Aufgaben:Exercise 4.4Z: Pointer Diagram for SSB-AM - Revision history

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	[[File:EN_Sig_Z_4_4_ML.png\|right\|frame\|Three different analytical signals]]		[[File:EN_Sig_Z_4_4_ML.png\|right\|frame\|Three different analytical signals]]
	'''(1)'''  The analytical signal is generally:		'''(1)'''  The analytical signal is generally:
	:$$s_{+}(t) = {\rm 1 \hspace{0.05cm} V} \cdot {\rm e}^{{\~~rmj~~}\hspace{0.05cm} \omega_{\rm 50}\hspace{0.05cm} t } - {\~~rmj~~}\cdot{\rm 1 \hspace{0.05cm} V} \cdot {\rm e}^{{\~~rmj~~}\hspace{0.05cm} \omega_{\rm 60} \hspace{0.05cm} t }.$$		:$$s_{+}(t) = {\rm 1 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} \omega_{\rm 50}\hspace{0.05cm} t } - {\rm j}\cdot{\rm 1 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm j}\hspace{0.05cm} \omega_{\rm 60} \hspace{0.05cm} t }.$$
	At time  $t = 0$  the complex exponential functions each take the value  $1$  and one obtains (see left graph):		At time  $t = 0$  the complex exponential functions each take the value  $1$  and one obtains (see left graph):
	*$\text{Re}[s_+(t = 0)] \; \underline{= +1\ \text{V}}$,		*$\text{Re}[s_+(t = 0)] \; \underline{= +1\ \text{V}}$,
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	[[Category:Signal Representation: Exercises\|^4.2 Analytical Signal and its Spectral Function^]]		[[Category:Signal Representation: Exercises\|^4.2 Analytical Signal and its Spectral Function^]]
	[[de:Aufgaben:4.~~4Z_Zeigerdiagramm_bei_ESB~~-AM]]		[[de:Aufgaben:Aufgabe 4.4Z: Zeigerdiagramm bei ESB-AM]]

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	<br clear=all>		<br clear=all>
	'''(2)'''  For the analytical signal it can also be written:		'''(2)'''  For the analytical signal it can also be written:
	:$$s_{+}(t) = {\rm 1 \hspace{0.05cm} V} \cdot \cos({ \omega_{\rm		:$$s_{+}(t) = {\rm 1 \hspace{0.05cm} V} \cdot \cos({ \omega_{\rm50}\hspace{0.05cm} t }) + {\rm j} \cdot{\rm 1 \hspace{0.05cm} V}\cdot \sin({ \omega_{\rm 50}\hspace{0.05cm} t }) - {\rm j}\cdot{\rm 1 \hspace{0.05cm} V} \cdot \cos({ \omega_{\rm60}\hspace{0.05cm} t }) + {\rm 1 \hspace{0.05cm} V} \cdot \sin({\omega_{\rm 60}\hspace{0.05cm} t }).$$
	50}\hspace{0.05cm} t }) + {\rm j} \cdot{\rm 1 \hspace{0.05cm} V}
	\cdot \sin({ \omega_{\rm 50}\hspace{0.05cm} t }) - {\rm j}
	\cdot
	{\rm 1 \hspace{0.05cm} V} \cdot \cos({ \omega_{\rm
	60}\hspace{0.05cm} t }) + {\rm 1 \hspace{0.05cm} V} \cdot \sin({
	\omega_{\rm 60}\hspace{0.05cm} t }).$$
	The real part of this describes the actual physical signal:		The real part of this describes the actual physical signal:
	:$$s(t) = {\rm 1 \hspace{0.05cm} V} \cdot \cos({ \omega_{\rm50}\hspace{0.05cm} t }) + {\rm 1 \hspace{0.05cm} V} \cdot \sin({\omega_{\rm 60}\hspace{0.05cm} t }).$$		:$$s(t) = {\rm 1 \hspace{0.05cm} V} \cdot \cos({ \omega_{\rm50}\hspace{0.05cm} t }) + {\rm 1 \hspace{0.05cm} V} \cdot \sin({\omega_{\rm 60}\hspace{0.05cm} t }).$$

