Aufgaben:Exercise 2.3: DSB-AM Realization - 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.~~3_ZSB–AM–Realisierung~~]]		[[de:Aufgaben:Aufgabe 2.3: ZSB–AM–Realisierung]]

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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:2.3_ZSB–AM–Realisierung]]

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

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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.

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

Guenter at 16:21, 29 November 2021

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Guenter at 16:10, 29 November 2021

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	[[File:EN_Mod_A_2_3.png\|right\|frame\|Nonlinear characteristic curve <br>~~forr~~ AM~~–~~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  "Double-Sideband Amplitude Modulation  $\text{(DSB-AM)}$  with carrier",  an amplifier with the following characteristic curve must be used:
	:$$y = g(x) = U \cdot \left( 1 -{\rm e} ^{-x/U}\right)$$		:$$y = g(x) = U \cdot \left( 1 -{\rm e} ^{-x/U}\right)$$
	~~must be used.~~ Here,  $x = x(t)$  and  $y = y(t)$ are time-dependent voltages at the input and output of the amplifier, respectively.   The parameter  $U = 3 \ \rm V$  indicates the saturation voltage of the amplifier.		*Here,  $x = x(t)$  and  $y = y(t)$  are time-dependent voltages at the input and output of the amplifier,  respectively.
			*The parameter  $U = 3 \ \rm V$  indicates the saturation voltage of the amplifier.

	This curve is ~~operated~~ at the operating point  $A_0 = 2\ \rm V$.  This is achieved, for example, by the input signal
			This curve is driven at the operating point  $A_0 = 2\ \rm V$.  This is achieved,  for example,  by the input signal
	:$$x(t) = A_0 + z(t) + q(t)\hspace{0.05cm}.$$		:$$x(t) = A_0 + z(t) + q(t)\hspace{0.05cm}.$$
	Assume cosine oscillations for both the carrier and the source signal:		Assume cosine oscillations for both the carrier and the source signal:
	:$$ z(t) = A_{\rm T} \cdot \cos (2 \pi f_{\rm T} t),\hspace{0.2cm} A_{\rm T} = 1\,{\rm V},\hspace{0.2cm} f_{\rm T} = 30\,{\rm kHz},$$		:$$ z(t) = A_{\rm T} \cdot \cos (2 \pi f_{\rm T} t),\hspace{0.2cm} A_{\rm T} = 1\,{\rm V},\hspace{0.2cm} f_{\rm T} = 30\,{\rm kHz},$$
	:$$ q(t) = A_{\rm N} \cdot \cos (2 \pi f_{\rm N} t),\hspace{0.2cm} A_{\rm N} = 1\,{\rm V},\hspace{0.2cm} f_{\rm N} = 3\,{\rm kHz}\hspace{0.05cm}.$$		:$$ q(t) = A_{\rm N} \cdot \cos (2 \pi f_{\rm N} t),\hspace{0.2cm} A_{\rm N} = 1\,{\rm V},\hspace{0.2cm} f_{\rm N} = 3\,{\rm kHz}\hspace{0.05cm}.$$
	In solving this problem, use the auxiliary quantity		In solving this problem,  use the auxiliary quantity
	:$$w(t) = x(t) - A_0 = z(t) + q(t)\hspace{0.05cm}.$$		:$$w(t) = x(t) - A_0 = z(t) + q(t)\hspace{0.05cm}.$$

	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  "Taylor series"  around the operating point:
	:$$y(x) = y(A_0) + \frac{1}{1!} \cdot y\hspace{0.08cm}{\rm '}(A_0) \cdot (x - A_0)+ \frac{1}{2!} \cdot y\hspace{0.08cm}''(A_0) \cdot (x - A_0)^2+		:$$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{ ...}$$		\frac{1}{3!} \cdot y\hspace{0.08cm}'''(A_0) \cdot (x - A_0)^3 + \text{ ...}$$
	The output signal can then also be represented as depending on the auxiliary quantity  $w(t)$  as follows:		The output signal can then also be represented as depending on the auxiliary quantity  $w(t)$  as follows:
	:$$y(t) = c_0 + c_1 \cdot w(t) + c_2 \cdot w^2(t)+ c_3 \cdot w^3(t) +\text{ ...}$$		:$$y(t) = c_0 + c_1 \cdot w(t) + c_2 \cdot w^2(t)+ c_3 \cdot w^3(t) +\text{ ...}$$
	The DSB–AM signal  $s(t)$  is obtained by band-limiting $y(t)$  to the frequency range from  $\text{23 kHz}$  to  $\text{37 kHz}$.  That is, all frequencies other than  $f_{\rm T}$,  $f_{\rm T}±f_{\rm N}$  and  $f_{\rm T}±2f_{\rm N}$  are removed by the ~~bandpass.~~		*The DSB–AM signal  $s(t)$  is obtained by band-limiting $y(t)$  to the frequency range from  $\text{23 kHz}$  to  $\text{37 kHz}$.
			*That is,  all frequencies other than  $f_{\rm T}$,  $f_{\rm T}±f_{\rm N}$  and  $f_{\rm T}±2f_{\rm N}$  are removed by the band-pass.
	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.



