Aufgaben:Exercise 4.15Z: MSK Basic Pulse and MSK Spectrum - Revision history

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	'''(2)'''  The spectral function of a rectangular pulse of height  $g_0$  and duration $ 2T$  is:		'''(2)'''  The spectral function of a rectangular pulse of height  $g_0$  and duration $ 2T$  is:
	:$$R(f) = g_0 \cdot 2 T \cdot {\rm si} ( \pi f \cdot 2T )\hspace{0.2cm}{\rm mit}\hspace{0.2cm}{\rm si} (x) = \sin(x)/x \hspace{0.3cm}		:$$R(f) = g_0 \cdot 2 T \cdot {\rm si} ( \pi f \cdot 2T )\hspace{0.2cm}{\rm mit}\hspace{0.2cm}{\rm si} (x) = \sin(x)/x \hspace{0.3cm}\Rightarrow \hspace{0.3cm}R(f = 0) \hspace{0.15cm}\underline {= 2} \cdot g_0 \cdot T\hspace{0.05cm}.$$
	\Rightarrow \hspace{0.3cm}R(f = 0) \hspace{0.15cm}\underline {= 2} \cdot g_0 \cdot T\hspace{0.05cm}.$$

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	[[Category:Modulation Methods: Exercises\|^4.4 Non-linear Digital Modulation^]]		[[Category:Modulation Methods: Exercises\|^4.4 Non-linear Digital Modulation^]]
			[[de:Aufgaben:4.14Z_MSK–Grundimpuls/Spektrum]]

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	[[File:P_ID1744__Mod_Z_4_14.png\|right\|frame\|MSK ~~fundamental~~ pulse and its spectrum]]		[[File:P_ID1744__Mod_Z_4_14.png\|right\|frame\|MSK basic pulse and its spectrum]]
	The fundamental pulse that is always required to  [[Modulation_Methods/Nonlinear_Digital_Modulation#Realizing_MSK_as_Offset.E2.80.93QPSK\|realize MSK as Offset–QPSK]]   has the form shown in the graph above:		The fundamental pulse that is always required to  [[Modulation_Methods/Nonlinear_Digital_Modulation#Realizing_MSK_as_Offset.E2.80.93QPSK\|realize MSK as Offset–QPSK]]   has the form shown in the graph above:
	:$$g_{\rm MSK}(t) = \left\{ \begin{array}{l} g_0 \cdot \cos (\pi/2 \cdot t/T) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} \| t \| \le T \hspace{0.05cm}, \\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$		:$$g_{\rm MSK}(t) = \left\{ \begin{array}{l} g_0 \cdot \cos (\pi/2 \cdot t/T) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} \| t \| \le T \hspace{0.05cm}, \\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$

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	''Hints:''		''Hints:''
	*This exercise belongs to the chapter  [[Modulation_Methods/Nonlinear_Digital_Modulation\|Nonlinear Digital Modulation]].		*This exercise belongs to the chapter  [[Modulation_Methods/Nonlinear_Digital_Modulation\|Nonlinear Digital Modulation]].
	*Particular reference is made ~~tot he~~ page  [[Modulation_Methods/Nonlinear_Digital_Modulation#Realizing_MSK_as_Offset.E2.80.93QPSK\|Realizing MSK as Offset–QPSK]].		*Particular reference is made to the page  [[Modulation_Methods/Nonlinear_Digital_Modulation#Realizing_MSK_as_Offset.E2.80.93QPSK\|Realizing MSK as Offset–QPSK]].

	*The result obtained here is also used in  [[Aufgaben:Exercise_4.15:_MSK_Compared_with_BPSK_and_QPSK\|Exercise 4.15]] .		*The result obtained here is also used in  [[Aufgaben:Exercise_4.15:_MSK_Compared_with_BPSK_and_QPSK\|Exercise 4.15]] .

