Aufgaben:Exercise 3.2Z: Bessel Spectrum - Revision history

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	[[Category:Modulation Methods: Exercises\|^3.1 Phase Modulation^]]		[[Category:Modulation Methods: Exercises\|^3.1 Phase Modulation^]]
	[[de:Aufgaben:3.~~2Z_Besselspektrum~~]]		[[de:Aufgaben:Aufgabe 3.2Z: Besselspektrum]]

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	[[Category:Modulation Methods: Exercises\|^3.1 Phase Modulation^]]		[[Category:Modulation Methods: Exercises\|^3.1 Phase Modulation^]]
			[[de:Aufgaben:3.2Z_Besselspektrum]]

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	{{quiz-Header\|Buchseite=~~Modulationsverfahren~~/~~Phasenmodulation~~ (PM)		{{quiz-Header\|Buchseite=Modulation_Methods/Phase_Modulation_(PM)
	}}		}}

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	*Particular reference is made to the page  [[Modulation_Methods/Phase_Modulation_(PM)#Equivalent_low-pass_signal_in_phase_modulation\|Equivalent low-pass signal in phase modulation]].		*Particular reference is made to the page  [[Modulation_Methods/Phase_Modulation_(PM)#Equivalent_low-pass_signal_in_phase_modulation\|Equivalent low-pass signal in phase modulation]].
	*The values of the Bessel functions can be found in collections of formulae in table form.		*The values of the Bessel functions can be found in collections of formulae in table form.
	*You can also use the interactive applet   [[Applets:~~Bessel_functions_of_the_first_kindt~~\| Bessel functions of the first kind]]  to solve this task.		*You can also use the interactive applet   [[Applets:Bessel_functions_of_the_first_kind\| Bessel functions of the first kind]]  to solve this task.

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	*Particular reference is made to the page  [[Modulation_Methods/Phase_Modulation_(PM)#Equivalent_low-pass_signal_in_phase_modulation\|Equivalent low-pass signal in phase modulation]].		*Particular reference is made to the page  [[Modulation_Methods/Phase_Modulation_(PM)#Equivalent_low-pass_signal_in_phase_modulation\|Equivalent low-pass signal in phase modulation]].
	*The values of the Bessel functions can be found in collections of formulae in table form.		*The values of the Bessel functions can be found in collections of formulae in table form.
	*You can also use the interactive applet   [[Applets:~~Besselfunktionen_erster_Art~~\| Bessel functions of the first kind]]  to solve this task.		*You can also use the interactive applet   [[Applets:Bessel_functions_of_the_first_kindt\| Bessel functions of the first kind]]  to solve this task.

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Solution

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Richtig ist nur der~~ <u>~~zweite Lösungsvorschlag~~</u>:		'''(1)'''  Only the <u>second answer</u> is correct:
	*$x(t)$  ~~ist ein komplexes Signal, das nur~~ in ~~Ausnahmefällen reell wird~~, ~~zum Beispiel zur Zeit~~  $t = 0$.		*$x(t)$  is a complex signal that only becomes real in exceptional cases, for example at time   $t = 0$.
	*~~Ein rein imaginärer Wert~~ (~~zu gewissen Zeiten~~) ~~kann sich nur dann ergeben, wenn ~~ $η ≥ π/2$  ~~ist~~   ⇒   ~~Antwort~~ 1 ~~ist falsch~~.		*A purely imaginary value (at certain times) can only result when $η ≥ π/2$    ⇒   Answer 1 is incorrect.
	*~~Mit~~  $T_0 = 2π/ω_0$  ~~gilt beispielsweise~~:		*For example, when  $T_0 = 2π/ω_0$ :
	:$$ x(t + k \cdot T_0) = {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} (t \hspace{0.05cm}+ \hspace{0.05cm} k \hspace{0.05cm}\cdot \hspace{0.05cm}T_0)) } =		:$$ x(t + k \cdot T_0) = {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} (t \hspace{0.05cm}+ \hspace{0.05cm} k \hspace{0.05cm}\cdot \hspace{0.05cm}T_0)) } =
	{\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} t \hspace{0.05cm} + \hspace{0.05cm} k \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi) } ={\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} t \hspace{0.05cm} ) } = x(t)\hspace{0.05cm}.$$		{\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} t \hspace{0.05cm} + \hspace{0.05cm} k \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi) } ={\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} t \hspace{0.05cm} ) } = x(t)\hspace{0.05cm}.$$
	*~~Dieses Signal ist periodisch~~.  ~~Zur Berechnung der Spektralfunktion muss die Fourierreihe und nicht das Fourierintegral herangezogen werden~~.		*This signal is periodic.  The Fourier series, not the Fourier integral, must be used to calculate the spectral function.



	'''(2)'''  ~~Die Fourierkoeffizienten lauten~~:		'''(2)'''  The Fourier coefficients are:
	:$$ D_n = \frac{1}{T_0}\cdot \int_{- T_0/2}^{+T_0/2}{\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm} \cdot \hspace{0.05cm} t) }\cdot{\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm 0} \hspace{0.05cm} \cdot \hspace{0.05cm} t }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$		:$$ D_n = \frac{1}{T_0}\cdot \int_{- T_0/2}^{+T_0/2}{\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm} \cdot \hspace{0.05cm} t) }\cdot{\rm e}^{\hspace{0.05cm}{-\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}n \hspace{0.05cm}\cdot \hspace{0.05cm}\omega_{\rm 0} \hspace{0.05cm} \cdot \hspace{0.05cm} t }\hspace{0.1cm}{\rm d}t \hspace{0.05cm}.$$
	*~~Durch Zusammenfassen der beiden Terme und nach der Substitution~~  $α = ω_0 · t$  ~~erhält man~~:		*Combining the two terms and after substituting   $α = ω_0 · t$ , we get:
	:$$D_n = \frac{1}{2\pi}\cdot \int_{-\pi}^{+\pi} {{\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin(\alpha) -\hspace{0.05cm} n \hspace{0.05cm}\cdot \hspace{0.05cm}\alpha)}}\hspace{0.1cm}{\rm d}\alpha \hspace{0.05cm} = {\rm J}_n (\eta) .$$		:$$D_n = \frac{1}{2\pi}\cdot \int_{-\pi}^{+\pi} {{\rm e}^{\hspace{0.05cm}{\rm j}\hspace{0.05cm}\cdot \hspace{0.05cm}(\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin(\alpha) -\hspace{0.05cm} n \hspace{0.05cm}\cdot \hspace{0.05cm}\alpha)}}\hspace{0.1cm}{\rm d}\alpha \hspace{0.05cm} = {\rm J}_n (\eta) .$$
	*~~Richtig ist also der~~ <u>~~zweite Lösungsvorschlag~~</u>.		*Thus, the <u>second answer</u> is correct.



