Aufgaben:Exercise 2.3Z: DSB-AM due to Nonlinearity - Revision history

Reed at 16:02, 24 March 2022

2022-03-24T16:02:22Z

Guenter at 17:16, 29 November 2021

2021-11-29T17:16:23Z

Guenter at 17:05, 29 November 2021

2021-11-29T17:05:17Z

Guenter at 17:02, 29 November 2021

2021-11-29T17:02:45Z

Guenter at 17:00, 29 November 2021

2021-11-29T17:00:33Z

Reed at 18:08, 24 November 2021

2021-11-24T18:08:01Z

Reed at 18:06, 24 November 2021

2021-11-24T18:06:46Z

Reed at 18:06, 24 November 2021

2021-11-24T18:06:24Z

Reed: /* Solution */

2021-11-24T18:01:14Z

Solution

Reed at 17:47, 24 November 2021

2021-11-24T17:47:17Z

← Older revision		Revision as of 16:02, 24 March 2022
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−	{{quiz-Header\|Buchseite=~~Modulationsverfahren~~/~~Zweiseitenband~~-~~Amplitudenmodulation~~	+	{{quiz-Header\|Buchseite=Modulation_Methods/Double-Sideband_Amplitude_Modulation
	}}		}}

@@ Line 84: / Line 84: @@
-'''(3)'''&nbsp; The quadratic term&nbsp; $z^2(t)$&nbsp;consists of a DC component &nbsp; $($at&nbsp; $f = 0)$&nbsp; and a component at &nbsp; $2f_{\rm T}$.
+'''(3)'''&nbsp; The quadratic term&nbsp; $z^2(t)$&nbsp; consists of a DC component &nbsp; $($at&nbsp; $f = 0)$&nbsp; and a component at &nbsp; $2f_{\rm T}$.
-*Additionally, all spectral components of &nbsp; $q^2(t)$&nbsp; are outside the bandpass.
+*Additionally,&nbsp; all spectral components of &nbsp; $q^2(t)$&nbsp; are outside the band-pass.
-*Therefore the <u>last answer</u> is correct.
+*Therefore the&nbsp; <u>last answer</u>&nbsp; is correct.
-'''(4)'''&nbsp;<u>Answers 1 and 3</u> are correct:
+'''(4)'''&nbsp;<u>Answers 1 and 3</u>&nbsp; are correct:
 *The term&nbsp; $\cos^3(ω_Tt)$&nbsp; has its largest signal component at&nbsp; $f = f_{\rm T}$.
 *The third answer&nbsp; $(3 · c_3 · z(t) · q^2(t))$&nbsp; lies between&nbsp; $100\ \rm  kHz ± 18 \ \rm  kHz $.
-*Some parts – namely the frequency components between &nbsp; $90\ \rm  kHz $ and $110 \ \rm  kHz$&nbsp; – are not removed by the bandpass and are thus also included in &nbsp; $s(t)$&nbsp;.
+*Some parts – namely the frequency components between &nbsp; $90\ \rm  kHz $ and $110 \ \rm  kHz$&nbsp; – are not removed by the band-pass and are thus also included in&nbsp; $s(t)$.
-[[File:P_ID993__Mod_Z_2_3_f.png|right|frame|Generated AM Spectrum]]
+[[File:P_ID993__Mod_Z_2_3_f.png|right|frame|Generated DSB-AM spectrum]]
 '''(5)'''&nbsp; The transmitted signal consists of a total of five frequencies:
 :$$s(t)  =  c_1 \cdot A_{\rm T} \cdot \cos(\omega_{\rm T} t)+ c_2 \cdot A_{\rm T} \cdot A_{\rm 1} \cdot \cos((\omega_{\rm T} \pm \omega_{\rm 1})t) + c_2 \cdot A_{\rm T} \cdot A_{\rm 2} \cdot \cos((\omega_{\rm T} \pm \omega_{\rm 2})t) \hspace{0.05cm}.$$
-*Note that the second and third terms each contain two signal frequencies::
+*Note that the second term and the third term each contain two signal frequencies::
 **$\text{99 kHz}$&nbsp; and &nbsp;$\text{101 kHz}$ and
 **$\text{91 kHz}$&nbsp; and&nbsp; $\text{109 kHz}$, respectively.
-*Given &nbsp; $A_{\rm T} = 4 \ \rm  V$,&nbsp; $A_1 = 1 V$,&nbsp; $A_9 = 2  \ \rm  V$,&nbsp; $c_1 = 1$&nbsp; and&nbsp; $c_2 = 1/A_{\rm T} = \rm 0.25/V$&nbsp;, it holds that:
