Aufgaben:Exercise 2.3: DSB-AM Realization - Revision history

Hwang at 14:18, 18 January 2023

2023-01-18T14:18:09Z

Reed at 15:41, 24 March 2022

2022-03-24T15:41:32Z

Reed at 15:40, 24 March 2022

2022-03-24T15:40:15Z

Guenter at 16:21, 29 November 2021

2021-11-29T16:21:15Z

Guenter at 16:10, 29 November 2021

2021-11-29T16:10:24Z

Reed at 16:46, 24 November 2021

2021-11-24T16:46:34Z

Reed at 16:43, 24 November 2021

2021-11-24T16:43:57Z

Reed: /* Solution */

2021-11-24T16:42:56Z

Solution

Reed at 16:30, 24 November 2021

2021-11-24T16:30:15Z

Reed at 16:29, 24 November 2021

2021-11-24T16:29:52Z

@@ Line 63: / Line 63: @@
 |type="[]"}
 - The weight of the spectral line at&nbsp;$f_{\rm T}$&nbsp; is unchanged.
-+ $s(t)$&nbsp; now includes Dirac lines at&nbsp;$f_{\rm T} ± 2f_{\rm N}$.
++ $s(t)$&nbsp; now includes Dirac delta lines at&nbsp;$f_{\rm T} ± 2f_{\rm N}$.
 + The cubic term leads to nonlinear distortions.
 - The cubic term leads to linear distortions.
@@ Line 108: / Line 108: @@
 *Accordingly, the third summand in the above equation leads to
 :$$3 \cdot c_3 \cdot z(t) \cdot q^2(t)= {3}/{2 }  \cdot A_{\rm T} \cdot A_{\rm N}^2 \cdot \left[ \cos(\omega_{\rm T} t) + \cos(\omega_{\rm T} t)\cdot \cos(2 \omega_{\rm N} t)\right] \hspace{0.05cm}.$$
-*Thus,&nbsp; within the frequency range from&nbsp; $\text{23 kHz}$&nbsp; to&nbsp; $\text{37 kHz}$,&nbsp; there is indeed a change in the spectral line at&nbsp; $f_{\rm T}$&nbsp; <br>and new Dirac lines are formed at &nbsp; $f_{\rm T} ± 2f_{\rm N}$,&nbsp; i.e.,&nbsp; at &nbsp; $\text{24 kHz}$&nbsp; and&nbsp; $\text{36 kHz}$.
+*Thus,&nbsp; within the frequency range from&nbsp; $\text{23 kHz}$&nbsp; to&nbsp; $\text{37 kHz}$,&nbsp; there is indeed a change in the spectral line at&nbsp; $f_{\rm T}$&nbsp; <br>and new Dirac delta lines are formed at &nbsp; $f_{\rm T} ± 2f_{\rm N}$,&nbsp; i.e.,&nbsp; at &nbsp; $\text{24 kHz}$&nbsp; and&nbsp; $\text{36 kHz}$.
 *The resulting distortions are thus nonlinear&nbsp; &rArr; &nbsp; Answer 3 ist correct and Answer 4 is wrong.

← Older revision		Revision as of 15:41, 24 March 2022
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−	The ~~graphic~~ shows the characteristic curve  $g(x)$  and the approximations  $g_1(x)$,  $g_2(x)$  and  $g_3(x)$, when the Taylor series is truncated after the first, second, or third term.  It can be seen that the approximation  $g_3(x)$  is indistinguishable from  $g(x)$  in the range shown.	+	The graph shows the characteristic curve  $g(x)$  and the approximations  $g_1(x)$,  $g_2(x)$  and  $g_3(x)$, when the Taylor series is truncated after the first, second, or third term.  It can be seen that the approximation  $g_3(x)$  is indistinguishable from  $g(x)$  in the range shown.

@@ Line 10: / Line 10: @@
-This curve is driven at the operating point &nbsp;$A_0 = 2\ \rm  V$.&nbsp; This is achieved,&nbsp; for example,&nbsp; by the input signal
+This curve is used at the operating point &nbsp;$A_0 = 2\ \rm  V$.&nbsp; This is achieved,&nbsp; for example,&nbsp; by the input signal
 :$$x(t) = A_0 + z(t) + q(t)\hspace{0.05cm}.$$
 Assume cosine oscillations for both the carrier and the source signal:

@@ Line 70: / Line 70: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; From&nbsp; $x(t) = A_0 + z(t) + q(t)$&nbsp;, with &nbsp; $A_0 = 2\ \rm  V$&nbsp; and&nbsp; $A_{\rm T} = A_{\rm N} = 1 \ \rm  V$&nbsp;, we get the possible range&nbsp; $0 \ {\rm V} ≤ x(t) ≤ 4\ \rm  V$.
+'''(1)'''&nbsp; From&nbsp; $x(t) = A_0 + z(t) + q(t)$,&nbsp; with &nbsp; $A_0 = 2\ \rm  V$&nbsp; and &nbsp; $A_{\rm T} = A_{\rm N} = 1 \ \rm  V$,&nbsp; we get the possible range&nbsp; $0 \ {\rm V} ≤ x(t) ≤ 4\ \rm  V$.
-*Thus, the auxiliary quantity&nbsp; $w(t)$&nbsp; can take values between&nbsp; $w_{\rm min}\hspace{0.15cm}\underline{ = -2 \ \rm V}$&nbsp; and&nbsp; $w_{\rm max}\hspace{0.15cm}\underline{ = +2 \ \rm V}$&nbsp;.
+*Thus,&nbsp; the auxiliary quantity&nbsp; $w(t)$&nbsp; can take values between&nbsp; $w_{\rm min}\hspace{0.15cm}\underline{ = -2 \ \rm V}$&nbsp; and&nbsp; $w_{\rm max}\hspace{0.15cm}\underline{ = +2 \ \rm V}$.
@@ Line 77: / Line 77: @@
 '''(2)'''&nbsp; The coefficient&nbsp; $c_0$&nbsp; is equal to the characteristic value at the operating point.&nbsp; Using&nbsp; $A_0 = 2 \ \rm V$&nbsp; and&nbsp; $U = 3 \ \rm V$&nbsp; we obtain:
 :$$c_0 = y(A_0) = U \cdot \left( 1 -{\rm e} ^{-A_0/U}\right) \hspace{0.15cm}\underline {= 1.460\,{\rm V}}\hspace{0.05cm}.$$
-*Accordingly, for the Taylor coefficient&nbsp; $c_1$:
+*Accordingly,&nbsp; for the Taylor coefficient&nbsp; $c_1$:
 :$$c_1 = y\hspace{0.06cm}'(A_0)= {\rm e} ^{-A_0/U}\hspace{0.15cm}\underline { = 0.513}\hspace{0.05cm}.$$
@@ Line 89: / Line 89: @@
-'''(4)'''&nbsp; Setting&nbsp; $c_3 = 0$, the output signal of the amplifier is:
+'''(4)'''&nbsp; Setting&nbsp; $c_3 = 0$,&nbsp; the output signal of the amplifier is:
 :$$y(t) = c_0 + c_1 \cdot (z(t) + q(t)) + c_2 \cdot (z^2(t) + q^2(t) + 2 \cdot z(t) \cdot q(t))\hspace{0.05cm}.$$
-*Thus, after the bandpass, the following signal components remain:
+*Thus,&nbsp; after the band-pass,&nbsp; the following signal components remain:
 :$$s(t)  =  c_1 \cdot z(t) + 2 \cdot c_2 \cdot z(t) \cdot q(t)
    =  \left[c_1 \cdot A_{\rm T} + 2 \cdot c_2 \cdot A_{\rm T} \cdot A_{\rm N} \cdot \cos(\omega_{\rm N} t)\right] \cdot \cos(\omega_{\rm T} t)\hspace{0.05cm}.$$