	'''(3)'''  ~~Mit dem Satz von~~ Euler ~~können die Fourierkoeffizienten wie folgt dargestellt werden~~:		'''(3)'''  Using Euler's theorem, the Fourier coefficients can be represented as follows:
	:$$D_n = \frac{1}{2\pi}\cdot \int_{-\pi}^{+\pi} {\cos( \eta \cdot \sin(\alpha) - n \cdot \alpha)}\hspace{0.1cm}{\rm d}\alpha +		:$$D_n = \frac{1}{2\pi}\cdot \int_{-\pi}^{+\pi} {\cos( \eta \cdot \sin(\alpha) - n \cdot \alpha)}\hspace{0.1cm}{\rm d}\alpha +
	\frac{\rm j}{2\pi}\cdot \int_{-\pi}^{+\pi} {\sin( \eta \cdot \sin(\alpha) - n \cdot \alpha)}\hspace{0.1cm}{\rm d}\alpha \hspace{0.05cm}.$$		\frac{\rm j}{2\pi}\cdot \int_{-\pi}^{+\pi} {\sin( \eta \cdot \sin(\alpha) - n \cdot \alpha)}\hspace{0.1cm}{\rm d}\alpha \hspace{0.05cm}.$$
	*~~Der Integrand des ersten Integrals ist eine gerade Funktion von~~  $\alpha$:		*The integrand of the first integral is an even function of  $\alpha$:
	:$$I_1 (-\alpha) = {\cos( \eta \cdot \sin(-\alpha) + n \cdot \alpha)} = {\cos( -\eta \cdot \sin(\alpha) + n \cdot \alpha)}=		:$$I_1 (-\alpha) = {\cos( \eta \cdot \sin(-\alpha) + n \cdot \alpha)} = {\cos( -\eta \cdot \sin(\alpha) + n \cdot \alpha)}=
	{\cos( \eta \cdot \sin(\alpha) - n \cdot \alpha)} = I_1 (\alpha) \hspace{0.05cm}.$$		{\cos( \eta \cdot \sin(\alpha) - n \cdot \alpha)} = I_1 (\alpha) \hspace{0.05cm}.$$
	*~~Dagegen ist der zweite Integrand eine ungerade Funktion~~:		*In contrast, the second integrand is an odd function:
	:$$I_2 (-\alpha) = {\sin( \eta \cdot \sin(-\alpha) + n \cdot \alpha)} = {\sin( -\eta \cdot \sin(\alpha) + n \cdot \alpha)}=		:$$I_2 (-\alpha) = {\sin( \eta \cdot \sin(-\alpha) + n \cdot \alpha)} = {\sin( -\eta \cdot \sin(\alpha) + n \cdot \alpha)}=
	-{\sin( \eta \cdot \sin(\alpha) - n \cdot \alpha)} = -I_2 (\alpha) \hspace{0.05cm}.$$		-{\sin( \eta \cdot \sin(\alpha) - n \cdot \alpha)} = -I_2 (\alpha) \hspace{0.05cm}.$$
	*~~Somit verschwindet das zweite Integral und man erhält unter Berücksichtigung der Symmetrie~~:		*Thus, the second integral vanishes and, taking symmetry into account, we obtain:
	:$$D_n = \frac{1}{\pi}\cdot \int_{0}^{\pi} {\cos( \eta \cdot \sin(\alpha) - n \cdot \alpha)}\hspace{0.1cm}{\rm d}\alpha \hspace{0.05cm}.$$		:$$D_n = \frac{1}{\pi}\cdot \int_{0}^{\pi} {\cos( \eta \cdot \sin(\alpha) - n \cdot \alpha)}\hspace{0.1cm}{\rm d}\alpha \hspace{0.05cm}.$$
	*~~Richtig ist somit der~~ <u>~~Lösungsvorschlag~~ 1</u>.		*Thus, the correct solution is <u>Answer 1</u>.



	'''(4)'''  ~~Entsprechend der iterativen Berechnungsformel gilt für~~  $η = 2$:		'''(4)'''  According to the formula for iterative calculation, when  $η = 2$:
	:$$ D_2 = D_1 - D_0 = 0.577 - 0.224 \hspace{0.15cm}\underline {= 0.353} \hspace{0.05cm},$$		:$$ D_2 = D_1 - D_0 = 0.577 - 0.224 \hspace{0.15cm}\underline {= 0.353} \hspace{0.05cm},$$
	:$$D_3 = 2 \cdot D_2 - D_1 = 2 \cdot 0.353 - 0.577 \hspace{0.15cm}\underline {= 0.129} \hspace{0.05cm}.$$		:$$D_3 = 2 \cdot D_2 - D_1 = 2 \cdot 0.353 - 0.577 \hspace{0.15cm}\underline {= 0.129} \hspace{0.05cm}.$$


	'''(5)'''  ~~Aufgrund der angegebenen Symmetriebeziehung gilt weiter~~:		'''(5)'''  Due to the given symmetry relation, it further holds that:
	:$$ D_{–2} = D_2\hspace{0.15cm}\underline {= 0.353} \hspace{0.05cm},$$		:$$ D_{–2} = D_2\hspace{0.15cm}\underline {= 0.353} \hspace{0.05cm},$$
	:$$D_{–3} = -D_3 \hspace{0.15cm}\underline {= -0.129} \hspace{0.05cm}.$$		:$$D_{–3} = -D_3 \hspace{0.15cm}\underline {= -0.129} \hspace{0.05cm}.$$

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	</quiz>		</quiz>

	===~~Musterlösung~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Richtig ist nur der <u>zweite Lösungsvorschlag</u>:		'''(1)'''  Richtig ist nur der <u>zweite Lösungsvorschlag</u>:

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Questions

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	\|type="()"}		\|type="()"}
	+ All  $D_n$  are purely real.		+ All  $D_n$  are purely real.
	- ~~Alle~~  $D_n$  are purely imaginary.		- All  $D_n$  are purely imaginary.

