Aufgaben:Exercise 2.1Z: DSB-AM without/with Carrier - Revision history

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	[[Category:Modulation Methods: Exercises\|^2.1 Double Sideband Amplitude Modulation^]]		[[Category:Modulation Methods: Exercises\|^2.1 Double Sideband Amplitude Modulation^]]
	[[de:Aufgaben:2.~~1Z_ZSB~~-~~AM_ohne~~/~~mit_Träger~~]]		[[de:Aufgaben:Aufgabe 2.1Z: ZSB-AM ohne/mit Träger]]

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	[[Category:Modulation Methods: Exercises\|^2.1 Double Sideband Amplitude Modulation^]]		[[Category:Modulation Methods: Exercises\|^2.1 Double Sideband Amplitude Modulation^]]
			[[de:Aufgaben:2.1Z_ZSB-AM_ohne/mit_Träger]]

Hwang at 14:17, 18 January 2023

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	{{quiz-Header\|Buchseite=~~Modulationsverfahren~~/~~Zweiseitenband~~-~~Amplitudenmodulation~~		{{quiz-Header\|Buchseite=Modulation_Methods/Double-Sideband_Amplitude_Modulation
	}}		}}

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

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

Guenter at 13:44, 29 November 2021

2021-11-29T13:44:10Z

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	'''(2)'''  From the graph, the period durations of   $200$ μs  and   $20$ μs   can be seen for $q(t)$  and $z(t)$, respectively.		'''(2)'''  From the graph,  the period durations of   $200$ μs  and   $20$ μs   can be seen for  $q(t)$  and $z(t)$,  respectively.
	*This gives the frequencies as   $f_{\rm N} \hspace{0.15cm}\underline { = 5}$  kHz and  $f_{\rm T} \hspace{0.15cm}\underline { = 50}$ kHz.		*This gives the frequencies as   $f_{\rm N} \hspace{0.15cm}\underline { = 5}$  kHz and  $f_{\rm T} \hspace{0.15cm}\underline { = 50}$ kHz.


	'''(3)'''  <u>Answers 1 and 2</u> are correct:
			'''(3)'''  <u>Answers 1 and 2</u>  are correct:
	*The zero crossings of  $z(t)$  at  $±5$ μs,  $±15$ μs,  $±25$ μs, ... ... are also present in the signal  $s(t)$    ⇒   Answer 1 is correct.		*The zero crossings of  $z(t)$  at  $±5$ μs,  $±15$ μs,  $±25$ μs, ... ... are also present in the signal  $s(t)$    ⇒   Answer 1 is correct.
	*Other zero intersects of  $s(t)$ – cause by  $q(t)$  – are present at  $±50$ μs,  $±150$ μs,  $±250$ μs, ....   ⇒   Answer 2 is also correct.		*Other zero intersects of  $s(t)$ – cause by  $q(t)$  – are present at  $±50$ μs,  $±150$ μs,  $±250$ μs, ....   ⇒   Answer 2 is also correct.
	*In contrast, the third statement is not true. Instead,   $ s(t) = a(t) \cdot \cos[\omega_{\rm T} t + \phi (t)] \hspace{0.05cm}.$		*In contrast, the third statement is not true. Instead,   $ s(t) = a(t) \cdot \cos[\omega_{\rm T} t + \phi (t)] \hspace{0.05cm}.$
			[[File:EN_Mod_Z_2_1_d.png\|right\|frame\|DSB–AM spectra  $Z(f)$,  $Q(f)$  and  $S(f)$]]

	*For  $q(t) > 0$  the phase function is  $ϕ(t) = 0$  and  $s(t)$  coincides with   $z(t)$.		*For  $q(t) > 0$  the phase function is  $ϕ(t) = 0$  and  $s(t)$  coincides with   $z(t)$.
	*n contrast, for  $q(t) < 0$:   $ϕ(t) = π = 180^\circ$.		*In contrast, for  $q(t) < 0$:   $ϕ(t) = π = 180^\circ$.
	*~~BAt~~ the zero crossings of  $q(t)$  , the modulated signal   $s(t)$  exhibits phase jumps.		*At the zero crossings of  $q(t)$,  the modulated signal   $s(t)$  exhibits phase jumps.