+*Given &nbsp; $A_{\rm T} = 4 \ \rm  V$,&nbsp; $A_1 = 1 V$,&nbsp; $A_9 = 2  \ \rm  V$,&nbsp; $c_1 = 1$&nbsp; and&nbsp; $c_2 = 1/A_{\rm T} = \rm 0.25/V$,&nbsp; it holds that:
 :$$s(t) = 4\,{\rm V} \cdot \cos(\omega_{\rm T} t) + 1\,{\rm V} \cdot \cos((\omega_{\rm T} \pm \omega_{\rm 1})t) + 2\,{\rm V}\cdot \cos((\omega_{\rm T} \pm \omega_{\rm 2})t) \hspace{0.05cm}.$$
@@ Line 117: / Line 117: @@
 * When&nbsp; $c_3 ≠ 0$&nbsp; the following additional spectral components arise:
 :$$c_3 \cdot z^3(t)= \frac{c_3 \cdot A_{\rm T}^3}{4} \cdot \left[ 3 \cdot \cos(\omega_{\rm T} t) + \cos(3\omega_{\rm T} t)\right] \hspace{0.05cm}.$$
-*The first component falls in the passband of the bandpass.  As a result, the Dirac weight at &nbsp; $f_{\rm T} = 100\ \rm kHz$&nbsp; is increased from the original&nbsp; $8 \ \rm V$&nbsp; to&nbsp; $\text{8 V + 0.75 · 0.01/V}^2 · 4^3 \text{ V}^3 = 8.48 \ \rm V$&nbsp;.
+*The first component falls in the pass-band of the band-pass.&nbsp;  As a result,&nbsp; the Dirac weight at &nbsp; $f_{\rm T} = 100\ \rm kHz$&nbsp; is increased from the original&nbsp; $8 \ \rm V$&nbsp; to&nbsp;
+*Furthermore,&nbsp; the third spectral component of subtask&nbsp; '''(4)'''&nbsp; provides an unwanted contribution to&nbsp; $S_+(f)$.&nbsp; In this case:
 :$$q^2(t)  =  \left[A_{\rm 1} \cdot \cos(\omega_{\rm 1} t)+A_{\rm 9} \cdot \cos(\omega_{\rm 9} t)\right]^2 = A_{\rm 1}^2 \cdot \cos^2(\omega_{\rm 1} t)+ A_{\rm 9}^2 \cdot \cos^2(\omega_{\rm 9}t) +
 \cdot A_{\rm 1} \cdot A_{\rm 9} \cdot \cos(\omega_{\rm 1} t)\cdot \cos(\omega_{\rm 9} t)$$
 :$$ \Rightarrow \hspace{0.2cm} q^2(t)  =  \frac{A_{\rm 1}^2}{2} +\frac{A_{\rm 1}^2}{2} \cdot \cos(\omega_{\rm 2} t)+ \frac{A_{\rm 9}^2}{2} + \frac{A_{\rm 9}^2}{2} \cdot \cos(\omega_{\rm 18} t) +  A_{\rm 1} \cdot A_{\rm 9} \cdot \cos(\omega_{\rm 8} t)+ A_{\rm 1} \cdot A_{\rm 9} \cdot \cos(\omega_{\rm 10} t).$$
-*After multiplying by &nbsp; $z(t)$&nbsp;, all but the fourth of these contributions fall within the &nbsp; $\text{90 kHz}$&nbsp; to&nbsp; $\text{110 kHz}$ range.&nbsp; The weight at&nbsp; $f_{\rm T} = 100\ \rm kHz$&nbsp;  is further increased by&nbsp; $3 · c_3 · A_{\rm T} · 0.5 (A_1^2 + A_9^2) = 0.6\ \rm  V$&nbsp; and is thus &nbsp; $9.08 \ \rm V$.
+*After multiplying by &nbsp; $z(t)$,&nbsp; all but the fourth of these contributions fall within the &nbsp; $\text{90 kHz}$&nbsp; to&nbsp; $\text{110 kHz}$ range.&nbsp; The weight at&nbsp; $f_{\rm T} = 100\ \rm kHz$&nbsp;  is further increased by&nbsp; $3 · c_3 · A_{\rm T} · 0.5 (A_1^2 + A_9^2) = 0.6\ \rm  V$&nbsp; and is thus &nbsp; $9.08 \ \rm V$.
@@ Line 133: / Line 133: @@
-The lower plot in the above graph shows the spectrum &nbsp; $S_+(f)$&nbsp; considering the cubic components. It can be seen that new frequencies have appeared, indicating nonlinear distortions.  The correct solutions are therefore <u>Answers 1 and 3</u>.
+The lower plot in the above graph shows the spectrum &nbsp; $S_+(f)$&nbsp; considering the cubic components.&nbsp; It can be seen that new frequencies have appeared, indicating nonlinear distortions.&nbsp;  The correct solutions are therefore <u>Answers 1 and 3</u>.
 {{ML-Fuß}}

