Aufgaben:Exercise 3.1: Phase Modulation Locus Curve - Revision history

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	[[Category:Modulation Methods: Exercises\|^3.1 Phase Modulation^]]		[[Category:Modulation Methods: Exercises\|^3.1 Phase Modulation^]]
	[[de:Aufgaben:3.~~1_Ortskurve_bei_Phasenmodulation~~]]		[[de:Aufgaben:Aufgabe 3.1: Ortskurve bei Phasenmodulation]]

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	Let the source signal be the same for both modulators:		Let the source signal be the same for both modulators:
	$$ q(t) = A_{\rm N} \cdot \cos(2 \pi f_{\rm N} \cdot t),\hspace{1cm}		$$ q(t) = A_{\rm N} \cdot \cos(2 \pi f_{\rm N} \cdot t),\hspace{1cm}{\rm with}\hspace{0.2cm} A_{\rm N} = 2\,{\rm V},\hspace{0.2cm}f_{\rm N} = 5\,{\rm kHz}\hspace{0.05cm}.$$
	{\rm with}\hspace{0.2cm} A_{\rm N} = 2\,{\rm V},\hspace{0.2cm}f_{\rm N} = 5\,{\rm kHz}\hspace{0.05cm}.$$
	One of the two modulators implements phase modulation, which is characterized by the following equations:		One of the two modulators implements phase modulation, which is characterized by the following equations:
	:$$ s(t) = A_{\rm T} \cdot \cos \hspace{-0.1cm} \big[\omega_{\rm T} \cdot t + \phi(t) \big]\hspace{0.05cm},$$		:$$ s(t) = A_{\rm T} \cdot \cos \hspace{-0.1cm} \big[\omega_{\rm T} \cdot t + \phi(t) \big]\hspace{0.05cm},$$

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	[[Category:Modulation Methods: Exercises\|^3.1 Phase Modulation^]]		[[Category:Modulation Methods: Exercises\|^3.1 Phase Modulation^]]
			[[de:Aufgaben:3.1_Ortskurve_bei_Phasenmodulation]]

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	*After a quarter of the period  $T_{\rm N} = 1/f_{\rm N} = 200 \ \rm µ s$ ,  $ϕ(t) = 0$  and  $s_{\rm TP}(t) = 1 \, \rm V$.		*After a quarter of the period  $T_{\rm N} = 1/f_{\rm N} = 200 \ \rm µ s$ ,  $ϕ(t) = 0$  and  $s_{\rm TP}(t) = 1 \, \rm V$.
	*At time  $t_1 = T_{\rm N}/2\hspace{0.15cm}\underline { = 100 \ \rm µ s}$ ,  $ϕ(t_1) = -π$  and  $s_{\rm TP}(t_1) = -1 \, \rm V$.		*At time  $t_1 = T_{\rm N}/2\hspace{0.15cm}\underline { = 100 \ \rm µ s}$ ,  $ϕ(t_1) = -π$  and  $s_{\rm TP}(t_1) = -1 \, \rm V$.
	*Afterwards, ~~move~~ counterclockwise along the arc.		*Afterwards, one moves counterclockwise along the arc.

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	{{quiz-Header\|Buchseite=~~Modulatio_Methods~~/Phase_Modulation_(PM)		{{quiz-Header\|Buchseite=Modulation_Methods/Phase_Modulation_(PM)
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	{{quiz-Header\|Buchseite=~~Modulationsverfahren~~/~~Phase Modulation~~ (PM)		{{quiz-Header\|Buchseite=Modulatio_Methods/Phase_Modulation_(PM)
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	{{quiz-Header\|Buchseite=Modulationsverfahren/~~Phasenmodulation~~ (PM)		{{quiz-Header\|Buchseite=Modulationsverfahren/Phase Modulation (PM)
	}}		}}

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	'''(5)'''  One moves clockwise along the circular arc:		'''(5)'''  One moves clockwise along the circular arc.
	*After a quarter of the period  $T_{\rm N} = 1/f_{\rm N} = 200 \ \rm µ s$ ,  $ϕ(t) = 0$  and  $s_{\rm TP}(t) = 1 \, \rm V$.		*After a quarter of the period  $T_{\rm N} = 1/f_{\rm N} = 200 \ \rm µ s$ ,  $ϕ(t) = 0$  and  $s_{\rm TP}(t) = 1 \, \rm V$.
	*At time  $t_1 = T_{\rm N}/2\hspace{0.15cm}\underline { = 100 \ \rm µ s}$ ,  $ϕ(t_1) = -π$  and  $s_{\rm TP}(t_1) = -1 \, \rm V$.		*At time  $t_1 = T_{\rm N}/2\hspace{0.15cm}\underline { = 100 \ \rm µ s}$ ,  $ϕ(t_1) = -π$  and  $s_{\rm TP}(t_1) = -1 \, \rm V$.

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Solution

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	'''(2)'''  ~~Hier handelt es sich um die Phasenmodulation~~   ⇒   <u>~~Antwort~~ 3</u>:		'''(2)'''  Here, we observe phase modulation   ⇒   <u>Answer 3</u>:
	*~~Die Einhüllende~~  $a(t) = A_{\rm T}$  ~~ist konstant~~,		*The envelope   $a(t) = A_{\rm T}$  is constant,
	*~~während die Phase~~  $ϕ(t)$  ~~entsprechend dem Quellensignal~~  $q(t)$  ~~cosinusförmig verläuft~~.		*while the phase  $ϕ(t)$  is cosinusoidal according to the source signal  $q(t)$ .




