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

Hwang at 14:17, 18 January 2023

2023-01-18T14:17:19Z

Reed at 15:32, 24 March 2022

2022-03-24T15:32:11Z

Guenter at 13:44, 29 November 2021

2021-11-29T13:44:10Z

Guenter at 13:30, 29 November 2021

2021-11-29T13:30:17Z

Reed at 11:25, 24 November 2021

2021-11-24T11:25:40Z

Reed: /* Solution */

2021-11-24T11:24:33Z

Solution

Reed at 10:59, 24 November 2021

2021-11-24T10:59:19Z

Reed at 10:54, 24 November 2021

2021-11-24T10:54:31Z

Reed at 10:54, 24 November 2021

2021-11-24T10:54:08Z

Reed at 10:42, 24 November 2021

2021-11-24T10:42:47Z

@@ Line 57: / Line 57: @@
 |type="[]"}
 +  $S(f)$&nbsp; now includes Dirac delta functions at &nbsp;$±f_{\rm T}$.
--  The weights of these Dirac lines are each &nbsp;$2\text{ V}$.
+-  The weights of these Dirac delta lines are each &nbsp;$2\text{ V}$.
 +  $q(t)$&nbsp; can be seen in the envelope of &nbsp;$s(t)$.
 -  Due to the additional carrier component,&nbsp; the power remains unchanged.
@@ Line 88: / Line 88: @@
 '''(4)'''&nbsp; The spectrum&nbsp; $S(f)$&nbsp; results from the convolution of the spectral functions &nbsp; $Z(f)$&nbsp; and&nbsp; $Q(f)$,&nbsp; each consisting of only two Dirac delta functions.&nbsp; The graph displays the result.
 *The Dirac delta functions plotted in red apply only to the&nbsp; "DSB-AM with carrier"&nbsp; and refer to subtask ('''6)'''.
-*Convolution of the two&nbsp; $Z(f)$&nbsp; Dirac delta functions at&nbsp; $f_{\rm T} = 50\text{ kHz}$&nbsp; with&nbsp; $Q(f)$&nbsp; leads to the Dirac lines at&nbsp; $f_{\rm T} - f_{\rm N}$&nbsp; and&nbsp; $f_{\rm T} + f_{\rm N}$,&nbsp; each with weight &nbsp; $0.5 · 0.5\text{ V}= 0.25\text{ V}$.
+*Convolution of the two&nbsp; $Z(f)$&nbsp; Dirac delta functions at&nbsp; $f_{\rm T} = 50\text{ kHz}$&nbsp; with&nbsp; $Q(f)$&nbsp; leads to the Dirac delta lines at&nbsp; $f_{\rm T} - f_{\rm N}$&nbsp; and&nbsp; $f_{\rm T} + f_{\rm N}$,&nbsp; each with weight &nbsp; $0.5 · 0.5\text{ V}= 0.25\text{ V}$.
 *Thus,&nbsp; the desired values are &nbsp; $f_1\hspace{0.15cm}\underline { = 45 \ \rm kHz}$&nbsp; and&nbsp; $f_1\hspace{0.15cm}\underline { = 55 \ \rm kHz}$.
-*The Dirac function&nbsp; $0.5 · δ(f + f_{\rm T})$&nbsp; with two markers leads to two more Dirac lines at &nbsp; $-f_1$&nbsp; and&nbsp; $-f_2$.
+*The Dirac function&nbsp; $0.5 · δ(f + f_{\rm T})$&nbsp; with two markers leads to two more Dirac delta lines at &nbsp; $-f_1$&nbsp; and&nbsp; $-f_2$.
@@ Line 99: / Line 99: @@
 '''(6)'''&nbsp; <u>Answers 1 and 3</u>&nbsp; are correct:
-*According to the sketch,&nbsp; Dirac lines result at &nbsp; $±f_{\rm T}$,&nbsp; both with impulse weight&nbsp; $A_{\rm T}/2 = 1\text{ V}$.
+*According to the sketch,&nbsp; Dirac delta lines result at &nbsp; $±f_{\rm T}$,&nbsp; both with impulse weight&nbsp; $A_{\rm T}/2 = 1\text{ V}$.
 *At&nbsp; $m ≤ 1$,&nbsp; $q(t)$&nbsp; is detectable in the envelope &nbsp; &rArr; &nbsp;  envelope demodulation is applicable.
 *However,&nbsp; this simpler receiver variant must be accounted for with a much larger transmission power. &nbsp;

