Aufgaben:Exercise 2.2: Modulation Depth - 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.~~2_Modulationsgrad~~]]		[[de:Aufgaben:Aufgabe 2.2: Modulationsgrad]]

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	:$$ A_{\rm max} = A_{\rm T}+A_{\rm N}=2\,\,{\rm V},\hspace{0.3cm} A_{\rm min} = A_{\rm T}-A_{\rm N}\hspace{0.05cm}$$		:$$ A_{\rm max} = A_{\rm T}+A_{\rm N}=2\,\,{\rm V},\hspace{0.3cm} A_{\rm min} = A_{\rm T}-A_{\rm N}\hspace{0.05cm}$$
	directly follows:		directly follows:
	:$$A_{\rm N} = (A_{\rm max} - A_{\rm min})/2,\hspace{0.3cm}		:$$A_{\rm N} = (A_{\rm max} - A_{\rm min})/2,\hspace{0.3cm}A_{\rm T} = (A_{\rm max} + A_{\rm min})/2\hspace{0.05cm}.$$
	A_{\rm T} = (A_{\rm max} + A_{\rm min})/2\hspace{0.05cm}.$$
	*Thus,  the modulation depth is		*Thus,  the modulation depth is
	:$$m = \frac{A_{\rm max} - A_{\rm min}}{A_{\rm max} + A_{\rm min}}\hspace{0.05cm}.$$		:$$m = \frac{A_{\rm max} - A_{\rm min}}{A_{\rm max} + A_{\rm min}}\hspace{0.05cm}.$$
	*With the given numerical values,  one obtains:		*With the given numerical values,  one obtains:
	:$$ m_1 = \frac{2\,{\rm V} - 0.667\,{\rm V}}{2\,{\rm V} + 0.667\,{\rm V}} \hspace{0.15cm}\underline {= 0.5}\hspace{0.05cm}, \hspace{0.5cm} m_2 = \frac{2\,{\rm V} - 0\,{\rm V}}{2\,{\rm V} + 0\,{\rm V}} \hspace{0.15cm}\underline {= 1.0}\hspace{0.05cm}, \hspace{0.5cm}		:$$ m_1 = \frac{2\,{\rm V} - 0.667\,{\rm V}}{2\,{\rm V} + 0.667\,{\rm V}} \hspace{0.15cm}\underline {= 0.5}\hspace{0.05cm}, \hspace{0.5cm} m_2 = \frac{2\,{\rm V} - 0\,{\rm V}}{2\,{\rm V} + 0\,{\rm V}} \hspace{0.15cm}\underline {= 1.0}\hspace{0.05cm}, \hspace{0.5cm}m_3 = \frac{2\,{\rm V} -(-1\,{\rm V})}{2\,{\rm V} + (-1\,{\rm V})} \hspace{0.15cm}\underline{=3.0}\hspace{0.05cm}.$$
	m_3 = \frac{2\,{\rm V} -(-1\,{\rm V})}{2\,{\rm V} + (-1\,{\rm V})} \hspace{0.15cm}\underline{=3.0}\hspace{0.05cm}.$$

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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:2.2_Modulationsgrad]]

Hwang at 14:17, 18 January 2023

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	[[File:P_ID990__Mod_A_2_2_c.png\|right\|frame\|Analytical signal's spectrum ]]		[[File:P_ID990__Mod_A_2_2_c.png\|right\|frame\|Analytical signal's spectrum ]]
	'''(3)'''  The spectrum  $S_+(f)$  is composed of three Dirac lines for each modulation depth   $m$  with the following weights:		'''(3)'''  The spectrum  $S_+(f)$  is composed of three Dirac delta lines for each modulation depth   $m$  with the following weights:
	*$A_{\rm T}$  $($at  $f = f_{\rm T})$,		*$A_{\rm T}$  $($at  $f = f_{\rm T})$,
	* $m/2 · A_{\rm T}$  $($at  $f = f_{\rm T} ± f_{\rm N})$.		* $m/2 · A_{\rm T}$  $($at  $f = f_{\rm T} ± f_{\rm N})$.