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	'''(2)'''  The spectral function of a rectangular pulse of height  $g_0$  ~~und~~ duration $ 2T$  is:		'''(2)'''  The spectral function of a rectangular pulse of height  $g_0$  and duration $ 2T$  is:
	:$$R(f) = g_0 \cdot 2 T \cdot {\rm si} ( \pi f \cdot 2T )\hspace{0.2cm}{\rm mit}\hspace{0.2cm}{\rm si} (x) = \sin(x)/x \hspace{0.3cm}		:$$R(f) = g_0 \cdot 2 T \cdot {\rm si} ( \pi f \cdot 2T )\hspace{0.2cm}{\rm mit}\hspace{0.2cm}{\rm si} (x) = \sin(x)/x \hspace{0.3cm}
	\Rightarrow \hspace{0.3cm}R(f = 0) \hspace{0.15cm}\underline {= 2} \cdot g_0 \cdot T\hspace{0.05cm}.$$		\Rightarrow \hspace{0.3cm}R(f = 0) \hspace{0.15cm}\underline {= 2} \cdot g_0 \cdot T\hspace{0.05cm}.$$

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Solution

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Die Periodendauer des Cosinussignals muss~~  $T_0 = 4T$  ~~sein~~.  ~~Damit ist die Frequenz~~ $f_0 = 1/T_0\hspace{0.15cm}\underline {= 0.25} · 1/T$.		'''(1)'''  The period of the cosine signal must be  $T_0 = 4T$ .  Thus, the frequency is $f_0 = 1/T_0\hspace{0.15cm}\underline {= 0.25} · 1/T$.


	'''(2)'''  ~~Die Spektralfunktion eines Rechteckimpulses der Höhe~~  $g_0$  und ~~der Dauer~~ $ 2T$  ~~lautet~~:		'''(2)'''  The spectral function of a rectangular pulse of height  $g_0$  und duration $ 2T$  is:
	:$$R(f) = g_0 \cdot 2 T \cdot {\rm si} ( \pi f \cdot 2T )\hspace{0.2cm}{\rm mit}\hspace{0.2cm}{\rm si} (x) = \sin(x)/x \hspace{0.3cm}		:$$R(f) = g_0 \cdot 2 T \cdot {\rm si} ( \pi f \cdot 2T )\hspace{0.2cm}{\rm mit}\hspace{0.2cm}{\rm si} (x) = \sin(x)/x \hspace{0.3cm}
	\Rightarrow \hspace{0.3cm}R(f = 0) \hspace{0.15cm}\underline {= 2} \cdot g_0 \cdot T\hspace{0.05cm}.$$		\Rightarrow \hspace{0.3cm}R(f = 0) \hspace{0.15cm}\underline {= 2} \cdot g_0 \cdot T\hspace{0.05cm}.$$


