Aufgaben:Exercise 2.11: Envelope Demodulation of an SSB Signal - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

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	[[Category:Modulation Methods: Exercises\|^2.4 Single Sideband Amplitude Modulation^]]		[[Category:Modulation Methods: Exercises\|^2.4 Single Sideband Amplitude Modulation^]]
	[[de:Aufgaben:2.~~10_Verzerrungen_durch_ESB/HKD~~]]		[[de:Aufgaben:Aufgabe 2.11: Hüllkurvendemodulation eines ESB-Signals]]

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	[[Category:Modulation Methods: Exercises\|^2.4 Single Sideband Amplitude Modulation^]]		[[Category:Modulation Methods: Exercises\|^2.4 Single Sideband Amplitude Modulation^]]
			[[de:Aufgaben:2.10_Verzerrungen_durch_ESB/HKD]]

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	'''(5)'''  A smaller sideband-to-carrier ratio also results in a smaller distortion factor:		'''(5)'''  A smaller sideband-to-carrier ratio also results in a smaller distortion factor:
	:$$K = \frac{0.058{\,\rm V}}{0.484\,{\rm V}}\hspace{0.15cm}\underline { \approx 12 \%}.$$		:$$K = \frac{0.058{\,\rm V}}{0.484\,{\rm V}}\hspace{0.15cm}\underline { \approx 12 \%}.$$
	*The simple approximation   $K ≈ μ/4$  yields ~~here~~  $12.5\%$.		*The simple approximation   $K ≈ μ/4$  here yields  $12.5\%$.
	*It can be concluded that the above rule of thumb is more accurate for smaller values of   $μ$.		*It can be concluded that the above rule of thumb is more accurate for smaller values of   $μ$.

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The maximum value   $a_{\rm max} = 2\ \rm V$  and the minimum value   $a_{\rm min} = 0$  can be read of the graph or calculated using the equation given:		'''(1)'''  The maximum value   $a_{\rm max} = 2\ \rm V$  and the minimum value   $a_{\rm min} = 0$  can be read off the graph or calculated using the equation given:
	:$$ a_{\rm max} = A_{\rm T} \cdot \sqrt{1+ \mu^2 + 2 \mu}= A_{\rm T} \cdot (1+ \mu) = 2\,{\rm V} \hspace{0.05cm},$$		:$$ a_{\rm max} = A_{\rm T} \cdot \sqrt{1+ \mu^2 + 2 \mu}= A_{\rm T} \cdot (1+ \mu) = 2\,{\rm V} \hspace{0.05cm},$$
	:$$a_{\rm min} = A_{\rm T} \cdot \sqrt{1+ \mu^2 - 2 \mu}= A_{\rm T} \cdot (1- \mu) = 0 \hspace{0.05cm}.$$		:$$a_{\rm min} = A_{\rm T} \cdot \sqrt{1+ \mu^2 - 2 \mu}= A_{\rm T} \cdot (1- \mu) = 0 \hspace{0.05cm}.$$

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	#a DSB-AM which has been extremely affected by the channel,		#a DSB-AM which has been extremely affected by the channel,

	:~~simultaneously provides~~  $K_{\rm max} ≈ μ/4$  an upper bound for the DSB-AM.		:  $K_{\rm max} ≈ μ/4$  simultaneously gives an upper bound for the DSB-AM.

	*A comparison of the parameters  $m = A_{\rm N}/A_{\rm T}$  and  $μ = A_{\rm N}/(2A_{\rm T})$  leads to the result:		*A comparison of the parameters  $m = A_{\rm N}/A_{\rm T}$  and  $μ = A_{\rm N}/(2A_{\rm T})$  leads to the result:

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	$K \ = \ $ { 22.3 3% } $\ \text{%}$		$K \ = \ $ { 22.3 3% } $\ \text{%}$

	{How can you recognize the nonlinear distortions in the ~~present~~ signal  $v(t)$?		{How can you recognize the nonlinear distortions in the signal  $v(t)$?
	\|type="[]"}		\|type="[]"}
	+ The lower cosine half-wave is more peaked than the upper one.		+ The lower cosine half-wave is more peaked than the upper one.

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	{{quiz-Header\|Buchseite=~~Modulationsverfahren~~/~~Einseitenbandmodulation~~		{{quiz-Header\|Buchseite=Modulation_Methods/Single-Sideband_Modulation
	}}		}}

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	The channel is assumed to be ideal such that the received signal  $r(t)$  is identical to the transmitted signal  $s(t)$ .  With the sideband-to-carrier ratio		The channel is assumed to be ideal such that the received signal  $r(t)$  is identical to the transmitted signal  $s(t)$ .  With the sideband-to-carrier ratio
	:$$ \mu = \frac{A_{\rm N}}{2 \cdot A_{\rm T}}$$		:$$ \mu = \frac{A_{\rm N}}{2 \cdot A_{\rm T}}$$
	the equivalent low-pass signal  (German:  "äquivalentes Tiefpass-Signal"   ⇒   subscript:  "TP") can be written as:		the equivalent low-pass signal  (German:  "äquivalentes Tiefpass-Signal"   ⇒   subscript:  "TP")  can be written as:
	:$$r_{\rm TP}(t) = A_{\rm T} \cdot \left( 1 + \mu \cdot {\rm e}^{{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\omega_{\rm N}\cdot \hspace{0.03cm}\hspace{0.03cm}t} \right) \hspace{0.05cm}$$		:$$r_{\rm TP}(t) = A_{\rm T} \cdot \left( 1 + \mu \cdot {\rm e}^{{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm}\omega_{\rm N}\cdot \hspace{0.03cm}\hspace{0.03cm}t} \right) \hspace{0.05cm}$$

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	'''(6)'''  Thus,  the distortion factor is largest when one of the sidebands is entirely cut out.		'''(6)'''  Thus,  the distortion factor is largest when one of the sidebands is entirely cut out.
	*However,  since the envelope demodulator has no information to distinguish between		*However,  since the envelope demodulator has no information to distinguish between
	**a SSB–AM, or		#a SSB–AM, or
	**a DSB-AM which has been extremely affected by the channel,		#a DSB-AM which has been extremely affected by the channel,

			:simultaneously provides  $K_{\rm max} ≈ μ/4$  an upper bound for the DSB-AM.
	~~⇒~~   $K_{\rm max} ≈ μ/4$  ~~simultaneously provides~~ an upper bound for the DSB-AM.

	*A comparison of the parameters  $m = A_{\rm N}/A_{\rm T}$  and  $μ = A_{\rm N}/(2A_{\rm T})$  leads to the result:		*A comparison of the parameters  $m = A_{\rm N}/A_{\rm T}$  and  $μ = A_{\rm N}/(2A_{\rm T})$  leads to the result: