Difference between revisions of "Aufgaben:Exercise 3.5: GMSK Modulation"

Revision as of 08:47, 30 June 2020

Verschiedene Signale bei GMSK-Modulation

The modulation method used for GSM is Gaussian Minimum Shift Keying, short GMSK. This is a special type of FSK (Frequency Shift Keying) with CP-FSK (Continuous Phase Matching), where

the modulation index has the smallest value that just satisfies the orthogonality condition: $h = 0.5$ ⇒ Minimum Shift Keying,
a Gaussian low pass with the impulse response $h_{\rm G}(t)$ is inserted before the FSK modulator, with the aim of saving even more bandwidth.

The graphic illustrates the situation:

The digital message is represented by the amplitude coefficients $a_{\mu} ∈ \{±1\}$ which are applied to a Dirac pulse. It should be noted that the sequence drawn in is assumed for the subtask (3).

The symmetrical rectangular pulse with duration $T = T_{\rm B}$ (GSM bit duration) is dimensionless:

$$g_{\rm R}(t) = \left\{ \begin{array}{c} 1 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{5}c} {\rm{{\rm{f\ddot{u}r}}}} \\ {\rm{{\rm{f\ddot{u}r}}}} \\ \end{array}\begin{array}{*{5}c} |\hspace{0.05cm} t \hspace{0.05cm}| < T/2 \hspace{0.05cm}, \\ |\hspace{0.05cm} t \hspace{0.05cm}| > T/2 \hspace{0.05cm}. \\ \end{array}$$

This results for the rectangular signal

$$q_{\rm R} (t) = q_{\rm \delta} (t) \star g_{\rm R}(t) = \sum_{\nu} a_{\nu}\cdot g_{\rm R}(t - \nu \cdot T)\hspace{0.05cm}.$$

The Gaussian low pass is given by its frequency response or impulse response:

$$H_{\rm G}(f) = {\rm e}^{-\pi\cdot (\frac{f}{2 f_{\rm G}})^2} \hspace{0.2cm}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \hspace{0.2cm} h_{\rm G}(t) = 2 f_{\rm G} \cdot {\rm e}^{-\pi\cdot (2 f_{\rm G}\cdot t)^2}\hspace{0.05cm},$$

where the system theoretical cut-off frequency $f_{\rm G}$ is used. In the GSM specification, however, the 3dB cut-off frequency is specified with $f_{\rm 3dB} = 0.3/T$ . From this, $f_{\rm G}$ can be calculated directly - see subtask (2).

The signal after the gauss low pass is thus

$$q_{\rm G} (t) = q_{\rm R} (t) \star h_{\rm G}(t) = \sum_{\nu} a_{\nu}\cdot g(t - \nu \cdot T)\hspace{0.05cm}.$$

Here $g(t)$ is referred to as frequency pulse. For this one:

$$g(t) = q_{\rm R} (t) \star h_{\rm G}(t) \hspace{0.05cm}.$$

With the low pass filtered signal $q_{\rm G}(t)$, the carrier frequency $f_{\rm T}$ and the frequency deviation $\Delta f_{\rm A}$ can thus be written for the instantaneous frequency at the output of the FSK modulator::$$f_{\rm A}(t) = f_{\rm T} + \Delta f_{\rm A} \cdot q_{\rm G} (t)\hspace{0.05cm}.$$

Notes:

This exercise belongs to the chapter Die Charakteristika von GSM.
Reference is also made to the chapter Funkschnittstelle in the book „Beispiele von Nachrichtensystemen”.

For your calculations use the exemplary values $f_{\rm T} = 900 \ \ \rm MHz$ and $\Delta f_{\rm A} = 68 \ \rm kHz$.
Use the Gaussian integral to solve the task (some numerical values are given in the table)

some values of the Gaussian integral

$$\phi(x) =\frac {1}{\sqrt{2 \pi}} \cdot \int^{x} _{-\infty} {\rm e}^{-u^2/2}\,{\rm d}u \hspace{0.05cm}.$$

Fragebogen

${\rm Max} \ \big[f_{\rm A}(t)\big] \ = \hspace{0.2cm} $

$\ \rm MHz$

${\rm Min} \ \big[f_{\rm A}(t)\big] \ = \hspace{0.28cm} $

$\ \rm MHz$

$f_{\rm G} \cdot T \ = \ $

$g(t = 0) \ = \ $

$q_{\rm G}(t = 3T) \ = \ $

$g(t = ±T) \ = \ $

${\rm Max} \ |q_{\rm G}(t)| \ = \ $

Sample solution

Solution

(1) If all amplitude coefficients $a_{\mu}$ are equal to $+1$, then $q_{\rm R}(t) = 1$ is a constant. Thus, the Gaussian low pass has no influence and $q_{\rm G}(t) = 1$ results.

