Aufgaben:Exercise 3.4Z: Continuous Phase Frequency Shift Keying - Revision history

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	[[Category:Examples of Communication Systems: Exercises\|^3.2 Radio Interface^]]		[[Category:Examples of Communication Systems: Exercises\|^3.2 Radio Interface^]]
	[[de:Aufgaben:3.~~4Z_Continuous_Phase_FSK~~]]		[[de:Aufgaben:Aufgabe 3.4Z: FSK mit kontinuierlicher Phase]]

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	[[Category:Examples of Communication Systems: Exercises\|^3.2 Radio Interface^]]		[[Category:Examples of Communication Systems: Exercises\|^3.2 Radio Interface^]]
			[[de:Aufgaben:3.4Z_Continuous_Phase_FSK]]

Guenter at 14:21, 22 January 2023

2023-01-22T14:21:17Z

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	'''(7)'''  Correct is the  <u>last proposed solution</u>,  since for this signal $h = 0.5$.		'''(7)'''  Correct is the  <u>last proposed solution</u>,  since for this signal   ⇒   $h = 0.5$.
	*This is the smallest possible modulation index for which there is orthogonality between  $f_{1}$  and  $f_{2}$  within the symbol duration  $T$.		*This is the smallest possible modulation index for which there is orthogonality between  $f_{1}$  and  $f_{2}$  within the symbol duration  $T$.

Guenter at 14:14, 22 January 2023

2023-01-22T14:14:44Z

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	$f_{2} \cdot T \ = \ $ { 3 3% }		$f_{2} \cdot T \ = \ $ { 3 3% }

	{What are the carrier frequency  $f_{\rm T}$,  the frequency deviation  $\~~delta~~ f_{\rm A}$  and the modulation index  $h$  for signal  $s_{\rm A}(t)$?		{What are the carrier frequency  $f_{\rm T}$,  the frequency deviation  $\Delta f_{\rm A}$  and the modulation index  $h$  for signal  $s_{\rm A}(t)$?
	\|type="{}"}		\|type="{}"}
	$f_{\rm T} \cdot T \ = \ $ { 4 3% }		$f_{\rm T} \cdot T \ = \ $ { 4 3% }

Guenter at 12:38, 19 January 2023

2023-01-19T12:38:30Z

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	\|type="{}"}		\|type="{}"}
	$f_{\rm T} \cdot T \ = \ $ { 4 3% }		$f_{\rm T} \cdot T \ = \ $ { 4 3% }
	$\~~delta~~ f_{\rm A} \cdot T \ = \ $ { 1 3% }		$\Delta f_{\rm A} \cdot T \ = \ $ { 1 3% }
	$h \ = \ $ { 2 3% }		$h \ = \ $ { 2 3% }

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	+ $s_{\rm C}(t)$.		+ $s_{\rm C}(t)$.

	{What signals describe  ''Minimum Shift Keying''  (MSK)?		{What signals describe  "Minimum Shift Keying"  $\rm (MSK)$?
	\|type="[]"}		\|type="[]"}
	- $s_{\rm A}(t)$,		- $s_{\rm A}(t)$,
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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>All statements except the third are true</u>:		'''(1)'''  <u>All statements except the third are true</u>:
	*Generally nonlinear FSK can only be demodulated coherently, while MSK can also use a noncoherent demodulation method.		*Generally nonlinear FSK can only be demodulated coherently,  while MSK can also use a noncoherent demodulation method.
	*Compared to QPSK with coherent demodulation, MSK requires $3 \ \rm dB$ more $E_{\rm B}/N_{0}$ (energy per bit related to the noise power density) for the same bit error rate.
	*The first zero in the power density spectrum occurs later in MSK than in QSPK, but it shows a faster asymptotic decay than in QSPK.		*Compared to QPSK with coherent demodulation,  MSK requires  $3 \ \rm dB$  more  $E_{\rm B}/N_{0}$  $($energy per bit related to the noise power density$)$  for the same bit error rate.
	*The constant envelope of MSK means that nonlinearities in the transmission line do not play a role. This allows the use of simple and inexpensive power amplifiers with lower power consumption and thus longer operating times of battery-powered devices.

			*The first zero in the power-spectral density occurs in MSK later than in QSPK,  but it shows a faster asymptotic decay than in QSPK.

			*The constant envelope of MSK means that nonlinearities in the transmission line do not play a role.  This allows the use of simple and inexpensive power amplifiers with lower power consumption and thus longer operating times of battery-powered devices.


	'''(2)'''  One can see from the graph five and three oscillations per symbol duration, respectively:
			'''(2)'''  One can see from the graph five and three oscillations per symbol duration,  respectively:
	:$$f_{\rm 1} \cdot T \hspace{0.15cm} \underline {= 5}\hspace{0.05cm},\hspace{0.2cm}f_{\rm 2} \cdot T \hspace{0.15cm} \underline { = 3}\hspace{0.05cm}.$$		:$$f_{\rm 1} \cdot T \hspace{0.15cm} \underline {= 5}\hspace{0.05cm},\hspace{0.2cm}f_{\rm 2} \cdot T \hspace{0.15cm} \underline { = 3}\hspace{0.05cm}.$$


	'''(3)'''  For FSK with rectangular pulse shape, only the two instantaneous frequencies $f_{1} = f_{\rm T} + \Delta f_{\rm A}$ and $f_{2} = f_{\rm T} - \Delta f_{\rm A}$ occur.		'''(3)'''  For FSK with rectangular pulse shape,  only the two instantaneous frequencies  $f_{1} = f_{\rm T} + \Delta f_{\rm A}$  and  $f_{2} = f_{\rm T} - \Delta f_{\rm A}$  occur.
	*With the result from '''(2)''' we thus obtain:		*With the result from subtask  '''(2)'''  we thus obtain:
	:$$f_{\rm T} \ = \ \frac{f_{\rm 1}+f_{\rm 2}}{2}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm T} \cdot T \hspace{0.15cm} \underline {= 4}\hspace{0.05cm},$$		:$$f_{\rm T} \ = \ \frac{f_{\rm 1}+f_{\rm 2}}{2}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm T} \cdot T \hspace{0.15cm} \underline {= 4}\hspace{0.05cm},$$
	:$$ \Delta f_{\rm A} \ = \ \frac{f_{\rm 1}-f_{\rm 2}}{2}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\Delta f_{\rm A} \cdot T \hspace{0.15cm} \underline { = 1}\hspace{0.05cm},$$		:$$ \Delta f_{\rm A} \ = \ \frac{f_{\rm 1}-f_{\rm 2}}{2}\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\Delta f_{\rm A} \cdot T \hspace{0.15cm} \underline { = 1}\hspace{0.05cm},$$
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	'''(4)'''  From the graph one can see the frequencies $f_{1} \cdot T = 4.5$ and $f_{2} \cdot T = 3.5$.		'''(4)'''  From the graph one can see the frequencies  $f_{1} \cdot T = 4.5$  and  $f_{2} \cdot T = 3.5$.
	*This results in the frequency deviation $\Delta f_{\rm A} \cdot T = 0.5$ and the modulation index $\underline{h = 1}$.		*This results in the frequency deviation  $\Delta f_{\rm A} \cdot T = 0.5$  and the modulation index  $\underline{h = 1}$.



	'''(5)'''  Here the two (normalized) frequencies $f_{1} \cdot T = 4.25$ and $f_{2} \cdot T = 3.75$ occur,		'''(5)'''  Here the two  $($normalized$)$  frequencies  $f_{1} \cdot T = 4.25$  and  $f_{2} \cdot T = 3.75$  occur,
	*which results in the frequency deviation $\Delta f_{\rm A} \cdot T = 0.25$ and the modulation index $\underline{h = 0.5}$ ~~can be calculated~~.		*which results in the frequency deviation  $\Delta f_{\rm A} \cdot T = 0.25$  and the modulation index  $\underline{h = 0.5}$.



	'''(6)'''  Correct are the <u>solutions 2 and 3</u>:		'''(6)'''  Correct are the  <u>solutions 2 and 3</u>:
	*Only at $s_{\rm A}(t)$ was no phase adjustment made.		*Only at  $s_{\rm A}(t)$  was no phase adjustment made.  Here,  the signal waveforms in the region of the first and second bit  $(a_{1} = a_{2} = +1)$  are each cosinusoidal like the carrier signal  $($with respect to the symbol boundary$)$.
	*Here, the signal waveforms in the region of the first and second bit ($a_{1} = a_{2} = +1$) are each cosinusoidal like the carrier signal (with respect to the symbol boundary).
	*In contrast, in the second symbol of $s_{\rm B}(t)$ a minus-cosine-shaped course (initial phase $\phi_{0} = π$ corresponding to $180^\circ$) can be seen and in the second symbol of $s_{\rm C}(t)$ a minus-sine-shaped course ($\phi_{0} = π /2$ or $90^\circ$).		*In contrast,  in the second symbol of  $s_{\rm B}(t)$  a minus-cosine-shaped course  $($initial phase  $\phi_{0} = π$,  corresponding to  $180^\circ)$  can be seen and in the second symbol of  $s_{\rm C}(t)$  a minus-sine-shaped course  $(\phi_{0} = π /2$  or  $90^\circ)$.
	*For $s_{\rm A}(t)$ the initial phase is always ~~$0$~~, for $s_{\rm B}(t)$ either ~~$0$~~ or $π$, while for the signal $s_{\rm C}(t)$ with modulation index $h = 0.5$ a total of four initial phases are possible: $0^\circ, \ 90^\circ, \ 180^\circ$ and $270^\circ$.
			*For $s_{\rm A}(t)$  the initial phase is always zero,  for  $s_{\rm B}(t)$  either zero or  $π$,  while for the signal $s_{\rm C}(t)$  with modulation index  $h = 0.5$  a total of four initial phases are possible:  $0^\circ, \ 90^\circ, \ 180^\circ$  and  $270^\circ$.



	'''(7)'''  Correct is the <u>last proposed solution</u>, since for this signal $h = 0.5$ ~~holds~~.		'''(7)'''  Correct is the  <u>last proposed solution</u>,  since for this signal $h = 0.5$.
	*This is the smallest possible modulation index for which there is orthogonality between $f_{1}$ and $f_{2}$ within the symbol duration $T$.		*This is the smallest possible modulation index for which there is orthogonality between  $f_{1}$  and  $f_{2}$  within the symbol duration  $T$.

	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 12:14, 19 January 2023

2023-01-19T12:14:48Z

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	[[File:P_ID1231__Bei_Z_3_4.png\|right\|frame\|Signals for  ~~"Continuous Phase~~ FSK"]]		[[File:P_ID1231__Bei_Z_3_4.png\|right\|frame\|Signals for  $\text{CP-FSK}$]]
	The graph shows three frequency shift keying  $\rm (FSK)$  transmitted signals which differ with respect to the frequency deviation  $\Delta f_{\rm A}$  distinguish and thus also by their modulation index		The graph shows three frequency shift keying  $\rm (FSK)$  transmitted signals which differ with respect to the frequency deviation  $\Delta f_{\rm A}$  distinguish and thus also by their modulation index
	:$$h = 2 \cdot \Delta f_{\rm A} \cdot T.$$		:$$h = 2 \cdot \Delta f_{\rm A} \cdot T.$$
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	<quiz display=simple>		<quiz display=simple>

	{Which statements are true for ~~the~~ FSK and specifically for ~~the~~ MSK?		{Which statements are true for FSK and specifically for MSK?
	\|type="[]"}		\|type="[]"}
	+ FSK is generally a nonlinear modulation method.		+ FSK is generally a nonlinear modulation method.
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	+ The MSK envelope is constant even with spectral shaping.		+ The MSK envelope is constant even with spectral shaping.

	{What frequencies  $f_{1}$  $($for amplitude coefficient  $a_{\nu} = +1)$  and $f_{2}$  $($for  $a_{\nu} = -1)$  does the signal  $s_{\rm A}(t)$ contain?		{What frequencies  $f_{1}$  $($for amplitude coefficient  $a_{\nu} = +1)$  and $f_{2}$  $($for  $a_{\nu} = -1)$  does the signal  $s_{\rm A}(t)$  contain?
	\|type="{}"}		\|type="{}"}
	$f_{1} \cdot T \ = \ $ { 5 3% }		$f_{1} \cdot T \ = \ $ { 5 3% }
	$f_{2} \cdot T \ = \ $ { 3 3% }		$f_{2} \cdot T \ = \ $ { 3 3% }

	{~~For the signal  $s_{\rm A}(t)$  what~~ are the carrier frequency  $f_{\rm T}$, frequency deviation  $\delta f_{\rm A}$  and modulation index  $h$?		{What are the carrier frequency  $f_{\rm T}$,  the frequency deviation  $\delta f_{\rm A}$  and the modulation index  $h$  for signal  $s_{\rm A}(t)$?
	\|type="{}"}		\|type="{}"}
	$f_{\rm T} \cdot T \ = \ $ { 4 3% }		$f_{\rm T} \cdot T \ = \ $ { 4 3% }
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	$h \ = \ $ { 2 3% }		$h \ = \ $ { 2 3% }

	{What is the modulation index at signal  $s_{\rm B}(t)$?		{What is the modulation index for signal  $s_{\rm B}(t)$?
	\|type="{}"}		\|type="{}"}
	$h \ = \ $ { 1 3% }		$h \ = \ $ { 1 3% }

	{What is the modulation index at signal  $s_{\rm C}(t)$?		{What is the modulation index for signal  $s_{\rm C}(t)$?
	\|type="{}"}		\|type="{}"}
	$h \ = \ $ { 0.5 3% }		$h \ = \ $ { 0.5 3% }

Guenter at 08:45, 19 January 2023

2023-01-19T08:45:14Z

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	[[File:P_ID1231__Bei_Z_3_4.png\|right\|frame\|Signals for  ~~$\text{CP-~~FSK}$]]		[[File:P_ID1231__Bei_Z_3_4.png\|right\|frame\|Signals for  "Continuous Phase FSK"]]
	The graph shows three frequency shift keying  $\rm (FSK)$  transmitted signals which differ with respect to the frequency deviation  $\Delta f_{\rm A}$  distinguish and thus also by their modulation index		The graph shows three frequency shift keying  $\rm (FSK)$  transmitted signals which differ with respect to the frequency deviation  $\Delta f_{\rm A}$  distinguish and thus also by their modulation index
	:$$h = 2 \cdot \Delta f_{\rm A} \cdot T.$$		:$$h = 2 \cdot \Delta f_{\rm A} \cdot T.$$

Noah at 18:53, 17 January 2023

2023-01-17T18:53:58Z

Show changes

Guenter at 15:38, 11 January 2023

2023-01-11T15:38:40Z

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	[[File:P_ID1231__Bei_Z_3_4.png\|right\|frame\|Signals for  $\text{CP-FSK$}]]		[[File:P_ID1231__Bei_Z_3_4.png\|right\|frame\|Signals for  $\text{CP-FSK}$]]
	The graph shows three frequency shift keying  $\rm (FSK)$  transmitted signals which differ with respect to the frequency deviation  $\Delta f_{\rm A}$  distinguish and thus also by their modulation index		The graph shows three frequency shift keying  $\rm (FSK)$  transmitted signals which differ with respect to the frequency deviation  $\Delta f_{\rm A}$  distinguish and thus also by their modulation index
	:$$h = 2 \cdot \Delta f_{\rm A} \cdot T.$$		:$$h = 2 \cdot \Delta f_{\rm A} \cdot T.$$

Guenter at 15:37, 11 January 2023

2023-01-11T15:37:54Z

← Older revision		Revision as of 17:37, 11 January 2023
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	[[File:P_ID1231__Bei_Z_3_4.png\|right\|frame\|Signals for  $\rm CP FSK$]]		[[File:P_ID1231__Bei_Z_3_4.png\|right\|frame\|Signals for  $\text{CP-FSK$}]]
	The graph shows three FSK ~~transmit~~ signals which differ with respect to the frequency deviation  $\Delta f_{\rm A}$  and thus also by their modulation index		The graph shows three frequency shift keying  $\rm (FSK)$  transmitted signals which differ with respect to the frequency deviation  $\Delta f_{\rm A}$  distinguish and thus also by their modulation index
	:$$h = 2 \cdot \Delta f_{\rm A} \cdot T$$		:$$h = 2 \cdot \Delta f_{\rm A} \cdot T.$$
	~~distinguish.~~

	The digital source signal  $q(t)$ underlying the signals  $s_{\rm A}(t),  s_{\rm B}(t)$  and  $s_{\rm C}(t)$  is shown above. All considered signals are normalized to amplitude  $1$  and time duration  $T$  and ~~are~~ based on a cosine carrier with frequency $f_{\rm T}$.		The digital source signal  $q(t)$ underlying the signals  $s_{\rm A}(t),  s_{\rm B}(t)$  and  $s_{\rm C}(t)$  is shown above.  All considered signals are normalized to amplitude  $1$  and time duration  $T$  and based on a cosine carrier with frequency  $f_{\rm T}$.

	With binary FSK (''Binary Frequency Shift Keying'') only two different frequencies occur.		With binary FSK  $($"Binary Frequency Shift Keying"$)$  only two different frequencies occur,  each of which remains constant over a bit duration:
	*$f_{1}$  $($if  $a_{\nu} = +1)$,		*$f_{1}$  $($if  $a_{\nu} = +1)$,
	*$f_{2}$  $($if  $a_{\nu} = -1)$

			*$f_{2}$  $($if  $a_{\nu} = -1)$.

	~~each of which remains constant over a bit duration.~~

	If the modulation index is not a multiple of  $2$, continuous phase adjustment is required to avoid phase jumps. This is called  ''Continuous Phase Frequency Shift Keying''  $ (\rm CP– FSK)$.

	An important special case is represented by binary FSK with modulation index  $h = 0.5$  which is also called  ''Minimum Shift Keying''  $(\rm MSK)$. This will be discussed in this exercise.

			If the modulation index is not a multiple of  $2$,  continuous phase adjustment is required to avoid phase jumps.  This is called  "Continuous Phase Frequency Shift Keying"   $(\text{CP-FSK)}$.

			An important special case is represented by binary FSK with modulation index  $h = 0.5$  which is also called  "Minimum Shift Keying"  $(\rm MSK)$.  This will be discussed in this exercise.




			<u>Hints:</u>

			*This exercise belongs to the chapter  [[Examples_of_Communication_Systems/Radio_Interface\|"Radio Interface"]].

	~~Hints:~~		*Reference is made in particular to the section  [[Examples_of_Communication_Systems/Radio_Interface#Continuous_phase_adjustment_with_FSK\|"Continuous phase adjustment with FSK]].
	*This exercise belongs to the chapter  [[Examples_of_Communication_Systems/Radio_Interface\|"Radio Interface"]].
	*Reference is made in particular to the ~~page~~  [[Examples_of_Communication_Systems/Radio_Interface#Continuous_phase_adjustment_with_FSK\|"Continuous phase adjustment with FSK]].

	*The interactive applet  [[Applets:Frequency_Shift_Keying_%26_Continuous_Phase_Modulation\|"Frequency Shift Keying & Continuous Phase Modulation"]]  illustrates the topic covered here.


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