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	The analytical signal  $s_+(t)$  with the line spectrum		The analytical signal  $s_+(t)$  with the line spectrum
	:$$S_{+}(f) = {\rm 1 \hspace{0.05cm} V} \cdot\delta (f - f_{\rm		:$$S_{+}(f) = {\rm 1 \hspace{0.05cm} V} \cdot\delta (f - f_{\rm50})- {\rm j} \cdot {\rm 1 \hspace{0.05cm} V} \cdot\delta (f -f_{\rm 60})$$
	50})- {\rm j} \cdot {\rm 1 \hspace{0.05cm} V} \cdot\delta (f -
	f_{\rm 60})$$
	is to be considered.		is to be considered.
	Here  $f_{50}$  and  $f_{60}$  are abbreviations for the frequencies  $50 \ \text{kHz}$  and  $60 \ \text{kHz}$, respectively.		Here  $f_{50}$  and  $f_{60}$  are abbreviations for the frequencies  $50 \ \text{kHz}$  and  $60 \ \text{kHz}$, respectively.
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	[[File:EN_Sig_Z_4_4_ML.png\|right\|frame\|Three different analytical signals]]		[[File:EN_Sig_Z_4_4_ML.png\|right\|frame\|Three different analytical signals]]
	'''(1)'''  The analytical signal is generally:		'''(1)'''  The analytical signal is generally:
	:$$s_{+}(t) = {\rm 1 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm		:$$s_{+}(t) = {\rm 1 \hspace{0.05cm} V} \cdot {\rm e}^{{\rmj}\hspace{0.05cm} \omega_{\rm 50}\hspace{0.05cm} t } - {\rmj}\cdot{\rm 1 \hspace{0.05cm} V} \cdot {\rm e}^{{\rmj}\hspace{0.05cm} \omega_{\rm 60} \hspace{0.05cm} t }.$$
	j}\hspace{0.05cm} \omega_{\rm 50}\hspace{0.05cm} t } - {\rm
	j}\cdot{\rm 1 \hspace{0.05cm} V} \cdot {\rm e}^{{\rm
	j}\hspace{0.05cm} \omega_{\rm 60} \hspace{0.05cm} t }.$$
	At time  $t = 0$  the complex exponential functions each take the value  $1$  and one obtains (see left graph):		At time  $t = 0$  the complex exponential functions each take the value  $1$  and one obtains (see left graph):
	*$\text{Re}[s_+(t = 0)] \; \underline{= +1\ \text{V}}$,		*$\text{Re}[s_+(t = 0)] \; \underline{= +1\ \text{V}}$,
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	\omega_{\rm 60}\hspace{0.05cm} t }).$$		\omega_{\rm 60}\hspace{0.05cm} t }).$$
	The real part of this describes the actual physical signal:		The real part of this describes the actual physical signal:
	:$$s(t) = {\rm 1 \hspace{0.05cm} V} \cdot \cos({ \omega_{\rm		:$$s(t) = {\rm 1 \hspace{0.05cm} V} \cdot \cos({ \omega_{\rm50}\hspace{0.05cm} t }) + {\rm 1 \hspace{0.05cm} V} \cdot \sin({\omega_{\rm 60}\hspace{0.05cm} t }).$$
	50}\hspace{0.05cm} t }) + {\rm 1 \hspace{0.05cm} V} \cdot \sin({
	\omega_{\rm 60}\hspace{0.05cm} t }).$$