@@ Line 76: / Line 76: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; The carrier frequency is most sensibly equal to the center frequency of the bandpass:&nbsp; $f_{\rm T}\hspace{0.15cm}\underline{ = 100\ \rm kHz}$.
-*If&nbsp;  $f_{\rm T}$&nbsp; does not deviate from this by more than&nbsp; $±1 \ \rm kHz$&nbsp;, this also results in a "DSB-AM".
+*If&nbsp;  $f_{\rm T}$&nbsp; does not deviate from this by more than&nbsp; $±1 \ \rm kHz$,&nbsp; this also results in a&nbsp; "DSB-AM".
-'''(2)'''&nbsp; $s_1(t)$&nbsp; includes only the carrier&nbsp; $z(t)$ &nbsp; &rArr; &nbsp; <u>Answer 1</u>. The source signal&nbsp; $q(t)$&nbsp;  is removed by the bandpass.
+'''(2)'''&nbsp; $s_1(t)$&nbsp; includes only the carrier&nbsp; $z(t)$ &nbsp; &rArr; &nbsp; <u>Answer 1</u>. The source signal&nbsp; $q(t)$&nbsp;  is removed by the band-pass.

@@ Line 45: / Line 45: @@
 {What signal components does &nbsp;$s_1(t)$&nbsp; include?
 |type="[]"}
-+ the term &nbsp;$c_1 \cdot z(t)$,
++ The term &nbsp;$c_1 \cdot z(t)$,
 - the term &nbsp;$c_1 \cdot q(t)$.
@@ Line 65: / Line 65: @@
 $m \ = \ $ { 0.75 3% }
-{Now calculate the transmitted signal &nbsp;$s(t)$&nbsp; given&nbsp; $c_3 = \rm 0.01/V^{2}$.&nbsp; Which of the following statements is/are true?
+{Now calculate the transmitted signal &nbsp;$s(t)$&nbsp; given&nbsp; $c_3 = \rm 0.01/V^{2}$.&nbsp; Which of the following statements are true?
 |type="[]"}
 + Because &nbsp;$c_3 ≠ 0$&nbsp; the spectral line changes at &nbsp;$f_{\rm T}$.

