Aufgaben:Exercise 4.6: Locality Curve for SSB-AM - Revision history

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← Older revision		Revision as of 15:48, 16 March 2026
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	We consider the analytical signal  $s_+(t)$  with the spectral function		We consider the analytical signal  $s_+(t)$  with the spectral function

	:$$S_{\rm +}(f) = 1 \cdot \delta (f - f_{\rm 50})- {\rm j} \cdot		:$$S_{\rm +}(f) = 1 \cdot \delta (f - f_{\rm 50})- {\rm j} \cdot\delta (f - f_{\rm 60}) .$$
	\delta (f - f_{\rm 60}) .$$

	Here  $f_{50}$  and  $f_{60}$  are abbreviations for the frequencies  $50 \ \rm kHz$  and  $60 \ \rm kHz$ respectively.		Here  $f_{50}$  and  $f_{60}$  are abbreviations for the frequencies  $50 \ \rm kHz$  and  $60 \ \rm kHz$ respectively.
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	'''(1)'''  The <u>proposed solution 2</u> is correct:		'''(1)'''  The <u>proposed solution 2</u> is correct:
	*The spectrum of the equivalent low-pass signal with carrier frequency  $f_{\rm T} = f_{50} = 50 \ \text{kHz}$:		*The spectrum of the equivalent low-pass signal with carrier frequency  $f_{\rm T} = f_{50} = 50 \ \text{kHz}$:
	:$$S_{\rm TP}(f ) = S_{\rm +}(f+ f_{\rm 50}) = 1 \cdot \delta (f)-		:$$S_{\rm TP}(f ) = S_{\rm +}(f+ f_{\rm 50}) = 1 \cdot \delta (f)-{\rm j} \cdot \delta (f - f_{\rm 10}) .$$
	{\rm j} \cdot \delta (f - f_{\rm 10}) .$$
	*This gives for the associated time signal:		*This gives for the associated time signal:
	:$$s_{\rm TP}(t) = {\rm 1 } - {\rm j} \cdot {\rm e}^{{\rm		:$$s_{\rm TP}(t) = {\rm 1 } - {\rm j} \cdot {\rm e}^{{\rmj}\hspace{0.05cm} \omega_{\rm 10} \hspace{0.05cm} t }.$$
	j}\hspace{0.05cm} \omega_{\rm 10} \hspace{0.05cm} t }.$$
	*Starting from the point  $(1, –{\rm j})$   ⇒   $s_{\rm TP}(t)$  runs on a circle with centre  $(1, 0)$  and radius  $1$.		*Starting from the point  $(1, –{\rm j})$   ⇒   $s_{\rm TP}(t)$  runs on a circle with centre  $(1, 0)$  and radius  $1$.
	*The period duration is equal to the reciprocal of the frequency:   $T_0 = 1/f_{10} = 100 \ µ \text{s}$.		*The period duration is equal to the reciprocal of the frequency:   $T_0 = 1/f_{10} = 100 \ µ \text{s}$.
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	'''(2)'''  Splitting the above equation into real and imaginary parts, we get:		'''(2)'''  Splitting the above equation into real and imaginary parts, we get:

	:$$s_{\rm TP}(t) = {\rm 1 } + \sin({ \omega_{\rm 10} \hspace{0.05cm} t })		:$$s_{\rm TP}(t) = {\rm 1 } + \sin({ \omega_{\rm 10} \hspace{0.05cm} t })-{\rm j}\cdot \cos({ \omega_{\rm 10} \hspace{0.05cm} t }).$$
	-{\rm j}\cdot \cos({ \omega_{\rm 10} \hspace{0.05cm} t }).$$

	*This leads to the magnitude function		*This leads to the magnitude function

	:$$a(t)= \|s_{\rm TP}(t)\|=\sqrt{{\rm Re}\left[s_{\rm		:$$a(t)= \|s_{\rm TP}(t)\|=\sqrt{{\rm Re}\left[s_{\rmTP}(t)\right]^2 + {\rm Im}\left[s_{\rm TP}(t)\right]^2 }=\sqrt{1 + 2 \sin(\omega_{\rm 10}\hspace{0.05cm} t)+\sin^2(\omega_{\rm 10}\hspace{0.05cm} t)+ \cos^2(\omega_{\rm10}\hspace{0.05cm} t)}= \sqrt{2 \cdot ( 1 + \sin(\omega_{\rm 10}\hspace{0.05cm} t))}.$$
	TP}(t)\right]^2 + {\rm Im}\left[s_{\rm TP}(t)\right]^2 }=
	\sqrt{1 + 2 \sin(\omega_{\rm 10}\hspace{0.05cm} t)+
	\sin^2(\omega_{\rm 10}\hspace{0.05cm} t)+ \cos^2(\omega_{\rm
	10}\hspace{0.05cm} t)}
	= \sqrt{2 \cdot ( 1 + \sin(\omega_{\rm 10}\hspace{0.05cm} t))}.$$