-*The modulation depth is then determined as the quotient of the "amplitude" of the message oscillation over the "amplitude" of the carrier:
+*The modulation depth is then determined as the quotient of the&nbsp; "amplitude of the message oscillation"&nbsp; over the "amplitude of the carrier":
 :$$m = \frac{2 \cdot |c_2| \cdot A_{\rm T} \cdot A_{\rm N}}{|c_1| \cdot A_{\rm T}} = \frac{2 \cdot |c_2| \cdot A_{\rm N}}{|c_1| }= \frac{2 \cdot 0.086 \cdot 1\,{\rm V}}{0.513 }\hspace{0.15cm}\underline { = 0.335}\hspace{0.05cm}.$$
-'''(5)'''&nbsp; <u>Answers 2 and 3</u> are correct:
+'''(5)'''&nbsp; <u>Answers 2 and 3</u>&nbsp; are correct:
 *Considering the cubic part,&nbsp; $y(t)$&nbsp; includes the following other components:
 :$$y_3(t)  =  c_3 \cdot (z(t) + q(t))^3
 =  c_3 \cdot z^3(t) + 3 \cdot c_3 \cdot z^2(t) \cdot q(t)+ 3 \cdot c_3 \cdot z(t) \cdot q^2(t) + c_3 \cdot q^3(t) \hspace{0.05cm}.$$
-*The first term results in components at&nbsp; $f_{\rm T}$&nbsp; and&nbsp; $3f_{\rm T}$, and the last term results in components at&nbsp; $f_{\rm N}$&nbsp; and&nbsp; $3f_{\rm N}$.&nbsp; The second term gives a component at &nbsp; $f_{\rm N}$&nbsp; and others at&nbsp; $2f_{\rm T} ± f_{\rm N}$:
+*The first term results in components at&nbsp; $f_{\rm T}$&nbsp; and&nbsp; $3f_{\rm T}$, and the last term results in components at&nbsp; $f_{\rm N}$&nbsp; and&nbsp; $3f_{\rm N}$.&nbsp;
 :$$3 \cdot c_3 \cdot z^2(t) \cdot q(t)= {3}/{2 } \cdot A_{\rm T}^2 \cdot A_{\rm N} \cdot \left[ \cos(\omega_{\rm N} t) + \cos(2\omega_{\rm T} t) \cdot \cos(\omega_{\rm N} t)\right] \hspace{0.05cm}.$$
 *Accordingly, the third summand in the above equation leads to
 :$$3 \cdot c_3 \cdot z(t) \cdot q^2(t)= {3}/{2 }  \cdot A_{\rm T} \cdot A_{\rm N}^2 \cdot \left[ \cos(\omega_{\rm T} t) + \cos(\omega_{\rm T} t)\cdot \cos(2 \omega_{\rm N} t)\right] \hspace{0.05cm}.$$
-*Thus, within the frequency range from&nbsp; $\text{23 kHz}$&nbsp; to&nbsp; $\text{37 kHz}$&nbsp;, there is indeed a change in the spectral line at&nbsp; $f_{\rm T}$&nbsp; and new Dirac lines are formed at &nbsp; $f_{\rm T} ± 2f_{\rm N}$, i.e., at &nbsp; $\text{24 kHz}$&nbsp; and&nbsp; $\text{36 kHz}$.
+*Thus,&nbsp; within the frequency range from&nbsp; $\text{23 kHz}$&nbsp; to&nbsp; $\text{37 kHz}$,&nbsp; there is indeed a change in the spectral line at&nbsp; $f_{\rm T}$&nbsp; <br>and new Dirac lines are formed at &nbsp; $f_{\rm T} ± 2f_{\rm N}$,&nbsp; i.e.,&nbsp; at &nbsp; $\text{24 kHz}$&nbsp; and&nbsp; $\text{36 kHz}$.
 *The resulting distortions are thus nonlinear&nbsp; &rArr; &nbsp; Answer 3 ist correct and Answer 4 is wrong.

