Aufgaben:Exercise 3.9: Circular Arc and Parabola - 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.~~8_Kreisbogen_und_Parabel~~]]		[[de:Aufgaben:Aufgabe 3.9: Kreisbogen und Parabel]]

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	:$$ s_{\rm TP}(t) = \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n\hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm}\cdot \hspace{0.05cm} t}.$$		:$$ s_{\rm TP}(t) = \sum_{n = - \infty}^{+\infty}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n\hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm}\cdot \hspace{0.05cm} t}.$$
	*This represents an arc.  Within the period $T_{\rm N} = 1/f_{\rm N} = 200 \ \rm µ s$  the following phase angles result:		*This represents an arc.  Within the period $T_{\rm N} = 1/f_{\rm N} = 200 \ \rm µ s$  the following phase angles result:
	:$$ \phi(0) = 0, \hspace{0.2cm}\phi(0.25 \cdot T_{\rm N}) = \eta, \hspace{0.2cm}\phi(0.5 \cdot T_{\rm N})= 0,\hspace{0.2cm}		:$$ \phi(0) = 0, \hspace{0.2cm}\phi(0.25 \cdot T_{\rm N}) = \eta, \hspace{0.2cm}\phi(0.5 \cdot T_{\rm N})= 0,\hspace{0.2cm}\phi(0.75 \cdot T_{\rm N})= -\eta,\hspace{0.2cm}\phi(T_{\rm N})= 0.$$
	\phi(0.75 \cdot T_{\rm N})= -\eta,\hspace{0.2cm}\phi(T_{\rm N})= 0.$$
	*Theoretically, the channel bandwidth required to transmit this signal is infinite.		*Theoretically, the channel bandwidth required to transmit this signal is infinite.

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	:$$r_{\rm TP}(t) = {\rm J}_0 + \left[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \gamma} - {\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \gamma} \right]\cdot {\rm J}_1 \hspace{0.27cm} +\left[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\gamma} + {\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\gamma} \right]\cdot {\rm J}_2 = {\rm J}_0 + 2 \cdot {\rm j} \cdot {\rm J}_1 \cdot \sin (\gamma)+ 2 \cdot {\rm J}_2 \cdot \cos (2\gamma)\hspace{0.05cm} .$$		:$$r_{\rm TP}(t) = {\rm J}_0 + \left[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \gamma} - {\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \gamma} \right]\cdot {\rm J}_1 \hspace{0.27cm} +\left[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\gamma} + {\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\gamma} \right]\cdot {\rm J}_2 = {\rm J}_0 + 2 \cdot {\rm j} \cdot {\rm J}_1 \cdot \sin (\gamma)+ 2 \cdot {\rm J}_2 \cdot \cos (2\gamma)\hspace{0.05cm} .$$
	*Thus, for the real part in general, including for   $η = 2.4$, that is,  ${\rm J}_0 = 0$  and  ${\rm J}_2 = 0.43$:		*Thus, for the real part in general, including for   $η = 2.4$, that is,  ${\rm J}_0 = 0$  and  ${\rm J}_2 = 0.43$:
	:$$ x(t) = {\rm J}_0 + 2 \cdot {\rm J}_2 \cdot \cos (2 \omega_{\rm N} t ) = 2 \cdot {\rm J}_2 \cdot \cos (2 \omega_{\rm N} t ) \hspace{0.3cm}		:$$ x(t) = {\rm J}_0 + 2 \cdot {\rm J}_2 \cdot \cos (2 \omega_{\rm N} t ) = 2 \cdot {\rm J}_2 \cdot \cos (2 \omega_{\rm N} t ) \hspace{0.3cm}\Rightarrow \hspace{0.3cm} x_{\rm max} = 2 \cdot {\rm J}_2\hspace{0.15cm}\underline { = 0.86}, \hspace{0.3cm} x_{\rm min} = -x_{\rm max} \hspace{0.15cm}\underline {= -0.86}\hspace{0.05cm}.$$
	\Rightarrow \hspace{0.3cm} x_{\rm max} = 2 \cdot {\rm J}_2\hspace{0.15cm}\underline { = 0.86}, \hspace{0.3cm} x_{\rm min} = -x_{\rm max} \hspace{0.15cm}\underline {= -0.86}\hspace{0.05cm}.$$