	*For the minimum value, considering  $\sin(\omega_{10} \cdot t) \geq -1$     ⇒    $a_{\text{min}} \; \underline{= 0}$.		*For the minimum value, considering  $\sin(\omega_{10} \cdot t) \geq -1$     ⇒    $a_{\text{min}} \; \underline{= 0}$.
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	'''(3)'''  According to the general definition:		'''(3)'''  According to the general definition:

	:$$\phi(t)= {\rm arctan} \hspace{0.1cm}\frac{{\rm Im}\left[s_{\rm		:$$\phi(t)= {\rm arctan} \hspace{0.1cm}\frac{{\rm Im}\left[s_{\rmTP}(t)\right]}{{\rm Re}\left[s_{\rm TP}(t)\right]}= {\rm arctan}\hspace{0.1cm}\frac{-\cos(\omega_{\rm 10}\hspace{0.05cm} t)}{1 +\sin(\omega_{\rm 10}\hspace{0.05cm} t)}.$$
	TP}(t)\right]}{{\rm Re}\left[s_{\rm TP}(t)\right]}= {\rm arctan}
	\hspace{0.1cm}\frac{-\cos(\omega_{\rm 10}\hspace{0.05cm} t)}{1 +
	\sin(\omega_{\rm 10}\hspace{0.05cm} t)}.$$

	*For  $t = 0$  ⇒   $\cos( \omega_{10} \cdot t ) = 1$  and  $\sin( \omega_{10} \cdot t ) = 0$.  It follows:		*For  $t = 0$  ⇒   $\cos( \omega_{10} \cdot t ) = 1$  and  $\sin( \omega_{10} \cdot t ) = 0$.  It follows:
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	*On the other hand, for  $t = T_0/4 =25 \ µ \text{s}$:		*On the other hand, for  $t = T_0/4 =25 \ µ \text{s}$:

	:$$\cos(\omega_{\rm 10}\hspace{0.05cm} t) = 0;		:$$\cos(\omega_{\rm 10}\hspace{0.05cm} t) = 0;\hspace{0.2cm}\sin(\omega_{\rm 10}\hspace{0.05cm} t) = 1\hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi(t = {\rm 25\hspace{0.05cm} µ s}) \hspace{0.15 cm}\underline{= 0}.$$
	\hspace{0.2cm}\sin(\omega_{\rm 10}\hspace{0.05cm} t) = 1
	\hspace{0.3cm} \Rightarrow \hspace{0.3cm}\phi(t = {\rm 25
	\hspace{0.05cm} µ s}) \hspace{0.15 cm}\underline{= 0}.$$

	*The two angles calculated so far can also be read from the graph above.		*The two angles calculated so far can also be read from the graph above.
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	*If one calculates the phase function for  $t =74 \ {\rm µ} \text{s}$, one obtains with  $\omega_{10} \cdot t = 1.48 \cdot \pi \; \Rightarrow \; 266.4^\circ$:		*If one calculates the phase function for  $t =74 \ {\rm µ} \text{s}$, one obtains with  $\omega_{10} \cdot t = 1.48 \cdot \pi \; \Rightarrow \; 266.4^\circ$:

	:$$\phi(t = {\rm 74 \hspace{0.05cm} {\rm µ} s})= {\rm arctan}		:$$\phi(t = {\rm 74 \hspace{0.05cm} {\rm µ} s})= {\rm arctan}\hspace{0.1cm}\frac{\cos(86.4^\circ)}{1 - \sin(86.4^\circ)} = {\rmarctan} \hspace{0.1cm}\frac{0.062}{1 - 0.998} \approx {\rmarctan}(31)\approx 88^\circ.$$
	\hspace{0.1cm}\frac{\cos(86.4^\circ)}{1 - \sin(86.4^\circ)} = {\rm
	~~arctan~~} \hspace{0.1cm}\frac{0.062}{1 - 0.998} \approx {\rm
	~~arctan~~}(31)\approx 88^\circ.$$

	*Accordingly, for  $t =76 \ {\rm µ} \text{s}$  with  $\omega_{10} \cdot t = 1.52 \cdot \pi \; \Rightarrow \; 273.6^\circ$ :		*Accordingly, for  $t =76 \ {\rm µ} \text{s}$  with  $\omega_{10} \cdot t = 1.52 \cdot \pi \; \Rightarrow \; 273.6^\circ$ :

	:$$\phi(t = {\rm 76 \hspace{0.05cm} {\rm µ} s})= {\rm arctan}		:$$\phi(t = {\rm 76 \hspace{0.05cm} {\rm µ} s})= {\rm arctan}\hspace{0.1cm}\frac{-\cos(86.4^\circ)}{1 - \sin(86.4^\circ)}\approx {\rm arctan}(-31)\approx -88^\circ.$$
	\hspace{0.1cm}\frac{-\cos(86.4^\circ)}{1 - \sin(86.4^\circ)}
	\approx {\rm arctan}(-31)\approx -88^\circ.$$

