Aufgaben:Exercise 2.12: Non-coherent Demodulation - Revision history

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	[[Category:Modulation Methods: Exercises\|^2.5 Other AM Variants^ ]]		[[Category:Modulation Methods: Exercises\|^2.5 Other AM Variants^ ]]
	[[de:Aufgaben:2.~~12Z_Inkohärente_Demodulationsverfahren~~]]		[[de:Aufgaben:Aufgabe 2.12: Zur nichtkohärenten Demodulation]]

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	[[Category:Modulation Methods: Exercises\|^2.5 Other AM Variants^ ]]		[[Category:Modulation Methods: Exercises\|^2.5 Other AM Variants^ ]]
			[[de:Aufgaben:2.12Z_Inkohärente_Demodulationsverfahren]]

Hwang at 12:46, 17 November 2022

2022-11-17T12:46:50Z

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	Hints:		Hints:
	*This exercise belongs to the chapter  [[Modulation_Methods/Further_AM_Variants\|Further AM Variants]].		*This exercise belongs to the chapter  [[Modulation_Methods/Further_AM_Variants\|Further AM Variants]].
	*Particular reference is made to the ~~page~~  [[Modulation_Methods/Further_AM_Variants#Incoherent_.28non-coherent.29_Demodulation\|Incoherent (non-coherent) Demodulation]].		*Particular reference is made to the section  [[Modulation_Methods/Further_AM_Variants#Incoherent_.28non-coherent.29_Demodulation\|Incoherent (non-coherent) Demodulation]].

	*The following trigonometric transformations are given:		*The following trigonometric transformations are given:

Reed at 14:26, 9 April 2022

2022-04-09T14:26:49Z

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	{{quiz-Header\|Buchseite=~~Modulationsverfahren~~/~~Weitere AM–Varianten~~		{{quiz-Header\|Buchseite=Modulation Methods/Further AM Variants
	}}		}}

Guenter at 15:38, 22 February 2022

2022-02-22T15:38:45Z

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	}}		}}

	[[File:~~P_ID1088__Mod_A_2_12~~.png\|right\|frame\|ASK Demodulation <br>(non-coherent) ]]		[[File:EN_Mod_A_2_12.png\|right\|frame\|ASK Demodulation <br>(non-coherent) ]]
	Consider an amplitude modulated signal:		Consider an amplitude modulated signal:
	:$$ s(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$		:$$ s(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$

Guenter at 16:16, 18 February 2022

2022-02-18T16:16:21Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Applying the trigonometric transformations given on the exercise page and taking into account the two ~~lowpass~~ filters (the components around twice the carrier frequency are removed), we obtain:		'''(1)'''  Applying the trigonometric transformations given on the exercise page and taking into account the two low-pass filters  (the components around twice the carrier frequency are removed),  we obtain:
	:$$b_1(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}) \cdot 2 \cdot \cos(\omega_{\rm T} \cdot t) = q(t) \cdot \cos(\Delta \phi_{\rm T})\hspace{0.05cm},$$		:$$b_1(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}) \cdot 2 \cdot \cos(\omega_{\rm T} \cdot t) = q(t) \cdot \cos(\Delta \phi_{\rm T})\hspace{0.05cm},$$
	:$$ b_2(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}) \cdot (-2) \cdot \sin(\omega_{\rm T} \cdot t) = q(t) \cdot \sin(\Delta \phi_{\rm T})\hspace{0.05cm}.$$		:$$ b_2(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}) \cdot (-2) \cdot \sin(\omega_{\rm T} \cdot t) = q(t) \cdot \sin(\Delta \phi_{\rm T})\hspace{0.05cm}.$$
	*Thus, <u>the first and fourth answers</u> are correct.		*Thus,  <u>the first and fourth answers</u>  are correct.


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	'''(4)'''  The result  $b(t) = q^2(t)$ – see subtask '''(2)'''  – here ~~leads~~ to the result:		'''(4)'''  The result  $b(t) = q^2(t)$ – see subtask '''(2)'''  – leads here to the result:
	:$$b_{\rm min}\hspace{0.15cm}\underline{ = 9},$$		:$$b_{\rm min}\hspace{0.15cm}\underline{ = 9},$$
	:$$b_{\rm max}\hspace{0.15cm}\underline{ =9}.$$		:$$b_{\rm max}\hspace{0.15cm}\underline{ =9}.$$
	This shows that the demodulator considered here only functions, if at all times   $q(t) ≥ 0$  or , $q(t) ≤ 0$  holds, and this is known at the receiver.
			This shows that the demodulator considered here only functions,
			*if at all times   $q(t) ≥ 0$   or   $q(t) ≤ 0$   holds,
			*and this is known at the receiver.

	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 16:07, 18 February 2022

2022-02-18T16:07:46Z

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	Consider an amplitude modulated signal:		Consider an amplitude modulated signal:
	:$$ s(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$		:$$ s(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$
	Reaching the receiver based on the channel propagation time, the signal is		Reaching the receiver based on the channel propagation time,  the signal is
	:$$ r(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}) \hspace{0.05cm}.$$		:$$ r(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}) \hspace{0.05cm}.$$
	The arrangement shown here allows perfect demodulation – that is,  $v(t) = q(t)$ – without knowledge of the phase  $Δϕ_T$, but only if the source signal  $q(t)$  satisfies certain conditions.		The arrangement shown here allows perfect demodulation – that is:  $v(t) = q(t)$ – without knowledge of the phase  $Δϕ_T$,  but only if the source signal  $q(t)$  satisfies certain conditions.

	The two receiver-side carrier signals are:		The two receiver-side carrier signals are:
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	:$$ z_{\rm 2, \hspace{0.08cm}E}(t) = -2 \cdot \sin(\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$		:$$ z_{\rm 2, \hspace{0.08cm}E}(t) = -2 \cdot \sin(\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$

	$\rm ~~TP_1~~$  and  $\rm ~~TP_2~~$  denote two ideal (rectangular) ~~lowpass~~ filters, each with cutoff frequency equal to the carrier frequency  $f_{\rm T}$~~ ~~.		$\rm LP_1$  and  $\rm LP_2$  denote two ideal  (rectangular)  low-pass filters,  each with cutoff frequency equal to the carrier frequency  $f_{\rm T}$.

	As (digital) source signals ~~we consider~~:		We consider as (digital) source signals:
	* the unipolar square wave  $q_1(t)$  with dimensionless amplitude values  $0$  and  $3$,		# the unipolar square wave signal  $q_1(t)$  with dimensionless amplitude values  $0$  and  $3$,
	* the bipolar square wave signal  $q_2(t)$  with the dimensionless amplitude values  $±3$.		# the bipolar square wave signal  $q_2(t)$  with the dimensionless amplitude values  $±3$.


	With respect to  $s(t)$ , these two signals result in an  [[Modulation_Methods/Linear_Digital_Modulation#ASK_.E2.80.93_Amplitude_Shift_Keying\|ASK signal]]  ~~and~~ a  [[Modulation_Methods/Linear_Digital_Modulation#BPSK_.E2.80.93_Binary_Phase_Shift_Keying\|BPSK signal]]~~, respectively~~.		With respect to  $s(t)$,  these two signals result in
			#an  [[Modulation_Methods/Linear_Digital_Modulation#ASK_.E2.80.93_Amplitude_Shift_Keying\|ASK signal]],
			#a  [[Modulation_Methods/Linear_Digital_Modulation#BPSK_.E2.80.93_Binary_Phase_Shift_Keying\|BPSK signal]].


	The nonlinear function  $v = g(b)$  is to be determined in this exercise.		The nonlinear function  $v = g(b)$  is to be determined in this exercise.
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			Hints:
			*This exercise belongs to the chapter  [[Modulation_Methods/Further_AM_Variants\|Further AM Variants]].


	''Hints:''
	*This exercise belongs to the chapter   [[Modulation_Methods/Further_AM_Variants\|Further AM Variants]].
	*Particular reference is made to the page  [[Modulation_Methods/Further_AM_Variants#Incoherent_.28non-coherent.29_Demodulation\|Incoherent (non-coherent) Demodulation]].		*Particular reference is made to the page  [[Modulation_Methods/Further_AM_Variants#Incoherent_.28non-coherent.29_Demodulation\|Incoherent (non-coherent) Demodulation]].

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	<quiz display=simple>		<quiz display=simple>
	{What are the signals  $b_1(t)$  and  $b_2(t)$  in both branches – after multiplier and ~~lowpass~~ respectively?  Which statements apply?		{What are the signals  $b_1(t)$  and  $b_2(t)$  in both branches – after multiplier and low-pass respectively?  Which statements apply?
	\|type="[]"}		\|type="[]"}
	+ $b_1(t) = q(t) · \cos(Δϕ_{\rm T})$.		+ $b_1(t) = q(t) · \cos(Δϕ_{\rm T})$.
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	- $b_1(t) = b_2(t) = q(t)$.		- $b_1(t) = b_2(t) = q(t)$.

	{What values of  $b_{\rm min}$  and  $b_{\rm max}$  does the signal  $b(t)$  take on, when the unipolar source signal  $q_1(t)$  is applied to the input?		{What values of  $b_{\rm min}$  and  $b_{\rm max}$  does the signal  $b(t)$  take on,  when the unipolar source signal  $q_1(t)$  is applied to the input?
	\|type="{}"}		\|type="{}"}
	$b_{\rm min} \ = \ $ { 0. }		$b_{\rm min} \ = \ $ { 0. }
	$b_{\rm max} \ = \ $ { 9 3% }		$b_{\rm max} \ = \ $ { 9 3% }

	{How should the characteristic curve  $v = g(b)$  be chosen, so that  $v(t) = q(t)$  holds?		{How should the characteristic curve  $v = g(b)$  be chosen,  so that  $v(t) = q(t)$  holds?
	\|type="()"}		\|type="()"}
	- $v=g(b) = b^2$.		- $v=g(b) = b^2$.
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	{What values of  $b_{\rm min}$  and  $b_{\rm max}$  does the signal  $b(t)$  take on, when the bipolar source signal  $q_2(t)$  is applied to the input?		{What values of  $b_{\rm min}$  and  $b_{\rm max}$  does the signal  $b(t)$  take on,  when the bipolar source signal  $q_2(t)$  is applied to the input?
	\|type="{}"}		\|type="{}"}
	$b_{\rm min} \ = \ $ { 9 3% }		$b_{\rm min} \ = \ $ { 9 3% }

Guenter at 15:51, 18 February 2022

2022-02-18T15:51:58Z

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	[[File:P_ID1088__Mod_A_2_12.png\|right\|frame\|~~Non-coherent~~ <br>~~ASK Demodulation~~]]		[[File:P_ID1088__Mod_A_2_12.png\|right\|frame\|ASK Demodulation <br>(non-coherent) ]]
	Consider an amplitude modulated signal:		Consider an amplitude modulated signal:
	:$$ s(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$		:$$ s(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t) \hspace{0.05cm}.$$

Reed at 19:54, 22 December 2021

2021-12-22T19:54:07Z

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	:$$b_{\rm min}\hspace{0.15cm}\underline{ = 9},$$		:$$b_{\rm min}\hspace{0.15cm}\underline{ = 9},$$
	:$$b_{\rm max}\hspace{0.15cm}\underline{ =9}.$$		:$$b_{\rm max}\hspace{0.15cm}\underline{ =9}.$$
	This shows, that the demodulator considered here only ~~function when for~~ all times   $q(t) ≥ 0$  or  $q(t) ≤ 0$  holds, and this is known at the receiver.		This shows that the demodulator considered here only functions, if at all times   $q(t) ≥ 0$  or , $q(t) ≤ 0$  holds, and this is known at the receiver.