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Mod_A_2_3.png|right|frame|Nonlinear characteristic curve <br>forr AM&ndash;realization]]
+[[File:EN_Mod_A_2_3.png|right|frame|Nonlinear characteristic curve <br>for DSB-AM realization]]
-In order to realize the so-called "DSB-AM with carrier", an amplifier with the characteristic curve
+In order to realize the so-called&nbsp; "Double-Sideband Amplitude Modulation&nbsp; $\text{(DSB-AM)}$&nbsp; with carrier",&nbsp; an amplifier with the following characteristic curve must be used:
 :$$y = g(x) = U \cdot \left( 1 -{\rm e} ^{-x/U}\right)$$
-must be used. Here, &nbsp;$x = x(t)$&nbsp; and &nbsp;$y = y(t)$&nbsp;are time-dependent voltages at the input and output of the amplifier, respectively. &nbsp; The parameter &nbsp;$U = 3 \ \rm V$&nbsp; indicates the saturation voltage of the amplifier.
+*Here, &nbsp;$x = x(t)$&nbsp; and &nbsp;$y = y(t)$&nbsp; are time-dependent voltages at the input and output of the amplifier,&nbsp; respectively. &nbsp;
-This curve is operated at the operating point &nbsp;$A_0 = 2\ \rm  V$.&nbsp; This is achieved, for example, by the input signal
 :$$x(t) = A_0 + z(t) + q(t)\hspace{0.05cm}.$$
 Assume cosine oscillations for both the carrier and the source signal:
 :$$ z(t)  = A_{\rm T} \cdot \cos (2 \pi f_{\rm T} t),\hspace{0.2cm} A_{\rm T} = 1\,{\rm V},\hspace{0.2cm} f_{\rm T} = 30\,{\rm kHz},$$
 :$$ q(t)  =  A_{\rm N} \cdot \cos (2 \pi f_{\rm N} t),\hspace{0.2cm} A_{\rm N} = 1\,{\rm V},\hspace{0.2cm} f_{\rm N} = 3\,{\rm kHz}\hspace{0.05cm}.$$
-In solving this problem, use the auxiliary quantity
+In solving this problem,&nbsp; use the auxiliary quantity
 :$$w(t) = x(t) - A_0 = z(t) + q(t)\hspace{0.05cm}.$$
-The nonlinear characteristic curve can be developed according to a ''Taylor series'' around the operating point:
+The nonlinear characteristic curve can be developed according to a&nbsp; "Taylor series"&nbsp; around the operating point:
 :$$y(x)  = y(A_0) + \frac{1}{1!} \cdot y\hspace{0.08cm}{\rm '}(A_0) \cdot (x - A_0)+ \frac{1}{2!} \cdot y\hspace{0.08cm}''(A_0) \cdot (x - A_0)^2+
    \frac{1}{3!} \cdot y\hspace{0.08cm}'''(A_0) \cdot (x - A_0)^3 + \text{ ...}$$
 The output signal can then also be represented as depending on the auxiliary quantity &nbsp;$w(t)$&nbsp; as follows:
 :$$y(t) = c_0 + c_1 \cdot w(t) + c_2 \cdot w^2(t)+ c_3 \cdot w^3(t) +\text{ ...}$$
-The DSB–AM signal &nbsp;$s(t)$&nbsp; is obtained by band-limiting&nbsp;$y(t)$&nbsp; to the frequency range from &nbsp;$\text{23 kHz}$&nbsp; to &nbsp;$\text{37 kHz}$.&nbsp; That is, all frequencies other than &nbsp;$f_{\rm T}$, &nbsp;$f_{\rm T}±f_{\rm N}$&nbsp; and &nbsp;$f_{\rm T}±2f_{\rm N}$&nbsp;  are removed by the bandpass.
