} \ = \ $ { 5 3% } $\ \cdot 1/T$
	$f_{–1} \hspace{0.18cm} = \ $ { 3 3% } $\ \cdot 1/T$		$f_{–1} \hspace{0.18cm} = \ $ { 3 3% } $\ \cdot 1/T$

	{~~Welche Aussagen treffen für das Signal~~  $b_{+1}(t)$  ~~im ersten Zeitintervall zu~~?		{Which statements are true for the signal  $b_{+1}(t)$  in the first time interval?
	\|type="[]"}		\|type="[]"}
	- ~~Das Signal beinhaltet die Frequenz~~  $2/T$.		- The signal contains the frequency  $2/T$.
	- ~~Das Signal beinhaltet die Frequenz~~  $4/T$.		- The signal contains the frequency  $4/T$.
	- ~~Das Signal beinhaltet die Frequenz~~  $6/T$.		- The signal contains the frequency  $6/T$.
	- ~~Das Signal beinhaltet die Frequenz~~  $8/T$.		- The signal contains the frequency  $8/T$.
	+ ~~Das Signal beinhaltet die Frequenz~~  $10/T$.		+ The signal contains the frequency  $10/T$.
	+ ~~Das Signal beinhaltet einen Gleichanteil~~.		+ The signal contains a DC component.

	{~~Welche Aussagen treffen für das Signal~~  $b_{–1}(t)$  ~~im ersten Zeitintervall zu~~?		{Which statements are true for the signal  $b_{-1}(t)$  in the first time interval?
	\|type="[]"}		\|type="[]"}
	+ ~~Das Signal beinhaltet die Frequenz~~  $2/T$.		+ The signal contains the frequency  $2/T$.
	- ~~Das Signal beinhaltet die Frequenz~~  $4/T$.		- The signal contains the frequency  $4/T$.
	- ~~Das Signal beinhaltet die Frequenz~~ $6/T$.		- The signal contains the frequency $6/T$.
	+ ~~Das Signal beinhaltet die Frequenz~~  $8/T$.		+ The signal contains the frequency  $8/T$.
	- ~~Das Signal beinhaltet die Frequenz~~  $10/T$.		- The signal contains the frequency  $10/T$.
	- ~~Das Signal beinhaltet einen Gleichanteil~~.		- The signal contains a DC component.

	{~~Welche normierten Entscheidungswerte ergeben sich für die einzelnen Symbole~~?		{What are the normalized decision values for each symbol?
	\|type="{}"}		\|type="{}"}
	$E_1 \hspace{0.2cm} = \ $ { 1 }		$E_1 \hspace{0.2cm} = \ $ { 1 }

Reed at 10:40, 21 March 2022

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	}}		}}

	[[File:P_ID1739__Mod_A_4_12.png\|right\|frame\|~~Signalverläufe der FSK–Demodulation~~]]		[[File:P_ID1739__Mod_A_4_12.png\|right\|frame\|Signal waveforms of FSK demodulation <b>KORREKTUR</b>]]
	The  [[Modulation_Methods/Nonlinear_Digital_Modulation#Coherent_demodulation_of_FSK\| block diagram]]  of the coherent FSK demodulator has already been given in the theory section, but in this exercise we will assume the lower system variant. Noise components are not considered here.		The  [[Modulation_Methods/Nonlinear_Digital_Modulation#Coherent_demodulation_of_FSK\| block diagram]]  of the coherent FSK demodulator has already been given in the theory section, but in this exercise we will assume the lower system variant. Noise components are not considered here.

Reed at 10:39, 21 March 2022

2022-03-21T10:39:47Z

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	* A technical hint: By enlarging the image, one can recognize the signal curves a little better.		* A technical hint: By enlarging the image, one can recognize the signal curves a little better.
	Please accept an apology on behalf of the author: blue on a black background is not great.		Please accept an apology on behalf of the author: blue on a black background is not great.
	The following trigonometric relation is given:		<br> The following trigonometric relation is given:
	:$$2 \cdot \sin(\alpha) \cdot \sin(\beta)= \cos(\alpha - \beta)- \cos(\alpha + \beta) \hspace{0.05cm}.$$		:$$2 \cdot \sin(\alpha) \cdot \sin(\beta)= \cos(\alpha - \beta)- \cos(\alpha + \beta) \hspace{0.05cm}.$$