The maximum frequency is thus

$${\rm Max}\ [f_{\rm A}(t)] = f_{\rm T} + \Delta f_{\rm A} \hspace{0.15cm} \underline {= 900.068\,{\rm MHz}} \hspace{0.05cm}.$$

The minimum instantaneous frequency

$${\rm Min}\ [f_{\rm A}(t)] = f_{\rm T} - \Delta f_{\rm A} \hspace{0.15cm} \underline { = 899.932\,{\rm MHz}} \hspace{0.05cm}$$

is obtained when all amplitude coefficients are negative. In this case $q_{\rm R}(t) = q_{\rm G}(t) = -1$.

(2) The frequency at which the logarithmic power transfer function is $3 \ \rm dB$ less than $f = 0$ is called the 3dB cut-off frequency.

This can also be expressed as follows:

$$\frac {|H(f = f_{\rm 3dB})|}{|H(f = 0)|}= \frac{1}{\sqrt{2}} \hspace{0.05cm}.$$

In particular the Gauss low pass because of $H(f = 0) = 1$:

$$ H(f = f_{\rm 3dB})= {\rm e}^{-\pi\cdot ({f_{\rm 3dB}}/{2 f_{\rm G}})^2} = \frac{1}{\sqrt{2}}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}(\frac{f_{\rm 3dB}}{2 f_{\rm G}})^2 = \frac{{\rm ln}\hspace{0.1cm}\sqrt{2}}{\pi} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm G} = \sqrt{\frac{\pi}{4 \cdot {\rm ln}\hspace{0.1cm}\sqrt{2}}}\cdot f_{\rm 3dB}\hspace{0.05cm}.$$

