Aufgaben:Exercise 2.6Z: Synchronous Demodulator - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:57:42Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:57, 16 March 2026
Line 81:		Line 81:
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The following holds for the band-pass signal after the second multiplier:		'''(1)'''  The following holds for the band-pass signal after the second multiplier:
	:$$b(t) = r(t) \cdot z_{\rm E}(t)= q(t) \cdot z(t) \cdot z_{\~~rmE~~}(t)= K \cdot q(t)\cdot\cos^2(\omega_{\rm T} t).$$		:$$b(t) = r(t) \cdot z_{\rm E}(t)= q(t) \cdot z(t) \cdot z_{\rm E}(t)= K \cdot q(t)\cdot\cos^2(\omega_{\rm T} t).$$

	*Using the trigonometric relation  $\cos^2(\omega_{\rm T} t) = {1}/{2} \cdot\big[ 1 +		*Using the trigonometric relation  $\cos^2(\omega_{\rm T} t) = {1}/{2} \cdot\big[ 1 +
Line 120:		Line 120:

	[[Category:Linear and Time-Invariant Systems: Exercises\|^2.3 Linear Distortions^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^2.3 Linear Distortions^]]
	[[de:Aufgaben:2.~~6Z_Synchrondemodulator~~]]		[[de:Aufgaben:Aufgabe 2.6Z: Synchrondemodulator]]

Maintenance script: Join split $$ formula lines for correct rendering

2026-03-16T13:30:40Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:30, 16 March 2026
Line 10:		Line 10:

	Let the source signal consist of two harmonic oscillations with frequencies  $f_2 = 2 \ \rm kHz$  and  $f_5 = 5 \ \rm kHz$:		Let the source signal consist of two harmonic oscillations with frequencies  $f_2 = 2 \ \rm kHz$  and  $f_5 = 5 \ \rm kHz$:
	:$$q(t) = {2 \, \rm V} \cdot {\rm cos}(\omega_2 t )+ {1 \, \rm V}		:$$q(t) = {2 \, \rm V} \cdot {\rm cos}(\omega_2 t )+ {1 \, \rm V}\cdot {\rm sin}(\omega_5 t ) .$$
	\cdot {\rm sin}(\omega_5 t ) .$$

	*This signal is multiplied by the dimensionless carrier signal  $z(t) = \cos(\omega_{\rm T} \cdot T)$  of carrier frequency  $f_{\rm T} = 50 \ \rm kHz$. <br>For DSB–AM, the dashed block is irrelevant so that the following holds for the transmission signal:		*This signal is multiplied by the dimensionless carrier signal  $z(t) = \cos(\omega_{\rm T} \cdot T)$  of carrier frequency  $f_{\rm T} = 50 \ \rm kHz$. <br>For DSB–AM, the dashed block is irrelevant so that the following holds for the transmission signal:
Line 27:		Line 26:

	For the transmitter, multiplication by the carrier signal  $z(t)$  generally results in two sidebands. In  [[Modulation_Methods/Single-Sideband_Modulation\|Single-Sideband Modulation]]  (ESB–AM), only one of the two bands is transmitted, for example the lower sideband (USB). Thus, the following is obtained with an ideal channel:		For the transmitter, multiplication by the carrier signal  $z(t)$  generally results in two sidebands. In  [[Modulation_Methods/Single-Sideband_Modulation\|Single-Sideband Modulation]]  (ESB–AM), only one of the two bands is transmitted, for example the lower sideband (USB). Thus, the following is obtained with an ideal channel:
	:$$r(t) = s(t)= {1 \, \rm V} \cdot {\rm cos}\big [(\omega_{\rm T} -		:$$r(t) = s(t)= {1 \, \rm V} \cdot {\rm cos}\big [(\omega_{\rm T} -\omega_2 )\cdot t \big ] - {0.5 \, \rm V} \cdot {\rm sin}\big [(\omega_{\rm T} -\omega_5 )\cdot t \big ] .$$
	\omega_2 )\cdot t \big ] - {0.5 \, \rm V} \cdot {\rm sin}\big [(\omega_{\rm T} -
	\omega_5 )\cdot t \big ] .$$