	{For  $η = 2$ , the coefficients are  $D_0 = 0.224$  and  $D_1 = 0.577$.  From this, calculate the coefficients  $D_2$  and  $D_3$.		{For  $η = 2$ , the coefficients are  $D_0 = 0.224$  and  $D_1 = 0.577$.  From this, calculate the coefficients  $D_2$  and  $D_3$.

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Questions

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	<quiz display=simple>		<quiz display=simple>

	{~~Welche Eigenschaften besitzt das Signal~~  $x(t)$?		{What are the properties of the signal  $x(t)$?
	\|type="[]"}		\|type="[]"}
	- $x(t)$  ~~ist für alle Zeiten~~  $t$  ~~imaginär~~.		- $x(t)$  is imaginary for all times  $t$ .
	+ $x(t)$  ~~ist periodisch~~.		+ $x(t)$  is periodic.
	- ~~Die Spektralfunktion~~  $X(f)$  ~~erhält man über das Fourierintegral~~.		- The spectral function $X(f)$  is obtained via the Fourier integral.

	{~~Schreiben Sie die Fourierkoeffizienten~~  $D_n$  ~~mit den Besselfunktionen erster Art~~   ⇒   ${\rm J}_n(η)$.  ~~Welche Zusammenhänge sind zu erkennen~~?		{Write the Fourier coefficients  $D_n$  together with the Bessel functions of the first kind   ⇒   ${\rm J}_n(η)$.  What relationships can be seen?
	\|type="[]"}		\|type="[]"}
	- ~~Alle~~  $D_n$  ~~sind gleich~~  ${\rm J}_η(0)$.		- All  $D_n$  are equal to  ${\rm J}_η(0)$.
	+ ~~Es gilt~~  $D_n = {\rm J}_n(η)$.		+  $D_n = {\rm J}_n(η)$ holds.
	- ~~Es gilt~~  $D_n = -{\rm J}_η(n)$.		-  $D_n = -{\rm J}_η(n)$ holds.

	{ ~~Welche Eigenschaften besitzen die Fourierkoeffizienten~~?		{ What are the properties of the Fourier coefficients?
	\|type="()"}		\|type="()"}
	+ ~~Alle~~  $D_n$  ~~sind rein reell~~.		+ All  $D_n$  are purely real.
	- Alle  $D_n$  ~~sind rein imaginär~~.		- Alle  $D_n$  are purely imaginary.

	{~~Für~~  $η = 2$  ~~lauten die Koeffizienten~~  $D_0 = 0.224$  ~~und~~  $D_1 = 0.577$.  ~~Berechnen Sie hieraus die Koeffizienten~~  $D_2$  ~~und~~  $D_3$.		{For  $η = 2$ , the coefficients are  $D_0 = 0.224$  and  $D_1 = 0.577$.  From this, calculate the coefficients  $D_2$  and  $D_3$.
	\|type="{}"}		\|type="{}"}
	$D_2 \ = \ $ { 0.353 3% }		$D_2 \ = \ $ { 0.353 3% }
	$D_3 \ = \ $ { 0.129 3% }		$D_3 \ = \ $ { 0.129 3% }

	{~~Wie lauten die Fourierkoeffizienten~~  $D_{-2}$  ~~und~~  $D_{-3}$ ?		{What are the Fourier coefficients  $D_{-2}$  and  $D_{-3}$ ?
	\|type="{}"}		\|type="{}"}
	$D_{-2} \ = \ $ { 0.353 3% }		$D_{-2} \ = \ $ { 0.353 3% }

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	*A purely imaginary value (at certain times) can only result when $η ≥ π/2$    ⇒   Answer 1 is incorrect.		*A purely imaginary value (at certain times) can only result when $η ≥ π/2$    ⇒   Answer 1 is incorrect.
	*For example, when  $T_0 = 2π/ω_0$ :		*For example, when  $T_0 = 2π/ω_0$ :
	:$$ x(t + k \cdot T_0) = {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} (t \hspace{0.05cm}+ \hspace{0.05cm} k \hspace{0.05cm}\cdot \hspace{0.05cm}T_0)) } =		:$$ x(t + k \cdot T_0) = {\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} (t \hspace{0.05cm}+ \hspace{0.05cm} k \hspace{0.05cm}\cdot \hspace{0.05cm}T_0)) } ={\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} t \hspace{0.05cm} + \hspace{0.05cm} k \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi) } ={\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} t \hspace{0.05cm} ) } = x(t)\hspace{0.05cm}.$$
	{\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} t \hspace{0.05cm} + \hspace{0.05cm} k \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi) } ={\rm e}^{\hspace{0.05cm}{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm}\eta \hspace{0.05cm}\cdot \hspace{0.05cm}\sin (\omega_{\rm 0} \hspace{0.05cm}\cdot \hspace{0.05cm} t \hspace{0.05cm} ) } = x(t)\hspace{0.05cm}.$$
	*This signal is periodic.  The Fourier series, not the Fourier integral, must be used to calculate the spectral function.		*This signal is periodic.  The Fourier series, not the Fourier integral, must be used to calculate the spectral function.

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	'''(3)'''  Using Euler's theorem, the Fourier coefficients can be represented as follows:		'''(3)'''  Using Euler's theorem, the Fourier coefficients can be represented as follows:
	:$$D_n = \frac{1}{2\pi}\cdot \int_{-\pi}^{+\pi} {\cos( \eta \cdot \sin(\alpha) - n \cdot \alpha)}\hspace{0.1cm}{\rm d}\alpha +		:$$D_n = \frac{1}{2\pi}\cdot \int_{-\pi}^{+\pi} {\cos( \eta \cdot \sin(\alpha) - n \cdot \alpha)}\hspace{0.1cm}{\rm d}\alpha +\frac{\rm j}{2\pi}\cdot \int_{-\pi}^{+\pi} {\sin( \eta \cdot \sin(\alpha) - n \cdot \alpha)}\hspace{0.1cm}{\rm d}\alpha \hspace{0.05cm}.$$
	\frac{\rm j}{2\pi}\cdot \int_{-\pi}^{+\pi} {\sin( \eta \cdot \sin(\alpha) - n \cdot \alpha)}\hspace{0.1cm}{\rm d}\alpha \hspace{0.05cm}.$$
	*The integrand of the first integral is an even function of  $\alpha$:		*The integrand of the first integral is an even function of  $\alpha$:
	:$$I_1 (-\alpha) = {\cos( \eta \cdot \sin(-\alpha) + n \cdot \alpha)} = {\cos( -\eta \cdot \sin(\alpha) + n \cdot \alpha)}=		:$$I_1 (-\alpha) = {\cos( \eta \cdot \sin(-\alpha) + n \cdot \alpha)} = {\cos( -\eta \cdot \sin(\alpha) + n \cdot \alpha)}={\cos( \eta \cdot \sin(\alpha) - n \cdot \alpha)} = I_1 (\alpha) \hspace{0.05cm}.$$
	{\cos( \eta \cdot \sin(\alpha) - n \cdot \alpha)} = I_1 (\alpha) \hspace{0.05cm}.$$
	*In contrast, the second integrand is an odd function:		*In contrast, the second integrand is an odd function:
	:$$I_2 (-\alpha) = {\sin( \eta \cdot \sin(-\alpha) + n \cdot \alpha)} = {\sin( -\eta \cdot \sin(\alpha) + n \cdot \alpha)}=		:$$I_2 (-\alpha) = {\sin( \eta \cdot \sin(-\alpha) + n \cdot \alpha)} = {\sin( -\eta \cdot \sin(\alpha) + n \cdot \alpha)}=-{\sin( \eta \cdot \sin(\alpha) - n \cdot \alpha)} = -I_2 (\alpha) \hspace{0.05cm}.$$
	-{\sin( \eta \cdot \sin(\alpha) - n \cdot \alpha)} = -I_2 (\alpha) \hspace{0.05cm}.$$
	*Thus, the second integral vanishes and, taking symmetry into account, we obtain:		*Thus, the second integral vanishes and, taking symmetry into account, we obtain:
	:$$D_n = \frac{1}{\pi}\cdot \int_{0}^{\pi} {\cos( \eta \cdot \sin(\alpha) - n \cdot \alpha)}\hspace{0.1cm}{\rm d}\alpha \hspace{0.05cm}.$$		:$$D_n = \frac{1}{\pi}\cdot \int_{0}^{\pi} {\cos( \eta \cdot \sin(\alpha) - n \cdot \alpha)}\hspace{0.1cm}{\rm d}\alpha \hspace{0.05cm}.$$