	~~[[File:EN_Mod_Z_2_1_d.png\|right\|frame\|DSB–AM Spectrum  $Z(f)$,  $Q(f)$  and  $S(f)$]]~~		'''(4)'''  The spectrum  $S(f)$  results from the convolution of the spectral functions   $Z(f)$  and  $Q(f)$,  each consisting of only two Dirac delta functions.  The graph displays the result.
	'''(4)'''  The spectrum  $S(f)$  results from the convolution of the spectral functions   $Z(f)$  and  $Q(f)$, each consisting of only two Dirac functions. The graph displays the result.		*The Dirac delta functions plotted in red apply only to the  "DSB-AM with carrier"  and refer to subtask ('''6)'''.
	*The Dirac functions plotted in red apply only to the "DSB-AM with carrier" and refer to subtask ('''6)'''.		*Convolution of the two  $Z(f)$  Dirac delta functions at  $f_{\rm T} = 50\text{ kHz}$  with  $Q(f)$  leads to the Dirac lines at  $f_{\rm T} - f_{\rm N}$  and  $f_{\rm T} + f_{\rm N}$,  each with weight   $0.5 · 0.5\text{ V}= 0.25\text{ V}$.
	*Convolution of the two  $Z(f)$~~–Dirac~~ functions at  $f_{\rm T} = 50\text{ kHz}$  with  $Q(f)$  leads to the Dirac lines at  $f_{\rm T} - f_{\rm N}$  and  $f_{\rm T} + f_{\rm N}$, each with weight   $0.5 · 0.5\text{ V}= 0.25\text{ V}$.		*Thus,  the desired values are   $f_1\hspace{0.15cm}\underline { = 45 \ \rm kHz}$  and  $f_1\hspace{0.15cm}\underline { = 55 \ \rm kHz}$.
	*Thus, the desired values are   $f_1\hspace{0.15cm}\underline { = 45 \ \rm kHz}$  and  $f_1\hspace{0.15cm}\underline { = 55 \ \rm kHz}$.
	*The Dirac function  $0.5 · δ(f + f_{\rm T})$  with two markers leads to two more Dirac lines at   $-f_1$  and  $-f_2$.		*The Dirac function  $0.5 · δ(f + f_{\rm T})$  with two markers leads to two more Dirac lines at   $-f_1$  and  $-f_2$.

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	'''(6)'''  <u>Answers 1 and 3</u> are correct:		'''(6)'''  <u>Answers 1 and 3</u>  are correct:
	*According to the sketch, Dirac lines result at   $±f_{\rm T}$, both with impulse weight  $A_{\rm T}/2 = 1\text{ V}$.		*According to the sketch,  Dirac lines result at   $±f_{\rm T}$,  both with impulse weight  $A_{\rm T}/2 = 1\text{ V}$.
	*At  $m ≤ 1$~~ ~~,  $q(t)$  is detectable in the envelope ~~and~~ envelope demodulation is applicable.		*At  $m ≤ 1$,  $q(t)$  is detectable in the envelope   ⇒   envelope demodulation is applicable.
	*However, this simpler receiver variant must be accounted for with a much larger transmission power.		*However,  this simpler receiver variant must be accounted for with a much larger transmission power.
	*In this example  $(m = 0.5)$  the addition of a carrier multiplies the transmission power by nine.		*In this example  $(m = 0.5)$  the addition of a carrier multiplies the transmission power by nine.

Guenter at 13:30, 29 November 2021

2021-11-29T13:30:17Z

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	}}		}}

	[[File:P_ID987__Mod_Z_2_1.png\|right\|frame\|The signals involved in AM]]		[[File:P_ID987__Mod_Z_2_1.png\|right\|frame\|The signals involved in this AM]]
	The red curve on the graph shows a section of the transmitted signal  $s(t) = q(t) · z(t)$  of a double-sideband amplitude modulation (abbreviated as DSB-AM) ~~without carrier.~~   ~~(abgekürzt mit ZSB-AM)  ohne Träger~~. The duration of the time interval is  $\rm 200 \ µ s$.		The red curve on the graph shows a section of the transmitted signal  $s(t) = q(t) · z(t)$  of a double-sideband amplitude modulation  (abbreviated as DSB-AM)  without carrier.  The duration of the time interval is  $\rm 200 \ µ s$.