	'''(3)'''  ~~Aus~~  $g(t) = c(t) · r(t)$  ~~folgt nach dem Faltungssatz~~:   $ G(f) = C(f) \star R(f)\hspace{0.05cm}.$		'''(3)'''  When  $g(t) = c(t) · r(t)$ , it follows from the convolution theorem that:   $ G(f) = C(f) \star R(f)\hspace{0.05cm}.$
	*~~Die Spektralfunktion~~  $C(f)$  ~~besteht aus zwei Diracfunktionen bei~~  $± f_0$, ~~jeweils mit dem Gewicht~~  $1/2$.  ~~Daraus folgt~~:		*The spectral function  $C(f)$  consists of two Dirac functions at  $± f_0$, each with weight  $1/2$.  From this follows:
	:$$ G(f) = 2 \cdot g_0 \cdot T \cdot \big [ 1/2 \cdot \delta (f - f_0 ) + 1/2 \cdot \delta (f + f_0 )\big ] \star {\rm si} ( 2 \pi f T )= g_0 \cdot T \cdot \big [ {\rm si} ( 2 \pi T \cdot (f - f_0 ) ) + {\rm si} ( 2 \pi T \cdot (f + f_0 ) ) \big ] \hspace{0.05cm}.$$		:$$ G(f) = 2 \cdot g_0 \cdot T \cdot \big [ 1/2 \cdot \delta (f - f_0 ) + 1/2 \cdot \delta (f + f_0 )\big ] \star {\rm si} ( 2 \pi f T )= g_0 \cdot T \cdot \big [ {\rm si} ( 2 \pi T \cdot (f - f_0 ) ) + {\rm si} ( 2 \pi T \cdot (f + f_0 ) ) \big ] \hspace{0.05cm}.$$
	*~~Mit dem Ergebnis~~  $f_0 = 1/(4T)$  ~~der Teilaufgabe~~  '''(1)'''  ~~gilt weiter~~:		*Using the result  $f_0 = 1/(4T)$  from question  '''(1)''' , it further holds that:
	:$$G(f) = g_0 \cdot T \cdot \big [ {\rm si} ( 2 \pi f T - \pi / 2 ) + {\rm si} ( 2 \pi f T + \pi / 2) \big ]$$		:$$G(f) = g_0 \cdot T \cdot \big [ {\rm si} ( 2 \pi f T - \pi / 2 ) + {\rm si} ( 2 \pi f T + \pi / 2) \big ]$$
	:$$\Rightarrow \hspace{0.3cm} G(f = 0) = g_0 \hspace{-0.02cm}\cdot\hspace{-0.02cm} T \hspace{-0.02cm}\cdot\hspace{-0.02cm} \big [ {\rm si} ( - \pi/2 ) + {\rm si} ( +\pi/2 ) \big ] = 2 \cdot g_0 \hspace{-0.02cm}\cdot\hspace{-0.02cm} T \hspace{-0.02cm}\cdot\hspace{-0.02cm} {\rm si} ( \pi/2 ) = 2 \hspace{-0.02cm}\cdot\hspace{-0.02cm} g_0 \hspace{-0.02cm}\cdot\hspace{-0.02cm} T \hspace{-0.02cm}\cdot\hspace{-0.02cm} \frac {{\rm sin}({\pi}/{2}) } { {\pi}/{2} } ={4}/{\pi} \hspace{-0.02cm}\cdot\hspace{-0.02cm} g_0 \hspace{-0.02cm}\cdot\hspace{-0.02cm} T \hspace{0.15cm}\underline {\approx 1.273} \hspace{-0.02cm}\cdot\hspace{-0.02cm} g_0 \hspace{-0.02cm}\cdot\hspace{-0.02cm} T .$$		:$$\Rightarrow \hspace{0.3cm} G(f = 0) = g_0 \hspace{-0.02cm}\cdot\hspace{-0.02cm} T \hspace{-0.02cm}\cdot\hspace{-0.02cm} \big [ {\rm si} ( - \pi/2 ) + {\rm si} ( +\pi/2 ) \big ] = 2 \cdot g_0 \hspace{-0.02cm}\cdot\hspace{-0.02cm} T \hspace{-0.02cm}\cdot\hspace{-0.02cm} {\rm si} ( \pi/2 ) = 2 \hspace{-0.02cm}\cdot\hspace{-0.02cm} g_0 \hspace{-0.02cm}\cdot\hspace{-0.02cm} T \hspace{-0.02cm}\cdot\hspace{-0.02cm} \frac {{\rm sin}({\pi}/{2}) } { {\pi}/{2} } ={4}/{\pi} \hspace{-0.02cm}\cdot\hspace{-0.02cm} g_0 \hspace{-0.02cm}\cdot\hspace{-0.02cm} T \hspace{0.15cm}\underline {\approx 1.273} \hspace{-0.02cm}\cdot\hspace{-0.02cm} g_0 \hspace{-0.02cm}\cdot\hspace{-0.02cm} T .$$