			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.




	''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]].
	*Reference will also be made to the chapter  [[Linear_and_Time_Invariant_Systems/Nonlinear_Distortions#Description_of_nonlinear_systems\|Description of nonlinear systems]]  in the book "Linear and Time Invariant Systems".		*Reference will also be made to the chapter  [[Linear_and_Time_Invariant_Systems/Nonlinear_Distortions#Description_of_nonlinear_systems\|Description of nonlinear systems]]  in the book "Linear and Time Invariant Systems".
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	<quiz display=simple>		<quiz display=simple>
	{In what range can the input signal  $x(t)$ vary~~? variieren~~? Give the minimum and maximum values of the auxiliary variable  $w(t) = x(t) - A_0$~~ ~~.		{In what range can the input signal  $x(t)$ vary? Give the minimum and maximum values of the auxiliary variable  $w(t) = x(t) - A_0$.
	\|type="{}"}		\|type="{}"}
	$w_{\rm min} \ = \ $ { -2.06--1.94 } $\ \text{V}$		$w_{\rm min} \ = \ $ { -2.06--1.94 } $\ \text{V}$
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	$c_1 \ = \ $ { 0.513 3% }		$c_1 \ = \ $ { 0.513 3% }

	{What are the coefficients  $c_2$  and  $c_3$&nbsp of the nonlinear characteristic curve?		{What are the coefficients  $c_2$  and  $c_3$  of the nonlinear characteristic curve?
	\|type="{}"}		\|type="{}"}
	$c_2\ = \ $ { -0.088--0.084 } $\ \rm V^{ -1 }$		$c_2\ = \ $ { -0.088--0.084 } $\ \rm V^{ -1 }$
	$c_3\ = \ $ { 0.0095 3% } $\ \rm V^{ -2 }$		$c_3\ = \ $ { 0.0095 3% } $\ \rm V^{ -2 }$

	{Show that a "DSB-AM with carrier" constellation results when  $c_3$  is considered negligibly small. What is the modulation depth  $m$?		{Show that a  "DSB-AM with carrier"  constellation results when  $c_3$  is considered negligibly small.  What is the modulation depth  $m$?
	\|type="{}"}		\|type="{}"}
	$m \ = \ $ { 0.335 3% }		$m \ = \ $ { 0.335 3% }

	{Assuming that  $c_3$  cannot be considered negligibly small, which of the following statements are true?		{Assuming that  $c_3$  cannot be considered negligibly small,  which of the following statements are true?
	\|type="[]"}		\|type="[]"}
	- The weight of the spectral line at $f_{\rm T}$  is unchanged.		- The weight of the spectral line at $f_{\rm T}$  is unchanged.
	+ $s(t)$  now includes Dirac lines at $f_{\rm T} ± 2f_{\rm N}$.		+ $s(t)$  now includes Dirac lines at $f_{\rm T} ± 2f_{\rm N}$.
	+ The cubic term leads to nonlinear distortions.		+ The cubic term leads to nonlinear distortions.
	- The cubic term leads to linear distortions..		- The cubic term leads to linear distortions.
	</quiz>		</quiz>

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	'''(5)'''  <u>Answers 2 and 3</u>are correct:		'''(5)'''  <u>Answers 2 and 3</u> are correct:
	*Considering the cubic part,  $y(t)$  includes the following other components:		*Considering the cubic part,  $y(t)$  includes the following other components:
	:$$y_3(t) = c_3 \cdot (z(t) + q(t))^3		:$$y_3(t) = c_3 \cdot (z(t) + q(t))^3