@@ Line 4: / Line 4: @@
 [[File:P_ID999__Mod_Z_2_3.png|right|frame|DSB–AM with nonlinearity]]
-In this task, we consider the realization of a double-sideband amplitude modulation using the nonlinear characteristic curve:
+In this task, we consider the realization of a double-sideband amplitude modulation&nbsp; $\text{DSB-AM}$&nbsp; using the nonlinear characteristic curve:
-$$y  = g(x)  =  c_1 \cdot x + c_2 \cdot x^2+ c_3 \cdot x^3\hspace{0.05cm}$$
+:$$y  = g(x)  =  c_1 \cdot x + c_2 \cdot x^2+ c_3 \cdot x^3\hspace{0.05cm}$$
-:$$ \Rightarrow   c_1 = 2,\hspace{0.2cm}c_2 = 0.25/{\rm V},\hspace{0.2cm}c_3 = 0 \hspace{0.1cm}{\rm bzw.}\hspace{0.1cm}c_3 = 0.01/{\rm V^2}\hspace{0.05cm}.$$
+:$$ \Rightarrow \hspace{0.5cm} c_1 = 2,\hspace{0.5cm}c_2 = 0.25/{\rm V},\hspace{0.5cm}c_3 = 0 \hspace{0.5cm}{\rm or}\hspace{0.5cm}c_3 = 0.01/{\rm V^2}\hspace{0.05cm}.$$
-The sum of the carrier signal and source signal is present at the input of this characteristic curve:
+The sum of the carrier signal&nbsp; $ z(t)$&nbsp; and the source signal&nbsp; $ q(t)$&nbsp; is present at the input of this characteristic curve:
-$$ x(t) = z(t) + q(t) = A_{\rm T} \cdot \cos(\omega_{\rm T} t)+ q(t),\hspace{0.2cm} A_{\rm T} = 4\,{\rm V}\hspace{0.05cm}.$$
+:$$ x(t) = z(t) + q(t) = A_{\rm T} \cdot \cos(\omega_{\rm T} t)+ q(t),\hspace{0.2cm} A_{\rm T} = 4\,{\rm V}\hspace{0.05cm}.$$
-*It is known that the source signal &nbsp;$q(t)$&nbsp; contains spectral components between &nbsp;$1 \ \rm kHz$&nbsp; and &nbsp;$9 \ \rm kHz$&nbsp; (up to and including these limits).
+*It is known that the source signal &nbsp;$q(t)$&nbsp; contains spectral components between &nbsp;$1 \ \rm kHz$&nbsp; and &nbsp;$9 \ \rm kHz$ &nbsp; (up to and including these limits).
 *From subtask &nbsp; '''(5)'''&nbsp; onwards, the following source signal should be assumed:
 :$$q(t) = A_{\rm 1} \cdot \cos(\omega_{\rm 1} t)+A_{\rm 9} \cdot \cos(\omega_{\rm 9} t) \hspace{0.05cm}.$$
@@ Line 20: / Line 20: @@
 :$$ y_2(t)  =  c_2 \cdot[z(t) + q(t)]^2,$$
 :$$y_3(t)  =  c_3 \cdot [z(t) + q(t)]^3 \hspace{0.05cm}.$$
-Thus, the transmitted signals &nbsp;$s(t)$&nbsp; i.e., &nbsp;$s_1(t)$, &nbsp;$s_2(t)$&nbsp; and &nbsp;$s_3(t)$&nbsp;, result from band-limiting to the range from &nbsp;$90 \ \rm kHz$&nbsp; to &nbsp;$110 \ \rm kHz$.
+Thus,&nbsp; the transmitted signals &nbsp;$s(t)$&nbsp; i.e., &nbsp;$s_1(t)$, &nbsp;$s_2(t)$&nbsp; and &nbsp;$s_3(t)$,&nbsp; result from band-limiting to the range from &nbsp;$90 \ \rm kHz$&nbsp; to &nbsp;$110 \ \rm kHz$.
@@ Line 26: / Line 26: @@
 Hints:
 *This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Double-Sideband_Amplitude_Modulation|Double-Sideband Amplitude Modulation]].
 *Particular reference is made to the page&nbsp;  [[Modulation_Methods/Double-Sideband_Amplitude_Modulation#Amplitude_modulation_with_a_quadratic_characteristic_curve|Amplitude modulation with a quadratic characteristic curve]].
@@ Line 47: / Line 43: @@
 $f_{\rm T} \ = \ $  { 100 3%  } $\ \text{kHz}$
-{What signal components does &nbsp;$s_1(t)$ include?
+{What signal components does &nbsp;$s_1(t)$&nbsp; include?
 |type="[]"}
 + the term &nbsp;$c_1 \cdot z(t)$,
 - the term &nbsp;$c_1 \cdot q(t)$.
-{What signal components does &nbsp;$s_2(t)$ include?
+{What signal components does &nbsp;$s_2(t)$&nbsp; include?
 |type="[]"}
-- the term &nbsp;$c_2 · z^2(t)$,
+- The term &nbsp;$c_2 · z^2(t)$,
 - the term &nbsp;$c_2 · q^2(t)$,
 + the term &nbsp;$2c_2 · z(t) · q(t)$.
-{What signal components does &nbsp;$s_3(t)$&nbsp; (at least partially) include?
+{What signal components does &nbsp;$s_3(t)$&nbsp; (at least partially)&nbsp; include?
 |type="[]"}
-+ the term &nbsp;$c_3 · z^3(t)$,
++ The term &nbsp;$c_3 · z^3(t)$,
-- the term&nbsp;$3 · c_3 · z^2(t) · q(t)$,
+- the term&nbsp; $3 · c_3 · z^2(t) · q(t)$,
 + the term &nbsp;$3 · c_3 · z(t) · q^2(t)$,
 - the term &nbsp;$c_3 · q^3(t)$.
-{Calculate &nbsp;$s(t)$, when &nbsp;$c_3 = 0$&nbsp; and the source signal &nbsp;$q(t)$&nbsp; is composed of two cosine oscillations.<br>What is the modulation depth &nbsp; $m$?
+{Calculate &nbsp;$s(t)$,&nbsp; when &nbsp;$c_3 = 0$&nbsp; and the source signal &nbsp;$q(t)$&nbsp; is composed of two cosine oscillations.&nbsp; What is the modulation depth&nbsp; $m$?
 |type="{}"}
 $m \ = \ $ { 0.75 3% }
-{Now calculate the transmitted signal &nbsp;$s(t)$&nbsp; given&nbsp;$c_3 = \rm 0.01/V^{2}$.&nbsp; Which of the following statements is true?
+{Now calculate the transmitted signal &nbsp;$s(t)$&nbsp; given&nbsp; $c_3 = \rm 0.01/V^{2}$.&nbsp; Which of the following statements is/are true?
 |type="[]"}
-+ Because &nbsp;$c_3 ≠ 0$&nbsp; the spectral line changes at &nbsp;$f_{\rm T}$&nbsp;.
++ Because &nbsp;$c_3 ≠ 0$&nbsp; the spectral line changes at &nbsp;$f_{\rm T}$.
-- Because &nbsp;$c_3 ≠ 0$&nbsp; linear distortions arise, which can be compensated for.
+- Because &nbsp;$c_3 ≠ 0$&nbsp; linear distortions arise,&nbsp; which can be compensated for.
-+ Because &nbsp;$c_3 ≠ 0$&nbsp; nonlinear (i.e. irreversible) distortions arise.
++ Because &nbsp;$c_3 ≠ 0$&nbsp; nonlinear&nbsp; (i.e. irreversible) distortions arise.
 </quiz>