	'''(4)'''  ~~Schreibt man die~~  $\rm si$~~–Funktion aus~~, ~~so erhält man mit~~  $\sin (α ± π/2) = ± \cos(α)$:		'''(4)'''  By writing out the  $\rm si$–function, with   $\sin (α ± π/2) = ± \cos(α)$, one gets:
	:$$G(f) = g_0 \cdot T \cdot \left [ \frac{{\rm sin} ( 2 \pi f T - \pi / 2 )}{2 \pi f T - \pi / 2 } + \frac{{\rm sin} ( 2 \pi f T + \pi / 2 )}{2 \pi f T + \pi / 2 } \right ]= g_0 \cdot T \cdot \frac {2}{\pi}\cdot\left [ \frac{-{\rm cos} ( 2 \pi f T )}{4 f T - 1 } + \frac{{\rm cos} ( 2 \pi f T )}{4 f T + 1 } \right ]$$		:$$G(f) = g_0 \cdot T \cdot \left [ \frac{{\rm sin} ( 2 \pi f T - \pi / 2 )}{2 \pi f T - \pi / 2 } + \frac{{\rm sin} ( 2 \pi f T + \pi / 2 )}{2 \pi f T + \pi / 2 } \right ]= g_0 \cdot T \cdot \frac {2}{\pi}\cdot\left [ \frac{-{\rm cos} ( 2 \pi f T )}{4 f T - 1 } + \frac{{\rm cos} ( 2 \pi f T )}{4 f T + 1 } \right ]$$
	:$$\Rightarrow \hspace{0.3cm} G(f) = g_0 \cdot T \cdot \frac {2}{\pi}\cdot \frac{(1+4 f T ) \cdot {\rm cos} ( 2 \pi f T )+ (1-4 f T ) \cdot {\rm cos} ( 2 \pi f T )}{1 - (4 f T)^2 } = \frac {4}{\pi}\cdot g_0 \cdot T \cdot \frac{ {\rm cos} ( 2 \pi f T )}{1 - (4 f T)^2 }\hspace{0.05cm}.$$		:$$\Rightarrow \hspace{0.3cm} G(f) = g_0 \cdot T \cdot \frac {2}{\pi}\cdot \frac{(1+4 f T ) \cdot {\rm cos} ( 2 \pi f T )+ (1-4 f T ) \cdot {\rm cos} ( 2 \pi f T )}{1 - (4 f T)^2 } = \frac {4}{\pi}\cdot g_0 \cdot T \cdot \frac{ {\rm cos} ( 2 \pi f T )}{1 - (4 f T)^2 }\hspace{0.05cm}.$$

	*~~Die Nullstellen von~~  $G(f)$  ~~werden allein durch die Cosinusfunktion im Zähler bestimmt und würden bei den Frequenzen~~  $f · T = 0.25,\ 0.75,\ 1.25,$  ..~~. liegen~~.		*The zeroes of  $G(f)$  are exclusively determined by the cosine function in the numerator, and are found at the frequencies  $f · T = 0.25,\ 0.75,\ 1.25,$  ...
	*~~Allerdings wird die erste Nullstelle bei~~  $f · T = 0.25$  ~~durch die gleichzeitige Nullstelle des Nenners aufgehoben~~.  ~~Deshalb gilt~~:		*However, the first zero at  $f · T = 0.25$  is cancelled out by the simultaneously occuring zero in the denominator.  Therefore:
	:$$f_1 \hspace{0.15cm}\underline {= 0.75} \cdot 1/T \hspace{0.05cm}.$$		:$$f_1 \hspace{0.15cm}\underline {= 0.75} \cdot 1/T \hspace{0.05cm}.$$

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	[[File:P_ID1744__Mod_Z_4_14.png\|right\|frame\|MSK fundamental pulse~~ ~~ and~~ ~~ its spectrum]]		[[File:P_ID1744__Mod_Z_4_14.png\|right\|frame\|MSK fundamental pulse and its spectrum]]
	The fundamental pulse that is always required to  [[Modulation_Methods/Nonlinear_Digital_Modulation#Realizing_MSK_as_Offset.E2.80.93QPSK\|realize MSK as Offset–QPSK]]   has the form shown in the graph above:		The fundamental pulse that is always required to  [[Modulation_Methods/Nonlinear_Digital_Modulation#Realizing_MSK_as_Offset.E2.80.93QPSK\|realize MSK as Offset–QPSK]]   has the form shown in the graph above:
	:$$g_{\rm MSK}(t) = \left\{ \begin{array}{l} g_0 \cdot \cos (\pi/2 \cdot t/T) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} \| t \| \le T \hspace{0.05cm}, \\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$		:$$g_{\rm MSK}(t) = \left\{ \begin{array}{l} g_0 \cdot \cos (\pi/2 \cdot t/T) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} \| t \| \le T \hspace{0.05cm}, \\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$