+*The DSB–AM signal &nbsp;$s(t)$&nbsp; is obtained by band-limiting&nbsp;$y(t)$&nbsp; to the frequency range from &nbsp;$\text{23 kHz}$&nbsp; to &nbsp;$\text{37 kHz}$.&nbsp;
+*That is,&nbsp; all frequencies other than &nbsp;$f_{\rm T}$, &nbsp;$f_{\rm T}±f_{\rm N}$&nbsp; and &nbsp;$f_{\rm T}±2f_{\rm N}$&nbsp;  are removed by the band-pass.
-''Hints:''
+Hints:
 *This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Double-Sideband_Amplitude_Modulation|Double-Sideband Amplitude Modulation]].
 *Reference will also be made to the chapter&nbsp;  [[Linear_and_Time_Invariant_Systems/Nonlinear_Distortions#Description_of_nonlinear_systems|Description of nonlinear systems]]&nbsp; in the book "Linear and Time Invariant Systems".
@@ Line 40: / Line 41: @@
 <quiz display=simple>
-{In what range can the input signal &nbsp;$x(t)$&nbsp;vary? variieren?&nbsp;Give the minimum and maximum values of the auxiliary variable  &nbsp;$w(t) = x(t) - A_0$&nbsp;.
+{In what range can the input signal &nbsp;$x(t)$&nbsp;vary?&nbsp;Give the minimum and maximum values of the auxiliary variable  &nbsp;$w(t) = x(t) - A_0$.
 |type="{}"}
 $w_{\rm min} \ = \ $ { -2.06--1.94 } $\ \text{V}$
@@ Line 50: / Line 51: @@
 $c_1 \ = \ $ { 0.513 3% }
-{What are the coefficients &nbsp;$c_2$&nbsp; and &nbsp;$c_3$&nbsp of the nonlinear characteristic curve?
+{What are the coefficients &nbsp;$c_2$&nbsp; and &nbsp;$c_3$&nbsp; of the nonlinear characteristic curve?
 |type="{}"}
 $c_2\ = \ $ { -0.088--0.084 }   $\ \rm V^{ -1 }$
 $c_3\ = \ $ { 0.0095 3% }  $\ \rm V^{ -2 }$
-{Show that a "DSB-AM with carrier" constellation results when &nbsp;$c_3$&nbsp; is considered negligibly small.  What is the modulation depth &nbsp;$m$?
+{Show that a&nbsp; "DSB-AM with carrier"&nbsp; constellation results when &nbsp;$c_3$&nbsp; is considered negligibly small.&nbsp;  What is the modulation depth &nbsp;$m$?
 |type="{}"}
 $m \ = \ $ { 0.335 3% }
-{Assuming that &nbsp;$c_3$&nbsp; cannot be considered negligibly small, which of the following statements are true?
+{Assuming that &nbsp;$c_3$&nbsp; cannot be considered negligibly small,&nbsp; which of the following statements are true?
 |type="[]"}
 - The weight of the spectral line at&nbsp;$f_{\rm T}$&nbsp; is unchanged.
 + $s(t)$&nbsp; now includes Dirac lines at&nbsp;$f_{\rm T} ± 2f_{\rm N}$.
 + The cubic term leads to nonlinear distortions.
-- The cubic term leads to linear distortions..
+- The cubic term leads to linear distortions.
 </quiz>