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Modulationsverfahren/Zweiseitenband-Amplitudenmodulation
+{{quiz-Header|Buchseite=Modulation_Methods/Double-Sideband_Amplitude_Modulation
 }}
@@ Line 26: / Line 26: @@
-===Fragebogen===
+===Questions===
 <quiz display=simple>

@@ Line 69: / Line 69: @@
-'''(2)'''&nbsp; From the graph, the period durations of &nbsp; $200$ μs&nbsp; and &nbsp; $20$ μs&nbsp;&nbsp;  can be seen for  $q(t)$&nbsp; and $z(t)$, respectively.
+'''(2)'''&nbsp; From the graph,&nbsp; the period durations of &nbsp; $200$ μs&nbsp; and &nbsp; $20$ μs&nbsp;&nbsp;  can be seen for&nbsp;  $q(t)$&nbsp; and $z(t)$,&nbsp; respectively.
 *This gives the frequencies as &nbsp; $f_{\rm N} \hspace{0.15cm}\underline { = 5}$&nbsp; kHz and&nbsp;  $f_{\rm T} \hspace{0.15cm}\underline { = 50}$  kHz.
 '''(3)'''&nbsp; <u>Answers 1 and 2</u> are correct:
 *The zero crossings of&nbsp; $z(t)$ &nbsp;at&nbsp; $±5$ μs,&nbsp; $±15$ μs,&nbsp; $±25$ μs, ... ... are also present in the signal&nbsp; $s(t)$&nbsp; &nbsp; &rArr; &nbsp; Answer 1 is correct.
 *Other zero intersects of&nbsp; $s(t)$ &ndash; cause by&nbsp; $q(t)$&nbsp; &ndash; are present at&nbsp; $±50$ μs,&nbsp; $±150$ μs,&nbsp; $±250$ μs, .... &nbsp; &rArr; &nbsp; Answer 2 is also correct.
 *In contrast, the third statement is not true. Instead, &nbsp; $ s(t) = a(t) \cdot \cos[\omega_{\rm T} t + \phi (t)] \hspace{0.05cm}.$
 *For&nbsp; $q(t) > 0$&nbsp; the phase function is&nbsp; $ϕ(t) = 0$&nbsp; and&nbsp; $s(t)$&nbsp; coincides with &nbsp; $z(t)$.
-*n contrast, for&nbsp; $q(t) < 0$: &nbsp; $ϕ(t) = π = 180^\circ$.
+*In contrast, for&nbsp; $q(t) < 0$: &nbsp; $ϕ(t) = π = 180^\circ$.
-*BAt the zero crossings of&nbsp; $q(t)$&nbsp; , the modulated signal &nbsp; $s(t)$&nbsp; exhibits phase jumps.
+*At the zero crossings of&nbsp; $q(t)$,&nbsp; the modulated signal &nbsp; $s(t)$&nbsp; exhibits phase jumps.
-[[File:EN_Mod_Z_2_1_d.png|right|frame|DSB–AM Spectrum&nbsp; $Z(f)$,&nbsp; $Q(f)$&nbsp; and&nbsp; $S(f)$]]
+'''(4)'''&nbsp; The spectrum&nbsp; $S(f)$&nbsp; results from the convolution of the spectral functions &nbsp; $Z(f)$&nbsp; and&nbsp; $Q(f)$,&nbsp; each consisting of only two Dirac delta functions.&nbsp; The graph displays the result.