	*The numerical values suggest that the limit values for  $t \; \rightarrow \; 75 \ {\rm µ} \text{s}$  result in  $\pm 90^\circ$  depending on whether this value is approached from above or below.		*The numerical values suggest that the limit values for  $t \; \rightarrow \; 75 \ {\rm µ} \text{s}$  result in  $\pm 90^\circ$  depending on whether this value is approached from above or below.
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	*With the carrier frequency $f_{\rm T} = f_{60} = 60 \ \text{ kHz}$  the equations for frequency and time domain are:		*With the carrier frequency $f_{\rm T} = f_{60} = 60 \ \text{ kHz}$  the equations for frequency and time domain are:

	:$$S_{\rm TP}(f ) = S_{\rm +}(f+ f_{\rm 60}) = -{\rm j} \cdot \delta		:$$S_{\rm TP}(f ) = S_{\rm +}(f+ f_{\rm 60}) = -{\rm j} \cdot \delta(f) + \delta (f + f_{\rm 10}) ;$$
	(f) + \delta (f + f_{\rm 10}) ;$$

	:$$s_{\rm TP}(t) = - {\rm j} + 1 \cdot {\rm e}^{-{\rm		:$$s_{\rm TP}(t) = - {\rm j} + 1 \cdot {\rm e}^{-{\rmj}\hspace{0.05cm} \omega_{\rm 10} \hspace{0.05cm} t }.$$
	j}\hspace{0.05cm} \omega_{\rm 10} \hspace{0.05cm} t }.$$

	In the graph,  $s_{\rm TP}(t)$  is shown. It can be seen:		In the graph,  $s_{\rm TP}(t)$  is shown. It can be seen:

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	[[Category:Signal Representation: Exercises\|^4.3 Equivalent LP Signal and its Spectral Function^]]		[[Category:Signal Representation: Exercises\|^4.3 Equivalent LP Signal and its Spectral Function^]]
			[[de:Aufgaben:4.6_Ortskurve_bei_ESB-AM]]

Guenter at 13:01, 24 May 2021

2021-05-24T13:01:55Z

← Older revision		Revision as of 15:01, 24 May 2021
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	{{quiz-Header\|Buchseite=Signal_Representation/~~Equivalent_Low_Pass_Signal_and_Its_Spectral_Function~~		{{quiz-Header\|Buchseite=Signal_Representation/Equivalent Low-Pass Signal and its Spectral Function
	}}		}}

Guenter at 09:36, 12 May 2021

2021-05-12T09:36:29Z

← Older revision		Revision as of 11:36, 12 May 2021
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	*For all times, the same pointer lengths result as for  $f_{\rm T} = f_{50}$.  The magnitude remains the same.		*For all times, the same pointer lengths result as for  $f_{\rm T} = f_{50}$.  The magnitude remains the same.
	*The phase function  $\phi(t)$  now yields values between  $-\pi$  and zero, while the phase function calculated in subtask  '''(3)'''  has assumed values between  $\pm \pi/2$.		*The phase function  $\phi(t)$  now yields values between  $-\pi$  and zero, while the phase function calculated in subtask  '''(3)'''  has assumed values between  $\pm \pi/2$.
	*It is valid for all times  $t$:		*It is valid for all times  $t$:     $\phi_{\rm subtask \ (4)}= -(\phi_{\rm subtask \ (3)} + 90^\circ).$

	:$$\phi_{\rm subtask \ (4)}= -(\phi_{\rm subtask \ (3)} + 90^\circ).$$

Guenter at 09:35, 12 May 2021

2021-05-12T09:35:17Z

Guenter at 08:56, 12 May 2021

2021-05-12T08:56:15Z

← Older revision		Revision as of 10:56, 12 May 2021
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	Here  $f_{50}$  and  $f_{60}$  are abbreviations for the frequencies  $50 \ \rm kHz$  and  $60 \ \rm kHz$ respectively.		Here  $f_{50}$  and  $f_{60}$  are abbreviations for the frequencies  $50 \ \rm kHz$  and  $60 \ \rm kHz$ respectively.