	*Here, synchronous demodulation results in the following distorted sink signal considering a phase shift  $\Delta \varphi$, the constant  $K = 4$  and the downstream low-pass filter:		*Here, synchronous demodulation results in the following distorted sink signal considering a phase shift  $\Delta \varphi$, the constant  $K = 4$  and the downstream low-pass filter:
	:$$v(t)= {1 \, \rm V} \cdot {1}/{2}\cdot 4 \cdot{\rm cos}(		:$$v(t)= {1 \, \rm V} \cdot {1}/{2}\cdot 4 \cdot{\rm cos}(\omega_2 t - \Delta \varphi)+ {0.5 \, \rm V} \cdot{1}/{2}\cdot 4 \cdot{\rm sin}( \omega_5 t - \Delta \varphi)$$
	\omega_2 t - \Delta \varphi)+ {0.5 \, \rm V} \cdot		:$$\Rightarrow \hspace{0.5cm}v(t)= {2 \, \rm V} \cdot{\rm cos}(\omega_2 t - \Delta \varphi)+ {1 \, \rm V} \cdot{\rm sin}(\omega_5 t - \Delta \varphi)$$
	{1}/{2}\cdot 4 \cdot{\rm sin}( \omega_5 t - \Delta \varphi)$$
	:$$\Rightarrow \hspace{0.5cm}v(t)= {2 \, \rm V} \cdot{\rm cos}(
	\omega_2 t - \Delta \varphi)+ {1 \, \rm V} \cdot{\rm sin}(
	\omega_5 t - \Delta \varphi)$$

	*In the ideal case of phase-synchronous demodulation  $(\Delta \varphi = 0)$,    $v(t) = q(t)$ holds again.		*In the ideal case of phase-synchronous demodulation  $(\Delta \varphi = 0)$,    $v(t) = q(t)$ holds again.
Line 48:		Line 41:
	*The topic  "amplitude modulation/synchronous demodulator"  is discussed in detail in the book  [[Modulation_Methods]].		*The topic  "amplitude modulation/synchronous demodulator"  is discussed in detail in the book  [[Modulation_Methods]].
	*The following trigonometric relationships are given:		*The following trigonometric relationships are given:
	:$$\cos^2(\alpha) = {1}/{2} \cdot \big [ 1 +		:$$\cos^2(\alpha) = {1}/{2} \cdot \big [ 1 +\cos(2\alpha) \big ] \hspace{0.05cm}, $$
	\cos(2\alpha) \big ] \hspace{0.05cm}, $$		:$$\cos(\alpha) \cdot \cos(\beta) = {1}/{2} \cdot \big[ \cos(\alpha -\beta)+ \cos(\alpha + \beta) \big],$$
	:$$\cos(\alpha) \cdot \cos(\beta) = {1}/{2} \cdot \big[ \cos(\alpha -		:$$ \sin(\alpha) \cdot \cos(\beta) = {1}/{2} \cdot \big[ \sin(\alpha -\beta)+ \sin(\alpha + \beta)\big] \hspace{0.05cm}.$$
	\beta)+ \cos(\alpha + \beta) \big],$$
	:$$ \sin(\alpha) \cdot \cos(\beta) = {1}/{2} \cdot \big[ \sin(\alpha -
	\beta)+ \sin(\alpha + \beta)
	\big] \hspace{0.05cm}.$$
	*The signal designations result from the German original of this exercise.  '''Here again as a listing'''   $q(t)$   ⇒   source signal,   $v(t)$   ⇒   sink signal,   $z(t)$   ⇒   transmit-site carrier signal,   $s(t)$   ⇒   transmission signal (BP),  $r(t)$   ⇒   received signal (BP),   $z_{\rm E}(t)$   ⇒   receive-site carrier signal,   $b(t)$   ⇒   BP signal before low-pass.		*The signal designations result from the German original of this exercise.  '''Here again as a listing'''   $q(t)$   ⇒   source signal,   $v(t)$   ⇒   sink signal,   $z(t)$   ⇒   transmit-site carrier signal,   $s(t)$   ⇒   transmission signal (BP),  $r(t)$   ⇒   received signal (BP),   $z_{\rm E}(t)$   ⇒   receive-site carrier signal,   $b(t)$   ⇒   BP signal before low-pass.