-'''(4)'''&nbsp; The spectrum&nbsp; $S(f)$&nbsp; results from the convolution of the spectral functions &nbsp; $Z(f)$&nbsp; and&nbsp; $Q(f)$, each consisting of only two Dirac functions. The graph displays the result.
+*The Dirac delta functions plotted in red apply only to the&nbsp; "DSB-AM with carrier"&nbsp; 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&nbsp; $Z(f)$&nbsp; Dirac delta functions at&nbsp; $f_{\rm T} = 50\text{ kHz}$&nbsp; with&nbsp; $Q(f)$&nbsp; leads to the Dirac lines at&nbsp; $f_{\rm T} - f_{\rm N}$&nbsp; and&nbsp; $f_{\rm T} + f_{\rm N}$,&nbsp; each with weight &nbsp; $0.5 · 0.5\text{ V}= 0.25\text{ V}$.
-*Convolution of the two&nbsp; $Z(f)$–Dirac functions at&nbsp; $f_{\rm T} = 50\text{ kHz}$&nbsp; with&nbsp; $Q(f)$&nbsp; leads to the Dirac lines at&nbsp; $f_{\rm T} - f_{\rm N}$&nbsp; and&nbsp; $f_{\rm T} + f_{\rm N}$, each with weight &nbsp; $0.5 · 0.5\text{ V}= 0.25\text{ V}$.
+*Thus,&nbsp; the desired values are &nbsp; $f_1\hspace{0.15cm}\underline { = 45 \ \rm kHz}$&nbsp; and&nbsp; $f_1\hspace{0.15cm}\underline { = 55 \ \rm kHz}$.
 *The Dirac function&nbsp; $0.5 · δ(f + f_{\rm T})$&nbsp; with two markers leads to two more Dirac lines at &nbsp; $-f_1$&nbsp; and&nbsp; $-f_2$.
@@ Line 96: / Line 98: @@
-'''(6)'''&nbsp; <u>Answers 1 and 3</u> are correct:
+'''(6)'''&nbsp; <u>Answers 1 and 3</u>&nbsp; are correct:
-*According to the sketch, Dirac lines result at &nbsp; $±f_{\rm T}$, both with impulse weight&nbsp; $A_{\rm T}/2 = 1\text{ V}$.
+*According to the sketch,&nbsp; Dirac lines result at &nbsp; $±f_{\rm T}$,&nbsp; both with impulse weight&nbsp; $A_{\rm T}/2 = 1\text{ V}$.
-*At&nbsp; $m ≤ 1$&nbsp;,&nbsp; $q(t)$&nbsp; is detectable in the envelope and envelope demodulation is applicable.
+*At&nbsp; $m ≤ 1$,&nbsp; $q(t)$&nbsp; is detectable in the envelope &nbsp; &rArr; &nbsp;  envelope demodulation is applicable.
-*However, this simpler receiver variant must be accounted for with a much larger transmission power. &nbsp;
+*However,&nbsp; this simpler receiver variant must be accounted for with a much larger transmission power. &nbsp;
 *In this example&nbsp; $(m = 0.5)$&nbsp; the addition of a carrier multiplies the transmission power by nine.