	Additionally plotted in the graph are:		Additionally plotted in the graph are:
	*the source signal (as a blue dashed curve):		*the source signal  (as a blue dashed curve):
	:$$q(t) = 1\,{\rm V} \cdot \cos(2 \pi f_{\rm N} t + \phi_{\rm N}),$$		:$$q(t) = 1\,{\rm V} \cdot \cos(2 \pi f_{\rm N} t + \phi_{\rm N}),$$
	*the carrier signal (as a grey dashed trace):		*the carrier signal  (as a grey dashed trace):
	:$$z(t) = 1 \cdot \cos(2 \pi f_{\rm T} t + \phi_{\rm T})$$		:$$z(t) = 1 \cdot \cos(2 \pi f_{\rm T} t + \phi_{\rm T}).$$

	From subtask '''(4)''' onwards, the "DSB-AM with carrier" is considered. In that case, with  $A_{\rm T} = 2\text{ V}$:		From subtask  '''(4)'''  onwards,  the "DSB-AM with carrier"  is considered.  In that case, with  $A_{\rm T} = 2\text{ V}$:
	:$$s(t) = \left(q(t) + A_{\rm T} \right) \cdot z(t) \hspace{0.05cm}.$$		:$$s(t) = \left(q(t) + A_{\rm T} \right) \cdot z(t) \hspace{0.05cm}.$$

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			Hints:


	''Hints:''
	*This exercise belongs to the chapter  [[Modulation_Methods/Double-Sideband_Amplitude_Modulation\|Double-Sideband Amplitude Modulation]].		*This exercise belongs to the chapter  [[Modulation_Methods/Double-Sideband_Amplitude_Modulation\|Double-Sideband Amplitude Modulation]].
	*Particlar reference is made to the pages  [[Modulation_Methods/Double-Sideband_Amplitude_Modulation#Description_in_the_time_domain\|Description in the time domain]]  and  [[Modulation_Methods/Double-Sideband_Amplitude_Modulation#Double-Sideband_Amplitude_Modulation_with_carrier\|Double-Sideband Amplitude Modulation with carrier]].		*Particlar reference is made to the pages  [[Modulation_Methods/Double-Sideband_Amplitude_Modulation#Description_in_the_time_domain\|Description in the time domain]]  and  [[Modulation_Methods/Double-Sideband_Amplitude_Modulation#Double-Sideband_Amplitude_Modulation_with_carrier\|Double-Sideband Amplitude Modulation with carrier]].
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	<quiz display=simple>		<quiz display=simple>
	{From the graph, determine the phase values of the ~~source~~ and carrier signals.		{From the graph,  determine the phase values of the message and carrier signals.
	\|type="{}"}		\|type="{}"}
	$\phi_{\rm N} \ = \ $ { 0. } $\ \text{degrees}$		$\phi_{\rm N} \ = \ $ { 0. } $\ \text{degrees}$
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	$f_{\rm T} \ = \ $ { 50 3% } $\ \text{kHz}$		$f_{\rm T} \ = \ $ { 50 3% } $\ \text{kHz}$

	{Analyze the zero crossings of  $s(t)$. Which statements are true?		{Analyze the zero crossings of  $s(t)$.  Which statements are true?
	\|type="[]"}		\|type="[]"}
	+ All zero crossings of  $z(t)$  are preserved in  $s(t)$~~ ~~.		+ All zero crossings of  $z(t)$  are preserved in  $s(t)$.
	+ There are additional zero crossings caused by  $q(t)$.		+ There are additional zero crossings caused by  $q(t)$.
	-  $s(t) = a(t) · \cos(ω_T · t)$  holds with  $a(t) = \|q(t)\|$.		-  $s(t) = a(t) · \cos(ω_T · t)$  holds with  $a(t) = \|q(t)\|$.