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  From  $x(t) = A_0 + z(t) + q(t)$ , t  $A_0 = 2\ \rm V$  and  $A_{\rm T} = A_{\rm N} = 1 \ \rm V$  we get the possible range  $0 \ {\rm V} ≤ x(t) ≤ 4\ \rm V$.		'''(1)'''  From  $x(t) = A_0 + z(t) + q(t)$ , with   $A_0 = 2\ \rm V$  and  $A_{\rm T} = A_{\rm N} = 1 \ \rm V$ , we get the possible range  $0 \ {\rm V} ≤ x(t) ≤ 4\ \rm V$.
	*Thus, the auxiliary quantity  $w(t)$  can take values between  $w_{\rm min}\hspace{0.15cm}\underline{ = -2 \ \rm V}$  and  $w_{\rm max}\hspace{0.15cm}\underline{ = +2 \ \rm V}$ .		*Thus, the auxiliary quantity  $w(t)$  can take values between  $w_{\rm min}\hspace{0.15cm}\underline{ = -2 \ \rm V}$  and  $w_{\rm max}\hspace{0.15cm}\underline{ = +2 \ \rm V}$ .

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	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  "Taylor series"  around the operating point:
	:$$y(x) = y(A_0) + \frac{1}{1!} \cdot y\hspace{0.08cm}{\rm '}(A_0) \cdot (x - A_0)+ \frac{1}{2!} \cdot y\hspace{0.08cm}''(A_0) \cdot (x - A_0)^2+		:$$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{ ...}$$
	\frac{1}{3!} \cdot y\hspace{0.08cm}'''(A_0) \cdot (x - A_0)^3 + \text{ ...}$$
	The output signal can then also be represented as depending on the auxiliary quantity  $w(t)$  as follows:		The output signal can then also be represented as depending on the auxiliary quantity  $w(t)$  as follows:
	:$$y(t) = c_0 + c_1 \cdot w(t) + c_2 \cdot w^2(t)+ c_3 \cdot w^3(t) +\text{ ...}$$		:$$y(t) = c_0 + c_1 \cdot w(t) + c_2 \cdot w^2(t)+ c_3 \cdot w^3(t) +\text{ ...}$$
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	:$$y(t) = c_0 + c_1 \cdot (z(t) + q(t)) + c_2 \cdot (z^2(t) + q^2(t) + 2 \cdot z(t) \cdot q(t))\hspace{0.05cm}.$$		:$$y(t) = c_0 + c_1 \cdot (z(t) + q(t)) + c_2 \cdot (z^2(t) + q^2(t) + 2 \cdot z(t) \cdot q(t))\hspace{0.05cm}.$$
	*Thus,  after the band-pass,  the following signal components remain:		*Thus,  after the band-pass,  the following signal components remain:
	:$$s(t) = c_1 \cdot z(t) + 2 \cdot c_2 \cdot z(t) \cdot q(t)		:$$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}.$$
	= \left[c_1 \cdot A_{\rm T} + 2 \cdot c_2 \cdot A_{\rm T} \cdot A_{\rm N} \cdot \cos(\omega_{\rm N} t)\right] \cdot \cos(\omega_{\rm T} t)\hspace{0.05cm}.$$
	*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  "amplitude of the message oscillation"  over the "amplitude of the carrier":
	:$$m = \frac{2 \cdot \|c_2\| \cdot A_{\rm T} \cdot A_{\rm N}}{\|c_1\| \cdot A_{\rm T}} = \frac{2 \cdot \|c_2\| \cdot A_{\rm N}}{\|c_1\| }= \frac{2 \cdot 0.086 \cdot 1\,{\rm V}}{0.513 }\hspace{0.15cm}\underline { = 0.335}\hspace{0.05cm}.$$		:$$m = \frac{2 \cdot \|c_2\| \cdot A_{\rm T} \cdot A_{\rm N}}{\|c_1\| \cdot A_{\rm T}} = \frac{2 \cdot \|c_2\| \cdot A_{\rm N}}{\|c_1\| }= \frac{2 \cdot 0.086 \cdot 1\,{\rm V}}{0.513 }\hspace{0.15cm}\underline { = 0.335}\hspace{0.05cm}.$$
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	'''(5)'''  <u>Answers 2 and 3</u>  are correct:		'''(5)'''  <u>Answers 2 and 3</u>  are correct:
	*Considering the cubic part,  $y(t)$  includes the following other components:		*Considering the cubic part,  $y(t)$  includes the following other components:
	:$$y_3(t) = c_3 \cdot (z(t) + q(t))^3		:$$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}.$$
	= c_3 \cdot z^3(t) + 3 \cdot c_3 \cdot z^2(t) \cdot q(t)+ 3 \cdot c_3 \cdot z(t) \cdot q^2(t) + c_3 \cdot q^3(t) \hspace{0.05cm}.$$
	*The first term results in components at  $f_{\rm T}$  and  $3f_{\rm T}$, and the last term results in components at  $f_{\rm N}$  and  $3f_{\rm N}$.		*The first term results in components at  $f_{\rm T}$  and  $3f_{\rm T}$, and the last term results in components at  $f_{\rm N}$  and  $3f_{\rm N}$.
	*The second term gives a component at   $f_{\rm N}$  and others at  $2f_{\rm T} ± f_{\rm N}$:		*The second term gives a component at   $f_{\rm N}$  and others at  $2f_{\rm T} ± f_{\rm N}$:

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  From  $x(t) = A_0 + z(t) + q(t)$ , with   $A_0 = 2\ \rm V$  and  $A_{\rm T} = A_{\rm N} = 1 \ \rm V$ , we get the possible range  $0 \ {\rm V} ≤ x(t) ≤ 4\ \rm V$.		'''(1)'''  From  $x(t) = A_0 + z(t) + q(t)$,  with   $A_0 = 2\ \rm V$  and   $A_{\rm T} = A_{\rm N} = 1 \ \rm V$,  we get the possible range  $0 \ {\rm V} ≤ x(t) ≤ 4\ \rm V$.
	*Thus, the auxiliary quantity  $w(t)$  can take values between  $w_{\rm min}\hspace{0.15cm}\underline{ = -2 \ \rm V}$  and  $w_{\rm max}\hspace{0.15cm}\underline{ = +2 \ \rm V}$~~ ~~.		*Thus,  the auxiliary quantity  $w(t)$  can take values between  $w_{\rm min}\hspace{0.15cm}\underline{ = -2 \ \rm V}$  and  $w_{\rm max}\hspace{0.15cm}\underline{ = +2 \ \rm V}$.


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	'''(2)'''  The coefficient  $c_0$  is equal to the characteristic value at the operating point.  Using  $A_0 = 2 \ \rm V$  and  $U = 3 \ \rm V$  we obtain:		'''(2)'''  The coefficient  $c_0$  is equal to the characteristic value at the operating point.  Using  $A_0 = 2 \ \rm V$  and  $U = 3 \ \rm V$  we obtain:
	:$$c_0 = y(A_0) = U \cdot \left( 1 -{\rm e} ^{-A_0/U}\right) \hspace{0.15cm}\underline {= 1.460\,{\rm V}}\hspace{0.05cm}.$$		:$$c_0 = y(A_0) = U \cdot \left( 1 -{\rm e} ^{-A_0/U}\right) \hspace{0.15cm}\underline {= 1.460\,{\rm V}}\hspace{0.05cm}.$$
	*Accordingly, for the Taylor coefficient  $c_1$:		*Accordingly,  for the Taylor coefficient  $c_1$:
	:$$c_1 = y\hspace{0.06cm}'(A_0)= {\rm e} ^{-A_0/U}\hspace{0.15cm}\underline { = 0.513}\hspace{0.05cm}.$$		:$$c_1 = y\hspace{0.06cm}'(A_0)= {\rm e} ^{-A_0/U}\hspace{0.15cm}\underline { = 0.513}\hspace{0.05cm}.$$

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	'''(4)'''  Setting  $c_3 = 0$, the output signal of the amplifier is:		'''(4)'''  Setting  $c_3 = 0$,  the output signal of the amplifier is:
	:$$y(t) = c_0 + c_1 \cdot (z(t) + q(t)) + c_2 \cdot (z^2(t) + q^2(t) + 2 \cdot z(t) \cdot q(t))\hspace{0.05cm}.$$		:$$y(t) = c_0 + c_1 \cdot (z(t) + q(t)) + c_2 \cdot (z^2(t) + q^2(t) + 2 \cdot z(t) \cdot q(t))\hspace{0.05cm}.$$
	*Thus, after the ~~bandpass~~, the following signal components remain:		*Thus,  after the band-pass,  the following signal components remain:
	:$$s(t) = c_1 \cdot z(t) + 2 \cdot c_2 \cdot z(t) \cdot q(t)		:$$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}.$$		= \left[c_1 \cdot A_{\rm T} + 2 \cdot c_2 \cdot A_{\rm T} \cdot A_{\rm N} \cdot \cos(\omega_{\rm N} t)\right] \cdot \cos(\omega_{\rm T} t)\hspace{0.05cm}.$$
	*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  "amplitude of the message oscillation"  over the "amplitude of the carrier":
	:$$m = \frac{2 \cdot \|c_2\| \cdot A_{\rm T} \cdot A_{\rm N}}{\|c_1\| \cdot A_{\rm T}} = \frac{2 \cdot \|c_2\| \cdot A_{\rm N}}{\|c_1\| }= \frac{2 \cdot 0.086 \cdot 1\,{\rm V}}{0.513 }\hspace{0.15cm}\underline { = 0.335}\hspace{0.05cm}.$$		:$$m = \frac{2 \cdot \|c_2\| \cdot A_{\rm T} \cdot A_{\rm N}}{\|c_1\| \cdot A_{\rm T}} = \frac{2 \cdot \|c_2\| \cdot A_{\rm N}}{\|c_1\| }= \frac{2 \cdot 0.086 \cdot 1\,{\rm V}}{0.513 }\hspace{0.15cm}\underline { = 0.335}\hspace{0.05cm}.$$