	{{ML-Fuß}}		{{ML-Fuß}}

Reed: /* Solution */

2021-12-22T19:52:57Z

Solution

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Durch Anwendung der auf der Angabenseite gegebenen trigonometrischen Umformungen erhält man unter Berücksichtigung der beiden Tiefpässe~~ (~~die Anteile um die doppelte Trägerfrequenz werden entfernt~~):		'''(1)'''  Applying the trigonometric transformations given on the exercise page and taking into account the two lowpass filters (the components around twice the carrier frequency are removed), we obtain:
	:$$b_1(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}) \cdot 2 \cdot \cos(\omega_{\rm T} \cdot t) = q(t) \cdot \cos(\Delta \phi_{\rm T})\hspace{0.05cm},$$		:$$b_1(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}) \cdot 2 \cdot \cos(\omega_{\rm T} \cdot t) = q(t) \cdot \cos(\Delta \phi_{\rm T})\hspace{0.05cm},$$
	:$$ b_2(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}) \cdot (-2) \cdot \sin(\omega_{\rm T} \cdot t) = q(t) \cdot \sin(\Delta \phi_{\rm T})\hspace{0.05cm}.$$		:$$ b_2(t) = q(t) \cdot \cos(\omega_{\rm T} \cdot t + \Delta \phi_{\rm T}) \cdot (-2) \cdot \sin(\omega_{\rm T} \cdot t) = q(t) \cdot \sin(\Delta \phi_{\rm T})\hspace{0.05cm}.$$
	*~~Richtig sind somit~~ <u>~~die erste und die vierte Antwort~~</u>.		*Thus, <u>the first and fourth answers</u> are correct.



	'''(2)'''  ~~Die Summe der Quadrate der beiden Teilsignale ergibt~~:		'''(2)'''  The sum of the squares of the two partial signals gives:
	:$$ b(t) = b_1^2(t) + b_2^2(t)= q^2(t) \cdot \left( \cos^2(\Delta \phi_{\rm T})+ \sin^2(\Delta \phi_{\rm T})\right) = q^2(t)\hspace{0.05cm}.$$		:$$ b(t) = b_1^2(t) + b_2^2(t)= q^2(t) \cdot \left( \cos^2(\Delta \phi_{\rm T})+ \sin^2(\Delta \phi_{\rm T})\right) = q^2(t)\hspace{0.05cm}.$$
	~~Die möglichen Amplitudenwerte sind somit~~: ~~ ~~		The possible amplitude values are thus:
	:$$b_{\rm min}\hspace{0.15cm}\underline{ = 0},$$		:$$b_{\rm min}\hspace{0.15cm}\underline{ = 0},$$
	:$$ b_{\rm max}\hspace{0.15cm}\underline{ =9}.$$		:$$ b_{\rm max}\hspace{0.15cm}\underline{ =9}.$$
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	'''(3)'''  ~~Richtig ist der~~ <u>~~zweite Lösungsvorschlag~~</u>:		'''(3)'''  The <u>second answer</u> is correct:
	:$$v=g(b) = \sqrt{b} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} v(t) = \sqrt{ q^2(t) } = q(t)\hspace{0.05cm}.$$		:$$v=g(b) = \sqrt{b} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} v(t) = \sqrt{ q^2(t) } = q(t)\hspace{0.05cm}.$$


	'''(4)'''  ~~Das Ergebnis~~  $b(t) = q^2(t)$ – ~~siehe Teilaufgabe ~~ '''(2)'''  – ~~führt hier zum Ergebnis~~:		'''(4)'''  The result  $b(t) = q^2(t)$ – see subtask '''(2)'''  – here leads to the result:
	:$$b_{\rm min}\hspace{0.15cm}\underline{ = 9},$$		:$$b_{\rm min}\hspace{0.15cm}\underline{ = 9},$$
	:$$b_{\rm max}\hspace{0.15cm}\underline{ =9}.$$		:$$b_{\rm max}\hspace{0.15cm}\underline{ =9}.$$
	~~Dies zeigt~~, ~~dass der hier betrachtete Demodulator nur dann funktioniert, wenn für alle Zeiten~~  $q(t) ≥ 0$  ~~oder~~  $q(t) ≤ 0$  ~~gilt und dies dem Empfänger auch bekannt ist~~.		This shows, that the demodulator considered here only function when for all times   $q(t) ≥ 0$  or  $q(t) ≤ 0$  holds, and this is known at the receiver.

	{{ML-Fuß}}		{{ML-Fuß}}