The numerical evaluation leads to $f_{\rm G} \approx 1.5 \cdot f_{\rm 3dB}$.
{\f_{\rm 3dB} \cdot T = 0.3$ is followed by $f_{\rm G} \cdot T \underline{\approx 0.45}$. '''(3)''' Der gesuchte Frequenzimpuls ${\rm g}(t)$ ergibt sich aus der Faltung von Rechteckfunktion $g_{\rm R}(t)$ mit der Impulsantwort $h_{\rm G}(t)$: :'"`UNIQ-MathJax54-QINU`"' *Mit der Substitution $u^{2} = 8π \cdot {f_{G}}^{2} \cdot \tau^{2}$ und der Funktion $\phi (x)$ kann man hierfür auch schreiben: :'"`UNIQ-MathJax55-QINU`"' *Für die Zeit $t = 0$ gilt unter Berücksichtigung von $\phi (-x) = 1 - \phi (x)$ und $f_{\rm G} \cdot T = 0.45$: :'"`UNIQ-MathJax56-QINU`"' '''(4)''' Mit $a_{3} = +1$ würde sich $q_{\rm G}(t = 3 T) = 1$ ergeben. Aufgrund der Linearität gilt somit: :'"`UNIQ-MathJax57-QINU`"' '''(5)''' Mit dem Ergebnis aus (3) und $f_{\rm G} \cdot T = 0.45$ erhält man: :'"`UNIQ-MathJax58-QINU`"' *Der Impulswert $g(t = -T)$ ist aufgrund der Symmetrie des Gaußtiefpasses genau so groß. '''(6)''' Bei alternierender Folge sind aus Symmetriegründen die Beträge $|q_{\rm G}(\mu \cdot T)|$ bei allen Vielfachen der Bitdauer $T$ alle gleich. *Alle Zwischenwerte bei $t \approx \mu \cdot T$ sind dagegen kleiner. *Unter Berücksichtigung von $g(t ≥ 2T) \approx 0$ wird jeder einzelne Impulswert $g(0)$ durch den vorangegangenen Impuls mit $g(t = T)$ verkleinert, ebenso vom folgenden Impuls mit $g(t = -T)$. *Es ergeben sich also Impulsinterferenzen und man erhält: :'"`UNIQ-MathJax59-QINU`"' '"`UNIQ--html-00000004-QINU`"' [[Category:Exercises for Mobile Communications|^3.3 Characteristics of GSM^]] '"`UNIQ--html-00000005-QINU`"' [[Mobile Kommunikation/Die Charakteristika von GSM | Return to book]] '"`UNIQ--html-00000006-QINU`"' [[File:EN_Mob_A_3_5.png|right|frame|Verschiedene Signale bei GMSK-Modulation]] The modulation method used for GSM is ''Gaussian Minimum Shift Keying'', short GMSK. This is a special type of FSK (''Frequency Shift Keying'') with CP-FSK (''Continuous Phase Matching''), where *the modulation index has the smallest value that just satisfies the orthogonality condition: $h = 0.5$ ⇒ ''Minimum Shift Keying'', *a Gaussian low pass with the impulse response $h_{\rm G}(t)$ is inserted before the FSK modulator, with the aim of saving even more bandwidth. The graphic illustrates the situation: *The digital message is represented by the amplitude coefficients $a_{\mu} ∈ \{±1\}$ which are applied to a Dirac pulse. It should be noted that the sequence drawn in is assumed for the subtask '''(3)'''. *The symmetrical rectangular pulse with duration $T = T_{\rm B}$ (GSM bit duration) is dimensionless: :'"`UNIQ-MathJax60-QINU`"' *This results for the rectangular signal :'"`UNIQ-MathJax61-QINU`"' *The Gaussian low pass is given by its frequency response or impulse response: :'"`UNIQ-MathJax62-QINU`"' :where the system theoretical cut-off frequency $f_{\rm G}$ is used. In the GSM specification, however, the 3dB cut-off frequency is specified with $f_{\rm 3dB} = 0.3/T$ . From this, $f_{\rm G}$ can be calculated directly - see subtask '''(2)'''. *The signal after the gauss low pass is thus :'"`UNIQ-MathJax63-QINU`"' Here $g(t)$ is referred to as ''frequency pulse''. For this one: :'"`UNIQ-MathJax64-QINU`"' *With the low pass filtered signal $q_{\rm G}(t)$, the carrier frequency $f_{\rm T}$ and the frequency deviation $\Delta f_{\rm A}$ can thus be written for the instantaneous frequency at the output of the FSK modulator::'"`UNIQ-MathJax65-QINU`"' ''Notes:'' *This exercise belongs to the chapter [[Mobile_Kommunikation/Die_Charakteristika_von_GSM|Die Charakteristika von GSM]]. *Reference is also made to the chapter [[Beispiele_von_Nachrichtensystemen/Funkschnittstelle|Funkschnittstelle]] in the book „Beispiele von Nachrichtensystemen”. *For your calculations use the exemplary values $f_{\rm T} = 900 \ \ \rm MHz$ and $\Delta f_{\rm A} = 68 \ \rm kHz$. *Use the Gaussian integral to solve the task (some numerical values are given in the table) [[File:P_ID2226__Bei_A_3_4b.png|right|frame|some values of the Gaussian integral]] :'"`UNIQ-MathJax66-QINU`"' <br clear="all"> ==='"`UNIQ--h-2--QINU`"'Fragebogen=== '"`UNIQ--quiz-00000007-QINU`"' ==='"`UNIQ--h-3--QINU`"'Sample solution=== '"`UNIQ--html-00000003-QINU`"' '''(1)''' If all amplitude coefficients $a_{\mu}$ are equal to $+1$, then $q_{\rm R}(t) = 1$ is a constant. Thus, the Gaussian low pass has no influence and $q_{\rm G}(t) = 1$ results. *The maximum frequency is thus :'"`UNIQ-MathJax67-QINU`"' *The minimum instantaneous frequency :'"`UNIQ-MathJax68-QINU`"' is obtained when all amplitude coefficients are negative. In this case $q_{\rm R}(t) = q_{\rm G}(t) = -1$. '''(2)''' The frequency at which the logarithmic power transfer function is $3 \ \rm dB$ less than $f = 0$ is called the 3dB cut-off frequency. *This can also be expressed as follows: :'"`UNIQ-MathJax69-QINU`"' *In particular the Gauss low pass because of $H(f = 0) = 1$: :'"`UNIQ-MathJax70-QINU`"' *The numerical evaluation leads to $f_{\rm G} \approx 1.5 \cdot f_{\rm 3dB}$.