@@ Line 3: / Line 3: @@
 }}
-[[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 &nbsp;$s(t) = q(t) · z(t)$&nbsp;  of a double-sideband amplitude modulation (abbreviated as DSB-AM) without carrier. &nbsp; (abgekürzt mit ZSB-AM)&nbsp; ohne Träger.&nbsp;The duration of the time interval is &nbsp;$\rm 200 \ &micro; s$.
+The red curve on the graph shows a section of the transmitted signal &nbsp;$s(t) = q(t) · z(t)$&nbsp;  of a double-sideband amplitude modulation&nbsp; (abbreviated as DSB-AM)&nbsp; without carrier.&nbsp; The duration of the time interval is &nbsp;$\rm 200 \ &micro; s$.
 Additionally plotted in the graph are:
-*the source signal (as a blue dashed curve):
+*the source signal&nbsp; (as a blue dashed curve):
 :$$q(t) = 1\,{\rm V} \cdot \cos(2 \pi f_{\rm N} t + \phi_{\rm N}),$$
-*the carrier signal (as a grey dashed trace):
+*the carrier signal&nbsp; (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 &nbsp;$A_{\rm T} = 2\text{ V}$:
+From subtask&nbsp; '''(4)'''&nbsp; onwards,&nbsp; the "DSB-AM with carrier"&nbsp; is considered.&nbsp;  In that case, with &nbsp;$A_{\rm T} = 2\text{ V}$:
 :$$s(t) = \left(q(t) + A_{\rm T} \right) \cdot z(t) \hspace{0.05cm}.$$
@@ Line 19: / Line 19: @@
 Hints:
 *This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Double-Sideband_Amplitude_Modulation|Double-Sideband Amplitude Modulation]].
 *Particlar reference is made to the pages &nbsp;[[Modulation_Methods/Double-Sideband_Amplitude_Modulation#Description_in_the_time_domain|Description in the time domain]]&nbsp; and &nbsp;[[Modulation_Methods/Double-Sideband_Amplitude_Modulation#Double-Sideband_Amplitude_Modulation_with_carrier|Double-Sideband Amplitude Modulation with carrier]].
@@ Line 32: / Line 29: @@
 <quiz display=simple>
-{From the graph, determine the phase values of the source and carrier signals.
+{From the graph,&nbsp; determine the phase values of the message and carrier signals.
 |type="{}"}
 $\phi_{\rm N} \ = \ $ { 0. } $\ \text{degrees}$
@@ Line 42: / Line 39: @@
 $f_{\rm T} \ = \ $  { 50 3% }  $\ \text{kHz}$
-{Analyze the zero crossings of &nbsp;$s(t)$. Which statements are true?
+{Analyze the zero crossings of &nbsp;$s(t)$.&nbsp; Which statements are true?
 |type="[]"}
-+ All zero crossings of  &nbsp;$z(t)$&nbsp; are preserved in &nbsp;$s(t)$&nbsp;.
++ All zero crossings of  &nbsp;$z(t)$&nbsp; are preserved in &nbsp;$s(t)$.
 + There are additional zero crossings caused by &nbsp;$q(t)$.
 - &nbsp;$s(t) = a(t) · \cos(ω_T · t)$&nbsp; holds with &nbsp;$a(t) = |q(t)|$.
-{Determine the spectral function &nbsp;$S(f)$&nbsp; by convolution.&nbsp;  Which (positive) frequencies  $f_1$&nbsp; and $f_2 > f_1$&nbsp; are included in the signal??
+{Determine the spectral function &nbsp;$S(f)$&nbsp; by convolution.&nbsp;  Which&nbsp; (positive)&nbsp; frequencies&nbsp;  $f_1$&nbsp; and&nbsp; $f_2 > f_1$&nbsp; are included in the signal??
 |type="{}"}
 $f_1 \ = \ $ { 45 3% } $\ \text{kHz}$
@@ Line 57: / Line 54: @@
 $m \ = \ $ { 0.5 3% }
-{WWhich of the statements are true for &nbsp; „DSB–AM with carrier”&nbsp; and &nbsp;$A_{\rm T} = 2\text{ V}$&nbsp;?
+{Which of the statements are true for &nbsp; "DSB–AM with carrier"&nbsp; and &nbsp;$A_{\rm T} = 2\text{ V}$&nbsp;?
 |type="[]"}
-+  $S(f)$&nbsp; now includes Dirac functions at &nbsp;$±f_{\rm T}$.
++  $S(f)$&nbsp; now includes Dirac delta functions at &nbsp;$±f_{\rm T}$.
 -  The weights of these Dirac lines are each &nbsp;$2\text{ V}$.
-+  $q(t)$&nbsp; can be seen in the envelope of &nbsp;$s(t)$&nbsp;.
++  $q(t)$&nbsp; can be seen in the envelope of &nbsp;$s(t)$.
--  Due to the additional carrier component, the power remains unchanged..
+-  Due to the additional carrier component,&nbsp; the power remains unchanged.
 </quiz>