	In this task, the course of the equivalent low-pass signal  $s_{\rm TP}(t)$  is computed. This course is also referred to as the &~~bdquo~~;~~locus curve~~&~~rdquo~~; in this tutorial.		In this task, the course of the equivalent low-pass signal  $s_{\rm TP}(t)$  is computed.  This course is also referred to as the  "Locality Curve"  in this tutorial.
	*In subtasks  '''(1)'''  to  '''(3)'''~~ ~~ , we assume that the signal  $s(t)$  is produced by a single sideband amplitude modulation of the sinusoidal ~~message~~ signal of frequency  $f_{\rm N} = 10 \ \text{ kHz}$  with a cosinusoidal carrier at  $f_{\rm T} = f_{50}$~~ ~~ , whereby only the upper sideband is transmitted (&~~bdquo~~;~~USB modulation~~&~~rdquo~~;).		*In subtasks  '''(1)'''  to  '''(3)''', we assume that the signal  $s(t)$  is produced by a single-sideband amplitude modulation of the sinusoidal source signal of frequency  $f_{\rm N} = 10 \ \text{ kHz}$  with a cosinusoidal carrier at  $f_{\rm T} = f_{50}$, whereby only the upper sideband is transmitted   ⇒   $\text{Upper Sideband Modulation}$.
	*In contrast, subtask  '''(4)'''  assumes the carrier frequency  $f_{\rm T} = f_{60}$ . This assumption presupposes that ~~LSB modulation~~ has taken place.		*In contrast, subtask  '''(4)'''  assumes the carrier frequency  $f_{\rm T} = f_{60}$ . This assumption presupposes that  $\text{Lower Sideband Modulation}$  has taken place.




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	''Hints:''		''Hints:''
	*This ~~task~~ belongs to the chapter  [[Signal_Representation/~~Equivalent_Low_Pass_Signal_and_Its_Spectral_Function~~\|Equivalent Low-Pass Signal and ~~Its~~ Spectral Function]].		*This exercise belongs to the chapter  [[Signal_Representation/Equivalent_Low-Pass_Signal_and_its_Spectral_Function\|Equivalent Low-Pass Signal and its Spectral Function]].

	*You can check your solution with the interactive applet  [[Applets:~~Physikalisches_Signal_~~%~~26_Äquivalentes_TP-Signal~~\|Physical ~~signal~~ & ~~equivalent LP signal~~]]   ⇒   ~~locus curve~~.		*You can check your solution with the interactive applet  [[Applets:Physical_Signal_%26_Equivalent_Lowpass_Signal\|Physical Signal & Equivalent Low-Pass Signal]]    ⇒   "Locality Curve".


	===Questions===		===Questions===

	<quiz display=simple>		<quiz display=simple>
	{Give the equivalent low-pass signal  $s_{\rm TP}(t)$  for the carrier frequency  $f_{\rm N} = 10 \ \text{ kHz}$ . Which of the following statements are true?		{Give the equivalent low-pass signal  $s_{\rm TP}(t)$  for the carrier frequency  $f_{\rm T} = 50 \ \text{ kHz}$.  Which of the following statements are true?
	\|type="()"}		\|type="()"}
	- The ~~locus~~ curve describes an ellipse.		- The locality curve describes an ellipse.
	+ The ~~locus~~ curve describes a circle.		+ The locality curve describes a circle.
	- The ~~locus curve~~ describes an arc.		- The localitycurve describes an arc.

	{Calculate the magnitude function  $a(t) = \|s_{\rm TP}(t)\|$. What is the value of  $a_0$  at  $t = 0$  and the minimum and maximum values of the magnitude?		{Calculate the magnitude function  $a(t) = \|s_{\rm TP}(t)\|$.  What is the value of  $a_0$  at  $t = 0$  and the minimum and maximum values of the magnitude?
	\|type="{}"}		\|type="{}"}
	$a_{\text{min}}\ = \ $ { 0. }		$a_{\text{min}}\ = \ $ { 0. }
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	{Calculate the phase function  $\phi(t)$. What are the phase values at  $t = 0$  and  $t=25 \ {\rm µ} \text{s}$? <br>Interpret  $\phi (t)$ in the range around  $t=75 \ {\rm µ} \text{s}$.		{Calculate the phase function  $\phi(t)$.  What are the phase values at  $t = 0$  and  $t=25 \ {\rm µ} \text{s}$? <br>Interpret  $\phi (t)$ in the range around  $t=75 \ {\rm µ} \text{s}$.
	\|type="{}"}		\|type="{}"}
	$\phi(t=0 \ {\rm µ} \text{s})\ = \ $ { -46--44 }  $\text{deg}$		$\phi(t=0 \ {\rm µ} \text{s})\ = \ $ { -46--44 }  $\text{deg}$
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	$\phi(t=75 \ {\rm µ} \text{s})\ = \ $ { 0. }  $\text{deg}$		$\phi(t=75 \ {\rm µ} \text{s})\ = \ $ { 0. }  $\text{deg}$

	{Give the equivalent low-pass signal  $s_{\rm TP}(t)$  ~~für~~  $f_{\rm T} = 60 \ \text{kHz} = f_{60}$  ~~an.~~ Which of the following statements are true?		{Give the equivalent low-pass signal  $s_{\rm TP}(t)$  for  $f_{\rm T} = 60 \ \text{kHz} = f_{60}$.  Which of the following statements are true?
	\|type="[]"}		\|type="[]"}
	+ The ~~locus~~ curve is a circle with radius  $1$  around the centre  $(0, –{\rm j})$ .		+ The locality curve is a circle with radius  $1$  around the centre  $(0, –{\rm j})$ .
	- Now  $s_{\rm TP}(t = 0) = 1 + {\rm j}$ applies.		- Now  $s_{\rm TP}(t = 0) = 1 + {\rm j}$  applies.
	+ The magnitude function  $a(t)$  is unchanged compared to   $f_{\rm T} = f_{50}$ .		+ The magnitude function  $a(t)$  is unchanged compared to  $f_{\rm T} = f_{50}$ .
	- The phase function  $\phi (t)$  is unchanged compared to   $f_{\rm T} = f_{50}$ .		- The phase function  $\phi (t)$  is unchanged compared to  $f_{\rm T} = f_{50}$ .