@@ Line 11: / Line 11: @@
 The graphic illustrates the situation:
-*The digital message is represented by the amplitude coefficients&nbsp; $a_{\mu} ∈ \{±1\}$&nbsp; which are applied to a Dirac pulse. It should be noted that the sequence drawn in is assumed for the subtask '''(3)''.
+*The digital message is represented by the amplitude coefficients&nbsp; $a_{\mu} ∈ \{±1\}$&nbsp; which are applied to a Dirac pulse. It should be noted that the sequence drawn in is assumed for the subtask '''(3)'''.
 *The symmetrical rectangular pulse with duration&nbsp; $T = T_{\rm B}$&nbsp; (GSM bit duration) is dimensionless:
@@ Line 19: / Line 19: @@
 *The Gaussian low pass is given by its frequency response or impulse response:
 :$$H_{\rm G}(f) = {\rm e}^{-\pi\cdot (\frac{f}{2 f_{\rm G}})^2} \hspace{0.2cm}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \hspace{0.2cm} h_{\rm G}(t) = 2 f_{\rm G} \cdot {\rm e}^{-\pi\cdot (2 f_{\rm G}\cdot t)^2}\hspace{0.05cm},$$
-:wobei die systemtheoretische Grenzfrequenz&nbsp; $f_{\rm G}$&nbsp; verwendet wird. In der GSM–Spezifikation wird aber die 3dB–Grenzfrequenz mit&nbsp; $f_{\rm 3dB} = 0.3/T$&nbsp; angegeben. Daraus kann&nbsp; $f_{\rm G}$&nbsp; direkt berechnet werden – siehe Teilaufgabe '''(2)'''.
+:where the system theoretical cut-off frequency&nbsp; $f_{\rm G}$&nbsp; is used. In the GSM specification, however, the 3dB cut-off frequency is specified with&nbsp; $f_{\rm 3dB} = 0.3/T$&nbsp;. From this,&nbsp; $f_{\rm G}$&nbsp; can be calculated directly - see subtask '''(2)'''.
-*Das Signal nach dem Gaußtiefpass lautet somit:
+*The signal after the gauss low pass is thus
 :$$q_{\rm G} (t) = q_{\rm R} (t) \star h_{\rm G}(t) = \sum_{\nu} a_{\nu}\cdot g(t - \nu \cdot T)\hspace{0.05cm}.$$
-:Hierbei wird&nbsp; $g(t)$&nbsp; als ''Frequenzimpuls'' bezeichnet. Für diesen gilt:
+Here&nbsp; $g(t)$&nbsp; is referred to as ''frequency pulse''. For this one:
 :$$g(t) = q_{\rm R} (t) \star h_{\rm G}(t) \hspace{0.05cm}.$$
-*Mit dem tiefpassgefilterten Signal&nbsp; $q_{\rm G}(t)$, der Trägerfrequenz&nbsp; $f_{\rm T}$&nbsp; und dem Frequenzhub&nbsp; $\Delta f_{\rm A}$&nbsp; kann somit für die Augenblicksfrequenz am Ausgang des FSK–Modulators geschrieben werden:
+*With the low pass filtered signal&nbsp; $q_{\rm G}(t)$, the carrier frequency&nbsp; $f_{\rm T}$&nbsp; and the frequency deviation&nbsp; $\Delta f_{\rm A}$&nbsp; can thus be written for the instantaneous frequency at the output of the FSK modulator::$$f_{\rm A}(t) = f_{\rm T} + \Delta f_{\rm A} \cdot q_{\rm G} (t)\hspace{0.05cm}.$$
-:$$f_{\rm A}(t) = f_{\rm T} + \Delta f_{\rm A} \cdot q_{\rm G} (t)\hspace{0.05cm}.$$
@@ Line 38: / Line 37: @@
 *This exercise belongs to the chapter&nbsp;  [[Mobile_Kommunikation/Die_Charakteristika_von_GSM|Die Charakteristika von GSM]].
-*Bezug genommen wird auch auf das Kapitel&nbsp;  [[Beispiele_von_Nachrichtensystemen/Funkschnittstelle|Funkschnittstelle]]&nbsp; im Buch „Beispiele von Nachrichtensystemen”.
+*Reference is also made to the chapter&nbsp;  [[Beispiele_von_Nachrichtensystemen/Funkschnittstelle|Funkschnittstelle]]&nbsp; in the book „Beispiele von Nachrichtensystemen”.
-*Verwenden Sie für Ihre Berechnungen die beispielhaften Werte&nbsp; $f_{\rm T} = 900 \ \rm  MHz$&nbsp; und&nbsp; $\Delta f_{\rm A} = 68 \ \rm kHz$.
+*For your calculations use the exemplary values&nbsp; $f_{\rm T} = 900 \ \ \rm MHz$&nbsp; and&nbsp; $\Delta f_{\rm A} = 68 \ \rm kHz$.
-*Verwenden Sie zur Lösung der Aufgabe das Gaußintegral (einige Zahlenwerte sind in der Tabelle angegeben):
+*Use the Gaussian integral to solve the task (some numerical values are given in the table)
-[[File:P_ID2226__Bei_A_3_4b.png|right|frame|Tabelle der Gaußschen Fehlerfunktion]]
+[[File:P_ID2226__Bei_A_3_4b.png|right|frame|some values of the Gaussian integral]]
 :$$\phi(x) =\frac {1}{\sqrt{2 \pi}} \cdot \int^{x} _{-\infty} {\rm e}^{-u^2/2}\,{\rm d}u \hspace{0.05cm}.$$
 <br clear=all>
@@ Line 52: / Line 51: @@
 <quiz display=simple>
-{In welchem Wertebereich kann die Augenblicksfrequenz&nbsp; $f_{\rm A}(t)$&nbsp; schwanken? Welche Voraussetzungen müssen dafür erfüllt sein?
+{In what range of values can the instantaneous frequency&nbsp; $f_{\rm A}(t)$&nbsp; fluctuate? Which requirements must be met?
 |type="{}"}
 ${\rm Max} \  \big[f_{\rm A}(t)\big] \ = \hspace{0.2cm} $ { 900.068 0.01% } $\ \rm MHz$
 ${\rm Min} \  \big[f_{\rm A}(t)\big] \ = \hspace{0.28cm} $ { 899.932 0.01% } $\ \rm MHz$
-{Welche (normierte) systemtheoretische Grenzfrequenz des Gaußtiefpasses ergibt sich aus der Forderung&nbsp; $f_{\rm 3dB} \cdot T = 0.3$?
+{Which (normalized) system-theoretical cut-off frequency of the Gaussian low pass results from the requirement&nbsp; $f_{\rm 3dB} \cdot T = 0.3$?
 |type="{}"}
 $f_{\rm G} \cdot T \ = \ $ { 0.45 3% }
-{Berechnen Sie den Frequenzimpuls&nbsp; $g(t)$&nbsp; unter Verwendung der Funktion&nbsp; $\phi (x)$. Wie groß ist der Impulswert&nbsp; $g(t = 0)$?
+{Calculate the frequency pulse&nbsp; $g(t)$&nbsp; using the function&nbsp; $\phi (x)$. How large is the pulse value&nbsp; $g(t = 0)$?
 |type="{}"}
 $g(t = 0) \ = \ $ { 0.737 3% }
-{Welcher Signalwert ergibt sich für&nbsp; $q_{\rm G}(t = 3T)$&nbsp; mit&nbsp; $a_{3} = –1$&nbsp; sowie&nbsp; $a_{\mu \ne 3} = +1$? Wie groß ist die Augenblicksfrequenz&nbsp; $f_{\rm A}(t = 3T)$?
+{Which signal value results for&nbsp; $q_{\rm G}(t = 3T)$&nbsp; with&nbsp; $a_{3} = -1$&nbsp; and&nbsp; $a_{\mu \ne 3} = +1$? What is the instantaneous frequency&nbsp; $f_{\rm A}(t = 3T)$?
 |type="{}"}
 $q_{\rm G}(t = 3T) \ = \ $ { -0.51822--0.42978 }
-{Berechnen Sie die Impulswerte&nbsp; $g(t = ±T)$&nbsp; des Frequenzimpulses.
+{Calculate the pulse values&nbsp; $g(t = ±T)$&nbsp; of the frequency pulse
 |type="{}"}
 $g(t = ±T) \ = \ $ { 0.131 3% }
-{Die Amplitudenkoeffizienten seien alternierend. Welcher maximale Betrag von&nbsp; $q_{G}(t)$&nbsp; ergibt sich? Berücksichtigen Sie&nbsp; $g(t ≥ 2 T) \approx 0$.
+{The amplitude coefficients are alternating. What is the maximum amount of&nbsp; $q_{G}(t)$&nbsp;? Consider&nbsp; $g(t ≥ 2 T) \approx 0$.
 |type="{}"}
 ${\rm Max} \ |q_{\rm G}(t)| \ = \ $ { 0.475 3% }
@@ Line 84: / Line 83: @@
 {{ML-Kopf}}
-'''(1)'''&nbsp; Sind alle Amplitudenkoeffizienten $a_{\mu}$ gleich $+1$, so ist $q_{\rm R}(t) = 1$ eine Konstante. Damit hat der Gaußtiefpass keinen Einfluss und es ergibt sich $q_{\rm G}(t) = 1$.
+'''(1)'''&nbsp; If all amplitude coefficients $a_{\mu}$ are equal to $+1$, then $q_{\rm R}(t) = 1$ is a constant. Thus, the Gaussian low pass has no influence and $q_{\rm G}(t) = 1$ results.
-*Die maximale Frequenz ist somit
+*The maximum frequency is thus
 :$${\rm Max}\ [f_{\rm A}(t)] = f_{\rm T} + \Delta f_{\rm A} \hspace{0.15cm} \underline {= 900.068\,{\rm MHz}} \hspace{0.05cm}.$$
-*Das Minimum der Augenblicksfrequenz
+*The minimum instantaneous frequency
 :$${\rm Min}\ [f_{\rm A}(t)] = f_{\rm T} - \Delta f_{\rm A} \hspace{0.15cm} \underline { = 899.932\,{\rm MHz}} \hspace{0.05cm}$$
-:ergibt sich, wenn alle Amplitudenkoeffizienten negativ sind. In diesem Fall ist $q_{\rm R}(t) = q_{\rm G}(t) = -1$.
+is obtained when all amplitude coefficients are negative. In this case $q_{\rm R}(t) = q_{\rm G}(t) = -1$.
-'''(2)'''&nbsp; Diejenige Frequenz, bei der die logarithmierte Leistungsübertragungsfunktion gegenüber $f = 0$ um $3 \ \rm  dB$ kleiner ist, bezeichnet man als die 3dB–Grenzfrequenz.
+'''(2)'''&nbsp; The frequency at which the logarithmic power transfer function is $3 \ \rm dB$ less than $f = 0$ is called the 3dB cut-off frequency.
-*Dies lässt sich auch wie folgt ausdrücken:
+*This can also be expressed as follows:
 :$$\frac {|H(f = f_{\rm 3dB})|}{|H(f = 0)|}= \frac{1}{\sqrt{2}} \hspace{0.05cm}.$$
-*Insbesondere gilt für den Gaußtiefpass wegen $H(f = 0) = 1$:
+*In particular the Gauss low pass because of $H(f = 0) = 1$:
 :$$ H(f = f_{\rm 3dB})= {\rm e}^{-\pi\cdot ({f_{\rm 3dB}}/{2 f_{\rm G}})^2} = \frac{1}{\sqrt{2}}\hspace{0.3cm}
 \Rightarrow \hspace{0.3cm}(\frac{f_{\rm 3dB}}{2 f_{\rm G}})^2 = \frac{{\rm ln}\hspace{0.1cm}\sqrt{2}}{\pi} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm G} = \sqrt{\frac{\pi}{4 \cdot {\rm ln}\hspace{0.1cm}\sqrt{2}}}\cdot f_{\rm 3dB}\hspace{0.05cm}.$$
-*Die numerische Auswertung führt auf $f_{\rm G} \approx 1.5 \cdot f_{\rm 3dB}$.
+*The numerical evaluation leads to  $f_{\rm G} \approx 1.5 \cdot f_{\rm 3dB}$.
-*Aus $f_{\rm 3dB} \cdot T = 0.3$ folgt somit $f_{\rm G} \cdot T \underline{\approx 0.45}$.
+*{\f_{\rm 3dB} \cdot T = 0.3$ is followed by $f_{\rm G} \cdot T \underline{\approx 0.45}$.
@@ Line 130: / Line 128: @@
 *Es ergeben sich also Impulsinterferenzen und man erhält:
 :$${\rm Max} \hspace{0.12cm}[q_{\rm G}(t)] = g(t = 0) - 2 \cdot g(t = T) = 0.737 - 2 \cdot 0.131 \hspace{0.15cm} \underline {= 0.475 }\hspace{0.05cm}.$$
+{{ML-Fuß}}
+[[Category:Exercises for Mobile Communications|^3.3 Characteristics of GSM^]]
+{{quiz-Header|Buchseite=Mobile Kommunikation/Die Charakteristika von GSM
+}}
+[[File:EN_Mob_A_3_5.png|right|frame|Verschiedene Signale bei GMSK-Modulation]]
+The modulation method used for GSM is&nbsp; ''Gaussian Minimum Shift Keying'', short GMSK. This is a special type of FSK (''Frequency Shift Keying'') with CP-FSK (''Continuous Phase Matching''), where
+*the modulation index has the smallest value that just satisfies the orthogonality condition: &nbsp; $h = 0.5$ &nbsp; &rArr; &nbsp; ''Minimum Shift Keying'',
+*a Gaussian low pass with the impulse response&nbsp; $h_{\rm G}(t)$&nbsp; is inserted before the FSK modulator, with the aim of saving even more bandwidth.
+The graphic illustrates the situation:
+*The digital message is represented by the amplitude coefficients&nbsp; $a_{\mu} ∈ \{±1\}$&nbsp; which are applied to a Dirac pulse. It should be noted that the sequence drawn in is assumed for the subtask '''(3)'''.
+*The symmetrical rectangular pulse with duration&nbsp; $T = T_{\rm B}$&nbsp; (GSM bit duration) is dimensionless:
+:$$g_{\rm R}(t) = \left\{ \begin{array}{c} 1 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{5}c} {\rm{{\rm{f\ddot{u}r}}}} \\ {\rm{{\rm{f\ddot{u}r}}}} \\ \end{array}\begin{array}{*{5}c} |\hspace{0.05cm} t \hspace{0.05cm}| < T/2 \hspace{0.05cm}, \\ |\hspace{0.05cm} t \hspace{0.05cm}| > T/2 \hspace{0.05cm}. \\ \end{array}$$
+*This results for the rectangular signal
+:$$q_{\rm R} (t) = q_{\rm \delta} (t) \star g_{\rm R}(t) = \sum_{\nu} a_{\nu}\cdot g_{\rm R}(t - \nu \cdot T)\hspace{0.05cm}.$$
+*The Gaussian low pass is given by its frequency response or impulse response:
+:$$H_{\rm G}(f) = {\rm e}^{-\pi\cdot (\frac{f}{2 f_{\rm G}})^2} \hspace{0.2cm}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \hspace{0.2cm} h_{\rm G}(t) = 2 f_{\rm G} \cdot {\rm e}^{-\pi\cdot (2 f_{\rm G}\cdot t)^2}\hspace{0.05cm},$$
+:where the system theoretical cut-off frequency&nbsp; $f_{\rm G}$&nbsp; is used. In the GSM specification, however, the 3dB cut-off frequency is specified with&nbsp; $f_{\rm 3dB} = 0.3/T$&nbsp;. From this,&nbsp; $f_{\rm G}$&nbsp; can be calculated directly - see subtask '''(2)'''.
+*The signal after the gauss low pass is thus
+:$$q_{\rm G} (t) = q_{\rm R} (t) \star h_{\rm G}(t) = \sum_{\nu} a_{\nu}\cdot g(t - \nu \cdot T)\hspace{0.05cm}.$$
+Here&nbsp; $g(t)$&nbsp; is referred to as ''frequency pulse''. For this one:
+:$$g(t) = q_{\rm R} (t) \star h_{\rm G}(t) \hspace{0.05cm}.$$
+*With the low pass filtered signal&nbsp; $q_{\rm G}(t)$, the carrier frequency&nbsp; $f_{\rm T}$&nbsp; and the frequency deviation&nbsp; $\Delta f_{\rm A}$&nbsp; can thus be written for the instantaneous frequency at the output of the FSK modulator::$$f_{\rm A}(t) = f_{\rm T} + \Delta f_{\rm A} \cdot q_{\rm G} (t)\hspace{0.05cm}.$$
+''Notes:''
+*This exercise belongs to the chapter&nbsp;  [[Mobile_Kommunikation/Die_Charakteristika_von_GSM|Die Charakteristika von GSM]].
+*Reference is also made to the chapter&nbsp;  [[Beispiele_von_Nachrichtensystemen/Funkschnittstelle|Funkschnittstelle]]&nbsp; in the book „Beispiele von Nachrichtensystemen”.
+*For your calculations use the exemplary values&nbsp; $f_{\rm T} = 900 \ \ \rm MHz$&nbsp; and&nbsp; $\Delta f_{\rm A} = 68 \ \rm kHz$.
+*Use the Gaussian integral to solve the task (some numerical values are given in the table)