@@ Line 3: / Line 3: @@
 }}
-[[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 &nbsp;$s(t) = q(t) · z(t)$&nbsp;  of a double-sideband amplitude modulation (abbreviated as DSB-AM) without carrier. &nbsp; (abgekürzt mit ZSB-AM)&nbsp; ohne Träger.&nbsp;The duration of the time interval is &nbsp;$\rm 200 \ &micro; s$.

@@ Line 68: / Line 68: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Beide Signale sind cosinusförmig &nbsp; &rArr; &nbsp;  $ϕ_{\rm N} \hspace{0.15cm}\underline { = 0}$&nbsp; und&nbsp;  $ϕ_{\rm T} \hspace{0.15cm}\underline { = 0}$.
+'''(1)'''&nbsp; Both signals are cosine &nbsp; &rArr; &nbsp;  $ϕ_{\rm N} \hspace{0.15cm}\underline { = 0}$&nbsp; and&nbsp;  $ϕ_{\rm T} \hspace{0.15cm}\underline { = 0}$.
-'''(2)'''&nbsp; Aus der Grafik können für&nbsp; $q(t)$&nbsp; und $z(t)$ die Periodendauern&nbsp; $200$ μs&nbsp; bzw.&nbsp; $20$ μs&nbsp; abgelesen werden.
+'''(2)'''&nbsp; From the graph, the period durations of &nbsp; $200$ μs&nbsp; and &nbsp; $20$ μs&nbsp;&nbsp;  can be seen for  $q(t)$&nbsp; and $z(t)$, respectively.
-*Daraus ergeben sich die Frequenzen zu&nbsp; $f_{\rm N} \hspace{0.15cm}\underline { = 5}$&nbsp; kHz und&nbsp;  $f_{\rm T} \hspace{0.15cm}\underline { = 50}$  kHz.
+*This gives the frequencies as &nbsp; $f_{\rm N} \hspace{0.15cm}\underline { = 5}$&nbsp; kHz and&nbsp;  $f_{\rm T} \hspace{0.15cm}\underline { = 50}$  kHz.
-'''(3)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 1 und 2</u>:
+'''(3)'''&nbsp; <u>Answers 1 and 2</u> are correct:
-*Die Nullstellen von&nbsp; $z(t)$ &nbsp;bei&nbsp; $±5$ μs,&nbsp; $±15$ μs,&nbsp; $±25$ μs, ... sind auch im Signal&nbsp; $s(t)$&nbsp; vorhanden &nbsp; &rArr; &nbsp; Aussage 1 ist richtig.
+*The zero crossings of&nbsp; $z(t)$ &nbsp;at&nbsp; $±5$ μs,&nbsp; $±15$ μs,&nbsp; $±25$ μs, ... ... are also present in the signal&nbsp; $s(t)$&nbsp; &nbsp; &rArr; &nbsp; Answer 1 is correct.
-*Weitere Nullstellen von&nbsp; $s(t)$ &ndash; verursacht durch&nbsp; $q(t)$&nbsp; &ndash; liegen bei&nbsp; $±50$ μs,&nbsp; $±150$ μs,&nbsp; $±250$ μs, .... &nbsp; &rArr; &nbsp; Aussage 2 ist richtig.
+*Other zero intersects of&nbsp; $s(t)$ &ndash; cause by&nbsp; $q(t)$&nbsp; &ndash; are present at&nbsp; $±50$ μs,&nbsp; $±150$ μs,&nbsp; $±250$ μs, .... &nbsp; &rArr; &nbsp; Answer 2 is also correct.
-*Die dritte Aussage trifft dagegen nicht zu, sondern es gilt: &nbsp; $ s(t) = a(t) \cdot \cos[\omega_{\rm T} t + \phi (t)] \hspace{0.05cm}.$
+*In contrast, the third statement is not true. Instead, &nbsp; $ s(t) = a(t) \cdot \cos[\omega_{\rm T} t + \phi (t)] \hspace{0.05cm}.$
-*Für&nbsp; $q(t) > 0$&nbsp; ist die Phasenfunktion&nbsp; $ϕ(t) = 0$&nbsp; und&nbsp; $s(t)$&nbsp; ist gleichlaufend mit&nbsp; $z(t)$.
+*For&nbsp; $q(t) > 0$&nbsp; the phase function is&nbsp; $ϕ(t) = 0$&nbsp; and&nbsp; $s(t)$&nbsp; coincides with &nbsp; $z(t)$.
-*Dagegen gilt für&nbsp; $q(t) < 0$: &nbsp; $ϕ(t) = π = 180^\circ$.
+*n contrast, for&nbsp; $q(t) < 0$: &nbsp; $ϕ(t) = π = 180^\circ$.
-*Bei den Nulldurchgängen von&nbsp; $q(t)$&nbsp; weist das modulierte Signal&nbsp; $s(t)$&nbsp; Phasensprünge auf.
+*BAt the zero crossings of&nbsp; $q(t)$&nbsp; , the modulated signal &nbsp; $s(t)$&nbsp; exhibits phase jumps.
-[[File:EN_Mod_Z_2_1_d.png|right|frame|ZSB–AM–Spektrum&nbsp; $Z(f)$,&nbsp; $Q(f)$&nbsp; und&nbsp; $S(f)$]]
+[[File:EN_Mod_Z_2_1_d.png|right|frame|DSB–AM Spectrum&nbsp; $Z(f)$,&nbsp; $Q(f)$&nbsp; and&nbsp; $S(f)$]]
-'''(4)'''&nbsp; Das Spektrum&nbsp; $S(f)$&nbsp; ergibt sich aus der Faltung der Spektralfunktionen&nbsp; $Z(f)$&nbsp; und&nbsp; $Q(f)$, die jeweils aus nur zwei Diracfunktionen bestehen.&nbsp; Die Grafik zeigt das Ergebnis.
+'''(4)'''&nbsp; The spectrum&nbsp; $S(f)$&nbsp; results from the convolution of the spectral functions &nbsp; $Z(f)$&nbsp; and&nbsp; $Q(f)$, each consisting of only two Dirac functions. The graph displays the result.
-*Die rot eingezeichneten Diracfunktionen gelten nur für die&nbsp; „ZSB–AM mit Träger”&nbsp; und beziehen sich auf die Teilaufgabe&nbsp; ('''6)'''.
+*The Dirac functions plotted in red apply only to the "DSB-AM with carrier" and refer to subtask ('''6)'''.
-*Die Faltung der beiden&nbsp; $Z(f)$–Diracfunktionen bei&nbsp; $f_{\rm T} = 50\text{ kHz}$&nbsp; mit&nbsp; $Q(f)$&nbsp; führt zu den Diraclinien bei&nbsp; $f_{\rm T} - f_{\rm N}$&nbsp; und&nbsp; $f_{\rm T} + f_{\rm N}$, jeweils mit Gewicht&nbsp; $0.5 · 0.5\text{ V}= 0.25\text{ V}$.
+*Convolution of the two&nbsp; $Z(f)$–Dirac functions at&nbsp; $f_{\rm T} = 50\text{ kHz}$&nbsp; with&nbsp; $Q(f)$&nbsp; leads to the Dirac lines at&nbsp; $f_{\rm T} - f_{\rm N}$&nbsp; and&nbsp; $f_{\rm T} + f_{\rm N}$, each with weight &nbsp; $0.5 · 0.5\text{ V}= 0.25\text{ V}$.
-*Die gesuchten Werte sind somit&nbsp; $f_1\hspace{0.15cm}\underline { = 45 \ \rm kHz}$&nbsp; und&nbsp; $f_1\hspace{0.15cm}\underline { = 55 \ \rm kHz}$.
+*Thus, the desired values are &nbsp; $f_1\hspace{0.15cm}\underline { = 45 \ \rm kHz}$&nbsp; and&nbsp; $f_1\hspace{0.15cm}\underline { = 55 \ \rm kHz}$.
-*Die mit zwei Markierungsstrichen versehene Diracfunktion&nbsp; $0.5 · δ(f + f_{\rm T})$&nbsp; führt zu zwei weiteren Diraclinien bei&nbsp; $-f_1$&nbsp; und&nbsp; $-f_2$.
+*The Dirac function&nbsp; $0.5 · δ(f + f_{\rm T})$&nbsp; with two markers leads to two more Dirac lines at &nbsp; $-f_1$&nbsp; and&nbsp; $-f_2$.
-'''(5)'''&nbsp; Der Modulationsgrad berechnet sich zu:
+'''(5)'''&nbsp; The modulation depth is calculated as:
 :$$ m = \frac{q_{\rm max}}{A_{\rm T}} = \frac{A_{\rm N}}{A_{\rm T}} \hspace{0.15cm}\underline {= 0.5} \hspace{0.05cm}.$$
-'''(6)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 1 und 3</u>:
+'''(6)'''&nbsp; <u>Answers 1 and 3</u> are correct:
-*Gemäß der Skizze ergeben sich Diraclinien bei&nbsp; $±f_{\rm T}$, beide mit dem Impulsgewicht&nbsp; $A_{\rm T}/2 = 1\text{ V}$.
+*According to the sketch, Dirac lines result at &nbsp; $±f_{\rm T}$, both with impulse weight&nbsp; $A_{\rm T}/2 = 1\text{ V}$.
-*Bei&nbsp; $m ≤ 1$&nbsp; ist&nbsp; $q(t)$&nbsp; in der Hüllkurve erkennbar und Hüllkurvendemodulation anwendbar.
+*At&nbsp; $m ≤ 1$&nbsp;,&nbsp; $q(t)$&nbsp; is detectable in the envelope and envelope demodulation is applicable.
-*Allerdings muss diese einfachere Empfängervariante durch eine sehr viel größere Sendeleistung erkauft werden.&nbsp;
+*However, this simpler receiver variant must be accounted for with a much larger transmission power. &nbsp;
-*In diesem Beispiel&nbsp; $(m = 0.5)$&nbsp; wird die Sendeleistung durch den Trägerzusatz verneunfacht.
+*In this example&nbsp; $(m = 0.5)$&nbsp; the addition of a carrier multiplies the transmission power by nine.
 {{ML-Fuß}}

@@ Line 34: / Line 34: @@
 {From the graph, determine the phase values of the source and carrier signals.
 |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  $f_{\rm N}$&nbsp; of the message signal &nbsp;$q(t)$&nbsp; and what is the frequency $f_{\rm T}$&nbsp; of the carrier signal &nbsp;$z(t)$?

@@ Line 4: / Line 4: @@
 [[File:P_ID987__Mod_Z_2_1.png|right|frame|Die bei der AM beteiligten Signale]]
-Die Grafik zeigt als rote Kurve einen Ausschnitt des Sendesignals &nbsp;$s(t) = q(t) · z(t)$&nbsp; einer Zweiseitenband–Amplitudenmodulation&nbsp; (abgekürzt mit ZSB-AM)&nbsp; ohne Träger.&nbsp; Die Dauer des Zeitausschnitts beträgt &nbsp;$\rm 200 \ &micro; s$.
+The red curve on the graph shows a section of the transmitted signal &nbsp;$s(t) = q(t) · z(t)$&nbsp;  of a double-sideband amplitude modulation (abbreviated as DSB-AM) without carrier. &nbsp; (abgekürzt mit ZSB-AM)&nbsp; ohne Träger.&nbsp;The duration of the time interval is &nbsp;$\rm 200 \ &micro; s$.
-Zusätzlich sind in der Grafik eingetragen:
+Additionally plotted in the graph are:
-*das Quellensignal (als blau–gestrichelte Kurve):
+*the source signal (as a blue dashed curve):
 :$$q(t) = 1\,{\rm V} \cdot \cos(2 \pi f_{\rm N} t + \phi_{\rm N}),$$
-*das Trägersignal (grau–gepunkteter Verlauf):
+*the carrier signal (gray-dotted trace):
 :$$z(t) = 1 \cdot \cos(2 \pi f_{\rm T} t + \phi_{\rm T})$$
-Ab der Teilaufgabe&nbsp; '''(4)'''&nbsp; wird die&nbsp; „ZSB–AM mit Träger”&nbsp; betrachtet.&nbsp; Dann gilt mit &nbsp;$A_{\rm T} = 2\text{ V}$:
+From subtask '''(4)''' onwards, the "DSB-AM with carrier" is considered.  In that case, with &nbsp;$A_{\rm T} = 2\text{ V}$:
 :$$s(t) = \left(q(t) + A_{\rm T} \right) \cdot z(t) \hspace{0.05cm}.$$
@@ Line 22: / Line 22: @@
-''Hinweise:''
+''Hints:''
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Modulation_Methods/Zweiseitenband-Amplitudenmodulation|Zweiseitenband-Amplitudenmodulation]].
+*This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Double-Sideband_Amplitude_Modulation|Double-Sideband Amplitude Modulation]].
-*Bezug genommen wird insbesondere auf die Seiten   &nbsp;[[Modulation_Methods/Zweiseitenband-Amplitudenmodulation#Beschreibung_im_Zeitbereich|Beschreibung im Zeitbereich]]&nbsp; und &nbsp;[[Modulation_Methods/Zweiseitenband-Amplitudenmodulation#ZSB-Amplitudenmodulation_mit_Tr.C3.A4ger|ZSB-Amplitudenmodulation mit Träger]].
+*Particlar reference is made to the pages &nbsp;[[Modulation_Methods/Double-Sideband_Amplitude_Modulation#Description_in_the_time_domain|Description in the time domain]]&nbsp; and &nbsp;[[Modulation_Methods/Double-Sideband_Amplitude_Modulation#Double-Sideband_Amplitude_Modulation_with_carrierr|Double-Sideband Amplitude Modulation with carrier]].