	{Determine the spectral function  $S(f)$  by convolution.  Which (positive) frequencies $f_1$  and $f_2 > f_1$  are included in the signal??		{Determine the spectral function  $S(f)$  by convolution.  Which  (positive)  frequencies  $f_1$  and  $f_2 > f_1$  are included in the signal??
	\|type="{}"}		\|type="{}"}
	$f_1 \ = \ $ { 45 3% } $\ \text{kHz}$		$f_1 \ = \ $ { 45 3% } $\ \text{kHz}$
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	$m \ = \ $ { 0.5 3% }		$m \ = \ $ { 0.5 3% }

	{~~WWhich~~ of the statements are true for   ~~„DSB–AM~~ with ~~carrier”~~  and  $A_{\rm T} = 2\text{ V}$ ?		{Which of the statements are true for   "DSB–AM with carrier"  and  $A_{\rm T} = 2\text{ V}$ ?
	\|type="[]"}		\|type="[]"}
	+ $S(f)$  now includes Dirac functions at  $±f_{\rm T}$.		+ $S(f)$  now includes Dirac delta functions at  $±f_{\rm T}$.
	- The weights of these Dirac lines are each  $2\text{ V}$.		- The weights of these Dirac lines are each  $2\text{ V}$.
	+ $q(t)$  can be seen in the envelope of  $s(t)$~~ ~~.		+ $q(t)$  can be seen in the envelope of  $s(t)$.
	- Due to the additional carrier component, the power remains unchanged..		- Due to the additional carrier component,  the power remains unchanged.

	</quiz>		</quiz>

Reed at 11:25, 24 November 2021

2021-11-24T11:25:40Z

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	}}		}}

	[[File:P_ID987__Mod_Z_2_1.png\|right\|frame\|~~Die bei der~~ AM ~~beteiligten Signale~~]]		[[File:P_ID987__Mod_Z_2_1.png\|right\|frame\|The signals involved in AM]]
	The red curve on the graph shows a section of the transmitted signal  $s(t) = q(t) · z(t)$  of a double-sideband amplitude modulation (abbreviated as DSB-AM) without carrier.   (abgekürzt mit ZSB-AM)  ohne Träger. The duration of the time interval is  $\rm 200 \ µ s$.		The red curve on the graph shows a section of the transmitted signal  $s(t) = q(t) · z(t)$  of a double-sideband amplitude modulation (abbreviated as DSB-AM) without carrier.   (abgekürzt mit ZSB-AM)  ohne Träger. The duration of the time interval is  $\rm 200 \ µ s$.

Reed: /* Solution */

2021-11-24T11:24:33Z

Solution

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Beide Signale sind cosinusförmig~~   ⇒   $ϕ_{\rm N} \hspace{0.15cm}\underline { = 0}$  ~~und~~  $ϕ_{\rm T} \hspace{0.15cm}\underline { = 0}$.		'''(1)'''  Both signals are cosine   ⇒   $ϕ_{\rm N} \hspace{0.15cm}\underline { = 0}$  and  $ϕ_{\rm T} \hspace{0.15cm}\underline { = 0}$.



	'''(2)'''  ~~Aus der Grafik können für~~  $~~q(t)~~$  ~~und $z(t)$ die Periodendauern~~  $~~200~~$ μs  ~~bzw.~~  $20$ μs  ~~abgelesen werden~~.		'''(2)'''  From the graph, the period durations of   $200$ μs  and   $20$ μs   can be seen for $q(t)$  and $z(t)$, respectively.
	*~~Daraus ergeben sich die Frequenzen zu~~  $f_{\rm N} \hspace{0.15cm}\underline { = 5}$  kHz ~~und~~  $f_{\rm T} \hspace{0.15cm}\underline { = 50}$ kHz.		*This gives the frequencies as   $f_{\rm N} \hspace{0.15cm}\underline { = 5}$  kHz and  $f_{\rm T} \hspace{0.15cm}\underline { = 50}$ kHz.


