Aufgaben:Exercise 3.3Z: Characteristics Determination - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:54:11Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:54, 16 March 2026
Line 96:		Line 96:

	[[Category:Modulation Methods: Exercises\|^3.1 Phase Modulation^]]		[[Category:Modulation Methods: Exercises\|^3.1 Phase Modulation^]]
	[[de:Aufgaben:3.~~3Z_Kenngrößenbestimmung~~]]		[[de:Aufgaben:Aufgabe 3.3Z: Kenngrößenbestimmung]]

Maintenance script: Add German interlanguage link

2026-03-16T12:32:17Z

Add German interlanguage link

← Older revision		Revision as of 14:32, 16 March 2026
Line 96:		Line 96:

	[[Category:Modulation Methods: Exercises\|^3.1 Phase Modulation^]]		[[Category:Modulation Methods: Exercises\|^3.1 Phase Modulation^]]
			[[de:Aufgaben:3.3Z_Kenngrößenbestimmung]]

Reed at 15:03, 9 April 2022

2022-04-09T15:03:39Z

← Older revision		Revision as of 17:03, 9 April 2022
Line 87:		Line 87:
	'''(5)'''  The given equation can be rewritten as follows:		'''(5)'''  The given equation can be rewritten as follows:
	:$$\eta = \frac{2 \cdot {\rm J}_{1}{(\eta)}}{{\rm J}_{0}(\eta) + {\rm J}_{2}(\eta)} \hspace{0.05cm}.$$		:$$\eta = \frac{2 \cdot {\rm J}_{1}{(\eta)}}{{\rm J}_{0}(\eta) + {\rm J}_{2}(\eta)} \hspace{0.05cm}.$$
	*~~Mit~~  ${\rm J}_0(η) = 0.512$,  ${\rm J}_1(η) = 0.558$  ~~und~~  ${\rm J}_2(η) = 0.232$  ~~erhält man somit~~:		*With  ${\rm J}_0(η) = 0.512$,  ${\rm J}_1(η) = 0.558$  and  ${\rm J}_2(η) = 0.232$  we thus get:
	:$$ \eta = \frac{2 \cdot 0.558}{0.512 + 0.232}\hspace{0.15cm}\underline { = 1.5}\hspace{0.05cm}.$$		:$$ \eta = \frac{2 \cdot 0.558}{0.512 + 0.232}\hspace{0.15cm}\underline { = 1.5}\hspace{0.05cm}.$$

Reed: /* Solution */

2022-04-09T15:03:02Z

Solution

← Older revision		Revision as of 17:03, 9 April 2022
Line 80:		Line 80:
	'''(4)'''  When   $n = 1$   ⇒   $f = 3 \ \rm kHz$  as in question   '''(2)''', the phase is:		'''(4)'''  When   $n = 1$   ⇒   $f = 3 \ \rm kHz$  as in question   '''(2)''', the phase is:
	:$$ \phi_{\rm N} + 90^\circ = 60^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} = -30^\circ\hspace{0.05cm}.$$		:$$ \phi_{\rm N} + 90^\circ = 60^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} = -30^\circ\hspace{0.05cm}.$$
	*~~Checkin~~ this result when   $n = 2$   ⇒   $f = 6 \ \rm kHz$  as in question  '''(3)'''  yields the same value:		*Checking this result when   $n = 2$   ⇒   $f = 6 \ \rm kHz$  as in question  '''(3)'''  yields the same value:
	:$$ 2\cdot (\phi_{\rm N} + 90^\circ) = 120^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} \hspace{0.15cm}\underline {= -30^\circ}\hspace{0.05cm}.$$		:$$ 2\cdot (\phi_{\rm N} + 90^\circ) = 120^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} \hspace{0.15cm}\underline {= -30^\circ}\hspace{0.05cm}.$$

Reed at 14:51, 9 April 2022

2022-04-09T14:51:48Z

← Older revision		Revision as of 16:51, 9 April 2022
Line 1:		Line 1:

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

Guenter at 08:33, 30 March 2022

2022-03-30T08:33:36Z

← Older revision		Revision as of 10:33, 30 March 2022
Line 3:		Line 3:
	}}		}}

	[[File:P_ID1085__Mod_Z_3_3.png\|right\|frame\|~~Spektrum des analytischen Signals~~]]		[[File:P_ID1085__Mod_Z_3_3.png\|right\|frame\|Spectrum of the analytical signal]]
	Let us consider the phase modulation of the harmonic oscillation		Let us consider the phase modulation of the harmonic oscillation
	:$$q(t) = A_{\rm N} \cdot \cos(2 \pi \cdot f_{\rm N} \cdot t + \phi_{\rm N}) \hspace{0.05cm},$$		:$$q(t) = A_{\rm N} \cdot \cos(2 \pi \cdot f_{\rm N} \cdot t + \phi_{\rm N}) \hspace{0.05cm},$$

Reed: /* Solution */

2022-03-14T17:49:46Z

Solution

← Older revision		Revision as of 19:49, 14 March 2022
Line 62:		Line 62:
	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Bezüglich~~  $\|S_+(f)\|$  ~~gibt es eine Symmetrie zur Trägerfrequenz~~  $f_{\rm T}\hspace{0.15cm}\underline { = 40 \ \rm kHz}$.  ~~Der Abstand zwischen den Spektrallinien beträgt~~  $f_{\rm N}\hspace{0.15cm}\underline { = 3 \ \rm kHz}$.		'''(1)'''  Regarding   $\|S_+(f)\|$  there is a symmetry with respect to the carrier frequency  $f_{\rm T}\hspace{0.15cm}\underline { = 40 \ \rm kHz}$.  The distance between the spectral lines is  $f_{\rm N}\hspace{0.15cm}\underline { = 3 \ \rm kHz}$.



	'''(2)'''  ~~Unter Berücksichtigung von~~  $S_{\rm TP}(f = 3{\ \rm kHz}) = S_+(f = 43 \ \rm kHz)$  ~~gilt~~:		'''(2)'''  Considering  $S_{\rm TP}(f = 3{\ \rm kHz}) = S_+(f = 43 \ \rm kHz)$ , it holds that:
	:$$\|S_{\rm TP}(f = 3\,{\rm kHz})\| = \sqrt{0.279^2 + 0.483^2} \hspace{0.15cm}\underline {= 0.558}\hspace{0.05cm},$$		:$$\|S_{\rm TP}(f = 3\,{\rm kHz})\| = \sqrt{0.279^2 + 0.483^2} \hspace{0.15cm}\underline {= 0.558}\hspace{0.05cm},$$
	:$$ {\rm arc}\hspace{0.15cm} S_{\rm TP}(f = 3\,{\rm kHz}) = \arctan \frac{0.483}{0.279} = \arctan 1.732\hspace{0.15cm}\underline { = 60^\circ} \hspace{0.05cm}.$$		:$$ {\rm arc}\hspace{0.15cm} S_{\rm TP}(f = 3\,{\rm kHz}) = \arctan \frac{0.483}{0.279} = \arctan 1.732\hspace{0.15cm}\underline { = 60^\circ} \hspace{0.05cm}.$$
Line 72:		Line 72:


	'''(3)'''  ~~In analoger Weise zur Teilaufgabe~~  '''(2)'''  ~~erhält man für die Frequenz~~  $f = 6 \ \rm kHz$:		'''(3)'''  Analogously to in question  '''(2)''' , at frequency   $f = 6 \ \rm kHz$ we obtain:
	:$$\|S_{\rm TP}(f = 6\,{\rm kHz})\| = \sqrt{(-0.116)^2 + 0.201^2} \hspace{0.15cm}\underline {= 0.232}\hspace{0.05cm},$$		:$$\|S_{\rm TP}(f = 6\,{\rm kHz})\| = \sqrt{(-0.116)^2 + 0.201^2} \hspace{0.15cm}\underline {= 0.232}\hspace{0.05cm},$$
	:$${\rm arc}\hspace{0.15cm} S_{\rm TP}(f = 6\,{\rm kHz}) = \arctan \frac{-0.116}{0.201} = 180^\circ - \arctan 1.732 \hspace{0.15cm}\underline {= 120^\circ} \hspace{0.05cm}.$$		:$${\rm arc}\hspace{0.15cm} S_{\rm TP}(f = 6\,{\rm kHz}) = \arctan \frac{-0.116}{0.201} = 180^\circ - \arctan 1.732 \hspace{0.15cm}\underline {= 120^\circ} \hspace{0.05cm}.$$
Line 78:		Line 78:


	'''(4)'''  ~~Die Phase lautet für~~  $n = 1$   ⇒   $f = 3 \ \rm kHz$  ~~entsprechend Teilaufgabe~~  '''(2)''':		'''(4)'''  When   $n = 1$   ⇒   $f = 3 \ \rm kHz$  as in question   '''(2)''', the phase is:
	:$$ \phi_{\rm N} + 90^\circ = 60^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} = -30^\circ\hspace{0.05cm}.$$		:$$ \phi_{\rm N} + 90^\circ = 60^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} = -30^\circ\hspace{0.05cm}.$$
	*~~Die Überprüfung dieses Ergebnisses mit~~  $n = 2$   ⇒   $f = 6 \ \rm kHz$  ~~entsprechend Teilaufgabe~~  '''(3)'''  ~~liefert den gleichen Wert~~:		*Checkin this result when   $n = 2$   ⇒   $f = 6 \ \rm kHz$  as in question  '''(3)'''  yields the same value:
	:$$ 2\cdot (\phi_{\rm N} + 90^\circ) = 120^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} \hspace{0.15cm}\underline {= -30^\circ}\hspace{0.05cm}.$$		:$$ 2\cdot (\phi_{\rm N} + 90^\circ) = 120^\circ \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi_{\rm N} \hspace{0.15cm}\underline {= -30^\circ}\hspace{0.05cm}.$$



	'''(5)'''  ~~Die angegebene Gleichung kann wie folgt umgeformt werden~~:		'''(5)'''  The given equation can be rewritten as follows:
	:$$\eta = \frac{2 \cdot {\rm J}_{1}{(\eta)}}{{\rm J}_{0}(\eta) + {\rm J}_{2}(\eta)} \hspace{0.05cm}.$$		:$$\eta = \frac{2 \cdot {\rm J}_{1}{(\eta)}}{{\rm J}_{0}(\eta) + {\rm J}_{2}(\eta)} \hspace{0.05cm}.$$
	*Mit  ${\rm J}_0(η) = 0.512$,  ${\rm J}_1(η) = 0.558$  und  ${\rm J}_2(η) = 0.232$  erhält man somit:		*Mit  ${\rm J}_0(η) = 0.512$,  ${\rm J}_1(η) = 0.558$  und  ${\rm J}_2(η) = 0.232$  erhält man somit:

Reed at 17:44, 14 March 2022

2022-03-14T17:44:52Z

← Older revision		Revision as of 19:44, 14 March 2022
Line 60:		Line 60:
	</quiz>		</quiz>

	===~~Musterlösung~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Bezüglich  $\|S_+(f)\|$  gibt es eine Symmetrie zur Trägerfrequenz  $f_{\rm T}\hspace{0.15cm}\underline { = 40 \ \rm kHz}$.  Der Abstand zwischen den Spektrallinien beträgt  $f_{\rm N}\hspace{0.15cm}\underline { = 3 \ \rm kHz}$.		'''(1)'''  Bezüglich  $\|S_+(f)\|$  gibt es eine Symmetrie zur Trägerfrequenz  $f_{\rm T}\hspace{0.15cm}\underline { = 40 \ \rm kHz}$.  Der Abstand zwischen den Spektrallinien beträgt  $f_{\rm N}\hspace{0.15cm}\underline { = 3 \ \rm kHz}$.