	</quiz>		</quiz>

Guenter: Guenter moved page Aufgaben:Exercise 4.6: Locus Curve for ESB-AM to Aufgaben:Exercise 4.6: Locality Curve for SSB-AM

2021-05-10T12:55:49Z

Guenter moved page Aufgaben:Exercise 4.6: Locus Curve for ESB-AM to Aufgaben:Exercise 4.6: Locality Curve for SSB-AM

← Older revision		Revision as of 14:55, 10 May 2021
(No difference)

Javier: Text replacement - "Low Pass" to "Low-Pass"

2021-05-04T07:33:28Z

Text replacement - "Low Pass" to "Low-Pass"

← Older revision		Revision as of 09:33, 4 May 2021
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	''Hints:''		''Hints:''
	*This task belongs to the chapter  [[Signal_Representation/Equivalent_Low_Pass_Signal_and_Its_Spectral_Function\|Equivalent Low Pass Signal and Its Spectral Function]].		*This task belongs to the chapter  [[Signal_Representation/Equivalent_Low_Pass_Signal_and_Its_Spectral_Function\|Equivalent Low-Pass Signal and Its Spectral Function]].

	*You can check your solution with the interactive applet  [[Applets:Physikalisches_Signal_%26_Äquivalentes_TP-Signal\|Physical signal & equivalent LP signal]]   ⇒   locus curve.		*You can check your solution with the interactive applet  [[Applets:Physikalisches_Signal_%26_Äquivalentes_TP-Signal\|Physical signal & equivalent LP signal]]   ⇒   locus curve.

Guenter at 08:54, 23 April 2021

2021-04-23T08:54:10Z

← Older revision		Revision as of 10:54, 23 April 2021
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	__NOEDITSECTION__		__NOEDITSECTION__
	[[Category:Signal Representation: Exercises\|^4.3 Equivalent ~~Low Pass~~ Signal and ~~Its~~ Spectral Function^]]		[[Category:Signal Representation: Exercises\|^4.3 Equivalent LP Signal and its Spectral Function^]]

← Older revision		Revision as of 17:58, 16 March 2026
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	:$$S_{\rm TP}(f ) = S_{\rm +}(f+ f_{\rm 50}) = 1 \cdot \delta (f)-{\rm j} \cdot \delta (f - f_{\rm 10}) .$$		:$$S_{\rm TP}(f ) = S_{\rm +}(f+ f_{\rm 50}) = 1 \cdot \delta (f)-{\rm j} \cdot \delta (f - f_{\rm 10}) .$$
	*This gives for the associated time signal:		*This gives for the associated time signal:
	:$$s_{\rm TP}(t) = {\rm 1 } - {\rm j} \cdot {\rm e}^{{\~~rmj~~}\hspace{0.05cm} \omega_{\rm 10} \hspace{0.05cm} t }.$$		:$$s_{\rm TP}(t) = {\rm 1 } - {\rm j} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} \omega_{\rm 10} \hspace{0.05cm} t }.$$
	*Starting from the point  $(1, –{\rm j})$   ⇒   $s_{\rm TP}(t)$  runs on a circle with centre  $(1, 0)$  and radius  $1$.		*Starting from the point  $(1, –{\rm j})$   ⇒   $s_{\rm TP}(t)$  runs on a circle with centre  $(1, 0)$  and radius  $1$.
	*The period duration is equal to the reciprocal of the frequency:   $T_0 = 1/f_{10} = 100 \ µ \text{s}$.		*The period duration is equal to the reciprocal of the frequency:   $T_0 = 1/f_{10} = 100 \ µ \text{s}$.
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	*This leads to the magnitude function		*This leads to the magnitude function

	:$$a(t)= \|s_{\rm TP}(t)\|=\sqrt{{\rm Re}\left[s_{\~~rmTP~~}(t)\right]^2 + {\rm Im}\left[s_{\rm TP}(t)\right]^2 }=\sqrt{1 + 2 \sin(\omega_{\rm 10}\hspace{0.05cm} t)+\sin^2(\omega_{\rm 10}\hspace{0.05cm} t)+ \cos^2(\omega_{\rm10}\hspace{0.05cm} t)}= \sqrt{2 \cdot ( 1 + \sin(\omega_{\rm 10}\hspace{0.05cm} t))}.$$		:$$a(t)= \|s_{\rm TP}(t)\|=\sqrt{{\rm Re}\left[s_{\rm TP}(t)\right]^2 + {\rm Im}\left[s_{\rm TP}(t)\right]^2 }=\sqrt{1 + 2 \sin(\omega_{\rm 10}\hspace{0.05cm} t)+\sin^2(\omega_{\rm 10}\hspace{0.05cm} t)+ \cos^2(\omega_{\rm10}\hspace{0.05cm} t)}= \sqrt{2 \cdot ( 1 + \sin(\omega_{\rm 10}\hspace{0.05cm} t))}.$$