+[[File:P_ID2226__Bei_A_3_4b.png|right|frame|some values of the Gaussian integral]]
+:$$\phi(x) =\frac {1}{\sqrt{2 \pi}} \cdot \int^{x} _{-\infty} {\rm e}^{-u^2/2}\,{\rm d}u \hspace{0.05cm}.$$
+<br clear=all>
+===Fragebogen===
+<quiz display=simple>
+{In what range of values can the instantaneous frequency&nbsp; $f_{\rm A}(t)$&nbsp; fluctuate? Which requirements must be met?
+|type="{}"}
+${\rm Max} \  \big[f_{\rm A}(t)\big] \ = \hspace{0.2cm} $ { 900.068 0.01% } $\ \rm MHz$
+${\rm Min} \  \big[f_{\rm A}(t)\big] \ = \hspace{0.28cm} $ { 899.932 0.01% } $\ \rm MHz$
+{Which (normalized) system-theoretical cut-off frequency of the Gaussian low pass results from the requirement&nbsp; $f_{\rm 3dB} \cdot T = 0.3$?
+|type="{}"}
+$f_{\rm G} \cdot T \ = \ $ { 0.45 3% }
+{Calculate the frequency pulse&nbsp; $g(t)$&nbsp; using the function&nbsp; $\phi (x)$. How large is the pulse value&nbsp; $g(t = 0)$?
+|type="{}"}
+$g(t = 0) \ = \ $ { 0.737 3% }
+{Which signal value results for&nbsp; $q_{\rm G}(t = 3T)$&nbsp; with&nbsp; $a_{3} = -1$&nbsp; and&nbsp; $a_{\mu \ne 3} = +1$? What is the instantaneous frequency&nbsp; $f_{\rm A}(t = 3T)$?
+|type="{}"}
+$q_{\rm G}(t = 3T) \ = \ $ { -0.51822--0.42978 }
+{Calculate the pulse values&nbsp; $g(t = ±T)$&nbsp; of the frequency pulse
+|type="{}"}
+$g(t = ±T) \ = \ $ { 0.131 3% }
+{The amplitude coefficients are alternating. What is the maximum amount of&nbsp; $q_{G}(t)$&nbsp;? Consider&nbsp; $g(t ≥ 2 T) \approx 0$.
+|type="{}"}
+${\rm Max} \ |q_{\rm G}(t)| \ = \ $ { 0.475 3% }
+</quiz>
+===Sample solution===
+{{ML-Kopf}}
+'''(1)'''&nbsp; If all amplitude coefficients $a_{\mu}$ are equal to $+1$, then $q_{\rm R}(t) = 1$ is a constant. Thus, the Gaussian low pass has no influence and $q_{\rm G}(t) = 1$ results.
+*The maximum frequency is thus
+:$${\rm Max}\ [f_{\rm A}(t)] = f_{\rm T} + \Delta f_{\rm A} \hspace{0.15cm} \underline {= 900.068\,{\rm MHz}} \hspace{0.05cm}.$$
+*The minimum instantaneous frequency
+:$${\rm Min}\ [f_{\rm A}(t)] = f_{\rm T} - \Delta f_{\rm A} \hspace{0.15cm} \underline { = 899.932\,{\rm MHz}} \hspace{0.05cm}$$
+is obtained when all amplitude coefficients are negative. In this case $q_{\rm R}(t) = q_{\rm G}(t) = -1$.
+'''(2)'''&nbsp; The frequency at which the logarithmic power transfer function is $3 \ \rm dB$ less than $f = 0$ is called the 3dB cut-off frequency.
+*This can also be expressed as follows:
+:$$\frac {|H(f = f_{\rm 3dB})|}{|H(f = 0)|}= \frac{1}{\sqrt{2}} \hspace{0.05cm}.$$
+*In particular the Gauss low pass because of $H(f = 0) = 1$:
+:$$ H(f = f_{\rm 3dB})= {\rm e}^{-\pi\cdot ({f_{\rm 3dB}}/{2 f_{\rm G}})^2} = \frac{1}{\sqrt{2}}\hspace{0.3cm}
+\Rightarrow \hspace{0.3cm}(\frac{f_{\rm 3dB}}{2 f_{\rm G}})^2 = \frac{{\rm ln}\hspace{0.1cm}\sqrt{2}}{\pi} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm G} = \sqrt{\frac{\pi}{4 \cdot {\rm ln}\hspace{0.1cm}\sqrt{2}}}\cdot f_{\rm 3dB}\hspace{0.05cm}.$$
+*The numerical evaluation leads to  $f_{\rm G} \approx 1.5 \cdot f_{\rm 3dB}$.