	'''(3)'''  ~~Richtig sind die~~ <u>~~Lösungsvorschläge~~ 1 ~~und~~ 2</u>:		'''(3)'''  <u>Answers 1 and 2</u> are correct:
	*~~Die Nullstellen von~~  $z(t)$  ~~bei~~  $±5$ μs,  $±15$ μs,  $±25$ μs, ... ~~sind auch im Signal~~  $s(t)$  ~~vorhanden~~   ⇒   ~~Aussage~~ 1 ~~ist richtig~~.		*The zero crossings of  $z(t)$  at  $±5$ μs,  $±15$ μs,  $±25$ μs, ... ... are also present in the signal  $s(t)$    ⇒   Answer 1 is correct.
	*~~Weitere Nullstellen von~~  $s(t)$ – ~~verursacht durch~~  $q(t)$  – ~~liegen bei~~  $±50$ μs,  $±150$ μs,  $±250$ μs, ....   ⇒   ~~Aussage~~ 2 ~~ist richtig~~.		*Other zero intersects of  $s(t)$ – cause by  $q(t)$  – are present at  $±50$ μs,  $±150$ μs,  $±250$ μs, ....   ⇒   Answer 2 is also correct.
	*~~Die dritte Aussage trifft dagegen nicht zu~~, ~~sondern es gilt:~~   $ s(t) = a(t) \cdot \cos[\omega_{\rm T} t + \phi (t)] \hspace{0.05cm}.$		*In contrast, the third statement is not true. Instead,   $ s(t) = a(t) \cdot \cos[\omega_{\rm T} t + \phi (t)] \hspace{0.05cm}.$
	*~~Für~~  $q(t) > 0$  ~~ist die Phasenfunktion~~  $ϕ(t) = 0$  ~~und~~  $s(t)$  ~~ist gleichlaufend mit~~  $z(t)$.		*For  $q(t) > 0$  the phase function is  $ϕ(t) = 0$  and  $s(t)$  coincides with   $z(t)$.
	*~~Dagegen gilt für~~  $q(t) < 0$:   $ϕ(t) = π = 180^\circ$.		*n contrast, for  $q(t) < 0$:   $ϕ(t) = π = 180^\circ$.
	*~~Bei den Nulldurchgängen von~~  $q(t)$  ~~weist das modulierte Signal~~  $s(t)$  ~~Phasensprünge auf~~.		*BAt the zero crossings of  $q(t)$  , the modulated signal   $s(t)$  exhibits phase jumps.



	[[File:EN_Mod_Z_2_1_d.png\|right\|frame\|~~ZSB–AM–Spektrum~~  $Z(f)$,  $Q(f)$  ~~und~~  $S(f)$]]		[[File:EN_Mod_Z_2_1_d.png\|right\|frame\|DSB–AM Spectrum  $Z(f)$,  $Q(f)$  and  $S(f)$]]
	'''(4)'''  ~~Das Spektrum~~  $S(f)$  ~~ergibt sich aus der Faltung der Spektralfunktionen~~  $Z(f)$  ~~und~~  $Q(f)$, ~~die jeweils aus nur zwei Diracfunktionen bestehen~~.~~  Die Grafik zeigt das Ergebnis~~.		'''(4)'''  The spectrum  $S(f)$  results from the convolution of the spectral functions   $Z(f)$  and  $Q(f)$, each consisting of only two Dirac functions. The graph displays the result.
	*~~Die rot eingezeichneten Diracfunktionen gelten nur für die  „ZSB–AM mit Träger”  und beziehen sich auf die Teilaufgabe ~~ ('''6)'''.		*The Dirac functions plotted in red apply only to the "DSB-AM with carrier" and refer to subtask ('''6)'''.
	*~~Die Faltung der beiden~~  $Z(f)$~~–Diracfunktionen bei~~  $f_{\rm T} = 50\text{ kHz}$  ~~mit~~  $Q(f)$  ~~führt zu den Diraclinien bei~~  $f_{\rm T} - f_{\rm N}$  ~~und~~  $f_{\rm T} + f_{\rm N}$, ~~jeweils mit Gewicht~~  $0.5 · 0.5\text{ V}= 0.25\text{ V}$.		*Convolution of the two  $Z(f)$–Dirac functions at  $f_{\rm T} = 50\text{ kHz}$  with  $Q(f)$  leads to the Dirac lines at  $f_{\rm T} - f_{\rm N}$  and  $f_{\rm T} + f_{\rm N}$, each with weight   $0.5 · 0.5\text{ V}= 0.25\text{ V}$.
	*~~Die gesuchten Werte sind somit~~  $f_1\hspace{0.15cm}\underline { = 45 \ \rm kHz}$  ~~und~~  $f_1\hspace{0.15cm}\underline { = 55 \ \rm kHz}$.		*Thus, the desired values are   $f_1\hspace{0.15cm}\underline { = 45 \ \rm kHz}$  and  $f_1\hspace{0.15cm}\underline { = 55 \ \rm kHz}$.
	*~~Die mit zwei Markierungsstrichen versehene Diracfunktion~~  $0.5 · δ(f + f_{\rm T})$  ~~führt zu zwei weiteren Diraclinien bei~~  $-f_1$  ~~und~~  $-f_2$.		*The Dirac function  $0.5 · δ(f + f_{\rm T})$  with two markers leads to two more Dirac lines at   $-f_1$  and  $-f_2$.