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	[[File:P_ID1744__Mod_Z_4_14.png\|right\|frame\|MSK~~–Grundimpuls~~  ~~und~~  ~~–Spektrum~~]]		[[File:P_ID1744__Mod_Z_4_14.png\|right\|frame\|MSK fundamental pulse  and  its spectrum]]
	The fundamental pulse that is always required to  [[Modulation_Methods/Nonlinear_Digital_Modulation#Realizing_MSK_as_Offset.E2.80.93QPSK\|realize MSK as Offset–QPSK]]   has the form shown in the graph above:		The fundamental pulse that is always required to  [[Modulation_Methods/Nonlinear_Digital_Modulation#Realizing_MSK_as_Offset.E2.80.93QPSK\|realize MSK as Offset–QPSK]]   has the form shown in the graph above:
	:$$g_{\rm MSK}(t) = \left\{ \begin{array}{l} g_0 \cdot \cos (\pi/2 \cdot t/T) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} \| t \| \le T \hspace{0.05cm}, \\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$		:$$g_{\rm MSK}(t) = \left\{ \begin{array}{l} g_0 \cdot \cos (\pi/2 \cdot t/T) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} \| t \| \le T \hspace{0.05cm}, \\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$

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	[[File:P_ID1744__Mod_Z_4_14.png\|right\|frame\|MSK–Grundimpuls  und  –Spektrum]]		[[File:P_ID1744__Mod_Z_4_14.png\|right\|frame\|MSK–Grundimpuls  und  –Spektrum]]
	The fundamental pulse that is always required to  [[Modulation_Methods/Nonlinear_Digital_Modulation#Realizing_MSK_as_Offset.E2.80.93QPSK\|realize MSK as Offset–QPSK]]   has the form shown in the graph above:		The fundamental pulse that is always required to  [[Modulation_Methods/Nonlinear_Digital_Modulation#Realizing_MSK_as_Offset.E2.80.93QPSK\|realize MSK as Offset–QPSK]]   has the form shown in the graph above:
	:$$g_{\rm MSK}(t) = \left\{ \begin{array}{l} g_0 \cdot \cos (\pi/2 \cdot t/T) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} \| t \| \le T \hspace{0.05cm}, \\ {\rm ~~sonst~~}\hspace{0.05cm}. \\ \end{array}$$		:$$g_{\rm MSK}(t) = \left\{ \begin{array}{l} g_0 \cdot \cos (\pi/2 \cdot t/T) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} \| t \| \le T \hspace{0.05cm}, \\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$
	The spectral function  $G(f)$ is drawn below, that is, the  [[Signal_Representation/Fourier_Transform_and_its_Inverse#The_first_Fourier_integral\|Fourier transform]]  of  $g(t)$.		The spectral function  $G(f)$ is drawn below, that is, the  [[Signal_Representation/Fourier_Transform_and_its_Inverse#The_first_Fourier_integral\|Fourier transform]]  of  $g(t)$.

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	[[File:P_ID1744__Mod_Z_4_14.png\|right\|frame\|MSK–Grundimpuls  und  –Spektrum]]		[[File:P_ID1744__Mod_Z_4_14.png\|right\|frame\|MSK–Grundimpuls  und  –Spektrum]]
	The fundamental pulse that is always required to  [[Modulation_Methods/Nonlinear_Digital_Modulation#Realizing_MSK_as_Offset.E2.80.93QPSK\|realize MSK as Offset–QPSK]]   has the form shown in the graph above:		The fundamental pulse that is always required to  [[Modulation_Methods/Nonlinear_Digital_Modulation#Realizing_MSK_as_Offset.E2.80.93QPSK\|realize MSK as Offset–QPSK]]   has the form shown in the graph above:
	:$$g_{\rm MSK}(t) = \left\{ \begin{array}{l} g_0 \cdot \cos (\pi/2 \cdot t/T) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} \| t \| \le T \hspace{0.05cm}, \\ {\rm ~~sonst~~}\hspace{0.05cm}. \\ \end{array}$$		:$$g_{\rm MSK}(t) = \left\{ \begin{array}{l} g_0 \cdot \cos (\pi/2 \cdot t/T) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} \| t \| \le T \hspace{0.05cm}, \\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$
	The spectral function  $G(f)$ is drawn below, that is, the  [[Signal_Representation/Fourier_Transform_and_its_Inverse#The_first_Fourier_integral\|Fourier transform]]  of  $g(t)$.		The spectral function  $G(f)$ is drawn below, that is, the  [[Signal_Representation/Fourier_Transform_and_its_Inverse#The_first_Fourier_integral\|Fourier transform]]  of  $g(t)$.

