Aufgaben:Exercise 4.5Z: Simple Phase Modulator - Revision history

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	*Due to the phase shift by  $\phi = 90^\circ$  the cosine function becomes the minus-sine function.		*Due to the phase shift by  $\phi = 90^\circ$  the cosine function becomes the minus-sine function.
	*With  $q(t) = \sin(\omega_{\rm N} t)$  holds:		*With  $q(t) = \sin(\omega_{\rm N} t)$  holds:
	:$${s(t)} = \cos({ \omega_{\rm T}\hspace{0.05cm} t }) - \sin({\omega_{\rm T}\hspace{0.05cm} t }) \cdot \sin({ \omega_{\~~rmN~~}\hspace{0.05cm} t }) = \cos({ \omega_{\rm T}\hspace{0.05cm} t }) - 0.5 \cdot \cos(({\omega_{\rm T}-\omega_{\rm N})\hspace{0.05cm} t }) + 0.5 \cdot\cos(({ \omega_{\rm T}+\omega_{\rm N})\hspace{0.05cm} t }).$$		:$${s(t)} = \cos({ \omega_{\rm T}\hspace{0.05cm} t }) - \sin({\omega_{\rm T}\hspace{0.05cm} t }) \cdot \sin({ \omega_{\rm N}\hspace{0.05cm} t }) = \cos({ \omega_{\rm T}\hspace{0.05cm} t }) - 0.5 \cdot \cos(({\omega_{\rm T}-\omega_{\rm N})\hspace{0.05cm} t }) + 0.5 \cdot\cos(({ \omega_{\rm T}+\omega_{\rm N})\hspace{0.05cm} t }).$$


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	:$$S_{\rm TP}(f) = \delta (f ) - 0.5 \cdot \delta (f + f_{\rm N})+0.5 \cdot \delta (f - f_{\rm N}) .$$		:$$S_{\rm TP}(f) = \delta (f ) - 0.5 \cdot \delta (f + f_{\rm N})+0.5 \cdot \delta (f - f_{\rm N}) .$$
	*This leads to the time function		*This leads to the time function
	:$$s_{\rm TP}(t) = {\rm 1 } - 0.5 \cdot {\rm e}^{{-\~~rmj~~}\hspace{0.05cm} \omega_{\rm N} \hspace{0.05cm} t }+ 0.5 \cdot{\rm e}^{{\rm j}\hspace{0.05cm} \omega_{\rm N} \hspace{0.05cm} t }= 1 + {\rm j} \cdot \sin(\omega_{\rm N} \hspace{0.05cm} t ).$$		:$$s_{\rm TP}(t) = {\rm 1 } - 0.5 \cdot {\rm e}^{{-\rm j}\hspace{0.05cm} \omega_{\rm N} \hspace{0.05cm} t }+ 0.5 \cdot{\rm e}^{{\rm j}\hspace{0.05cm} \omega_{\rm N} \hspace{0.05cm} t }= 1 + {\rm j} \cdot \sin(\omega_{\rm N} \hspace{0.05cm} t ).$$
	*At time  $t = 0$   ⇒   $s_{\rm TP}(t) = 1$, is real. Thus:		*At time  $t = 0$   ⇒   $s_{\rm TP}(t) = 1$, is real. Thus:

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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.~~5Z_Einfacher_Phasenmodulator~~]]		[[de:Aufgaben:Aufgabe 4.5Z: Einfacher Phasenmodulator]]

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	*Due to the phase shift by  $\phi = 90^\circ$  the cosine function becomes the minus-sine function.		*Due to the phase shift by  $\phi = 90^\circ$  the cosine function becomes the minus-sine function.
	*With  $q(t) = \sin(\omega_{\rm N} t)$  holds:		*With  $q(t) = \sin(\omega_{\rm N} t)$  holds:
	:$${s(t)} = \cos({ \omega_{\rm T}\hspace{0.05cm} t }) - \sin({		:$${s(t)} = \cos({ \omega_{\rm T}\hspace{0.05cm} t }) - \sin({\omega_{\rm T}\hspace{0.05cm} t }) \cdot \sin({ \omega_{\rmN}\hspace{0.05cm} t }) = \cos({ \omega_{\rm T}\hspace{0.05cm} t }) - 0.5 \cdot \cos(({\omega_{\rm T}-\omega_{\rm N})\hspace{0.05cm} t }) + 0.5 \cdot\cos(({ \omega_{\rm T}+\omega_{\rm N})\hspace{0.05cm} t }).$$
	\omega_{\rm T}\hspace{0.05cm} t }) \cdot \sin({ \omega_{\rm
	N}\hspace{0.05cm} t }) = \cos({ \omega_{\rm T}\hspace{0.05cm} t }) - 0.5 \cdot \cos(({
	\omega_{\rm T}-\omega_{\rm N})\hspace{0.05cm} t }) + 0.5 \cdot
	\cos(({ \omega_{\rm T}+\omega_{\rm N})\hspace{0.05cm} t }).$$



	'''(2)'''  The spectrum of the analytical signal is:		'''(2)'''  The spectrum of the analytical signal is:
	:$$S_{\rm +}(f) = \delta (f - f_{\rm T}) - 0.5 \cdot \delta (f -		:$$S_{\rm +}(f) = \delta (f - f_{\rm T}) - 0.5 \cdot \delta (f -f_{\rm \Delta})+ 0.5 \cdot \delta (f - f_{\rm \Sigma}) .$$
	f_{\rm \Delta})+ 0.5 \cdot \delta (f - f_{\rm \Sigma}) .$$
	*By shitfing  $f_{\rm T}$  one arrives at the spectrum of the equivalent low-pass signal:		*By shitfing  $f_{\rm T}$  one arrives at the spectrum of the equivalent low-pass signal:
	:$$S_{\rm TP}(f) = \delta (f ) - 0.5 \cdot \delta (f + f_{\rm N})+		:$$S_{\rm TP}(f) = \delta (f ) - 0.5 \cdot \delta (f + f_{\rm N})+0.5 \cdot \delta (f - f_{\rm N}) .$$
	0.5 \cdot \delta (f - f_{\rm N}) .$$
	*This leads to the time function		*This leads to the time function
	:$$s_{\rm TP}(t) = {\rm 1 } - 0.5 \cdot {\rm e}^{{-\rm		:$$s_{\rm TP}(t) = {\rm 1 } - 0.5 \cdot {\rm e}^{{-\rmj}\hspace{0.05cm} \omega_{\rm N} \hspace{0.05cm} t }+ 0.5 \cdot{\rm e}^{{\rm j}\hspace{0.05cm} \omega_{\rm N} \hspace{0.05cm} t }= 1 + {\rm j} \cdot \sin(\omega_{\rm N} \hspace{0.05cm} t ).$$
	j}\hspace{0.05cm} \omega_{\rm N} \hspace{0.05cm} t }+ 0.5 \cdot
	{\rm e}^{{\rm j}\hspace{0.05cm} \omega_{\rm N} \hspace{0.05cm} t }
	= 1 + {\rm j} \cdot \sin(\omega_{\rm N} \hspace{0.05cm} t ).$$
	*At time  $t = 0$   ⇒   $s_{\rm TP}(t) = 1$, is real. Thus:		*At time  $t = 0$   ⇒   $s_{\rm TP}(t) = 1$, is real. Thus:

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	[[File:P_ID762__Sig_Z_4_5_a.png\|right\|frame\|Locality curve of a simple phase modulator]]		[[File:P_ID762__Sig_Z_4_5_a.png\|right\|frame\|Locality curve of a simple phase modulator]]
	'''(3)'''  The locality curve is a vertical straight line   ⇒   <u>Proposition 3</u> with the following values:		'''(3)'''  The locality curve is a vertical straight line   ⇒   <u>Proposition 3</u> with the following values:
	:$$s_{\rm TP}(t = 0) = s_{\rm TP}(t = {\rm 50 \hspace{0.05cm} µ s})		:$$s_{\rm TP}(t = 0) = s_{\rm TP}(t = {\rm 50 \hspace{0.05cm} µ s})= \text{ ...} = 1,$$
	= \text{ ...} = 1,$$		:$$s_{\rm TP}(t = {\rm 25 \hspace{0.05cm} µ s}) = s_{\rm TP}(t ={\rm 125 \hspace{0.05cm} \mu s}) = \text{ ...} = 1 + {\rm j},$$
	:$$s_{\rm TP}(t = {\rm 25 \hspace{0.05cm} µ s}) = s_{\rm TP}(t =		:$$s_{\rm TP}(t = {\rm 75 \hspace{0.05cm} µ s}) = s_{\rm TP}(t ={\rm 175 \hspace{0.05cm} \mu s}) = \text{ ...} = 1 - {\rm j}.$$
	{\rm 125 \hspace{0.05cm} \mu s}) = \text{ ...} = 1 + {\rm j},$$
	:$$s_{\rm TP}(t = {\rm 75 \hspace{0.05cm} µ s}) = s_{\rm TP}(t =
	{\rm 175 \hspace{0.05cm} \mu s}) = \text{ ...} = 1 - {\rm j}.$$


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	'''(5)'''  The real part is always  $1$, the imaginary part equal to  $\sin(\omega_{\rm N} \cdot t) $.		'''(5)'''  The real part is always  $1$, the imaginary part equal to  $\sin(\omega_{\rm N} \cdot t) $.
	*From this follows the phase function:		*From this follows the phase function:
	:$$\phi(t)= {\rm arctan} \hspace{0.1cm}{\left(\sin(\omega_{\rm N}		:$$\phi(t)= {\rm arctan} \hspace{0.1cm}{\left(\sin(\omega_{\rm N}\hspace{0.05cm} t )\right)}.$$
	\hspace{0.05cm} t )\right)}.$$
	*The maximum value of the sine function is  $1$. From this follows:		*The maximum value of the sine function is  $1$. From this follows:
	:$$\phi_{\rm max} = \arctan (1) \; \underline{= \pi /4 } \; \Rightarrow \; \underline{45^\circ}.$$		:$$\phi_{\rm max} = \arctan (1) \; \underline{= \pi /4 } \; \Rightarrow \; \underline{45^\circ}.$$

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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:4.5Z_Einfacher_Phasenmodulator]]

Guenter at 14:16, 24 May 2021

2021-05-24T14:16:26Z

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	}}		}}

	[[File:P_ID757__Sig_Z_4_5.png\|right\|frame\|Model of the phase modulator ~~under consideration~~]]		[[File:P_ID757__Sig_Z_4_5.png\|right\|frame\|Model of the considered phase modulator]]
	The diagram shows a quite simple arrangement for approximating a phase modulator.  All signals are dimensionless.		The diagram shows a quite simple arrangement for approximating a phase modulator.  All signals are dimensionless.

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	[[File:P_ID762__Sig_Z_4_5_a.png\|right\|frame\|~~Locus~~ curve of a simple phase modulator]]		[[File:P_ID762__Sig_Z_4_5_a.png\|right\|frame\|Locality curve of a simple phase modulator]]
	'''(3)'''  The locality curve is a vertical straight line   ⇒   <u>Proposition 3</u> with the following values:		'''(3)'''  The locality curve is a vertical straight line   ⇒   <u>Proposition 3</u> with the following values:
	:$$s_{\rm TP}(t = 0) = s_{\rm TP}(t = {\rm 50 \hspace{0.05cm} µ s})		:$$s_{\rm TP}(t = 0) = s_{\rm TP}(t = {\rm 50 \hspace{0.05cm} µ s})

Guenter at 13:01, 24 May 2021

2021-05-24T13:01:14Z

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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 16:08, 11 May 2021

2021-05-11T16:08:42Z

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	{Which of the following statements are true for the "Locality Curve"  $s_{\rm TP}(t)$ zu?		{Which of the following statements are true for the "Locality Curve"  $s_{\rm TP}(t)$ zu?
	\|type="[]"}		\|type="()"}
	- The locality curve is a circular arc.		- The locality curve is a circular arc.
	- The locality curve is a horizontal straight line.		- The locality curve is a horizontal straight line.
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	{What is the phase function  $\phi(t)$.  What is its maximum value?		{What is the phase function  $\phi(t)$.  What is its maximum value?
			\|type="{}"}
	$\phi_{\rm max}\ = \ $ { 45 3% }  $\text{deg}$		$\phi_{\rm max}\ = \ $ { 45 3% }  $\text{deg}$

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	{\rm e}^{{\rm j}\hspace{0.05cm} \omega_{\rm N} \hspace{0.05cm} t }		{\rm e}^{{\rm j}\hspace{0.05cm} \omega_{\rm N} \hspace{0.05cm} t }
	= 1 + {\rm j} \cdot \sin(\omega_{\rm N} \hspace{0.05cm} t ).$$		= 1 + {\rm j} \cdot \sin(\omega_{\rm N} \hspace{0.05cm} t ).$$
	*At time  $t = 0$ ~~ist~~ $s_{\rm TP}(t) = 1$, is real. Thus:		*At time  $t = 0$   ⇒   $s_{\rm TP}(t) = 1$, is real. Thus:

	:* $s_{\rm I}(t = 0) = \text{Re}[s_{\rm TP}(t = 0)]\; \underline{= 1}$,		:* $s_{\rm I}(t = 0) = \text{Re}[s_{\rm TP}(t = 0)]\; \underline{= 1}$,
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	[[File:P_ID762__Sig_Z_4_5_a.png\|right\|frame\|Locus curve of a simple phase modulator]]		[[File:P_ID762__Sig_Z_4_5_a.png\|right\|frame\|Locus curve of a simple phase modulator]]
	'''(3)'''  The ~~locus~~ curve is a vertical straight line   ⇒   <u>Proposition 3</u> with the following values:		'''(3)'''  The locality curve is a vertical straight line   ⇒   <u>Proposition 3</u> with the following values:
	:$$s_{\rm TP}(t = 0) = s_{\rm TP}(t = {\rm 50 \hspace{0.05cm} µ s})		:$$s_{\rm TP}(t = 0) = s_{\rm TP}(t = {\rm 50 \hspace{0.05cm} µ s})
	= \text{ ...} = 1,$$		= \text{ ...} = 1,$$
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	'''(5)'''  The real part is always  $1$, the imaginary part equal to  $\sin(\omega_{\rm N} \cdot t) $. From this follows the phase function:		'''(5)'''  The real part is always  $1$, the imaginary part equal to  $\sin(\omega_{\rm N} \cdot t) $.
			*From this follows the phase function:
	:$$\phi(t)= {\rm arctan} \hspace{0.1cm}{\left(\sin(\omega_{\rm N}		:$$\phi(t)= {\rm arctan} \hspace{0.1cm}{\left(\sin(\omega_{\rm N}
	\hspace{0.05cm} t )\right)}.$$		\hspace{0.05cm} t )\right)}.$$

Guenter at 16:00, 11 May 2021

2021-05-11T16:00:35Z

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	[[File:P_ID757__Sig_Z_4_5.png\|right\|frame\|Model of the phase modulator under consideration]]		[[File:P_ID757__Sig_Z_4_5.png\|right\|frame\|Model of the phase modulator under consideration]]
	The diagram shows a quite simple arrangement for approximating a phase modulator. All signals are dimensionless ~~quantities~~.		The diagram shows a quite simple arrangement for approximating a phase modulator.  All signals are dimensionless.

	The sinusoidal ~~message~~ signal  $q(t)$  of frequency  $f_{\rm N} = 10 \ \text{kHz}$  is multiplied by the signal  $m(t)$~~ ~~, which results from the cosinusoidal carrier signal  $z(t)$  by phase shifting by  $\phi = 90^\circ$ :		The sinusoidal source signal  $q(t)$  of frequency  $f_{\rm N} = 10 \ \text{kHz}$  is multiplied by the signal  $m(t)$, which results from the cosinusoidal carrier signal  $z(t)$  by phase shifting by  $\phi = 90^\circ$ :
	:$$m(t) = {\cos} ( \omega_{\rm T} \cdot t + 90^\circ).$$		:$$m(t) = {\cos} ( \omega_{\rm T} \cdot t + 90^\circ).$$

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	*the difference frequency  $f_{\rm \Delta} = f_{\rm T} - f_{\rm N} = 0.99 \ \text{MHz}$,		*the difference frequency  $f_{\rm \Delta} = f_{\rm T} - f_{\rm N} = 0.99 \ \text{MHz}$,
	*the sum frequency  $f_{\rm \Sigma} = f_{\rm T} + f_{\rm N} = 1.01\ \text{MHz}$,		*the sum frequency  $f_{\rm \Sigma} = f_{\rm T} + f_{\rm N} = 1.01\ \text{MHz}$,
	*the two ~~angular~~ frequencies  $\omega_{\rm \Delta} = 2\pi \cdot f_{\rm \Delta}$  and  $\omega_{\rm \Sigma} = 2\pi \cdot f_{\rm \Sigma}$.		*the two circular frequencies  $\omega_{\rm \Delta} = 2\pi \cdot f_{\rm \Delta}$  and  $\omega_{\rm \Sigma} = 2\pi \cdot f_{\rm \Sigma}$.




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	''Hints:''		''Hints:''
	*This exercise 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]].

	*Consider the trigonomic transformations		*Consider the trigonomic transformations
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	{Calculate the equivalent low-pass signal  $s_{\rm TP}(t)$. What are the inphase and quadrature components at time  $t = 0$?		{Calculate the equivalent low-pass signal  $s_{\rm TP}(t)$.  What are the inphase and quadrature components at time  $t = 0$?
	\|type="{}"}		\|type="{}"}
	$s_{\rm I}(t = 0)\ = \ $ { 1 3% }		$s_{\rm I}(t = 0)\ = \ $ { 1 3% }
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	{Which of the following statements are true for the ~~locus curve~~  $s_{\rm TP}(t)$ zu?		{Which of the following statements are true for the "Locality Curve"  $s_{\rm TP}(t)$ zu?
	\|type="[]"}		\|type="[]"}
	- The ~~locus~~ curve is a circular arc.		- The locality curve is a circular arc.
	- The ~~locus~~ curve is a horizontal straight line.		- The locality curve is a horizontal straight line.
	+ The ~~locus~~ curve is a vertical straight line.		+ The locality curve is a vertical straight line.


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	{What is the phase function  $\phi(t)$. What is its maximum value?		{What is the phase function  $\phi(t)$.  What is its maximum value?

	$\phi_{\rm max}\ = \ $ { 45 3% }  $\text{deg}$		$\phi_{\rm max}\ = \ $ { 45 3% }  $\text{deg}$

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 exercise 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]].

	*Consider the trigonomic transformations		*Consider the trigonomic transformations

Javier: Text replacement - "low pass" to "low-pass"

2021-05-03T15:10:39Z

Text replacement - "low pass" to "low-pass"

← Older revision		Revision as of 17:10, 3 May 2021
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	{Calculate the equivalent low pass signal  $s_{\rm TP}(t)$. What are the inphase and quadrature components at time  $t = 0$?		{Calculate the equivalent low-pass signal  $s_{\rm TP}(t)$. What are the inphase and quadrature components at time  $t = 0$?
	\|type="{}"}		\|type="{}"}
	$s_{\rm I}(t = 0)\ = \ $ { 1 3% }		$s_{\rm I}(t = 0)\ = \ $ { 1 3% }
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	:$$S_{\rm +}(f) = \delta (f - f_{\rm T}) - 0.5 \cdot \delta (f -		:$$S_{\rm +}(f) = \delta (f - f_{\rm T}) - 0.5 \cdot \delta (f -
	f_{\rm \Delta})+ 0.5 \cdot \delta (f - f_{\rm \Sigma}) .$$		f_{\rm \Delta})+ 0.5 \cdot \delta (f - f_{\rm \Sigma}) .$$
	*By shitfing  $f_{\rm T}$  one arrives at the spectrum of the equivalent low pass signal:		*By shitfing  $f_{\rm T}$  one arrives at the spectrum of the equivalent low-pass signal:
	:$$S_{\rm TP}(f) = \delta (f ) - 0.5 \cdot \delta (f + f_{\rm N})+		:$$S_{\rm TP}(f) = \delta (f ) - 0.5 \cdot \delta (f + f_{\rm N})+
	0.5 \cdot \delta (f - f_{\rm N}) .$$		0.5 \cdot \delta (f - f_{\rm N}) .$$

Guenter at 08:53, 23 April 2021

2021-04-23T08:53:43Z

← Older revision		Revision as of 10:53, 23 April 2021
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	[[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^]]