Reed at 15:32, 24 March 2022

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

Reed at 15:27, 24 March 2022

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	'''(4)'''  The power (~~root~~ mean square) of a harmonic oscillation with amplitude   $A_{\rm T} = 1 \ \rm V$  referenced to the   $1 \ Ω$  resistor is:		'''(4)'''  The power (mean square) of a harmonic oscillation with amplitude   $A_{\rm T} = 1 \ \rm V$  referenced to the   $1 \ Ω$  resistor is:
	:$$P_{\rm T} ={A_{\rm T}^2}/{2} = 0.5\,{\rm V}^2 \hspace{0.05cm}.$$		:$$P_{\rm T} ={A_{\rm T}^2}/{2} = 0.5\,{\rm V}^2 \hspace{0.05cm}.$$
	*In the same way,  for the powers of the lower and the upper sideband we obtain:		*In the same way,  for the powers of the lower and the upper sideband we obtain:

Guenter at 15:13, 29 November 2021

2021-11-29T15:13:27Z

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	:$$m = \frac{A_{\rm max} - A_{\rm min}}{A_{\rm max} + A_{\rm min}}\hspace{0.05cm}.$$		:$$m = \frac{A_{\rm max} - A_{\rm min}}{A_{\rm max} + A_{\rm min}}\hspace{0.05cm}.$$
	*With the given numerical values,  one obtains:		*With the given numerical values,  one obtains:
	$$ m_1 = \frac{2\,{\rm V} - 0.667\,{\rm V}}{2\,{\rm V} + 0.667\,{\rm V}} \hspace{0.15cm}\underline {= 0.5}\hspace{0.05cm}, \hspace{0.5cm} m_2 = \frac{2\,{\rm V} - 0\,{\rm V}}{2\,{\rm V} + 0\,{\rm V}} \hspace{0.15cm}\underline {= 1.0}\hspace{0.05cm}, \hspace{0.5cm}		:$$ m_1 = \frac{2\,{\rm V} - 0.667\,{\rm V}}{2\,{\rm V} + 0.667\,{\rm V}} \hspace{0.15cm}\underline {= 0.5}\hspace{0.05cm}, \hspace{0.5cm} m_2 = \frac{2\,{\rm V} - 0\,{\rm V}}{2\,{\rm V} + 0\,{\rm V}} \hspace{0.15cm}\underline {= 1.0}\hspace{0.05cm}, \hspace{0.5cm}
	m_3 = \frac{2\,{\rm V} -(-1\,{\rm V})}{2\,{\rm V} + (-1\,{\rm V})} \hspace{0.15cm}\underline{=3.0}\hspace{0.05cm}.$$		m_3 = \frac{2\,{\rm V} -(-1\,{\rm V})}{2\,{\rm V} + (-1\,{\rm V})} \hspace{0.15cm}\underline{=3.0}\hspace{0.05cm}.$$

Guenter at 14:58, 29 November 2021

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	:$$A_{\rm N} = (A_{\rm max} - A_{\rm min})/2,\hspace{0.3cm}		:$$A_{\rm N} = (A_{\rm max} - A_{\rm min})/2,\hspace{0.3cm}
	A_{\rm T} = (A_{\rm max} + A_{\rm min})/2\hspace{0.05cm}.$$		A_{\rm T} = (A_{\rm max} + A_{\rm min})/2\hspace{0.05cm}.$$
	*Thus, the modulation depth is		*Thus,  the modulation depth is
	:$$m = \frac{A_{\rm max} - A_{\rm min}}{A_{\rm max} + A_{\rm min}}\hspace{0.05cm}.$$		:$$m = \frac{A_{\rm max} - A_{\rm min}}{A_{\rm max} + A_{\rm min}}\hspace{0.05cm}.$$
	*With the given numerical values, one obtains:		*With the given numerical values,  one obtains:
	$$ m_1 = \frac{2\,{\rm V} - 0.667\,{\rm V}}{2\,{\rm V} + 0.667\,{\rm V}} \hspace{0.15cm}\underline {= 0.5}\hspace{0.05cm}, \hspace{0.5cm} m_2 = \frac{2\,{\rm V} - 0\,{\rm V}}{2\,{\rm V} + 0\,{\rm V}} \hspace{0.15cm}\underline {= 1.0}\hspace{0.05cm}, \hspace{0.5cm}		$$ m_1 = \frac{2\,{\rm V} - 0.667\,{\rm V}}{2\,{\rm V} + 0.667\,{\rm V}} \hspace{0.15cm}\underline {= 0.5}\hspace{0.05cm}, \hspace{0.5cm} m_2 = \frac{2\,{\rm V} - 0\,{\rm V}}{2\,{\rm V} + 0\,{\rm V}} \hspace{0.15cm}\underline {= 1.0}\hspace{0.05cm}, \hspace{0.5cm}
	m_3 = \frac{2\,{\rm V} -(-1\,{\rm V})}{2\,{\rm V} + (-1\,{\rm V})} \hspace{0.15cm}\underline{=3.0}\hspace{0.05cm}.$$		m_3 = \frac{2\,{\rm V} -(-1\,{\rm V})}{2\,{\rm V} + (-1\,{\rm V})} \hspace{0.15cm}\underline{=3.0}\hspace{0.05cm}.$$
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	'''(2)'''  <u>Answers 1 and 3</u> are correct:		'''(2)'''  <u>Answers 1 and 3</u>  are correct:
	*In this case,  $A_{\rm T} = 0$, which means it is indeed "DSB-AM without a carrier".		*In this case,  $A_{\rm T} = 0$,  which means it is indeed  "DSB-AM without carrier".
	*The modulation depth  $m = A_{\rm N}/A_{\rm T}$  is infinitely large.		*The modulation depth  $m = A_{\rm N}/A_{\rm T}$  is infinitely large.