	'''(5)'''  <u>Answers 2 and 3</u> are correct:		'''(5)'''  <u>Answers 2 and 3</u>  are correct:
	*Considering the cubic part,  $y(t)$  includes the following other components:		*Considering the cubic part,  $y(t)$  includes the following other components:
	:$$y_3(t) = c_3 \cdot (z(t) + q(t))^3		:$$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}.$$		= c_3 \cdot z^3(t) + 3 \cdot c_3 \cdot z^2(t) \cdot q(t)+ 3 \cdot c_3 \cdot z(t) \cdot q^2(t) + c_3 \cdot q^3(t) \hspace{0.05cm}.$$
	*The first term results in components at  $f_{\rm T}$  and  $3f_{\rm T}$, and the last term results in components at  $f_{\rm N}$  and  $3f_{\rm N}$.  The second term gives a component at   $f_{\rm N}$  and others at  $2f_{\rm T} ± f_{\rm N}$:		*The first term results in components at  $f_{\rm T}$  and  $3f_{\rm T}$, and the last term results in components at  $f_{\rm N}$  and  $3f_{\rm N}$.
			*The second term gives a component at   $f_{\rm N}$  and others at  $2f_{\rm T} ± f_{\rm N}$:
	:$$3 \cdot c_3 \cdot z^2(t) \cdot q(t)= {3}/{2 } \cdot A_{\rm T}^2 \cdot A_{\rm N} \cdot \left[ \cos(\omega_{\rm N} t) + \cos(2\omega_{\rm T} t) \cdot \cos(\omega_{\rm N} t)\right] \hspace{0.05cm}.$$		:$$3 \cdot c_3 \cdot z^2(t) \cdot q(t)= {3}/{2 } \cdot A_{\rm T}^2 \cdot A_{\rm N} \cdot \left[ \cos(\omega_{\rm N} t) + \cos(2\omega_{\rm T} t) \cdot \cos(\omega_{\rm N} t)\right] \hspace{0.05cm}.$$
	*Accordingly, the third summand in the above equation leads to		*Accordingly, the third summand in the above equation leads to
	:$$3 \cdot c_3 \cdot z(t) \cdot q^2(t)= {3}/{2 } \cdot A_{\rm T} \cdot A_{\rm N}^2 \cdot \left[ \cos(\omega_{\rm T} t) + \cos(\omega_{\rm T} t)\cdot \cos(2 \omega_{\rm N} t)\right] \hspace{0.05cm}.$$		:$$3 \cdot c_3 \cdot z(t) \cdot q^2(t)= {3}/{2 } \cdot A_{\rm T} \cdot A_{\rm N}^2 \cdot \left[ \cos(\omega_{\rm T} t) + \cos(\omega_{\rm T} t)\cdot \cos(2 \omega_{\rm N} t)\right] \hspace{0.05cm}.$$
	*Thus, within the frequency range from  $\text{23 kHz}$  to  $\text{37 kHz}$ , there is indeed a change in the spectral line at  $f_{\rm T}$  and new Dirac lines are formed at   $f_{\rm T} ± 2f_{\rm N}$, i.e., at   $\text{24 kHz}$  and  $\text{36 kHz}$.		*Thus,  within the frequency range from  $\text{23 kHz}$  to  $\text{37 kHz}$,  there is indeed a change in the spectral line at  $f_{\rm T}$  <br>and new Dirac lines are formed at   $f_{\rm T} ± 2f_{\rm N}$,  i.e.,  at   $\text{24 kHz}$  and  $\text{36 kHz}$.
	*The resulting distortions are thus nonlinear  ⇒   Answer 3 ist correct and Answer 4 is wrong.		*The resulting distortions are thus nonlinear  ⇒   Answer 3 ist correct and Answer 4 is wrong.