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	[[Category:Modulation Methods: Exercises\|^3.2 Frequency Modulation^]]		[[Category:Modulation Methods: Exercises\|^3.2 Frequency Modulation^]]
			[[de:Aufgaben:3.8_Kreisbogen_und_Parabel]]

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Solution

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	:$$x = {\rm J}_0 + 4 \cdot {\rm J}_2 \cdot \cos^2 (\gamma) - 2 \cdot {\rm J}_2\hspace{0.05cm},\hspace{0.3cm} y = 2 \cdot {\rm J}_1 \cdot \sin (\gamma) \hspace{0.05cm}.$$		:$$x = {\rm J}_0 + 4 \cdot {\rm J}_2 \cdot \cos^2 (\gamma) - 2 \cdot {\rm J}_2\hspace{0.05cm},\hspace{0.3cm} y = 2 \cdot {\rm J}_1 \cdot \sin (\gamma) \hspace{0.05cm}.$$

	*These equations can be ~~rearranegd~~:		*These equations can be rearranged:
	:$$\cos^2 (\gamma) =\frac{x-{\rm J}_0 + 2 \cdot {\rm J}_2 }{4 \cdot {\rm J}_2} \hspace{0.05cm},\hspace{0.3cm} \sin^2 (\gamma) = \frac{y^2 }{4 \cdot {\rm J}_1^2}$$		:$$\cos^2 (\gamma) =\frac{x-{\rm J}_0 + 2 \cdot {\rm J}_2 }{4 \cdot {\rm J}_2} \hspace{0.05cm},\hspace{0.3cm} \sin^2 (\gamma) = \frac{y^2 }{4 \cdot {\rm J}_1^2}$$

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	However, if the bandwidth is limited to  $B_{\rm K} = 25 \ \rm kHz$, for example,the equivalent low-pass signal of the received signal can be described as follows:		However, if the bandwidth is limited to  $B_{\rm K} = 25 \ \rm kHz$, for example, the equivalent low-pass signal of the received signal can be described as follows:
	:$$r_{\rm TP}(t) = \sum_{n = - 2}^{+2}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n\hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm}\cdot \hspace{0.05cm} t}.$$		:$$r_{\rm TP}(t) = \sum_{n = - 2}^{+2}{\rm J}_n (\eta) \cdot {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n\hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm N} \hspace{0.05cm}\cdot \hspace{0.05cm} t}.$$
	In this case, the result is a parabolic locus curve		In this case, the result is a parabolic locus curve

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

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Solution

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Bei Frequenzmodulation eines Cosinussignals gilt für den Modulationsindex~~:		'''(1)'''  In the frequency modulation of a cosine signal, the modulation index is:
	:$$ \eta = \frac{K_{\rm FM} \cdot A_{\rm N}}{ \omega_{\rm N}} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} K_{\rm FM} = \frac{2 \pi \cdot f_{\rm N }\cdot \eta}{ A_{\rm N}} = \frac{2 \pi \cdot 5 \cdot 10^3 \,{\rm Hz}\cdot 2.4}{ 1\,{\rm V}} \hspace{0.15cm}\underline {\approx 7.54 }\cdot 10^4 \,\,{\rm V}^{-1}{\rm s}^{-1}\hspace{0.05cm}.$$		:$$ \eta = \frac{K_{\rm FM} \cdot A_{\rm N}}{ \omega_{\rm N}} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} K_{\rm FM} = \frac{2 \pi \cdot f_{\rm N }\cdot \eta}{ A_{\rm N}} = \frac{2 \pi \cdot 5 \cdot 10^3 \,{\rm Hz}\cdot 2.4}{ 1\,{\rm V}} \hspace{0.15cm}\underline {\approx 7.54 }\cdot 10^4 \,\,{\rm V}^{-1}{\rm s}^{-1}\hspace{0.05cm}.$$



	'''(2)'''  ~~Die angegebene Gleichung für das äquivalente TP–Signal lautet in ausgeschriebener Form mit~~  $γ = ω_{\rm N} · t$  ~~unter Berücksichtigung von~~  ${\rm J}_{–1} = –{\rm J}_1$  ~~und~~  ${\rm J}_{–2} = {\rm J}_2$:		'''(2)'''  The equation given for the equivalent low-pass signal when  $γ = ω_{\rm N} · t$  and considering   ${\rm J}_{–1} = –{\rm J}_1$  and  ${\rm J}_{–2} = {\rm J}_2$, is written out as:
	:$$r_{\rm TP}(t) = {\rm J}_0 + \left[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \gamma} - {\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \gamma} \right]\cdot {\rm J}_1 \hspace{0.27cm} +\left[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\gamma} + {\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\gamma} \right]\cdot {\rm J}_2 = {\rm J}_0 + 2 \cdot {\rm j} \cdot {\rm J}_1 \cdot \sin (\gamma)+ 2 \cdot {\rm J}_2 \cdot \cos (2\gamma)\hspace{0.05cm} .$$		:$$r_{\rm TP}(t) = {\rm J}_0 + \left[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \gamma} - {\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \gamma} \right]\cdot {\rm J}_1 \hspace{0.27cm} +\left[ {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\gamma} + {\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2\gamma} \right]\cdot {\rm J}_2 = {\rm J}_0 + 2 \cdot {\rm j} \cdot {\rm J}_1 \cdot \sin (\gamma)+ 2 \cdot {\rm J}_2 \cdot \cos (2\gamma)\hspace{0.05cm} .$$
	*~~Somit ergibt sich für den Realteil allgemein bzw. für~~  $η = 2.4$, ~~das heißt~~  ${\rm J}_0 = 0$  ~~und~~  ${\rm J}_2 = 0.43$:		*Thus, for the real part in general, including for   $η = 2.4$, that is,  ${\rm J}_0 = 0$  and  ${\rm J}_2 = 0.43$:
	:$$ x(t) = {\rm J}_0 + 2 \cdot {\rm J}_2 \cdot \cos (2 \omega_{\rm N} t ) = 2 \cdot {\rm J}_2 \cdot \cos (2 \omega_{\rm N} t ) \hspace{0.3cm}		:$$ x(t) = {\rm J}_0 + 2 \cdot {\rm J}_2 \cdot \cos (2 \omega_{\rm N} t ) = 2 \cdot {\rm J}_2 \cdot \cos (2 \omega_{\rm N} t ) \hspace{0.3cm}
	\Rightarrow \hspace{0.3cm} x_{\rm max} = 2 \cdot {\rm J}_2\hspace{0.15cm}\underline { = 0.86}, \hspace{0.3cm} x_{\rm min} = -x_{\rm max} \hspace{0.15cm}\underline {= -0.86}\hspace{0.05cm}.$$		\Rightarrow \hspace{0.3cm} x_{\rm max} = 2 \cdot {\rm J}_2\hspace{0.15cm}\underline { = 0.86}, \hspace{0.3cm} x_{\rm min} = -x_{\rm max} \hspace{0.15cm}\underline {= -0.86}\hspace{0.05cm}.$$
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	'''(3)'''  ~~Entsprechend dem Ergebnis der Teilaufgabe~~  '''(2)'''  ~~erhält man für den Imaginärteil~~  $({\rm J}_1 = 0.52)$:		'''(3)'''  According to the result of question  '''(2)'''  we get an imginary part   $({\rm J}_1 = 0.52)$ of:
	:$$y(t) = 2 \cdot {\rm J}_1 \cdot \sin ( \omega_{\rm N} t )\hspace{0.3cm}\Rightarrow \hspace{0.3cm} y_{\rm max} =2 \cdot {\rm J}_1\hspace{0.15cm}\underline { = 1.04}\hspace{0.05cm}, \hspace{0.3cm} y_{\rm min} = -y_{\rm max}\hspace{0.15cm}\underline { = -1.04}\hspace{0.05cm}.$$		:$$y(t) = 2 \cdot {\rm J}_1 \cdot \sin ( \omega_{\rm N} t )\hspace{0.3cm}\Rightarrow \hspace{0.3cm} y_{\rm max} =2 \cdot {\rm J}_1\hspace{0.15cm}\underline { = 1.04}\hspace{0.05cm}, \hspace{0.3cm} y_{\rm min} = -y_{\rm max}\hspace{0.15cm}\underline { = -1.04}\hspace{0.05cm}.$$