			[[File:P_ID990__Mod_A_2_2_c.png\|right\|frame\|Analytical signal's spectrum ]]
	[[File:P_ID990__Mod_A_2_2_c.png\|right\|frame\|~~Spectrum: analytical~~ signal]]
	'''(3)'''  The spectrum  $S_+(f)$  is composed of three Dirac lines for each modulation depth   $m$  with the following weights:		'''(3)'''  The spectrum  $S_+(f)$  is composed of three Dirac lines for each modulation depth   $m$  with the following weights:
	*$A_{\rm T}$  $($at  $f = f_{\rm T})$,		*$A_{\rm T}$  $($at  $f = f_{\rm T})$,
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	For $m = 1$, the weights are obtained according to the graph:		For  $m = 1$,  the weights are obtained according to the graph:
	*$S_+(f_{\rm T}) = 1\ \rm V$,		*$S_+(f_{\rm T}) = 1\ \rm V$,
	*$S_+(f_{\rm T} ± f_{\rm T}) = 0.5\ \rm V$.		*$S_+(f_{\rm T} ± f_{\rm T}) = 0.5\ \rm V$.



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	'''(4)'''  The power (root mean square) of a harmonic oscillation with amplitude   $A_{\rm T} = 1 \ \rm V$  referenced to the   $1 \ Ω$  resistor is:		'''(4)'''  The power (root mean square) of a harmonic oscillation with amplitude   $A_{\rm T} = 1 \ \rm V$  referenced to the   $1 \ Ω$  resistor is:
	:$$P_{\rm T} ={A_{\rm T}^2}/{2} = 0.5\,{\rm V}^2 \hspace{0.05cm}.$$		:$$P_{\rm T} ={A_{\rm T}^2}/{2} = 0.5\,{\rm V}^2 \hspace{0.05cm}.$$
	*In the same way, for the powers of the lower and the upper sideband we obtain:		*In the same way,  for the powers of the lower and the upper sideband we obtain:
	:$$P_{\rm ~~USB~~} = P_{\rm ~~OSB~~} =({A_{\rm N}}/{2})^2/2 = 0.125\,{\rm V}^2 \hspace{0.05cm}.$$		:$$P_{\rm LSB} = P_{\rm USB} =({A_{\rm N}}/{2})^2/2 = 0.125\,{\rm V}^2 \hspace{0.05cm}.$$
	*Thus, for  $m=1$, the ratio we are looking for is:		*Thus,  for  $m=1$,  the ratio we are looking for is:
	:$${P_{\rm T}}/{P_{\rm S}}= \frac{P_{\rm T}}{P_{\rm USB} + P_{\rm T}+ P_{\rm OSB}}= \frac{0.5\,{\rm V}^2}{0.125\,{\rm V}^2 + 0.5\,{\rm V}^2+ 0.125\,{\rm V}^2}= 2/3\hspace{0.15cm}\underline { = 0.667}\hspace{0.05cm}.$$		:$${P_{\rm T}}/{P_{\rm S}}= \frac{P_{\rm T}}{P_{\rm USB} + P_{\rm T}+ P_{\rm OSB}}= \frac{0.5\,{\rm V}^2}{0.125\,{\rm V}^2 + 0.5\,{\rm V}^2+ 0.125\,{\rm V}^2}= 2/3\hspace{0.15cm}\underline { = 0.667}\hspace{0.05cm}.$$



	'''(5)'''  Using the Dirac weights  $m/2 · A_{\rm T}$  of the two sidebands corresponding to subtask  '''(3)''' , we get:		'''(5)'''  Using the Dirac weights  $m/2 · A_{\rm T}$  of the two sidebands corresponding to subtask  '''(3)''',  we get:
	:$${P_{\rm T}}/{P_{\rm S}}= \frac{A_{\rm T}^2/2}{A_{\rm T}^2/2 + 2 \cdot (m/2)^2 \cdot A_{\rm T}^2/2}= \frac{2}{2 + m^2}\hspace{0.05cm}.$$		:$${P_{\rm T}}/{P_{\rm S}}= \frac{A_{\rm T}^2/2}{A_{\rm T}^2/2 + 2 \cdot (m/2)^2 \cdot A_{\rm T}^2/2}= \frac{2}{2 + m^2}\hspace{0.05cm}.$$
	*This leads to the numerical values   $8/9 = 0.889$  $($for  $m = 0.5)$,     $2/11 = 0.182$  $($for  $m = 3)$~~  und~~     $0$  $($for  $m \to ∞$).		*This leads to the numerical values   $8/9 = 0.889$  $($for  $m = 0.5)$,     $2/11 = 0.182$  $($for  $m = 3)$,     $0$  $($for  $m \to ∞$).



	'''(6)'''  <u>Both statements</u> are true:		'''(6)'''  <u>Both statements</u>  are true:
	*The addition of the carrier only makes sense in order to use the simpler envelope demodulator. This is only possible for  $m \le 1$.		*The addition of the carrier only makes sense in order to use the simpler envelope demodulator.  This is only possible for  $m \le 1$.
	*However, should the modulation depth be  $m > 1$  and the use of a synchronous demodulator therefore be required, the carrier should be (almost) completely omitted for energy reasons.		*However,  should the modulation depth be  $m > 1$  and the use of a synchronous demodulator therefore be required,  the carrier should be (almost) completely omitted for energy reasons.
	*Similarly due to energy concerns, if an envelope demodulator is used, the largest possible modulation depth  $m < 1$    ⇒   $m \to 1$  should be aimed for.		*Similarly due to energy concerns,  if an envelope demodulator is used,  the largest possible modulation depth  $m < 1$    ⇒   $m \to 1$  should be aimed for.
	*However, a small residual carrier can facilitate carrier recovery, which is needed in a synchronous demodulator for frequency and phase synchronization. Thus, the second statement is only conditionally correct.		*However,  a small residual carrier can facilitate carrier recovery,  which is needed in a synchronous demodulator for frequency and phase synchronization.  Thus,  the second statement is only conditionally correct.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 14:46, 29 November 2021

2021-11-29T14:46:50Z

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	[[File:P_ID989__Mod_A_2_2.png\|right\|frame\|Definition of modulation depth for DSB–AM]]		[[File:P_ID989__Mod_A_2_2.png\|right\|frame\|Definition of modulation depth for DSB–AM]]
	The graph shows  ~~2DSB~~ amplitude-modulated signals"   $s_1(t)$  to  $s_4(t)$  with differing modulation depth  $m$.  Let the message signal  $q(t)$  and the carrier signal  $z(t)$  each be cosine:		The graph shows  "DSB amplitude-modulated signals"   $s_1(t)$  to  $s_4(t)$  with differing modulation depth  $m$.  Let the message signal  $q(t)$  and the carrier signal  $z(t)$  each be cosine:
	:$$q(t) = A_{\rm N} \cdot \cos (2 \pi f_{\rm N} t),\hspace{0.2cm} f_{\rm N} = 4\,{\rm kHz}\hspace{0.05cm},$$		:$$q(t) = A_{\rm N} \cdot \cos (2 \pi f_{\rm N} t),\hspace{0.2cm} f_{\rm N} = 4\,{\rm kHz}\hspace{0.05cm},$$
	:$$ z(t) = \hspace{0.2cm}1 \hspace{0.15cm} \cdot \cos (2 \pi f_{\rm T} t),\hspace{0.2cm} f_{\rm T} = 50\,{\rm kHz}\hspace{0.05cm}.$$		:$$ z(t) = \hspace{0.2cm}1 \hspace{0.15cm} \cdot \cos (2 \pi f_{\rm T} t),\hspace{0.2cm} f_{\rm T} = 50\,{\rm kHz}\hspace{0.05cm}.$$

Guenter at 14:46, 29 November 2021

2021-11-29T14:46:26Z

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	[[File:P_ID989__Mod_A_2_2.png\|right\|frame\|Definition of modulation depth for DSB–AM]]		[[File:P_ID989__Mod_A_2_2.png\|right\|frame\|Definition of modulation depth for DSB–AM]]
	The graph shows ~~DSB~~ amplitude-modulated signals  $s_1(t)$  to  $s_4(t)$  with differing modulation depth  $m$. Let message signal  $q(t)$  and carrier signal  $z(t)$  each be cosine:		The graph shows  2DSB amplitude-modulated signals"   $s_1(t)$  to  $s_4(t)$  with differing modulation depth  $m$.  Let the message signal  $q(t)$  and the carrier signal  $z(t)$  each be cosine:
	:$$q(t) = A_{\rm N} \cdot \cos (2 \pi f_{\rm N} t),\hspace{0.2cm} f_{\rm N} = 4\,{\rm kHz}\hspace{0.05cm},$$		:$$q(t) = A_{\rm N} \cdot \cos (2 \pi f_{\rm N} t),\hspace{0.2cm} f_{\rm N} = 4\,{\rm kHz}\hspace{0.05cm},$$
	:$$ z(t) = \hspace{0.2cm}1 \hspace{0.15cm} \cdot \cos (2 \pi f_{\rm T} t),\hspace{0.2cm} f_{\rm T} = 50\,{\rm kHz}\hspace{0.05cm}.$$		:$$ z(t) = \hspace{0.2cm}1 \hspace{0.15cm} \cdot \cos (2 \pi f_{\rm T} t),\hspace{0.2cm} f_{\rm T} = 50\,{\rm kHz}\hspace{0.05cm}.$$
	The modulated signal (transmitted signal) with the DC component added in the modulator is  $A_{\rm T}$:		The modulated signal  (transmitted signal)  with the DC component added in the modulator is  $A_{\rm T}$:
	:$$s(t ) = A(t) \cdot z(t), \hspace{0.2cm} A(t) = q(t) + A_{\rm T}\hspace{0.05cm}.$$		:$$s(t ) = A(t) \cdot z(t), \hspace{0.2cm} A(t) = q(t) + A_{\rm T}\hspace{0.05cm}.$$
	In the graphs, the chosen normalization was:		In the graphs,  the chosen normalization was:
	:$$A_{\rm T}+ A_{\rm N} = 2\,{\rm V}\hspace{0.05cm}.$$		:$$A_{\rm T}+ A_{\rm N} = 2\,{\rm V}\hspace{0.05cm}.$$
	*If the modulation depth is  $m ≤ 1$, then  $A(t)= q(t) + A_{\rm T}$  is equal to the envelope  $a(t)$.		*If the modulation depth is  $m ≤ 1$,  then  $A(t)= q(t) + A_{\rm T}$  is equal to the envelope  $a(t)$.
	*In contrast, for a modulation depth  $m > 1$:		*In contrast,  for a modulation depth  $m > 1$:
	:$$a(t ) = \|A(t)\|\hspace{0.05cm}.$$		:$$a(t ) = \|A(t)\|\hspace{0.05cm}.$$
			*The cosine curve  $A(t)$  varies between  $A_{\rm max}$  and  $A_{\rm min}$; because of normalization,   ⇒   $A_{\rm max} = 2 \ \rm V$.
	*The cosine curve  $A(t)$  varies between  $A_{\rm max}$  and  $A_{\rm min}$; because of normalization,  $A_{\rm max} = 2 \ \rm V$ ~~is always the case~~.		*The minimum values of  $A(t)$  occur at half the period of the source signal  $($i.e., for  $t = 125 \ \rm µ s)$:
	*The minimum values of  $A(t)$  occur at half the period of the source signal $($i.e., for  $t = 125 \ \rm µ s)$~~ ~~:
	:$$A_{\rm min} = q(T_0/2)+ A_{\rm T} = A_{\rm T}-A_{\rm N}.$$		:$$A_{\rm min} = q(T_0/2)+ A_{\rm T} = A_{\rm T}-A_{\rm N}.$$
	*The numerical values are given in the graph.		*The numerical values are given in the graph.
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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]].
	*Particular reference is made to the page  [[Modulation_Methods/Double-Sideband_Amplitude_Modulation#Double-Sideband_Amplitude_Modulation_with_carrier\|DSB-AM with carrier]].		*Particular reference is made to the page  [[Modulation_Methods/Double-Sideband_Amplitude_Modulation#Double-Sideband_Amplitude_Modulation_with_carrier\|DSB-AM with carrier]].
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	<quiz display=simple>		<quiz display=simple>