	Correct is the <u>proposed solution 3</u>:		Correct is the <u>proposed solution 3</u>:
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	This case is reached for the first time when the faster pointer with circular velocity  $\omega_{60}$  has caught up its "lag" of  $90^{\circ} \; (\pi /2)$  with the slower pointer  ($\omega_{50}$) :		This case is reached for the first time when the faster pointer with circular velocity  $\omega_{60}$  has caught up its "lag" of  $90^{\circ} \; (\pi /2)$  with the slower pointer  ($\omega_{50}$) :
	:$$\omega_{\rm 60} \cdot t_2 - \omega_{\rm		:$$\omega_{\rm 60} \cdot t_2 - \omega_{\rm50}\cdot t_2 = \frac{\pi}{2} \hspace{0.3cm}\Rightarrow\hspace{0.3cm}t_2 = \frac{\pi/2}{2\pi (f_{\rm 60}-f_{\rm 50})} = \frac{1}{4\cdot(f_{\rm 60}- f_{\rm 50})}\hspace{0.15 cm}\underline{= {\rm 25 \hspace{0.05cm} {\rm µ s}}}.$$
	50}\cdot t_2 = \frac{\pi}{2} \hspace{0.3cm}
	\Rightarrow\hspace{0.3cm}t_2 = \frac{\pi/2}{2\pi (f_{\rm 60}-
	f_{\rm 50})} = \frac{1}{4
	\cdot(f_{\rm 60}- f_{\rm 50})}\hspace{0.15 cm}\underline{= {\rm 25 \hspace{0.05cm} {\rm µ s}}}.$$
	*At this point, the two pointers have made  $5/4$  and  $6/4$  rotations respectively and both point in the direction of the imaginary axis (see right graph).		*At this point, the two pointers have made  $5/4$  and  $6/4$  rotations respectively and both point in the direction of the imaginary axis (see right graph).
	*The actual physical signal  $s(t)$ – i.e. the real part of  $s_+(t)$ – is therefore zero at this moment.		*The actual physical signal  $s(t)$ – i.e. the real part of  $s_+(t)$ – is therefore zero at this moment.
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	*Analogous to the sample solution of sub-task  '''(3)''' , therefore the following applies:		*Analogous to the sample solution of sub-task  '''(3)''' , therefore the following applies:
	:$$t_3 = \frac{3\pi/2}{2\pi (f_{\rm 60}- f_{\rm 50})} \hspace{0.15 cm}\underline{=		:$$t_3 = \frac{3\pi/2}{2\pi (f_{\rm 60}- f_{\rm 50})} \hspace{0.15 cm}\underline{={\rm 75 \hspace{0.05cm} {\rm µ s}}}.$$
	{\rm 75 \hspace{0.05cm} {\rm µ s}}}.$$
	{{ML-Fuß}}		{{ML-Fuß}}

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	[[Category:Signal Representation: Exercises\|^4.2 Analytical Signal and its Spectral Function^]]		[[Category:Signal Representation: Exercises\|^4.2 Analytical Signal and its Spectral Function^]]
			[[de:Aufgaben:4.4Z_Zeigerdiagramm_bei_ESB-AM]]

Hwang at 03:15, 18 September 2022

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	'''(3)'''  The maximum value of  $\|s_+(t)\|$  is reached when both pointers point in the same direction. The ~~amount~~ of the sum pointer is then equal to the sum of the two individual pointers; i.e.  $\underline {2\ \text{V}}$.		'''(3)'''  The maximum value of  $\|s_+(t)\|$  is reached when both pointers point in the same direction. The magnitude of the sum pointer is then equal to the sum of the two individual pointers; i.e.  $\underline {2\ \text{V}}$.

	This case is reached for the first time when the faster pointer with circular velocity  $\omega_{60}$  has caught up its "lag" of  $90^{\circ} \; (\pi /2)$  with the slower pointer  ($\omega_{50}$) :		This case is reached for the first time when the faster pointer with circular velocity  $\omega_{60}$  has caught up its "lag" of  $90^{\circ} \; (\pi /2)$  with the slower pointer  ($\omega_{50}$) :

Guenter at 14:37, 7 May 2021

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Guenter at 14:26, 7 May 2021

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	Here  $f_{50}$  and  $f_{60}$  are abbreviations for the frequencies  $50 \ \text{kHz}$  and  $60 \ \text{kHz}$, respectively.		Here  $f_{50}$  and  $f_{60}$  are abbreviations for the frequencies  $50 \ \text{kHz}$  and  $60 \ \text{kHz}$, respectively.