@@ Line 118: / Line 118: @@
-'''(6)'''&nbsp; The graph above shows the spectrum &nbsp; $S_+(f)$&nbsp; – that is, only positive frequencies – with&nbsp; $c_3 = 0$.&nbsp;
+'''(6)'''&nbsp; The graph above shows the spectrum &nbsp; $S_+(f)$&nbsp; – that is, only positive frequencies – where&nbsp; $c_3 = 0$.&nbsp;
 * When&nbsp; $c_3 ≠ 0$&nbsp; the following additional spectral components arise:
 :$$c_3 \cdot z^3(t)= \frac{c_3 \cdot A_{\rm T}^3}{4} \cdot \left[ 3 \cdot \cos(\omega_{\rm T} t) + \cos(3\omega_{\rm T} t)\right] \hspace{0.05cm}.$$

@@ Line 124: / Line 124: @@
-*Weiterhin liefert der dritte Spektralanteil von Teilaufgabe&nbsp; '''(4)'''&nbsp; einen unerwünschten Beitrag zu&nbsp; $S_+(f)$.&nbsp; Dabei gilt:
+*Furthermore, the third spectral component of subtask&nbsp; '''(4)'''&nbsp; provides an unwanted contribution to&nbsp; $S_+(f)$.&nbsp; In this case:
 :$$q^2(t)  =  \left[A_{\rm 1} \cdot \cos(\omega_{\rm 1} t)+A_{\rm 9} \cdot \cos(\omega_{\rm 9} t)\right]^2 = A_{\rm 1}^2 \cdot \cos^2(\omega_{\rm 1} t)+ A_{\rm 9}^2 \cdot \cos^2(\omega_{\rm 9}t) +
 \cdot A_{\rm 1} \cdot A_{\rm 9} \cdot \cos(\omega_{\rm 1} t)\cdot \cos(\omega_{\rm 9} t)$$
 :$$ \Rightarrow \hspace{0.2cm} q^2(t)  =  \frac{A_{\rm 1}^2}{2} +\frac{A_{\rm 1}^2}{2} \cdot \cos(\omega_{\rm 2} t)+ \frac{A_{\rm 9}^2}{2} + \frac{A_{\rm 9}^2}{2} \cdot \cos(\omega_{\rm 18} t) +  A_{\rm 1} \cdot A_{\rm 9} \cdot \cos(\omega_{\rm 8} t)+ A_{\rm 1} \cdot A_{\rm 9} \cdot \cos(\omega_{\rm 10} t).$$
-*Nach der Multiplikation mit&nbsp; $z(t)$&nbsp; fallen alle diese Beiträge bis auf den vierten in den Bereich von&nbsp; $\text{90 kHz}$&nbsp; bis&nbsp; $\text{110 kHz}$.&nbsp; Das Gewicht bei&nbsp; $f_{\rm T} = 100\ \rm kHz$&nbsp;  wird um&nbsp; $3 · c_3 · A_{\rm T} · 0.5 (A_1^2 + A_9^2) = 0.6\ \rm  V$&nbsp; weiter erhöht und ist somit&nbsp; $9.08 \ \rm V$.
+*After multiplying by &nbsp; $z(t)$&nbsp;, all but the fourth of these contributions fall within the &nbsp; $\text{90 kHz}$&nbsp; to&nbsp; $\text{110 kHz}$ range.&nbsp; The weight at&nbsp; $f_{\rm T} = 100\ \rm kHz$&nbsp;  is further increased by&nbsp; $3 · c_3 · A_{\rm T} · 0.5 (A_1^2 + A_9^2) = 0.6\ \rm  V$&nbsp; and is thus &nbsp; $9.08 \ \rm V$.
-Weitere Anteile ergeben sich bei:
+Further components are obtained at:
-*$98 \ \rm kHz$&nbsp; und&nbsp; $102 \ \rm kHz$&nbsp; mit den Gewichten&nbsp; $c_3 · A_{\rm T}/2 · A_1^2/2 = 0.03\ \rm  V$,
+*$98 \ \rm kHz$&nbsp; and&nbsp; $102 \ \rm kHz$&nbsp; with weights&nbsp; $c_3 · A_{\rm T}/2 · A_1^2/2 = 0.03\ \rm  V$,
-* $92 \ \rm kHz$&nbsp; und&nbsp; $108 \ \rm kHz$&nbsp; mit den Gewichten&nbsp; $3c_3 · A_{\rm T}/2 · A_1 · A_9 = 0.12\ \rm  V$,
+* $92 \ \rm kHz$&nbsp; and&nbsp; $108 \ \rm kHz$&nbsp; with weights&nbsp; $3c_3 · A_{\rm T}/2 · A_1 · A_9 = 0.12\ \rm  V$,
-* $90 \ \rm kHz$&nbsp; und&nbsp; $110 \ \rm kHz$&nbsp; mit den Gewichten&nbsp; $3c_3 · A_{\rm T}/2 · A_1 · A_9 = 0.12\ \rm  V$.
+* $90 \ \rm kHz$&nbsp; and&nbsp; $110 \ \rm kHz$&nbsp; with weights&nbsp; $3c_3 · A_{\rm T}/2 · A_1 · A_9 = 0.12\ \rm  V$.
-Die untereSkizze in obiger  Grafik zeigt das Spektrum&nbsp; $S_+(f)$&nbsp; unter Berücksichtigung der kubischen Anteile.&nbsp; Man erkennt, dass neue Frequenzen entstanden sind, was auf nichtlineare Verzerrungen hindeutet.&nbsp; Richtig sind somit die <u>Lösungsvorschläge 1 und 3</u>.
+The lower plot in the above graph shows the spectrum &nbsp; $S_+(f)$&nbsp; considering the cubic components. It can be seen that new frequencies have appeared, indicating nonlinear distortions.  The correct solutions are therefore <u>Answers 1 and 3</u>.
 {{ML-Fuß}}