	'''(5)'''  ~~Der Modulationsgrad berechnet sich zu~~:		'''(5)'''  The modulation depth is calculated as:
	:$$ m = \frac{q_{\rm max}}{A_{\rm T}} = \frac{A_{\rm N}}{A_{\rm T}} \hspace{0.15cm}\underline {= 0.5} \hspace{0.05cm}.$$		:$$ m = \frac{q_{\rm max}}{A_{\rm T}} = \frac{A_{\rm N}}{A_{\rm T}} \hspace{0.15cm}\underline {= 0.5} \hspace{0.05cm}.$$


	'''(6)'''  ~~Richtig sind die~~ <u>~~Lösungsvorschläge~~ 1 ~~und~~ 3</u>:		'''(6)'''  <u>Answers 1 and 3</u> are correct:
	*~~Gemäß der Skizze ergeben sich Diraclinien bei~~  $±f_{\rm T}$, ~~beide mit dem Impulsgewicht~~  $A_{\rm T}/2 = 1\text{ V}$.		*According to the sketch, Dirac lines result at   $±f_{\rm T}$, both with impulse weight  $A_{\rm T}/2 = 1\text{ V}$.
	*~~Bei~~  $m ≤ 1$  ~~ist~~  $q(t)$  in ~~der Hüllkurve erkennbar und Hüllkurvendemodulation anwendbar~~.		*At  $m ≤ 1$ ,  $q(t)$  is detectable in the envelope and envelope demodulation is applicable.
	*~~Allerdings muss diese einfachere Empfängervariante durch eine sehr viel größere Sendeleistung erkauft werden~~.		*However, this simpler receiver variant must be accounted for with a much larger transmission power.
	*In ~~diesem Beispiel~~  $(m = 0.5)$  ~~wird die Sendeleistung durch den Trägerzusatz verneunfacht~~.		*In this example  $(m = 0.5)$  the addition of a carrier multiplies the transmission power by nine.

	{{ML-Fuß}}		{{ML-Fuß}}

Reed at 10:59, 24 November 2021

2021-11-24T10:59:19Z

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	*the source signal (as a blue dashed curve):		*the source signal (as a blue dashed curve):
	:$$q(t) = 1\,{\rm V} \cdot \cos(2 \pi f_{\rm N} t + \phi_{\rm N}),$$		:$$q(t) = 1\,{\rm V} \cdot \cos(2 \pi f_{\rm N} t + \phi_{\rm N}),$$
	*the carrier signal (~~gray-dotted~~ trace):		*the carrier signal (as a grey dashed trace):
	:$$z(t) = 1 \cdot \cos(2 \pi f_{\rm T} t + \phi_{\rm T})$$		:$$z(t) = 1 \cdot \cos(2 \pi f_{\rm T} t + \phi_{\rm T})$$

Reed at 10:54, 24 November 2021

2021-11-24T10:54:31Z

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	{From the graph, determine the phase values of the source and carrier signals.		{From the graph, determine the phase values of the source and carrier signals.
	\|type="{}"}		\|type="{}"}
	$\phi_{\rm N} \ = \ $ { 0. } $\ \text{~~Grad~~}$		$\phi_{\rm N} \ = \ $ { 0. } $\ \text{degrees}$
	$\phi_{\rm T} \ = \ $ { 0. } $\ \text{~~Grad~~}$		$\phi_{\rm T} \ = \ $ { 0. } $\ \text{degrees}$

	{What is the frequency $f_{\rm N}$  of the message signal  $q(t)$  and what is the frequency $f_{\rm T}$  of the carrier signal  $z(t)$?		{What is the frequency $f_{\rm N}$  of the message signal  $q(t)$  and what is the frequency $f_{\rm T}$  of the carrier signal  $z(t)$?