	This analytical signal could occur, for example, with the  [[Modulation_Methods/Einseitenbandmodulation\|Single Sideband Amplitude Modulation]]  $\text{(SSB-AM)}$  of a sinusoidal message signal  $($Frequenz  $f_{\rm N} = 10 \ \text{kHz})$  with a cosinusoidal carrier signal  $(f_{\rm T} = 50 \ \text{kHz})$ , whereby only the upper sideband is transmitted   ⇒ &nbsp $\text{Upper Sideband Modulation}$.		This analytical signal could occur, for example, with the  [[Modulation_Methods/Einseitenbandmodulation\|Single Sideband Amplitude Modulation]]  $\text{(SSB–AM)}$  of a sinusoidal message signal  $($Frequenz  $f_{\rm N} = 10 \ \text{kHz})$  with a cosinusoidal carrier signal  $(f_{\rm T} = 50 \ \text{kHz})$ , whereby only the upper sideband is transmitted   ⇒   $\text{Upper Sideband Modulation}$.

	However, the analytical signal could also result from a  $\text{(Lower Sideband Modulation)}$  of the same sinusoidal signal if a sinusoidal carrier with frequency  $f_{\rm T} = 60 \ \text{kHz}$  is used.		However, the analytical signal could also result from a  $\text{Lower Sideband Modulation}$  of the same sinusoidal signal if a sinusoidal carrier with frequency  $f_{\rm T} = 60 \ \text{kHz}$  is used.


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	<quiz display=simple>		<quiz display=simple>
	{Give the analytical signal  $s_+(t)$  as a formula. What value results at the starting time  $t = 0$?		{Give the analytical signal  $s_+(t)$  as a formula.  What value results at the starting time  $t = 0$?
	\|type="{}"}		\|type="{}"}
	$\text{Re}[s_+(t = 0)]\ = \ $ { 1 3% }  $\text{V}$		$\text{Re}[s_+(t = 0)]\ = \ $ { 1 3% }  $\text{V}$
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	{What is the maximum value of  $\|s_+(t)\|$ ? At what time  $t_2$  is this maximum value reached for the first time?		{What is the maximum value of  $\|s_+(t)\|$?  At what time  $t_2$  is this maximum value reached for the first time?
	\|type="{}"}		\|type="{}"}
	$\|s_+(t)\|_{\rm max}\ = \ ${ 2 3% }  $\text{V}$		$\|s_+(t)\|_{\rm max}\ = \ ${ 2 3% }  $\text{V}$

Guenter at 14:23, 7 May 2021

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	~~''Hints:''~~
	*This exercise belongs to the chapter  [[Signal_Representation/Analytical_Signal_and_Its_Spectral_Function\|Analytical Signal and its Spectral Function]].
	*The interactive applet  [[Applets:Physical_Signal_%26_Analytic_Signal\|Physical and Analytical Signal]]  illustrates the topic covered here.

	''Hints:''		''Hints:''
	*This exercise belongs to the chapter  [[Signal_Representation/Analytical_Signal_and_Its_Spectral_Function\|Analytical Signal and its Spectral Function]].		*This exercise belongs to the chapter  [[Signal_Representation/Analytical_Signal_and_Its_Spectral_Function\|Analytical Signal and its Spectral Function]].

	*You can check your solution with the interaction module  [[Applets:~~Physikalisches_Signal_~~%~~26_Analytisches_Signal~~\|Physical and Analytical Signal]].		*You can check your solution with the interaction module  [[Applets:Physical_Signal_%26_Analytic_Signal\|Physical and Analytical Signal]].

	~~$\text{Please note:}$     In this task we use the following nomenclature because of the German original:~~
	*The index  $\rm N$  stands for "source signal"   ⇒   (German:  "Nachrichtenignal").
	*The index  $\rm T$  stands for  "carrier"   ⇒   (German:  "Trägersignal").
	*$\rm OSB$  denotes the  "upper sideband"   ⇒   (German:  "oberes Seitenband").
	*$\rm USB$  denotes the  "lower sideband"   ⇒   (German:  "unteres Seitenband").