	*For the minimum value, considering  $\sin(\omega_{10} \cdot t) \geq -1$     ⇒    $a_{\text{min}} \; \underline{= 0}$.		*For the minimum value, considering  $\sin(\omega_{10} \cdot t) \geq -1$     ⇒    $a_{\text{min}} \; \underline{= 0}$.
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	'''(3)'''  According to the general definition:		'''(3)'''  According to the general definition:

	:$$\phi(t)= {\rm arctan} \hspace{0.1cm}\frac{{\rm Im}\left[s_{\~~rmTP~~}(t)\right]}{{\rm Re}\left[s_{\rm TP}(t)\right]}= {\rm arctan}\hspace{0.1cm}\frac{-\cos(\omega_{\rm 10}\hspace{0.05cm} t)}{1 +\sin(\omega_{\rm 10}\hspace{0.05cm} t)}.$$		:$$\phi(t)= {\rm arctan} \hspace{0.1cm}\frac{{\rm Im}\left[s_{\rm TP}(t)\right]}{{\rm Re}\left[s_{\rm TP}(t)\right]}= {\rm arctan}\hspace{0.1cm}\frac{-\cos(\omega_{\rm 10}\hspace{0.05cm} t)}{1 +\sin(\omega_{\rm 10}\hspace{0.05cm} t)}.$$

	*For  $t = 0$  ⇒   $\cos( \omega_{10} \cdot t ) = 1$  and  $\sin( \omega_{10} \cdot t ) = 0$.  It follows:		*For  $t = 0$  ⇒   $\cos( \omega_{10} \cdot t ) = 1$  and  $\sin( \omega_{10} \cdot t ) = 0$.  It follows:
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	*If one calculates the phase function for  $t =74 \ {\rm µ} \text{s}$, one obtains with  $\omega_{10} \cdot t = 1.48 \cdot \pi \; \Rightarrow \; 266.4^\circ$:		*If one calculates the phase function for  $t =74 \ {\rm µ} \text{s}$, one obtains with  $\omega_{10} \cdot t = 1.48 \cdot \pi \; \Rightarrow \; 266.4^\circ$:

	:$$\phi(t = {\rm 74 \hspace{0.05cm} {\rm µ} s})= {\rm arctan}\hspace{0.1cm}\frac{\cos(86.4^\circ)}{1 - \sin(86.4^\circ)} = {\~~rmarctan~~} \hspace{0.1cm}\frac{0.062}{1 - 0.998} \approx {\~~rmarctan~~}(31)\approx 88^\circ.$$		:$$\phi(t = {\rm 74 \hspace{0.05cm} {\rm µ} s})= {\rm arctan}\hspace{0.1cm}\frac{\cos(86.4^\circ)}{1 - \sin(86.4^\circ)} = {\rm arctan} \hspace{0.1cm}\frac{0.062}{1 - 0.998} \approx {\rm arctan}(31)\approx 88^\circ.$$

	*Accordingly, for  $t =76 \ {\rm µ} \text{s}$  with  $\omega_{10} \cdot t = 1.52 \cdot \pi \; \Rightarrow \; 273.6^\circ$ :		*Accordingly, for  $t =76 \ {\rm µ} \text{s}$  with  $\omega_{10} \cdot t = 1.52 \cdot \pi \; \Rightarrow \; 273.6^\circ$ :
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	:$$S_{\rm TP}(f ) = S_{\rm +}(f+ f_{\rm 60}) = -{\rm j} \cdot \delta(f) + \delta (f + f_{\rm 10}) ;$$		:$$S_{\rm TP}(f ) = S_{\rm +}(f+ f_{\rm 60}) = -{\rm j} \cdot \delta(f) + \delta (f + f_{\rm 10}) ;$$

	:$$s_{\rm TP}(t) = - {\rm j} + 1 \cdot {\rm e}^{-{\~~rmj~~}\hspace{0.05cm} \omega_{\rm 10} \hspace{0.05cm} t }.$$		:$$s_{\rm TP}(t) = - {\rm j} + 1 \cdot {\rm e}^{-{\rm j}\hspace{0.05cm} \omega_{\rm 10} \hspace{0.05cm} t }.$$

	In the graph,  $s_{\rm TP}(t)$  is shown. It can be seen:		In the graph,  $s_{\rm TP}(t)$  is shown. It can be seen:
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	[[de:Aufgaben:4.~~6_Ortskurve_bei_ESB~~-AM]]		[[de:Aufgaben:Aufgabe 4.6: Ortskurve bei ESB-AM]]

← Older revision		Revision as of 11:35, 12 May 2021
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	*You can check your solution with the interactive applet  [[Applets:Physical_Signal_%26_Equivalent_Lowpass_Signal\|Physical Signal & Equivalent Low-Pass Signal]]    ⇒   "Locality Curve".		*You can check your solution with the interactive applet  [[Applets:Physical_Signal_%26_Equivalent_Lowpass_Signal\|Physical Signal & Equivalent Low-Pass Signal]]    ⇒   "Locality Curve".