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	\|type="[]"}		\|type="[]"}
	+ $S(f)$  now includes Dirac delta functions at  $±f_{\rm T}$.		+ $S(f)$  now includes Dirac delta functions at  $±f_{\rm T}$.
	- The weights of these Dirac lines are each  $2\text{ V}$.		- The weights of these Dirac delta lines are each  $2\text{ V}$.
	+ $q(t)$  can be seen in the envelope of  $s(t)$.		+ $q(t)$  can be seen in the envelope of  $s(t)$.
	- Due to the additional carrier component,  the power remains unchanged.		- Due to the additional carrier component,  the power remains unchanged.
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	'''(4)'''  The spectrum  $S(f)$  results from the convolution of the spectral functions   $Z(f)$  and  $Q(f)$,  each consisting of only two Dirac delta functions.  The graph displays the result.		'''(4)'''  The spectrum  $S(f)$  results from the convolution of the spectral functions   $Z(f)$  and  $Q(f)$,  each consisting of only two Dirac delta functions.  The graph displays the result.
	*The Dirac delta functions plotted in red apply only to the  "DSB-AM with carrier"  and refer to subtask ('''6)'''.		*The Dirac delta functions plotted in red apply only to the  "DSB-AM with carrier"  and refer to subtask ('''6)'''.
	*Convolution of the two  $Z(f)$  Dirac delta functions at  $f_{\rm T} = 50\text{ kHz}$  with  $Q(f)$  leads to the Dirac lines at  $f_{\rm T} - f_{\rm N}$  and  $f_{\rm T} + f_{\rm N}$,  each with weight   $0.5 · 0.5\text{ V}= 0.25\text{ V}$.		*Convolution of the two  $Z(f)$  Dirac delta functions at  $f_{\rm T} = 50\text{ kHz}$  with  $Q(f)$  leads to the Dirac delta lines at  $f_{\rm T} - f_{\rm N}$  and  $f_{\rm T} + f_{\rm N}$,  each with weight   $0.5 · 0.5\text{ V}= 0.25\text{ V}$.
	*Thus,  the desired values are   $f_1\hspace{0.15cm}\underline { = 45 \ \rm kHz}$  and  $f_1\hspace{0.15cm}\underline { = 55 \ \rm kHz}$.		*Thus,  the desired values are   $f_1\hspace{0.15cm}\underline { = 45 \ \rm kHz}$  and  $f_1\hspace{0.15cm}\underline { = 55 \ \rm kHz}$.
	*The Dirac function  $0.5 · δ(f + f_{\rm T})$  with two markers leads to two more Dirac lines at   $-f_1$  and  $-f_2$.		*The Dirac function  $0.5 · δ(f + f_{\rm T})$  with two markers leads to two more Dirac delta lines at   $-f_1$  and  $-f_2$.


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	'''(6)'''  <u>Answers 1 and 3</u>  are correct:		'''(6)'''  <u>Answers 1 and 3</u>  are correct:
	*According to the sketch,  Dirac lines result at   $±f_{\rm T}$,  both with impulse weight  $A_{\rm T}/2 = 1\text{ V}$.		*According to the sketch,  Dirac delta lines result at   $±f_{\rm T}$,  both with impulse weight  $A_{\rm T}/2 = 1\text{ V}$.
	*At  $m ≤ 1$,  $q(t)$  is detectable in the envelope   ⇒   envelope demodulation is applicable.		*At  $m ≤ 1$,  $q(t)$  is detectable in the envelope   ⇒   envelope demodulation is applicable.
	*However,  this simpler receiver variant must be accounted for with a much larger transmission power.		*However,  this simpler receiver variant must be accounted for with a much larger transmission power.