Guenter at 14:22, 7 May 2021

2021-05-07T14:22:05Z

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	}}		}}

	[[File:P_ID732__Sig_Z_4_4_neu.png\|right\|frame\|Given spectrum  $S_+(f)$]]		[[File:P_ID732__Sig_Z_4_4_neu.png\|right\|frame\|Given analytical spectrum  $S_+(f)$]]

	~~$\text{Please note:}$     In this task we use the following nomenclature because of the German original:~~
	*The index  $\rm N$  stands for "source signal"   ⇒   (German:  "Nachrichtenignal").
	*The index  $\rm T$  stands for  "carrier"   ⇒   (German:  "Trägersignal").
	*$\rm OSB$  denotes the  "upper sideband"   ⇒   (German:  "oberes Seitenband").
	*$\rm USB$  denotes the  "lower sideband"   ⇒   (German:  "unteres Seitenband").


	The analytical signal  $s_+(t)$  with the line spectrum		The analytical signal  $s_+(t)$  with the line spectrum
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	f_{\rm 60})$$		f_{\rm 60})$$
	is to be considered.		is to be considered.
	Here  $f_{50}$  and  $f_{60}$  are abbreviations for the frequencies  $50 \ \text{kHz}$  and  $60 \ \text{kHz}$, respectively..		Here  $f_{50}$  and  $f_{60}$  are abbreviations for the frequencies  $50 \ \text{kHz}$  and  $60 \ \text{kHz}$, respectively.

	This analytical signal could occur, for example, with the  [[Modulation_Methods/Einseitenbandmodulation\|single sideband amplitude modulation]]  (SSB-AM) of a sinusoidal message signal  $($Frequenz  $f_{\rm N} = 10 \ \text{kHz})$  with a cosinusoidal carrier signal  $(f_{\rm T} = 50 \ \text{kHz})$ , whereby only the <u>upper sideband</u> is transmitted (''USB-Modulation'').

	However, the analytical signal could also result from a  ''LSB-Modulation''  modulation of the same sinusoidal signal if a sinusoidal carrier signal with carrier frequency  $f_{\rm T} = 60 \ \text{kHz}$  is used.


			This analytical signal could occur, for example, with the  [[Modulation_Methods/Einseitenbandmodulation\|Single Sideband Amplitude Modulation]]  $\text{(SSB-AM)}$  of a sinusoidal message signal  $($Frequenz  $f_{\rm N} = 10 \ \text{kHz})$  with a cosinusoidal carrier signal  $(f_{\rm T} = 50 \ \text{kHz})$ , whereby only the upper sideband is transmitted   ⇒ &nbsp $\text{Upper Sideband Modulation}$.

			However, the analytical signal could also result from a  $\text{(Lower Sideband Modulation)}$  of the same sinusoidal signal if a sinusoidal carrier with frequency  $f_{\rm T} = 60 \ \text{kHz}$  is used.


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	''Hints:''		''Hints:''
	*This exercise belongs to the chapter  [[Signal_Representation/Analytical_Signal_and_Its_Spectral_Function\|Analytical Signal and ~~Its~~ Spectral Function]].		*This exercise belongs to the chapter  [[Signal_Representation/Analytical_Signal_and_Its_Spectral_Function\|Analytical Signal and its Spectral Function]].

	*You can check your solution with the interaction module  [[Applets:Physikalisches_Signal_%26_Analytisches_Signal\|Physical ~~signal &~~ Analytical Signal]] .		*You can check your solution with the interaction module  [[Applets:Physikalisches_Signal_%26_Analytisches_Signal\|Physical and Analytical Signal]].