Line 92:		Line 81:
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The following holds for the band-pass signal after the second multiplier:		'''(1)'''  The following holds for the band-pass signal after the second multiplier:
	:$$b(t) = r(t) \cdot z_{\rm E}(t)= q(t) \cdot z(t) \cdot z_{\rm		:$$b(t) = r(t) \cdot z_{\rm E}(t)= q(t) \cdot z(t) \cdot z_{\rmE}(t)= K \cdot q(t)\cdot\cos^2(\omega_{\rm T} t).$$
	E}(t)= K \cdot q(t)\cdot
	\cos^2(\omega_{\rm T} t).$$

	*Using the trigonometric relation  $\cos^2(\omega_{\rm T} t) = {1}/{2} \cdot\big[ 1 +		*Using the trigonometric relation  $\cos^2(\omega_{\rm T} t) = {1}/{2} \cdot\big[ 1 +
	\cos(2\omega_{\rm T} t)\big]$,  one obtains:		\cos(2\omega_{\rm T} t)\big]$,  one obtains:
	:$$b(t) = {K}/{2} \cdot q(t) + {K}/{2} \cdot q(t)\cdot		:$$b(t) = {K}/{2} \cdot q(t) + {K}/{2} \cdot q(t)\cdot\cos(2\omega_{\rm T} t).$$
	\cos(2\omega_{\rm T} t).$$

	*The second component is located at around twice the carrier frequency  ⇒   $2 f_{\rm T}$.		*The second component is located at around twice the carrier frequency  ⇒   $2 f_{\rm T}$.
Line 109:		Line 95:

	'''(2)'''  Considering the relation		'''(2)'''  Considering the relation
	:$$\cos(\omega_{\rm T} t) \cdot \cos(\omega_{\rm T} t - \Delta \varphi) = {1}/{2} \cdot		:$$\cos(\omega_{\rm T} t) \cdot \cos(\omega_{\rm T} t - \Delta \varphi) = {1}/{2} \cdot\big[ \cos(\Delta \varphi)+ \cos(2\omega_{\rm T} t - \Delta \varphi) \big]$$
	\big[ \cos(\Delta \varphi)+ \cos(2\omega_{\rm T} t - \Delta \varphi) \big]$$

	and the downstream low-pass filter,  which removes the component at around twice the carrier frequency, the following is obtained here with  $ {K = 2}$:		and the downstream low-pass filter,  which removes the component at around twice the carrier frequency, the following is obtained here with  $ {K = 2}$:
Line 123:		Line 108:
	'''(3)'''  Here, <u>proposed solution 4</u>  is correct.		'''(3)'''  Here, <u>proposed solution 4</u>  is correct.
	*Exactly the same phase shift  $\Delta \varphi$  occurs for both summands,  and phase distortions occur here:		*Exactly the same phase shift  $\Delta \varphi$  occurs for both summands,  and phase distortions occur here:
	:$$v(t)= {2 \, \rm V} \cdot{\rm cos}\big[ \omega_2 \cdot (t - \tau_2) \big]+		:$$v(t)= {2 \, \rm V} \cdot{\rm cos}\big[ \omega_2 \cdot (t - \tau_2) \big]+{1 \, \rm V} \cdot{\rm sin}\big[ \omega_5 t \cdot (t - \tau_5)\big],$$
	{1 \, \rm V} \cdot{\rm sin}\big[ \omega_5 t \cdot (t - \tau_5)\big],$$		:$${\rm where}\hspace{0.5cm}\tau_2 = \frac{\Delta \varphi}{\omega_2}\hspace{0.5cm}\ne \hspace{0.5cm} \tau_5 = \frac{\Delta\varphi}{\omega_5}.$$
	:$${\rm where}\hspace{0.5cm}\tau_2 = \frac{\Delta \varphi}{\omega_2}
	\hspace{0.5cm}\ne \hspace{0.5cm} \tau_5 = \frac{\Delta
	\varphi}{\omega_5}.$$

	*A phase shift of  $\varphi =60^\circ$ corresponding to  $\pi/3$  leads to the following delay times here:		*A phase shift of  $\varphi =60^\circ$ corresponding to  $\pi/3$  leads to the following delay times here:
	:$$\tau_2 = \frac{\pi/3}{2 \pi \cdot 2\,\,{\rm kHz }} \~~approx~~		:$$\tau_2 = \frac{\pi/3}{2 \pi \cdot 2\,\,{\rm kHz }} \approx83.3\,{\rm µ s }, \hspace{0.5cm}\tau_5 = \frac{\pi/3}{2 \pi \cdot 5\,\,{\rm kHz }} \approx33.3\,{\rm µ s }.$$
	83.3\,{\rm µ s }, \hspace{0.5cm}
	\tau_5 = \frac{\pi/3}{2 \pi \cdot 5\,\,{\rm kHz }} \~~approx~~
	33.3\,{\rm µ s }.$$

	*The lower-frequency signal is thus delayed more.		*The lower-frequency signal is thus delayed more.

Maintenance script: Add German interlanguage link

2026-03-16T12:31:32Z

Add German interlanguage link

← Older revision		Revision as of 14:31, 16 March 2026
Line 141:		Line 141:

	[[Category:Linear and Time-Invariant Systems: Exercises\|^2.3 Linear Distortions^]]		[[Category:Linear and Time-Invariant Systems: Exercises\|^2.3 Linear Distortions^]]
			[[de:Aufgaben:2.6Z_Synchrondemodulator]]

Guenter at 13:29, 6 October 2021

2021-10-06T13:29:31Z

Guenter at 13:22, 6 October 2021

2021-10-06T13:22:59Z

← Older revision		Revision as of 15:22, 6 October 2021
Line 55:		Line 55:
	\beta)+ \sin(\alpha + \beta)		\beta)+ \sin(\alpha + \beta)
	\big] \hspace{0.05cm}.$$		\big] \hspace{0.05cm}.$$
	*The signal designations result from the German original of this exercise.  '''Here again as a listing'''   $q(t)$   ⇒   source signal,   $v(t)$   ⇒   sink signal,   $z(t)$   ⇒   transmit-site carrier signal,   $s(t)$   ⇒   transmission signal (HF),  $r(t)$   ⇒   received signal,   $z_{\rm E}(t)$   ⇒   receive-site carrier signal,   $b(t)$   ⇒   HF signal before low-pass		*The signal designations result from the German original of this exercise.  '''Here again as a listing'''   $q(t)$   ⇒   source signal,   $v(t)$   ⇒   sink signal,   $z(t)$   ⇒   transmit-site carrier signal,   $s(t)$   ⇒   transmission signal (BP),  $r(t)$   ⇒   received signal (BP),   $z_{\rm E}(t)$   ⇒   receive-site carrier signal,   $b(t)$   ⇒   BP signal before low-pass.

Guenter at 13:19, 6 October 2021

2021-10-06T13:19:48Z

← Older revision		Revision as of 15:19, 6 October 2021
Line 62:		Line 62:

	<quiz display=simple>		<quiz display=simple>
	{What is the sink signal $v(t)$  for ~~ZSB~~-AM and phase-~~synchronous~~ synchronous modulation  ⇒   $\Delta \varphi = 0$? <br>How should  $K$  be chosen such that  $v(t) = q(t)$  holds?		{What is the sink signal  $v(t)$  for DSB-AM and phase-synchronous modulation  ⇒   $\Delta \varphi = 0$? <br>How should  $K$  be chosen such that  $v(t) = q(t)$  holds?
	\|type="{}"}		\|type="{}"}
	$K \ = \ $ { 2 3% }		$K \ = \ $ { 2 3% }


	{The following holds:  $K = 2$. Specify the sink signal $v(t)$  considering a phase shift $\Delta \varphi$ ~~. <br>~~Which of the following statements are true?		{The following holds:  $K = 2$.  Specify the sink signal  $v(t)$  considering a phase shift $\Delta \varphi$.  Which of the following statements are true?
	\|type="[]"}		\|type="[]"}
	- $v(t) = q(t)$ holds independently of $\Delta \varphi$ .		- $v(t) = q(t)$  holds independently of  $\Delta \varphi$ .
	+ $\Delta \varphi \ne 0$  results in frequency-independent attenuation.		+ $\Delta \varphi \ne 0$  results in frequency-independent attenuation.
	- A phase shift $\Delta \varphi \ne 0$  results in attenuation distortions.		- A phase shift  $\Delta \varphi \ne 0$  results in attenuation distortions.
	- A phase shift $\Delta \varphi \ne 0$  results in phase distortions.		- A phase shift  $\Delta \varphi \ne 0$  results in phase distortions.
	+ $v(t) = q(t)/2$ holds with $\Delta \varphi = \hspace{-0.05cm}-\hspace{0.05cm}60^\circ$~~ ~~.		+ $v(t) = q(t)/2$  holds with  $\Delta \varphi = \hspace{-0.05cm}-\hspace{0.05cm}60^\circ$.


	{Which statements hold for synchronous demodulation of the ~~ESB–~~signal if a phase shift of $\Delta \varphi$  is considered?		{Which statements hold for synchronous demodulation of the SSB signal if a phase shift of $\Delta \varphi$  is considered?
	\|type="[]"}		\|type="[]"}
	- Regardless of $\Delta \varphi$,~~ ~~  $v(t) = q(t)$ holds.		- Regardless of  $\Delta \varphi$,  $v(t) = q(t)$  holds.
	- $\Delta \varphi \ne 0$  results in frequency-independent attenuation.		- $\Delta \varphi \ne 0$  results in frequency-independent attenuation.
	- A phase shift $\Delta \varphi \ne 0$  results in attenuation distortions.		- A phase shift  $\Delta \varphi \ne 0$  results in attenuation distortions.
	+ A phase shift $\Delta \varphi \ne 0$  results in phase distortions.		+ A phase shift  $\Delta \varphi \ne 0$  results in phase distortions.
	- $v(t) = q(t)/2$ holds with $\Delta \varphi = \hspace{-0.05cm}-\hspace{0.05cm}60^\circ$ .		- $v(t) = q(t)/2$  holds with  $\Delta \varphi = \hspace{-0.05cm}-\hspace{0.05cm}60^\circ$ .

Guenter at 13:13, 6 October 2021

2021-10-06T13:13:17Z

← Older revision		Revision as of 15:13, 6 October 2021
Line 3:		Line 3:
	}}		}}

	[[File:P_ID913__LZI_Z_2_6_neu.png\|right\|frame\|~~AM–~~modulator (top) ~~and~~ synchronous demodulator ~~(bottom)~~]]		[[File:P_ID913__LZI_Z_2_6_neu.png\|right\|frame\|Amplitude modulator (top), <br>synchronous demodulator]]
	The depicted block diagram shows a transmission system		The depicted block diagram shows a transmission system
	*with  [[Modulation_Methods/Double-Sideband_Amplitude_Modulation\|Double-Sideband Amplitude Modulation]]  (~~ZSB~~-AM)		*with  [[Modulation_Methods/Double-Sideband_Amplitude_Modulation\|Double-Sideband Amplitude Modulation]]  $\rm(DSB\hspace{0.03cm}–\hspace{-0.1cm}AM)$
	*and  [[Modulation_Methods/Synchronous_Demodulation\|Synchronous Demodulation]] (SD).		*and  [[Modulation_Methods/Synchronous_Demodulation\|Synchronous Demodulation]]  $\rm (SD)$.