	===Questions===		===Questions===
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	{{ML-Kopf}}		{{ML-Kopf}}

	[[File:P_ID766__Sig_A_4_6_a.png\|250px\|right\|frame\|~~Ortskurve für OSB~~]]		[[File:P_ID766__Sig_A_4_6_a.png\|250px\|right\|frame\|Locality curve for <br>Upper Sideband Modulation]]
	'''(1)'''  <u>~~Proposed~~ solution 2</u> is correct:		'''(1)'''  The <u>proposed solution 2</u> is correct:
	*The spectrum of the equivalent ~~LP signal~~ signal with carrier frequency  $f_{\rm T} = f_{50} = 50 \ \text{kHz}$:		*The spectrum of the equivalent low-pass signal with carrier frequency  $f_{\rm T} = f_{50} = 50 \ \text{kHz}$:
	:$$S_{\rm TP}(f ) = S_{\rm +}(f+ f_{\rm 50}) = 1 \cdot \delta (f)-		:$$S_{\rm TP}(f ) = S_{\rm +}(f+ f_{\rm 50}) = 1 \cdot \delta (f)-
	{\rm j} \cdot \delta (f - f_{\rm 10}) .$$		{\rm j} \cdot \delta (f - f_{\rm 10}) .$$
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	:$$s_{\rm TP}(t) = {\rm 1 } - {\rm j} \cdot {\rm e}^{{\rm		:$$s_{\rm TP}(t) = {\rm 1 } - {\rm j} \cdot {\rm e}^{{\rm
	j}\hspace{0.05cm} \omega_{\rm 10} \hspace{0.05cm} t }.$$		j}\hspace{0.05cm} \omega_{\rm 10} \hspace{0.05cm} t }.$$
	*Starting from the point  $(1, –{\rm j})$    $s_{\rm TP}(t)$  runs on a circle with centre  $(1, 0)$  and radius  $1$.		*Starting from the point  $(1, –{\rm j})$   ⇒   $s_{\rm TP}(t)$  runs on a circle with centre  $(1, 0)$  and radius  $1$.
	*The period is equal to the reciprocal of the frequency:   $T_0 = 1/f_{10} = 100 \ µ \text{s}$ ~~   ⇒    <u>Answer 2</u>~~.		*The period duration is equal to the reciprocal of the frequency:   $T_0 = 1/f_{10} = 100 \ µ \text{s}$.


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	= \sqrt{2 \cdot ( 1 + \sin(\omega_{\rm 10}\hspace{0.05cm} t))}.$$		= \sqrt{2 \cdot ( 1 + \sin(\omega_{\rm 10}\hspace{0.05cm} t))}.$$

	*For the minimum value, considering  $\sin(\omega_{10} \cdot t) \geq -1$     ⇒    $a_{\text{min}} \; \underline{= 0}$ ~~is obtained.~~.		*For the minimum value, considering  $\sin(\omega_{10} \cdot t) \geq -1$     ⇒    $a_{\text{min}} \; \underline{= 0}$.
	*The maximum value is obtained from  $\sin(\omega_{10} \cdot t \leq 1$ )    ⇒    $a_{\text{max}} \; \underline{= 2}$.		*The maximum value is obtained from  $\sin(\omega_{10} \cdot t \leq 1$ )    ⇒    $a_{\text{max}} \; \underline{= 2}$.
	*At  $t = 0$ , the magnitude is equal to  $a_0 = \sqrt{2 }\; \underline{\approx 1.414}$.		*At  $t = 0$ , the magnitude is equal to  $a_0 = \sqrt{2 }\; \underline{\approx 1.414}$.
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	\sin(\omega_{\rm 10}\hspace{0.05cm} t)}.$$		\sin(\omega_{\rm 10}\hspace{0.05cm} t)}.$$

	*For  $t = 0$    $\cos( \omega_{10} \cdot t ) = 1$  and  $\sin( \omega_{10} \cdot t ) = 0$. It follows:		*For  $t = 0$  ⇒   $\cos( \omega_{10} \cdot t ) = 1$  and  $\sin( \omega_{10} \cdot t ) = 0$.  It follows:

	:$$\phi(t = 0)= {\rm arctan} (-1) \hspace{0.15 cm}\underline{= -45^\circ}.$$		:$$\phi(t = 0)= {\rm arctan} (-1) \hspace{0.15 cm}\underline{= -45^\circ}.$$

	*On the other hand, for  $t = T_0/4 =25 \ µ \text{s}$ :		*On the other hand, for  $t = T_0/4 =25 \ µ \text{s}$:

	:$$\cos(\omega_{\rm 10}\hspace{0.05cm} t) = 0;		:$$\cos(\omega_{\rm 10}\hspace{0.05cm} t) = 0;
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	The phase value at  $t =75 \ µ \text{s}$~~ ~~, on the other hand, must be determined by boundary crossing, since here both the real and imaginary parts become zero and thus the argument of the arctan function is indeterminate. One obtains  $\phi(t=75 \ µ \text{s}) \; \underline{= 0}.$ This result is to be verified numerically here:		The phase value at  $t =75 \ µ \text{s}$, on the other hand, must be determined by boundary crossing, since here both the real and imaginary parts become zero and thus the argument of the arctan function is indeterminate.  One obtains  $\phi(t=75 \ µ \text{s}) \; \underline{= 0}.$  This result is to be verified numerically here:
	*If one calculates the phase function for  $t =74 \ {\rm µ} \text{s}$, ~~so erhält man mit~~  $\omega_{10} \cdot t = 1.48 \cdot \pi \; \Rightarrow \; 266.4^\circ$:		*If one calculates the phase function for  $t =74 \ {\rm µ} \text{s}$, one obtains with  $\omega_{10} \cdot t = 1.48 \cdot \pi \; \Rightarrow \; 266.4^\circ$:

	:$$\phi(t = {\rm 74 \hspace{0.05cm} {\rm µ} s})= {\rm arctan}		:$$\phi(t = {\rm 74 \hspace{0.05cm} {\rm µ} s})= {\rm arctan}
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	[[File:P_ID767__Sig_A_4_6_d.png\|250px\|right\|frame\|~~Locus~~ curve for ~~LSB~~]]		[[File:P_ID767__Sig_A_4_6_d.png\|250px\|right\|frame\|Locality curve for <br>Lower Sideband Modulation]]
	'''(4)'''  With the carrier frequency $f_{\rm T} = f_{60} = 60 \ \text{ kHz}$  the equations for time ~~and frequency~~ domain are:		'''(4)'''  <u>The first and third suggested solutions</u> are correct:
			*With the carrier frequency $f_{\rm T} = f_{60} = 60 \ \text{ kHz}$  the equations for frequency and time domain are:

	:$$S_{\rm TP}(f ) = S_{\rm +}(f+ f_{\rm 60}) = -{\rm j} \cdot \delta		:$$S_{\rm TP}(f ) = S_{\rm +}(f+ f_{\rm 60}) = -{\rm j} \cdot \delta
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	In the graph,  $s_{\rm TP}(t)$  is shown. It can be seen:		In the graph,  $s_{\rm TP}(t)$  is shown. It can be seen:
	*The ~~locus~~ curve is again a circle with radius  $1$, but now with centre  $(0, –{\rm j})$.		*The locality curve is again a circle with radius  $1$, but now with centre  $(0, –{\rm j})$.
	*Here, too  $s_{\rm TP}(t = 0) = 1 - {\rm j}$ ~~applies~~.		*Here, too:  $s_{\rm TP}(t = 0) = 1 - {\rm j}$.
	*You now move clockwise on the ~~locus~~ curve..		*You now move clockwise on the locality curve.
	*The period continues to be  $T_0 = 1/f_{10} = 100 \ µ \text{s}$.		*The period continues to be  $T_0 = 1/f_{10} = 100 \ µ \text{s}$.
	*The ~~locus~~ curve is now rotated by $90^\circ$  in the complex plane compared to sub-task  '''(1)'''~~  nun ~~.		*The locality curve is now rotated by $90^\circ$  in the complex plane compared to sub-task  '''(1)'''.
	*For all times, the same pointer lengths result as for  $f_{\rm T} = f_{50}$. The magnitude remains the same.		*For all times, the same pointer lengths result as for  $f_{\rm T} = f_{50}$.  The magnitude remains the same.
	*The phase function  $\phi(t)$  now yields values between  $-\pi$  and zero, while the phase function calculated in subtask  '''(3)'''  has assumed values between  $-\~~pi/2$  and  $+~~\pi /2$ . It is valid for all times  $t$:		*The phase function  $\phi(t)$  now yields values between  $-\pi$  and zero, while the phase function calculated in subtask  '''(3)'''  has assumed values between  $\pm \pi/2$.
			*It is valid for all times  $t$:

	:$$\phi_{\rm subtask \ (4)}= -(\phi_{\rm subtask \ (3)} + 90^\circ).$$		:$$\phi_{\rm subtask \ (4)}= -(\phi_{\rm subtask \ (3)} + 90^\circ).$$

	~~<u>The first and third suggested solutions</u> are therefore correct.~~
	{{ML-Fuß}}		{{ML-Fuß}}

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