	'''(3)'''  ~~Bei der Phasenmodulation gilt~~:		'''(3)'''  In the case of phase modulation:
	:$$s_{\rm TP}(t) = A_{\rm T} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\phi(t) }\hspace{0.05cm}.$$		:$$s_{\rm TP}(t) = A_{\rm T} \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\phi(t) }\hspace{0.05cm}.$$
	*~~Aus der Grafik kann man die Trägeramplitude~~  $A_{\rm T}\hspace{0.15cm}\underline{ = 1 \ \rm V}$  ~~als den Kreisradius ablesen~~.		*From the graph, we can read the carrier amplitude   $A_{\rm T}\hspace{0.15cm}\underline{ = 1 \ \rm V}$  as the radius of the circle.




	'''(4)'''  ~~Das Quellensignal~~  $q(t)$  ~~ist zum Zeitpunkt~~  $t = 0$  ~~maximal und damit auch die Phasenfunktion~~:		'''(4)'''  The source signal  $q(t)$  is at its maximum at time  $t = 0$  and therefore so is the phase function:
	:$$ \eta = \phi_{\rm max} = \phi( t =0) = \pi\hspace{0.15cm}\underline { = 3.1415} \hspace{0.05cm}.$$		:$$ \eta = \phi_{\rm max} = \phi( t =0) = \pi\hspace{0.15cm}\underline { = 3.1415} \hspace{0.05cm}.$$
	*~~Daraus erhält man für die Modulatorkonstante~~:		*This gives the modulator constant:
	$$K_{\rm PM} = \frac{\eta}{A_{\rm N}} = \frac{\pi}{2\,{\rm V}}\hspace{0.15cm}\underline {= 1.571\,{\rm V}^{-1}}\hspace{0.05cm}.$$		$$K_{\rm PM} = \frac{\eta}{A_{\rm N}} = \frac{\pi}{2\,{\rm V}}\hspace{0.15cm}\underline {= 1.571\,{\rm V}^{-1}}\hspace{0.05cm}.$$

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	'''(5)'''  ~~Man bewegt sich auf dem Kreis(bogen) im Uhrzeigersinn.~~		'''(5)'''  One moves clockwise along the circular arc:
	*~~Nach einem Viertel der Periodendauer~~  $T_{\rm N} = 1/f_{\rm N} = 200 \ \rm µ s$  ~~ist~~  $ϕ(t) = 0$  ~~und~~  $s_{\rm TP}(t) = 1 \, \rm V$.		*After a quarter of the period  $T_{\rm N} = 1/f_{\rm N} = 200 \ \rm µ s$ ,  $ϕ(t) = 0$  and  $s_{\rm TP}(t) = 1 \, \rm V$.
	*~~Zur Zeit~~  $t_1 = T_{\rm N}/2\hspace{0.15cm}\underline { = 100 \ \rm µ s}$  ~~gilt~~  $ϕ(t_1) = -π$  ~~und~~  $s_{\rm TP}(t_1) = -1 \, \rm V$.		*At time  $t_1 = T_{\rm N}/2\hspace{0.15cm}\underline { = 100 \ \rm µ s}$ ,  $ϕ(t_1) = -π$  and  $s_{\rm TP}(t_1) = -1 \, \rm V$.
	*~~Danach bewegt man sich auf dem Kreisbogen entgegen dem Uhrzeigersinn~~.		*Afterwards, move counterclockwise along the arc.

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Solution

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	===Solution===		===Solution===
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	'''(1)'''  ~~Es handelt sich um eine ESB–AM mit dem Seitenband–zu–Träger–Verhältnis ~~ $μ = 1$   ⇒   <u>~~Antwort~~ 2</u>:		'''(1)'''  We are dealing with SSB-AM with a sideband-to-carrier ratio $μ = 1$   ⇒   <u>Answer 2</u>:
	*~~Bewegt man sich auf dem Kreis~~ in ~~mathematisch~~ positive ~~Richtung~~, ~~so liegt speziell eine OSB–AM vor~~, ~~andernfalls eine USB–AM~~.		*If one moves in the mathematically positive direction on the circle, it is specifically an USB–AM, otherwise it is a LSB–AM.
	*~~Die Phasenfunktion~~  $ϕ(t)$  ~~als der Winkel eines Punktes~~  $s_{\rm TP}(t)$  ~~auf dem Kreis~~(~~bogen~~) ~~bezogen auf den Koordinatenursprung kann Werte zwischen~~  $±π/2$  ~~annehmen und zeigt keinen Cosinusverlauf~~.		*The phase function  $ϕ(t)$  as the angle of a point  $s_{\rm TP}(t)$  on the circle (arc) with respect to the coordinate origin can take values between  $±π/2$  and does not show a cosine progression.
	*~~Aber auch die Hüllkurve~~  $a(t) = \|s_{\rm TP}(t)\|$  ~~ist nicht cosinusförmig~~.		*The envelope   $a(t) = \|s_{\rm TP}(t)\|$  is also not cosine.
	*~~Würde man beim Empfänger für~~  $\rm M_1$  ~~einen Hüllkurvendemodulator einsetzen~~, ~~so käme es zu nichtlinearen Verzerrungen im Gegensatz zur ZSB–AM~~, ~~deren Ortskurve eine horizontale Gerade ist~~.		*If an envelope demodulator were used for  $\rm M_1$  at the receiver, nonlinear distortions would occur, in contrast to DSB–AM, which has a horizontal straight line for a locus curve.