	{Determine the modulation depth for each of the signals  $s_1(t)$,  $s_2(t)$,  $s_3(t)$~~ ~~.		{Determine the modulation depth for each of the signals  $s_1(t)$,  $s_2(t)$,  $s_3(t)$.
	\|type="{}"}		\|type="{}"}
	$m_1 \ = \ $ { 0.5 3% }		$m_1 \ = \ $ { 0.5 3% }
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	$m_3 \ = \ $ { 3 3% }		$m_3 \ = \ $ { 3 3% }

	{Which statements are true for the signal  $s_4(t)$~~ ~~?		{Which statements are true for the signal  $s_4(t)$?
	\|type="[]"}		\|type="[]"}
	+ This is a case of "DSB–AM without a carrier".		+ This is a case of  "DSB–AM without carrier".
	- The modulation depth is  $m = 0$.		- The modulation depth is  $m = 0$.
	+ The modulation depth  $m$  is infinite.		+ The modulation depth  $m$  is infinite.


	{Let  $A_{\rm T} = A_{\rm N} = 1\ \rm V$~~, so~~  $m = 1$.  What is the spectrum  $S_+(f)$  of the analytical signal? ~~<br>~~Which Dirac weights occur at  $f_{\rm T}$  as well as at  $f_{\rm T}± f_{\rm N}$~~ ~~?		{Let  $A_{\rm T} = A_{\rm N} = 1\ \rm V$   ⇒   $m = 1$.  What is the spectrum  $S_+(f)$  of the analytical signal?  Which Dirac weights occur at  $f_{\rm T}$  as well as at  $f_{\rm T}± f_{\rm N}$?
	\|type="{}"}		\|type="{}"}
	$S_+(f_{\rm T}) \ = \ $ { 1 3% } $\ \text{V}$		$S_+(f_{\rm T}) \ = \ $ { 1 3% } $\ \text{V}$
	$S_+(f_{\rm T} ± f_{\rm N}) \ = \ $ { 0.5 3% } $\ \text{V}$		$S_+(f_{\rm T} ± f_{\rm N}) \ = \ $ { 0.5 3% } $\ \text{V}$

	{Now let  $m = 1$.  Which fraction  $P_{\rm T}/P_{\rm S}$  of the total transmission power  $P_{\rm S}$  is due to the carrier alone, and thus cannot be used for demodulation??		{Now let  $m = 1$.  Which fraction  $P_{\rm T}/P_{\rm S}$  of the total transmission power  $P_{\rm S}$  is due to the carrier alone,  and thus cannot be used for demodulation??
	\|type="{}"}		\|type="{}"}
	$P_{\rm T}/P_{\rm S} \ = \ $ { 0.667 3% }		$P_{\rm T}/P_{\rm S} \ = \ $ { 0.667 3% }

	{Generalize the result from   '''(4)'''  for an arbitrary modulation depth $m$. ~~<br>~~What are the power ratios for  $m = 0.5$,  $m = 3$  and  $m → ∞$ ?		{Generalize the result from   '''(4)'''  for an arbitrary modulation depth  $m$.  What are the power ratios for  $m = 0.5$,  $m = 3$  and  $m → ∞$?
	\|type="{}"}		\|type="{}"}
	$m = 0.5\text{:}\hspace{0.3cm} P_{\rm T}/P_{\rm S} \ = \ $ { 0.889 3% }		$m = 0.5\text{:}\hspace{0.3cm} P_{\rm T}/P_{\rm S} \ = \ $ { 0.889 3% }
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	$m → ∞ \text{:}\hspace{0.3cm} P_{\rm T}/P_{\rm S} \ = \ $ { 0. }		$m → ∞ \text{:}\hspace{0.3cm} P_{\rm T}/P_{\rm S} \ = \ $ { 0. }

	{Based on the calculations so far, which of the following ~~assessments~~ seem reasonable to you?		{Based on the calculations so far,  which of the following statements seem reasonable to you?
	\|type="[]"}		\|type="[]"}
	+ $m ≈ 1$  s more favorable than a small  $m$ for energy reasons.		+ $m ≈ 1$  is more favorable than a small  $m$  for energy reasons.
	+ The carrier is only useful for envelope demodulation.		+ The carrier is only useful for envelope demodulation.
	</quiz>		</quiz>