Line 13:		Line 13:
	\cdot {\rm sin}(\omega_5 t ) .$$		\cdot {\rm sin}(\omega_5 t ) .$$

	*This signal is multiplied by the dimensionless carrier signal  $z(t) = \cos(\omega_{\rm T} \cdot T)$  of frequency  $f_{\rm T} = 50 \ \rm kHz$. For ~~ZSB~~–AM, the dashed block is irrelevant so that the following holds for the ~~transmit~~ signal:		*This signal is multiplied by the dimensionless carrier signal  $z(t) = \cos(\omega_{\rm T} \cdot T)$  of carrier frequency  $f_{\rm T} = 50 \ \rm kHz$. <br>For DSB–AM, the dashed block is irrelevant so that the following holds for the transmission signal:
	:$$s(t) = q(t) \cdot {\rm cos}(\omega_{\rm T} t ) .$$		:$$s(t) = q(t) \cdot {\rm cos}(\omega_{\rm T} t ) .$$

	*In the synchronous demodulator, the ~~receiver~~ signal  $r(t)$  – identical to the ~~transmit~~ signal  $s(t)$  ~~in an ideal channel~~ – is multiplied by the receive-site carrier signal  $z_{\rm E}(t)$  where the following applies:		*In the synchronous demodulator, the received signal  $r(t)$  – in an ideal channel identical to the signal  $s(t)$  – is multiplied by the receive-site carrier signal  $z_{\rm E}(t)$  where the following applies:
	:$$z_{\rm E}(t) = K \cdot {\rm cos}(\omega_{\rm T} t - \Delta \varphi ) .$$		:$$z_{\rm E}(t) = K \cdot {\rm cos}(\omega_{\rm T} t - \Delta \varphi ) .$$

	*This signal should not only be frequency-synchronous with  $z(t)$  but also phase-synchronous &~~ndash~~; hence the name "synchronous demodulator".		*This signal should not only be frequency-synchronous with  $z(t)$  but also phase-synchronous   ⇒   hence the name "synchronous demodulator".
	*The above approach takes into account a phase shift between  $z(t)$  and  $z_{\rm E}(t)$, which should ideally be  $\Delta \varphi = 0$  but often cannot be avoided in real systems.		*The above approach takes into account a phase shift between  $z(t)$  and  $z_{\rm E}(t)$, which should ideally be  $\Delta \varphi = 0$  but often cannot be avoided in real systems.

	*The output signal  $b(t)$  of the second multiplier includes components around twice the carrier frequency in addition to the desired NF component.		*The output signal  $b(t)$  of the second multiplier includes components around twice the carrier frequency in addition to the desired low-frequency component.
	*Using an ideal low-pass filter &~~ndash~~; ~~for example~~ with cut-off frequency  $f_{\rm T}$&nbsp~~; &ndash~~; the sink signal  $v(t)$,  which ideally should be equal to the source signal  $q(t)$,  can be obtained.		*Using an ideal low-pass filter  $\rm (LP)$  (e.g. with cut-off frequency  $f_{\rm T}$)  the sink signal  $v(t)$,  which ideally should be equal to the source signal  $q(t)$,  can be obtained.


Line 44:		Line 44:


			Please note:
			*The exercise belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Linear_Distortions\|Linear Distortions]].
			*The topic  "amplitude modulation/synchronous demodulator"  is discussed in detail in the book  [[Modulation_Methods]].


	''Please note:''
	*The ~~task~~ belongs to the chapter  [[Linear_and_Time_Invariant_Systems/Linear_Distortions\|Linear Distortions]].
	*The topic "amplitude modulation/synchronous demodulator" is discussed in detail in the book  [[Modulation_Methods]]~~ ~~.

	*The following trigonometric relationships are given:		*The following trigonometric relationships are given:
	:$$\cos^2(\alpha) = {1}/{2} \cdot \big [ 1 +		:$$\cos^2(\alpha) = {1}/{2} \cdot \big [ 1 +
Line 61:		Line 55:
	\beta)+ \sin(\alpha + \beta)		\beta)+ \sin(\alpha + \beta)
	\big] \hspace{0.05cm}.$$		\big] \hspace{0.05cm}.$$
			*The signal designations result from the German original of this exercise.  '''Here again as a listing'''   $q(t)$   ⇒   source signal,   $v(t)$   ⇒   sink signal,   $z(t)$   ⇒   transmit-site carrier signal,   $s(t)$   ⇒   transmission signal (HF),  $r(t)$   ⇒   received signal,   $z_{\rm E}(t)$   ⇒   receive-site carrier signal,   $b(t)$   ⇒   HF signal before low-pass