	'''(4)'''  ~~Der Imaginärteil ist zu diesen Zeitpunkten jeweils Null und damit auch die Phasenfunktion~~: ~~ ~~		'''(4)'''  The imaginary part is zero at each of these time points and so is the phase function:
	:$$ϕ(t = n · T_{\rm N}/2)\hspace{0.15cm}\underline{= 0}.$$		:$$ϕ(t = n · T_{\rm N}/2)\hspace{0.15cm}\underline{= 0}.$$
	*~~Diesen Sachverhalt erkennt man auch aus der Skizze auf der Angabenseite~~.		*This fact can also be seen from the sketch on the exercise page.




	'''(5)'''  ~~Aus der Skizze ist bereits zu erkennen, dass der Phasenwinkel beispielsweise für~~  $t = T_{\rm N}/4$  ~~seinen Maximalwert erreicht~~.		'''(5)'''  From the sketch it can already be seen that the phase angle reaches its maximum value at  $t = T_{\rm N}/4$ , for example.
	*~~Dieser kann mit~~  $y_{\rm max} = 1.04$  ~~und~~  $x_{\rm min} = -0.86$  ~~wie folgt berechnet werden~~:		*This can be calculated with  $y_{\rm max} = 1.04$  and  $x_{\rm min} = -0.86$  as follows:
	:$$\phi_{\rm max} = \arctan \frac{y_{\rm max}}{x_{\rm min}} = \arctan (-1.21) = 180^\circ - 50.4^\circ \hspace{0.15cm}\underline {= 129.6^\circ} \hspace{0.05cm}.$$		:$$\phi_{\rm max} = \arctan \frac{y_{\rm max}}{x_{\rm min}} = \arctan (-1.21) = 180^\circ - 50.4^\circ \hspace{0.15cm}\underline {= 129.6^\circ} \hspace{0.05cm}.$$
	*~~Ohne Bandbegrenzung würde sich hier der Phasenwinkel ~~ $ϕ(t = T_{\rm N}/4) = η = 2.4 = 137.5^\circ$  ~~ergeben~~.		*Without band-limiting, the phase angle here would be $ϕ(t = T_{\rm N}/4) = η = 2.4 = 137.5^\circ$ .
	*~~Die maximale Abweichung des Sinkensignals vom Quellensignal tritt somit beispielsweise zur Zeit~~  $t = T_{\rm N}/4$  ~~auf~~.		*Thus, the maximum deviation of the sink signal from the source signal occurs at time  $t = T_{\rm N}/4$ , for example.



	[[File:P_ID1113__Mod_A_3_8_f.png\|right\|frame\|~~Parabelverlauf~~]]		[[File:P_ID1113__Mod_A_3_8_f.png\|right\|frame\|Parabolic progression]]
	'''(6)'''  ~~Mit~~  $γ = ω_{\rm N} · t$  ~~und~~  $\cos(2γ) = 1 - 2 · \cos^2(γ)$  ~~kann für Real– und Imaginärteil geschrieben werden~~:		'''(6)'''  Using  $γ = ω_{\rm N} · t$  and  $\cos(2γ) = 1 - 2 · \cos^2(γ)$  the real and imaginary parts can be written as:

	:$$x = {\rm J}_0 + 4 \cdot {\rm J}_2 \cdot \cos^2 (\gamma) - 2 \cdot {\rm J}_2\hspace{0.05cm},\hspace{0.3cm} y = 2 \cdot {\rm J}_1 \cdot \sin (\gamma) \hspace{0.05cm}.$$		:$$x = {\rm J}_0 + 4 \cdot {\rm J}_2 \cdot \cos^2 (\gamma) - 2 \cdot {\rm J}_2\hspace{0.05cm},\hspace{0.3cm} y = 2 \cdot {\rm J}_1 \cdot \sin (\gamma) \hspace{0.05cm}.$$

	*~~Diese Gleichungen können wie folgt umgeformt werden~~:		*These equations can be rearranegd:
	:$$\cos^2 (\gamma) =\frac{x-{\rm J}_0 + 2 \cdot {\rm J}_2 }{4 \cdot {\rm J}_2} \hspace{0.05cm},\hspace{0.3cm} \sin^2 (\gamma) = \frac{y^2 }{4 \cdot {\rm J}_1^2}$$		:$$\cos^2 (\gamma) =\frac{x-{\rm J}_0 + 2 \cdot {\rm J}_2 }{4 \cdot {\rm J}_2} \hspace{0.05cm},\hspace{0.3cm} \sin^2 (\gamma) = \frac{y^2 }{4 \cdot {\rm J}_1^2}$$

	$$\Rightarrow \hspace{0.3cm} \frac{y^2 }{4 \cdot {\rm J}_1^2} + \frac{x-{\rm J}_0 + 2 \cdot {\rm J}_2 }{4 \cdot {\rm J}_2} =1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {y^2 } + \frac{{\rm J}_1^2}{ {\rm J}_2} \cdot x + {{\rm J}_1^2} \cdot \left ( 2 - \frac{{\rm J}_0}{ {\rm J}_2} \right ) =0\hspace{0.05cm}.$$		$$\Rightarrow \hspace{0.3cm} \frac{y^2 }{4 \cdot {\rm J}_1^2} + \frac{x-{\rm J}_0 + 2 \cdot {\rm J}_2 }{4 \cdot {\rm J}_2} =1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {y^2 } + \frac{{\rm J}_1^2}{ {\rm J}_2} \cdot x + {{\rm J}_1^2} \cdot \left ( 2 - \frac{{\rm J}_0}{ {\rm J}_2} \right ) =0\hspace{0.05cm}.$$

	*~~Damit lauten die Parabelparameter für~~  $ {\rm J}_0 = 0$:		*Thus, the parabolic parameters for  $ {\rm J}_0 = 0$ are:
	:$$a = \frac{{\rm J}_1^2}{ {\rm J}_2} \hspace{0.15cm}\underline {\approx 0.629}, \hspace{0.3cm} b = 2 \cdot {\rm J}_1^2 \hspace{0.15cm}\underline {\approx 0.541} \hspace{0.05cm}.$$		:$$a = \frac{{\rm J}_1^2}{ {\rm J}_2} \hspace{0.15cm}\underline {\approx 0.629}, \hspace{0.3cm} b = 2 \cdot {\rm J}_1^2 \hspace{0.15cm}\underline {\approx 0.541} \hspace{0.05cm}.$$
	*~~Zur Kontrolle verwenden wir~~  $y = 0$   ⇒   $ x_{\rm max} = {b}/{a} = 2 \cdot {\rm J}_2 = 0.86 \hspace{0.05cm}.$		*To double-check, we use   $y = 0$   ⇒   $ x_{\rm max} = {b}/{a} = 2 \cdot {\rm J}_2 = 0.86 \hspace{0.05cm}.$
	*~~Die Werte bei~~  $x = 0$  ~~sind somit:~~     $y_0 = \pm \sqrt{2} \cdot {\rm J}_1 \approx 0.735 \hspace{0.05cm}.$		*Thus, the values at  $x = 0$  are     $y_0 = \pm \sqrt{2} \cdot {\rm J}_1 \approx 0.735 \hspace{0.05cm}.$

	{{ML-Fuß}}		{{ML-Fuß}}

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	</quiz>		</quiz>

	===~~Musterlösung~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Bei Frequenzmodulation eines Cosinussignals gilt für den Modulationsindex:		'''(1)'''  Bei Frequenzmodulation eines Cosinussignals gilt für den Modulationsindex:

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Question

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	<quiz display=simple>		<quiz display=simple>
	{~~Wie groß ist die Modulatorkonstante~~  $K_{\rm FM}$?		{What is the modulator constant  $K_{\rm FM}$?
	\|type="{}"}		\|type="{}"}
	$K_{\rm FM} \ = \ $ { 7.54 3% } $\ \cdot 10^4 \ \rm (Vs)^{-1}$		$K_{\rm FM} \ = \ $ { 7.54 3% } $\ \cdot 10^4 \ \rm (Vs)^{-1}$


	{~~Berechnen Sie den Realteil~~  $x(t) = {\rm Re}\big[r_{\rm TP}(t)\big]$  ~~des äquivalenten Tiefpass~~-~~Signals und geben Sie dessen Maximum und Minimum an~~.		{Calculate the real part $x(t) = {\rm Re}\big[r_{\rm TP}(t)\big]$  of the equivalent low-pass signal and give its maximum and minimum.
	\|type="{}"}		\|type="{}"}
	$x_{\rm max} \ = \ $ { 0.86 3% }		$x_{\rm max} \ = \ $ { 0.86 3% }
	$x_{\rm min} \ = \ $ { -0.89--0.83 }		$x_{\rm min} \ = \ $ { -0.89--0.83 }

	{~~Wie groß ist das Maximum und Minimum des Imaginärteils~~  $y(t) = {\rm Im}\big[r_{\rm TP}(t)\big]$?		{What is the maximum and minimum of the imaginary part  $y(t) = {\rm Im}\big[r_{\rm TP}(t)\big]$?
	\|type="{}"}		\|type="{}"}
	$y_{\rm max} \ = \ $ { 1.04 3% }		$y_{\rm max} \ = \ $ { 1.04 3% }
	$y_{\rm min} \ = \ $ { -1.07--1.01 }		$y_{\rm min} \ = \ $ { -1.07--1.01 }

	{~~Welche Phasenwerte ergeben sich bei allen Vielfachen von~~  $T_{\rm N}/2$?		{What are the phase values for all multiples of  $T_{\rm N}/2$?
	\|type="{}"}		\|type="{}"}
	$ϕ(t = n · T_{\rm N}/2) \ = \ $ { 0. } $\ \rm ~~Grad~~$		$ϕ(t = n · T_{\rm N}/2) \ = \ $ { 0. } $\ \rm degrees$

	{~~Wie groß ist der maximale Phasenwinkel~~  $ϕ_{\rm max}$?  ~~Interpretieren Sie das Ergebnis~~.		{What is the maximum phase angle  $ϕ_{\rm max}$?  Interpret the result.
	\|type="{}"}		\|type="{}"}
	$ϕ_{\rm max} \ = \ $ { 129.6 3% } $\ \rm ~~Grad~~$		$ϕ_{\rm max} \ = \ $ { 129.6 3% } $\ \rm degrees$

	{~~Zeigen Sie, dass man die Ortskurve~~ in ~~der Form~~  $y^2 + a · x + b = 0$  ~~angeben kann~~.  ~~Bestimmen Sie die Parabelparameter~~  $a$  ~~und~~  $b$.		{Show that the locus curve can be given in the form  $y^2 + a · x + b = 0$ .  Determine the parabolic parameters  $a$  and  $b$.
	\|type="{}"}		\|type="{}"}
	$a\ = \ $ { 0.629 3% }		$a\ = \ $ { 0.629 3% }

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	'''Hints:''		''Hints:''
	*This exercise belongs to the chapter  [[Modulation_Methods/Frequency_Modulation_(FM)\|Frequency Modulation]].		*This exercise belongs to the chapter  [[Modulation_Methods/Frequency_Modulation_(FM)\|Frequency Modulation]].
	*Reference is also made to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)\|Phase Modulation]]  and particularly to the section  [[Modulation_Methods/Frequency_Modulation_(FM)#Signal_characteristics_with_frequency_modulation\|Signal characteristics with frequency modulation]].		*Reference is also made to the chapter  [[Modulation_Methods/Phase_Modulation_(PM)\|Phase Modulation]]  and particularly to the section  [[Modulation_Methods/Frequency_Modulation_(FM)#Signal_characteristics_with_frequency_modulation\|Signal characteristics with frequency modulation]].