@@ Line 79: / Line 79: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Die Trägerfrequenz ist sinnvollerweise gleich der Mittenfrequenz des Bandpasses:&nbsp; $f_{\rm T}\hspace{0.15cm}\underline{ = 100\ \rm kHz}$.
+'''(1)'''&nbsp; The carrier frequency is most sensibly equal to the center frequency of the bandpass:&nbsp; $f_{\rm T}\hspace{0.15cm}\underline{ = 100\ \rm kHz}$.
-*Weicht&nbsp;  $f_{\rm T}$&nbsp; um nicht mehr als&nbsp; $±1 \ \rm kHz$&nbsp; davon ab, ergibt sich ebenfalls eine „ZSB–AM”.
+*If&nbsp;  $f_{\rm T}$&nbsp; does not deviate from this by more than&nbsp; $±1 \ \rm kHz$&nbsp; this also results in a "DSB-AM".
-'''(2)'''&nbsp; $s_1(t)$&nbsp; beinhaltet nur den Träger&nbsp; $z(t)$ &nbsp; &rArr; &nbsp; <u>Antwort 1</u>. Das Quellensignal&nbsp; $q(t)$&nbsp; wird durch den Bandpass entfernt.
+'''(2)'''&nbsp; $s_1(t)$&nbsp; includes only the carrier&nbsp; $z(t)$ &nbsp; &rArr; &nbsp; <u>Answer 1</u>. The source signal&nbsp; $q(t)$&nbsp;  is removed by the bandpass.
-'''(3)'''&nbsp; Der quadratische Term&nbsp; $z^2(t)$&nbsp; besteht aus einem Gleichanteil&nbsp; $($bei&nbsp; $f = 0)$&nbsp; sowie einem Anteil bei&nbsp; $2f_{\rm T}$.
+'''(3)'''&nbsp; The quadratic term&nbsp; $z^2(t)$&nbsp;consists of a DC component &nbsp; $($at&nbsp; $f = 0)$&nbsp; and a component at &nbsp; $2f_{\rm T}$.
-*Auch alle Spektralanteile von&nbsp; $q^2(t)$&nbsp; liegen außerhalb des Bandpasses.
+*Additionally, all spectral components of &nbsp; $q^2(t)$&nbsp; are outside the bandpass.
-*Richtig ist somit die <u>letzte Antwort</u>.
+*Therefore the <u>last answer</u> is correct.
-'''(4)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 1 und 3</u>:
+'''(4)'''&nbsp;<u>Answers 1 and 3</u> are correct:
-*Der Term&nbsp; $\cos^3(ω_Tt)$&nbsp; hat seinen größten Signalanteil bei&nbsp; $f = f_{\rm T}$.
+*The term&nbsp; $\cos^3(ω_Tt)$&nbsp; has its largest signal component at&nbsp; $f = f_{\rm T}$.
-*Der dritte Lösungsvorschlag&nbsp; $(3 · c_3 · z(t) · q^2(t))$&nbsp; liegt zwischen&nbsp; $100\ \rm  kHz ± 18 \ \rm  kHz $.
+*The third answer&nbsp; $(3 · c_3 · z(t) · q^2(t))$&nbsp; lies between&nbsp; $100\ \rm  kHz ± 18 \ \rm  kHz $.
-*Teile davon – nämlich die Frequenzanteile zwischen&nbsp; $90\ \rm  kHz $ und $110 \ \rm  kHz$&nbsp; – werden durch den Bandpass nicht entfernt und sind somit auch in&nbsp; $s(t)$&nbsp; enthalten.
+*Some parts – namely the frequency components between &nbsp; $90\ \rm  kHz $ and $110 \ \rm  kHz$&nbsp; – are not removed by the bandpass and are thus also included in &nbsp; $s(t)$&nbsp;.
-[[File:P_ID993__Mod_Z_2_3_f.png|right|frame|Erzeugtes AM–Spektrum]]
+[[File:P_ID993__Mod_Z_2_3_f.png|right|frame|Generated AM Spectrum]]