Oezer at 14:48, 20 September 2021

2021-09-20T14:48:58Z

← Older revision		Revision as of 16:48, 20 September 2021
Line 129:		Line 129:
	:$$v(t)= {2 \, \rm V} \cdot{\rm cos}\big[ \omega_2 \cdot (t - \tau_2) \big]+		:$$v(t)= {2 \, \rm V} \cdot{\rm cos}\big[ \omega_2 \cdot (t - \tau_2) \big]+
	{1 \, \rm V} \cdot{\rm sin}\big[ \omega_5 t \cdot (t - \tau_5)\big],$$		{1 \, \rm V} \cdot{\rm sin}\big[ \omega_5 t \cdot (t - \tau_5)\big],$$
	:$${\rm ~~wobei~~}\hspace{0.5cm}\tau_2 = \frac{\Delta \varphi}{\omega_2}		:$${\rm where}\hspace{0.5cm}\tau_2 = \frac{\Delta \varphi}{\omega_2}
	\hspace{0.5cm}\ne \hspace{0.5cm} \tau_5 = \frac{\Delta		\hspace{0.5cm}\ne \hspace{0.5cm} \tau_5 = \frac{\Delta
	\varphi}{\omega_5}.$$		\varphi}{\omega_5}.$$

Oezer at 14:46, 20 September 2021

2021-09-20T14:46:44Z

← Older revision		Revision as of 16:46, 20 September 2021
Line 71:		Line 71:


	{The following holds:  $K = 2$. Specify the sink signal $v(t)$  considering a phase shift $\Delta \varphi$  an. <br>Which of the following statements are true?		{The following holds:  $K = 2$. Specify the sink signal $v(t)$  considering a phase shift $\Delta \varphi$ . <br>Which of the following statements are true?
	\|type="[]"}		\|type="[]"}
	- $v(t) = q(t)$ holds independently of $\Delta \varphi$ .		- $v(t) = q(t)$ holds independently of $\Delta \varphi$ .

Oezer at 14:45, 20 September 2021

2021-09-20T14:45:19Z

← Older revision		Revision as of 16:45, 20 September 2021
Line 26:		Line 26:


	For the transmitter, multiplication by the carrier signal  $z(t)$  generally results in two sidebands. In  [[Modulation_Methods/Single-Sideband_Modulation\|Single-Sideband Modulation]]  (ESB–AM) only one of the two bands is transmitted, for example the lower sideband (USB). Thus, the following is obtained with an ideal channel:		For the transmitter, multiplication by the carrier signal  $z(t)$  generally results in two sidebands. In  [[Modulation_Methods/Single-Sideband_Modulation\|Single-Sideband Modulation]]  (ESB–AM), only one of the two bands is transmitted, for example the lower sideband (USB). Thus, the following is obtained with an ideal channel:
	:$$r(t) = s(t)= {1 \, \rm V} \cdot {\rm cos}\big [(\omega_{\rm T} -		:$$r(t) = s(t)= {1 \, \rm V} \cdot {\rm cos}\big [(\omega_{\rm T} -
	\omega_2 )\cdot t \big ] - {0.5 \, \rm V} \cdot {\rm sin}\big [(\omega_{\rm T} -		\omega_2 )\cdot t \big ] - {0.5 \, \rm V} \cdot {\rm sin}\big [(\omega_{\rm T} -

← Older revision		Revision as of 15:29, 6 October 2021
Line 97:		Line 97:

	*Using the trigonometric relation  $\cos^2(\omega_{\rm T} t) = {1}/{2} \cdot\big[ 1 +		*Using the trigonometric relation  $\cos^2(\omega_{\rm T} t) = {1}/{2} \cdot\big[ 1 +
	\cos(2\omega_{\rm T} t)\big]$,		\cos(2\omega_{\rm T} t)\big]$,  one obtains:
	:$$b(t) = {K}/{2} \cdot q(t) + {K}/{2} \cdot q(t)\cdot		:$$b(t) = {K}/{2} \cdot q(t) + {K}/{2} \cdot q(t)\cdot
	\cos(2\omega_{\rm T} t)$$ ~~is obtained.~~		\cos(2\omega_{\rm T} t).$$