@@ Line 69: / Line 69: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Aus&nbsp; $x(t) = A_0 + z(t) + q(t)$&nbsp; erhält man mit&nbsp; $A_0 = 2\ \rm  V$&nbsp; und&nbsp; $A_{\rm T} = A_{\rm N} = 1 \ \rm  V$&nbsp; den möglichen Bereich&nbsp; $0 \ {\rm V} ≤ x(t) ≤ 4\ \rm  V$.
+'''(1)'''&nbsp; From&nbsp; $x(t) = A_0 + z(t) + q(t)$&nbsp;, t&nbsp; $A_0 = 2\ \rm  V$&nbsp; and&nbsp; $A_{\rm T} = A_{\rm N} = 1 \ \rm  V$&nbsp; we get the possible range&nbsp; $0 \ {\rm V} ≤ x(t) ≤ 4\ \rm  V$.
-*Die Hilfsgröße&nbsp; $w(t)$&nbsp; kann somit Werte zwischen&nbsp; $w_{\rm min}\hspace{0.15cm}\underline{ = -2 \ \rm V}$&nbsp; und&nbsp; $w_{\rm max}\hspace{0.15cm}\underline{ = +2 \ \rm V}$&nbsp; annehmen.
+*Thus, the auxiliary quantity&nbsp; $w(t)$&nbsp; can take values between&nbsp; $w_{\rm min}\hspace{0.15cm}\underline{ = -2 \ \rm V}$&nbsp; and&nbsp; $w_{\rm max}\hspace{0.15cm}\underline{ = +2 \ \rm V}$&nbsp;.
-'''(2)'''&nbsp; Der Koeffizient&nbsp; $c_0$&nbsp; ist gleich dem Kennlinienwert im Arbeitspunkt.&nbsp; Mit&nbsp; $A_0 = 2 \ \rm V$&nbsp; und&nbsp; $U = 3 \ \rm V$&nbsp; erhält man:
+'''(2)'''&nbsp; The coefficient&nbsp; $c_0$&nbsp; is equal to the characteristic value at the operating point.&nbsp; Using&nbsp; $A_0 = 2 \ \rm V$&nbsp; and&nbsp; $U = 3 \ \rm V$&nbsp; we obtain:
 :$$c_0 = y(A_0) = U \cdot \left( 1 -{\rm e} ^{-A_0/U}\right) \hspace{0.15cm}\underline {= 1.460\,{\rm V}}\hspace{0.05cm}.$$
-*Entsprechend gilt für den Taylorkoeffizienten&nbsp; $c_1$:
+*Accordingly, for the Taylor coefficient&nbsp; $c_1$:
 :$$c_1 = y\hspace{0.06cm}'(A_0)= {\rm e} ^{-A_0/U}\hspace{0.15cm}\underline { = 0.513}\hspace{0.05cm}.$$
-'''(3)'''&nbsp; Die weiteren Ableitungen&nbsp; $(n ≥ 2)$&nbsp; lauten:
+'''(3)'''&nbsp; The further derivatives &nbsp; $(n ≥ 2)$&nbsp; are:
 :$$y^{(n)}(A_0)= \frac{(-1)^{n-1}}{U^{n-1}} \cdot {\rm e} ^{-A_0/U} \hspace{0.05cm}.$$
-*Daraus ergeben sich folgende Koeffizienten:
+*This results in the following coefficients:
 :$$ c_2  =  \frac{1}{2!} \cdot y^{(2)}(A_0)= \frac{1}{2U} \cdot {\rm e}^{-A_0/U} \hspace{0.15cm}\underline {= -0.086\,{\rm V^{-1}}}\hspace{0.05cm},$$
 :$$c_3  =  \frac{1}{3!} \cdot y^{(3)}(A_0)= \frac{1}{6U^2} \cdot {\rm e}^{-A_0/U}\hspace{0.15cm}\underline { = 0.0095\,{\rm V^{-2}}}\hspace{0.05cm}.$$
-'''(4)'''&nbsp; Setzt man&nbsp; $c_3 = 0$, so lautet das Ausgangssignal des Verstärkers:
+'''(4)'''&nbsp; Setting&nbsp; $c_3 = 0$, the output signal of the amplifier is:
 :$$y(t) = c_0 + c_1 \cdot (z(t) + q(t)) + c_2 \cdot (z^2(t) + q^2(t) + 2 \cdot z(t) \cdot q(t))\hspace{0.05cm}.$$
-*Nach dem Bandpass verbleiben somit noch folgende Signalanteile:
+*Thus, after the bandpass, the following signal components remain:
 :$$s(t)  =  c_1 \cdot z(t) + 2 \cdot c_2 \cdot z(t) \cdot q(t)
    =  \left[c_1 \cdot A_{\rm T} + 2 \cdot c_2 \cdot A_{\rm T} \cdot A_{\rm N} \cdot \cos(\omega_{\rm N} t)\right] \cdot \cos(\omega_{\rm T} t)\hspace{0.05cm}.$$