-'''(5)'''&nbsp; Das Sendesignal besteht aus insgesamt fünf Frequenzen:
+'''(5)'''&nbsp; The transmitted signal consists of a total of five frequencies:
 :$$s(t)  =  c_1 \cdot A_{\rm T} \cdot \cos(\omega_{\rm T} t)+ c_2 \cdot A_{\rm T} \cdot A_{\rm 1} \cdot \cos((\omega_{\rm T} \pm \omega_{\rm 1})t) + c_2 \cdot A_{\rm T} \cdot A_{\rm 2} \cdot \cos((\omega_{\rm T} \pm \omega_{\rm 2})t) \hspace{0.05cm}.$$
-*Beachten Sie hierbei, dass der zweite und dritte Term jeweils zwei Signalfrequenzen beinhaltet:
+*Note that the second and third terms each contain two signal frequencies::
-**$\text{99 kHz}$&nbsp; und &nbsp;$\text{101 kHz}$ bzw.
+**$\text{99 kHz}$&nbsp; and &nbsp;$\text{101 kHz}$ and
-**$\text{91 kHz}$&nbsp; und&nbsp; $\text{109 kHz}$.
+**$\text{91 kHz}$&nbsp; and&nbsp; $\text{109 kHz}$, respectively.
-*Mit&nbsp; $A_{\rm T} = 4 \ \rm  V$,&nbsp; $A_1 = 1 V$,&nbsp; $A_9 = 2  \ \rm  V$,&nbsp; $c_1 = 1$&nbsp; und&nbsp; $c_2 = 1/A_{\rm T} = \rm 0.25/V$&nbsp; gilt auch:
+*Given &nbsp; $A_{\rm T} = 4 \ \rm  V$,&nbsp; $A_1 = 1 V$,&nbsp; $A_9 = 2  \ \rm  V$,&nbsp; $c_1 = 1$&nbsp; and&nbsp; $c_2 = 1/A_{\rm T} = \rm 0.25/V$&nbsp;, it holds that:
 :$$s(t) = 4\,{\rm V} \cdot \cos(\omega_{\rm T} t) + 1\,{\rm V} \cdot \cos((\omega_{\rm T} \pm \omega_{\rm 1})t) + 2\,{\rm V}\cdot \cos((\omega_{\rm T} \pm \omega_{\rm 2})t) \hspace{0.05cm}.$$
-*Daran erkennt man, dass für den  Modulationsgrad gilt:
+*From this it can be seen that for the modulation depth:
 :$$m =\frac{A_1 + A_9}{A_{\rm T}} = \rm \frac{1\ V + 2 \ V}{4 \ V}  \hspace{0.15cm}\underline{=0.75}.$$
-'''(6)'''&nbsp; Die Grafik zeigt oben das Spektrum&nbsp; $S_+(f)$&nbsp; – also nur positive Frequenzen – mit&nbsp; $c_3 = 0$.&nbsp;
+'''(6)'''&nbsp; The graph above shows the spectrum &nbsp; $S_+(f)$&nbsp; – that is, only positive frequencies – with&nbsp; $c_3 = 0$.&nbsp;
-* Mit&nbsp; $c_3 ≠ 0$&nbsp; fallen folgende zusätzliche Spektralanteile an:
+* When&nbsp; $c_3 ≠ 0$&nbsp; the following additional spectral components arise:
 :$$c_3 \cdot z^3(t)= \frac{c_3 \cdot A_{\rm T}^3}{4} \cdot \left[ 3 \cdot \cos(\omega_{\rm T} t) + \cos(3\omega_{\rm T} t)\right] \hspace{0.05cm}.$$
-*Der erste Anteil fällt in den Durchlassbereich des Bandpasses.&nbsp; Das Diracgewicht bei&nbsp; $f_{\rm T} = 100\ \rm kHz$&nbsp; wird dadurch von ursprünglich&nbsp; $8 \ \rm V$&nbsp; auf&nbsp; $\text{8 V + 0.75 · 0.01/V}^2 · 4^3 \text{ V}^3 = 8.48 \ \rm V$&nbsp; erhöht.
+*The first component falls in the passband of the bandpass.  As a result, the Dirac weight at &nbsp; $f_{\rm T} = 100\ \rm kHz$&nbsp; is increased from the original&nbsp; $8 \ \rm V$&nbsp; to&nbsp; $\text{8 V + 0.75 · 0.01/V}^2 · 4^3 \text{ V}^3 = 8.48 \ \rm V$&nbsp;.