			$\text{Please note:}$     In this task we use the following nomenclature because of the German original:
			*The index  $\rm N$  stands for "source signal"   ⇒   (German:  "Nachrichtenignal").
			*The index  $\rm T$  stands for  "carrier"   ⇒   (German:  "Trägersignal").
			*$\rm OSB$  denotes the  "upper sideband"   ⇒   (German:  "oberes Seitenband").
			*$\rm USB$  denotes the  "lower sideband"   ⇒   (German:  "unteres Seitenband").

Guenter at 16:54, 6 May 2021

2021-05-06T16:54:40Z

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	[[File:P_ID732__Sig_Z_4_4_neu.png\|right\|frame\|Given spectrum  $S_+(f)$]]		[[File:P_ID732__Sig_Z_4_4_neu.png\|right\|frame\|Given spectrum  $S_+(f)$]]

			$\text{Please note:}$     In this task we use the following nomenclature because of the German original:
			*The index  $\rm N$  stands for "source signal"   ⇒   (German:  "Nachrichtenignal").
			*The index  $\rm T$  stands for  "carrier"   ⇒   (German:  "Trägersignal").
			*$\rm OSB$  denotes the  "upper sideband"   ⇒   (German:  "oberes Seitenband").
			*$\rm USB$  denotes the  "lower sideband"   ⇒   (German:  "unteres Seitenband").


	The analytical signal  $s_+(t)$  with the line spectrum		The analytical signal  $s_+(t)$  with the line spectrum
	:$$S_{+}(f) = {\rm 1 \hspace{0.05cm} V} \cdot\delta (f - f_{\rm		:$$S_{+}(f) = {\rm 1 \hspace{0.05cm} V} \cdot\delta (f - f_{\rm
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			''Hints:''
			*This exercise belongs to the chapter  [[Signal_Representation/Analytical_Signal_and_Its_Spectral_Function\|Analytical Signal and its Spectral Function]].
			*The interactive applet  [[Applets:Physical_Signal_%26_Analytic_Signal\|Physical and Analytical Signal]]  illustrates the topic covered here.

	''Hints:''		''Hints:''

← Older revision		Revision as of 16:37, 7 May 2021
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	60}\hspace{0.05cm} t }) + {\rm 1 \hspace{0.05cm} V} \cdot \sin({		60}\hspace{0.05cm} t }) + {\rm 1 \hspace{0.05cm} V} \cdot \sin({
	\omega_{\rm 60}\hspace{0.05cm} t }).$$		\omega_{\rm 60}\hspace{0.05cm} t }).$$
	The real part of this describes the actual, physical signal:		The real part of this describes the actual physical signal:
	:$$s(t) = {\rm 1 \hspace{0.05cm} V} \cdot \cos({ \omega_{\rm		:$$s(t) = {\rm 1 \hspace{0.05cm} V} \cdot \cos({ \omega_{\rm
	50}\hspace{0.05cm} t }) + {\rm 1 \hspace{0.05cm} V} \cdot \sin({		50}\hspace{0.05cm} t }) + {\rm 1 \hspace{0.05cm} V} \cdot \sin({
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	Correct is the <u>proposed solution 3</u>:		Correct is the <u>proposed solution 3</u>:
	*Considering the  $50 \ \text{kHz~~-Cosinussignals~~}$  cosine signal alone, the first zero crossing would occur at  $t_1 = T_0/4$  , i.e. after  $5 \ {\rm µ s}$, where  $T_0 = 1/f_{50} = 20 \ {\rm µ s}$  denotes the period duration of this signal.		*Considering the  $50 \ \text{kHz}$  cosine signal alone, the first zero crossing would occur at  $t_1 = T_0/4$  , i.e. after  $5 \ {\rm µ s}$, where  $T_0 = 1/f_{50} = 20 \ {\rm µ s}$  denotes the period duration of this signal.
	*The sinusoidal signal with the frequency  $60 \ \text{kHz}$  is positive during the entire first half-wave  $(0 \, \text{...} \, 8.33\ {\rm µ s})$ .		*The sinusoidal signal with the frequency  $60 \ \text{kHz}$  is positive during the entire first half-wave  $(0 \, \text{...} \, 8.33\ {\rm µ s})$ .
	*Due to the plus sign, the first zero crossing of  $s(t) \ \Rightarrow \ t_1 > 5\ {\rm µ s}$ is delayed.		*Due to the plus sign, the first zero crossing of  $s(t) \ \Rightarrow \ t_1 > 5\ {\rm µ s}$ is delayed.
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	'''(3)'''  The maximum value of  $\|s_+(t)\|$  is reached when both pointers point in the same direction. The amount of the sum pointer is then equal to the sum of the two individual pointers; i.e.  $\underline {2\ \text{ V}}$.		'''(3)'''  The maximum value of  $\|s_+(t)\|$  is reached when both pointers point in the same direction. The amount of the sum pointer is then equal to the sum of the two individual pointers; i.e.  $\underline {2\ \text{V}}$.