-*Der Modulationsgrad ist dann als Quotient der „Amplitude” der Nachrichtenschwingung zur „Amplitude” des Trägers zu bestimmen:
+*The modulation depth is then determined as the quotient of the "amplitude" of the message oscillation over the "amplitude" of the carrier:
 :$$m = \frac{2 \cdot |c_2| \cdot A_{\rm T} \cdot A_{\rm N}}{|c_1| \cdot A_{\rm T}} = \frac{2 \cdot |c_2| \cdot A_{\rm N}}{|c_1| }= \frac{2 \cdot 0.086 \cdot 1\,{\rm V}}{0.513 }\hspace{0.15cm}\underline { = 0.335}\hspace{0.05cm}.$$
-'''(5)'''&nbsp; Richtig sind die <u>Aussagen 2 und 3</u>:
+'''(5)'''&nbsp; <u>Answers 2 and 3</u>are correct:
-*Unter Berücksichtigung des kubischen Anteils beinhaltet&nbsp; $y(t)$&nbsp; noch folgende weitere Anteile:
+*Considering the cubic part,&nbsp; $y(t)$&nbsp; includes the following other components:
 :$$y_3(t)  =  c_3 \cdot (z(t) + q(t))^3
 =  c_3 \cdot z^3(t) + 3 \cdot c_3 \cdot z^2(t) \cdot q(t)+ 3 \cdot c_3 \cdot z(t) \cdot q^2(t) + c_3 \cdot q^3(t) \hspace{0.05cm}.$$
-*Der erste Term führt zu Anteilen bei&nbsp; $f_{\rm T}$&nbsp; und&nbsp; $3f_{\rm T}$, der letzte bei&nbsp; $f_{\rm N}$&nbsp; und&nbsp; $3f_{\rm N}$.&nbsp; Der zweite Term ergibt einen Anteil bei&nbsp; $f_{\rm N}$&nbsp; und weitere bei&nbsp; $2f_{\rm T} ± f_{\rm N}$:
+*The first term results in components at&nbsp; $f_{\rm T}$&nbsp; and&nbsp; $3f_{\rm T}$, and the last term results in components at&nbsp; $f_{\rm N}$&nbsp; and&nbsp; $3f_{\rm N}$.&nbsp; The second term gives a component at &nbsp; $f_{\rm N}$&nbsp; and others at&nbsp; $2f_{\rm T} ± f_{\rm N}$:
 :$$3 \cdot c_3 \cdot z^2(t) \cdot q(t)= {3}/{2 } \cdot A_{\rm T}^2 \cdot A_{\rm N} \cdot \left[ \cos(\omega_{\rm N} t) + \cos(2\omega_{\rm T} t) \cdot \cos(\omega_{\rm N} t)\right] \hspace{0.05cm}.$$
-*Entsprechend führt der dritte Summand in obiger Gleichung zu
+*Accordingly, the third summand in the above equation leads to
 :$$3 \cdot c_3 \cdot z(t) \cdot q^2(t)= {3}/{2 }  \cdot A_{\rm T} \cdot A_{\rm N}^2 \cdot \left[ \cos(\omega_{\rm T} t) + \cos(\omega_{\rm T} t)\cdot \cos(2 \omega_{\rm N} t)\right] \hspace{0.05cm}.$$
-*Innerhalb des Frequenzbereichs von&nbsp; $\text{23 kHz}$&nbsp; bis&nbsp; $\text{37 kHz}$&nbsp; kommt es also tatsächlich zu einer Veränderung der Spektrallinie bei&nbsp; $f_{\rm T}$&nbsp; und es entstehen neue Diraclinien bei&nbsp; $f_{\rm T} ± 2f_{\rm N}$, also bei&nbsp; $\text{24 kHz}$&nbsp; und&nbsp; $\text{36 kHz}$.
+*Thus, within the frequency range from&nbsp; $\text{23 kHz}$&nbsp; to&nbsp; $\text{37 kHz}$&nbsp;, there is indeed a change in the spectral line at&nbsp; $f_{\rm T}$&nbsp; and new Dirac lines are formed at &nbsp; $f_{\rm T} ± 2f_{\rm N}$, i.e., at &nbsp; $\text{24 kHz}$&nbsp; and&nbsp; $\text{36 kHz}$.
-*Die dadurch verbundenen Verzerrungen sind somit nichtlinear &nbsp; &rArr; &nbsp; Antwort 3 ist richtig und Antwort 4 ist falsch.
+*The resulting distortions are thus nonlinear&nbsp; &rArr; &nbsp; Answer 3 ist correct and Answer 4 is wrong.
 {{ML-Fuß}}