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	[[File:P_ID1744__Mod_Z_4_14.png\|right\|frame\|MSK basic pulse and its spectrum]]		[[File:P_ID1744__Mod_Z_4_14.png\|right\|frame\|MSK basic pulse and its spectrum]]
	The fundamental pulse that is always required to  [[Modulation_Methods/~~Nonlinear_Digital_Modulation~~#~~Realizing_MSK_as_Offset.E2.80.93QPSK~~\|realize MSK as Offset–QPSK]]   has the form shown in the graph above:		The fundamental pulse that is always required to  [[Modulation_Methods/Non-Linear_Digital_Modulation#Realizing_MSK_as_Offset–QPSK\|realize MSK as Offset–QPSK]]   has the form shown in the graph above:
	:$$g_{\rm MSK}(t) = \left\{ \begin{array}{l} g_0 \cdot \cos (\pi/2 \cdot t/T) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} \| t \| \le T \hspace{0.05cm}, \\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$		:$$g_{\rm MSK}(t) = \left\{ \begin{array}{l} g_0 \cdot \cos (\pi/2 \cdot t/T) \\ 0 \\ \end{array} \right.\quad \begin{array}{*{10}c} \| t \| \le T \hspace{0.05cm}, \\ {\rm otherwise}\hspace{0.05cm}. \\ \end{array}$$
	The spectral function  $G(f)$ is drawn below, that is, the  [[Signal_Representation/~~Fourier_Transform_and_its_Inverse~~#The_first_Fourier_integral\|Fourier transform]]  of  $g(t)$.		The spectral function  $G(f)$ is drawn below, that is, the  [[Signal_Representation/The_Fourier_Transform_and_its_Inverse#The_first_Fourier_integral\|Fourier transform]]  of  $g(t)$.

	The corresponding equation is to be determined in this task, by considering:		The corresponding equation is to be determined in this task, by considering:
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	''Hints:''		''Hints:''
	*This exercise belongs to the chapter  [[Modulation_Methods/Nonlinear_Digital_Modulation\|Nonlinear Digital Modulation]].		*This exercise belongs to the chapter  [[Modulation_Methods/Nonlinear_Digital_Modulation\|Nonlinear Digital Modulation]].
	*Particular reference is made to the page  [[Modulation_Methods/~~Nonlinear_Digital_Modulation~~#~~Realizing_MSK_as_Offset.E2.80.93QPSK~~\|Realizing MSK as Offset–QPSK]].		*Particular reference is made to the page  [[Modulation_Methods/Non-Linear_Digital_Modulation#Realizing_MSK_as_Offset–QPSK\|Realizing MSK as Offset–QPSK]].

	*The result obtained here is also used in  [[Aufgaben:Exercise_4.15:_MSK_Compared_with_BPSK_and_QPSK\|Exercise 4.15]] .		*The result obtained here is also used in  [[Aufgaben:Exercise_4.15:_MSK_Compared_with_BPSK_and_QPSK\|Exercise 4.15]] .
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	[[Category:Modulation Methods: Exercises\|^4.4 Non-linear Digital Modulation^]]		[[Category:Modulation Methods: Exercises\|^4.4 Non-linear Digital Modulation^]]
	[[de:Aufgaben:4.~~14Z_MSK–Grundimpuls/~~Spektrum]]		[[de:Aufgaben:Aufgabe 4.15Z: MSK–Grundimpuls und MSK-Spektrum]]