	*The second component is located at around twice the carrier frequency  ⇒   $2 f_{\rm T}$.		*The second component is located at around twice the carrier frequency  ⇒   $2 f_{\rm T}$.
	*This is removed by the low-pass filter  $($with the cut-off frequency  $ f_{\rm G} = f_{\rm T})$ .		*This is removed by the low-pass filter  $($with the cut-off frequency  $ f_{\rm G} = f_{\rm T})$ .
	*Hence, the following is obtained:   $v(t) = {K}/{2} \cdot q(t) .$		*Hence, the following is obtained:   $v(t) = {K}/{2} \cdot q(t) .$
	*An ideal demodulation  ⇒   $v(t) = q(t)$ is obtained with  $\underline {K = 2}$ .		*An ideal demodulation  ⇒   $v(t) = q(t)$  is obtained with  $\underline {K = 2}$ .


Line 112:		Line 112:
	\big[ \cos(\Delta \varphi)+ \cos(2\omega_{\rm T} t - \Delta \varphi) \big]$$		\big[ \cos(\Delta \varphi)+ \cos(2\omega_{\rm T} t - \Delta \varphi) \big]$$

	and the downstream low-pass filter, which ~~again~~ removes the component at around twice the carrier frequency, the following is obtained here with  $ {K = 2}$:		and the downstream low-pass filter,  which removes the component at around twice the carrier frequency, the following is obtained here with  $ {K = 2}$:
	:$$v(t) = q(t) \cdot \cos(\Delta \varphi).$$		:$$v(t) = q(t) \cdot \cos(\Delta \varphi).$$

	<u>Proposed solutions 2 and 5</u> are correct:		<u>Proposed solutions 2 and 5</u>  are correct:
	*A phase shift $\Delta \varphi$  only results in frequency-independent attenuation and not in attenuation or phase distortions.		*A phase shift $\Delta \varphi$  only results in frequency-independent attenuation and not in attenuation distortions or phase distortions.
	*A phase shift by $\varphi =\pm 60^\circ$  results in halving of the signal ~~respectively~~.		*A phase shift by $\varphi =\pm 60^\circ$  results in halving of the signal amplitude.



	'''(3)'''  Here, <u>proposed solution 4</u> is correct.		'''(3)'''  Here, <u>proposed solution 4</u>  is correct.
	*Exactly the same phase shift  $\Delta \varphi$  occurs for both summands, and phase distortions occur here:		*Exactly the same phase shift  $\Delta \varphi$  occurs for both summands,  and phase distortions occur here:
	:$$v(t)= {2 \, \rm V} \cdot{\rm cos}\big[ \omega_2 \cdot (t - \tau_2) \big]+		:$$v(t)= {2 \, \rm V} \cdot{\rm cos}\big[ \omega_2 \cdot (t - \tau_2) \big]+
	{1 \, \rm V} \cdot{\rm sin}\big[ \omega_5 t \cdot (t - \tau_5)\big],$$		{1 \, \rm V} \cdot{\rm sin}\big[ \omega_5 t \cdot (t - \tau_5)\big],$$
Line 129:		Line 129:
	\varphi}{\omega_5}.$$		\varphi}{\omega_5}.$$

	*A phase shift of $\varphi =60^\circ$ corresponding to  $\pi/3$  leads to the following delay times here:		*A phase shift of  $\varphi =60^\circ$ corresponding to  $\pi/3$  leads to the following delay times here:
	:$$\tau_2 = \frac{\pi/3}{2 \pi \cdot 2\,\,{\rm kHz }} \approx		:$$\tau_2 = \frac{\pi/3}{2 \pi \cdot 2\,\,{\rm kHz }} \approx
	83.3\,{\rm µ s }, \hspace{0.5cm}		83.3\,{\rm µ s }, \hspace{0.5cm}