	This case is reached for the first time when the faster pointer with ~~angular~~ velocity  $\omega_{60}$  has caught up its "lag" of  $90^{\circ} \; (\pi /2)$  with the slower pointer  ($\omega_{50}$) :		This case is reached for the first time when the faster pointer with circular velocity  $\omega_{60}$  has caught up its "lag" of  $90^{\circ} \; (\pi /2)$  with the slower pointer  ($\omega_{50}$) :
	:$$\omega_{\rm 60} \cdot t_2 - \omega_{\rm		:$$\omega_{\rm 60} \cdot t_2 - \omega_{\rm
	50}\cdot t_2 = \frac{\pi}{2} \hspace{0.3cm}		50}\cdot t_2 = \frac{\pi}{2} \hspace{0.3cm}
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	f_{\rm 50})} = \frac{1}{4		f_{\rm 50})} = \frac{1}{4
	\cdot(f_{\rm 60}- f_{\rm 50})}\hspace{0.15 cm}\underline{= {\rm 25 \hspace{0.05cm} {\rm µ s}}}.$$		\cdot(f_{\rm 60}- f_{\rm 50})}\hspace{0.15 cm}\underline{= {\rm 25 \hspace{0.05cm} {\rm µ s}}}.$$
	*At this point, the two pointers have made  $5/4$  ~~bzw.~~  $6/4$  ~~revolutions~~ respectively and both point in the direction of the imaginary axis (see right graph).		*At this point, the two pointers have made  $5/4$  and  $6/4$  rotations respectively and both point in the direction of the imaginary axis (see right graph).
	*The actual, physical signal  $s(t)$ – i.e. the real part of  $s_+(t)$ – is therefore zero at this moment.		*The actual physical signal  $s(t)$ – i.e. the real part of  $s_+(t)$ – is therefore zero at this moment.



	'''(4)'''  The condition for  $\|s_+(t_3)\| = 0$  is that there is a phase offset of  $180^\circ$  between the two equally long pointers so that they cancel each other out.		'''(4)'''  The condition for  $\|s_+(t_3)\| = 0$  is that there is a phase offset of  $180^\circ$  between the two equally long pointers so that they cancel each other out.
	*This further means that the faster pointer has rotated  $3\pi /2$  further than the  $50 \ \text{kHz~~-component~~}$.		*This further means that the faster pointer has rotated  $3\pi /2$  further than the  $50 \ \text{kHz}$  component.

	*Analogous to the sample solution of sub-task  '''(3)''' , the following ~~therefore~~ applies:		*Analogous to the sample solution of sub-task  '''(3)''' , therefore the following applies:
	:$$t_3 = \frac{3\pi/2}{2\pi (f_{\rm 60}- f_{\rm 50})} \hspace{0.15 cm}\underline{=		:$$t_3 = \frac{3\pi/2}{2\pi (f_{\rm 60}- f_{\rm 50})} \hspace{0.15 cm}\underline{=
	{\rm 75 \hspace{0.05cm} {\rm µ s}}}.$$		{\rm 75 \hspace{0.05cm} {\rm µ s}}}.$$