Reed: /* Questions */

2022-03-14T17:44:29Z

Questions

← Older revision		Revision as of 19:44, 14 March 2022
Line 33:		Line 33:

	<quiz display=simple>		<quiz display=simple>
	{~~Wie groß sind die Frequenzen~~  $f_{\rm T}$  ~~und~~  $f_{\rm N}$?		{What are the frequencies  $f_{\rm T}$  and  $f_{\rm N}$?
	\|type="{}"}		\|type="{}"}
	$f_{\rm T} \ = \ $ { 40 3% } $\ \rm kHz$		$f_{\rm T} \ = \ $ { 40 3% } $\ \rm kHz$
	$f_{\rm N} \ = \ $ { 3 3% } $\ \rm kHz$		$f_{\rm N} \ = \ $ { 3 3% } $\ \rm kHz$

	{~~Berechnen Sie den Betrag und die Phase von~~  $S_{\rm TP}(f = 3 \ \rm kHz)$.		{Calculate the magnitude and the phase of  $S_{\rm TP}(f = 3 \ \rm kHz)$.
	\|type="{}"}		\|type="{}"}
	$\|S_{\rm TP}(f = 3 \ \rm kHz)\| \ = \ $ { 0.558 3% }		$\|S_{\rm TP}(f = 3 \ \rm kHz)\| \ = \ $ { 0.558 3% }
	${\rm arc} \ S_{\rm TP}(f = 3\ \rm kHz) \ = \ $ { 60 3% } $\ \rm Grad$		${\rm arc} \ S_{\rm TP}(f = 3\ \rm kHz) \ = \ $ { 60 3% } $\ \rm Grad$

	{~~Berechnen Sie den Betrag und die Phase von~~  $S_{\rm TP}(f = 6 \ \rm kHz)$.		{Calculate the magnitude and the phase of  $S_{\rm TP}(f = 6 \ \rm kHz)$.
	\|type="{}"}		\|type="{}"}
	$\|S_{\rm TP}(f = 6 \ \rm kHz)\| \ = \ $ { 0.232 3% }		$\|S_{\rm TP}(f = 6 \ \rm kHz)\| \ = \ $ { 0.232 3% }
	${\rm arc} \ S_{\rm TP}(f = 6\ \rm kHz) \ = \ $ { 120 3% } $\ \rm Grad$		${\rm arc} \ S_{\rm TP}(f = 6\ \rm kHz) \ = \ $ { 120 3% } $\ \rm Grad$

	{~~Wie groß ist die Phase des Quellensignals~~  $q(t)$?		{What is the phase of the source signal  $q(t)$?
	\|type="{}"}		\|type="{}"}
	$ϕ_{\rm N} \ = \ $ { -30.9--29.1 } $\ \rm Grad$		$ϕ_{\rm N} \ = \ $ { -30.9--29.1 } $\ \rm Grad$

	{~~Wie groß ist der Modulationsindex~~  $η$ ?		{What is the magnitude of the modulation index  $η$ ?
	\|type="{}"}		\|type="{}"}
	$η \ = \ $ { 1.5 3% }		$η \ = \ $ { 1.5 3% }

Reed at 17:15, 14 March 2022

2022-03-14T17:15:21Z

← Older revision		Revision as of 19:15, 14 March 2022
Line 24:		Line 24:
	''Hints:''		''Hints:''
	*This exercise belongs to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)\|Phase Modulation]].		*This exercise belongs to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)\|Phase Modulation]].
	*Particular reference is made to the page  [[Modulation_Methods/~~Phasenmodulation_~~(PM)#~~.C3.84quivalentes_TP.E2.80.93Signal_bei_Phasenmodulation~~\|~~Äquivalentes Tiefpass~~-~~Signal bei Phasenmodulation~~]].		*Particular reference is made to the page  [[Modulation_Methods/Phase_Modulation_(PM)#Equivalent_low-pass_signal_in_phase_modulation\|Equivalent low-pass signal in phase modulation]].

	*For the calculation of the modulation index, you can take advantage of the following property of the Bessel function:		*For the calculation of the modulation index, you can take advantage of the following property of the Bessel function: