Aufgaben:Exercise 3.5: PM and FM for Rectangular Signals - Revision history

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	[[Category:Modulation Methods: Exercises\|^3.2 Frequency Modulation^]]		[[Category:Modulation Methods: Exercises\|^3.2 Frequency Modulation^]]
	[[de:Aufgaben:3.~~5_PM_und_FM_bei_Rechtecken~~]]		[[de:Aufgaben:Aufgabe 3.5: PM und FM bei Rechtecksignalen]]

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	[[Category:Modulation Methods: Exercises\|^3.2 Frequency Modulation^]]		[[Category:Modulation Methods: Exercises\|^3.2 Frequency Modulation^]]
			[[de:Aufgaben:3.5_PM_und_FM_bei_Rechtecken]]

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	Assume a bipolar and rectangular source signal $q(t)$ , as shown in the upper diagram.  This signal can only take on the two signal values  $±A = ±2 \ \rm V$  and the duration of the positive and negative rectangles are each $T = 1 \ \rm ms$.  The period of  $q(t)$  is therefore  $T_0 = 2 \ \rm ms$.		Assume a bipolar and rectangular source signal $q(t)$ , as shown in the upper diagram.  This signal can only take on the two signal values  $±A = ±2 \ \rm V$  and the duration of the positive and negative rectangles are each $T = 1 \ \rm ms$.  The period of  $q(t)$  is therefore  $T_0 = 2 \ \rm ms$.

	The signals $s_1(t)$  and  $s_2(t)$  display two ~~transmit~~ signals with angle modulation  $\rm (WM)$, each of which can be represented as		The signals $s_1(t)$  and  $s_2(t)$  display two transmitted signals with angle modulation  $\rm (WM)$, each of which can be represented as
	:$$s(t) = A_{\rm T} \cdot \cos \hspace{-0.05cm}\big [\psi (t) \big ]$$		:$$s(t) = A_{\rm T} \cdot \cos \hspace{-0.05cm}\big [\psi (t) \big ]$$
	Here, we distinguish between phase modulation  $\rm (PM)$  with the angular function		Here, we distinguish between phase modulation  $\rm (PM)$  with the angular function

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	{{quiz-Header\|Buchseite=~~Modulationsverfahren~~/~~Frequenzmodulation~~ (FM)		{{quiz-Header\|Buchseite=Modulation_Methods/Frequency_Modulation_(FM)
	}}		}}

	[[File:P_ID1099__Mod_A_3_5.png\|right\|frame\|~~Zwei Signalverläufe bei Winkelmodulation~~ ]]		[[File:P_ID1099__Mod_A_3_5.png\|right\|frame\|Two signal waveforms in angle modulation]]
	Assume a bipolar and rectangular source signal $q(t)$ , as shown in the upper diagram.  This signal can only take on the two signal values  $±A = ±2 \ \rm V$  and the duration of the positive and negative rectangles are each $T = 1 \ \rm ms$.  The period of  $q(t)$  is therefore  $T_0 = 2 \ \rm ms$.		Assume a bipolar and rectangular source signal $q(t)$ , as shown in the upper diagram.  This signal can only take on the two signal values  $±A = ±2 \ \rm V$  and the duration of the positive and negative rectangles are each $T = 1 \ \rm ms$.  The period of  $q(t)$  is therefore  $T_0 = 2 \ \rm ms$.

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	'''(6)'''  ~~Der Frequency~~ deviation can also be represented as follows:		'''(6)'''  The frequency deviation can also be represented as follows:
	:$$\Delta f_{\rm A} = \frac {K_{\rm FM}}{2\pi}\cdot A \hspace{0.05cm}.$$		:$$\Delta f_{\rm A} = \frac {K_{\rm FM}}{2\pi}\cdot A \hspace{0.05cm}.$$
	*~~Mit~~  $Δf_{\rm A} · {\rm A} = 2$  ~~erhält man somit~~:		*With   $Δf_{\rm A} · {\rm A} = 2$  we thus get:
	:$$K_{\rm FM} = \frac {2 \cdot 2\pi}{A \cdot T}= \frac {4\pi}{2\,{\rm V} \cdot 1\,{\rm ms}}\hspace{0.15cm}\underline {= 6283 \,{\rm V}^{-1}{\rm s}^{-1}} \hspace{0.05cm}.$$		:$$K_{\rm FM} = \frac {2 \cdot 2\pi}{A \cdot T}= \frac {4\pi}{2\,{\rm V} \cdot 1\,{\rm ms}}\hspace{0.15cm}\underline {= 6283 \,{\rm V}^{-1}{\rm s}^{-1}} \hspace{0.05cm}.$$

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	*By counting the oscillations of  $s_2(t)$  in the time interval  $T$ , it can be seen that  $f_{\rm T} · T\hspace{0.15cm}\underline{ = 6}$  was used.		*By counting the oscillations of  $s_2(t)$  in the time interval  $T$ , it can be seen that  $f_{\rm T} · T\hspace{0.15cm}\underline{ = 6}$  was used.
	*When frequency modulating a bipolar source signal,   $f_{\rm T}$  does not occur directly.		*When frequency modulating a bipolar source signal,   $f_{\rm T}$  does not occur directly.
	*However, the graphs ~~Grafiken~~ do indicate that   $f_{\rm T} · T = 6$  is also used here.		*However, the graphs do indicate that   $f_{\rm T} · T = 6$  is also used here.

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	'''(1)'''  <u>Answer 2</u> is correct:		'''(1)'''  <u>Answer 2</u> is correct:
	*For a rectangular (digital) source signal, phase modulation (PM) can be recognised by the typical phase jumps – see the signal waveform  $s_2(t)$.		*For a rectangular (digital) source signal, phase modulation (PM) can be recognised by the typical phase jumps – see the signal waveform  $s_2(t)$.
	*Frequency modulation (FM), on the other hand, has diverse instantaneous frequencies at different times as in  $s_1(t)$.		*Frequency modulation (FM), on the other hand, has diverse instantaneous frequencies at different times, as in  $s_1(t)$.

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	{Which of the signals is due to phase modulation and which is due to frquency modulation?		{Which of the signals is due to phase modulation and which is due to frquency modulation?
	\|type="()"}		\|type="()"}
	- $s_1(t)$  ~~describes~~ a phase modulation.		- $s_1(t)$  represents a phase modulation.
	+ $s_1(t)$  ~~describes~~ a frequency modulation.		+ $s_1(t)$  represents a frequency modulation.

	{What is the carrier phase  $ϕ_{\rm T}$ that could be measured without a message signal   ⇒   $q(t) \equiv 0$ ?		{What is the carrier phase  $ϕ_{\rm T}$ that could be measured without a message signal   ⇒   $q(t) \equiv 0$ ?

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	*By counting the oscillations of  $s_2(t)$  in the time interval  $T$ , it can be seen that  $f_{\rm T} · T\hspace{0.15cm}\underline{ = 6}$  was used.		*By counting the oscillations of  $s_2(t)$  in the time interval  $T$ , it can be seen that  $f_{\rm T} · T\hspace{0.15cm}\underline{ = 6}$  was used.
	*When frequency modulating a bipolar source signal,   $f_{\rm T}$  does not occur directly.		*When frequency modulating a bipolar source signal,   $f_{\rm T}$  does not occur directly.
	*~~however~~, the graphs Grafiken do indicate that   $f_{\rm T} · T = 6$  is also used here.		*However, the graphs Grafiken do indicate that   $f_{\rm T} · T = 6$  is also used here.

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	'''(1)'''  <u>Answer 2</u> is correct:		'''(1)'''  <u>Answer 2</u> is correct:
	*For a rectangular (digital) source signal, phase modulation (PM) can be recognised by the typical phase jumps – see the signal waveform  $s_2(t)$.		*For a rectangular (digital) source signal, phase modulation (PM) can be recognised by the typical phase jumps – see the signal waveform  $s_2(t)$.
	*Frequency modulation (FM), on the other hand, has a diverse instantaneous frequencies at different times as in  $s_1(t)$.		*Frequency modulation (FM), on the other hand, has diverse instantaneous frequencies at different times as in  $s_1(t)$.