Exercise 3.5: GMSK Modulation

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-MathJax55-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-MathJax56-QINU`"' *Für die Zeit $t = 0$ gilt unter Berücksichtigung von $\phi (-x) = 1 - \phi (x)$ und $f_{\rm G} \cdot T = 0.45$: :'"`UNIQ-MathJax57-QINU`"' '''(4)''' Mit $a_{3} = +1$ würde sich $q_{\rm G}(t = 3 T) = 1$ ergeben. Aufgrund der Linearität gilt somit: :'"`UNIQ-MathJax58-QINU`"' '''(5)''' Mit dem Ergebnis aus (3) und $f_{\rm G} \cdot T = 0.45$ erhält man: :'"`UNIQ-MathJax59-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-MathJax60-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-MathJax61-QINU`"' *This results for the rectangular signal :'"`UNIQ-MathJax62-QINU`"' *The Gaussian low pass is given by its frequency response or impulse response: :'"`UNIQ-MathJax63-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-MathJax64-QINU`"' Here $g(t)$ is referred to as ''frequency pulse''. For this one: :'"`UNIQ-MathJax65-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-MathJax66-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-MathJax67-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-MathJax68-QINU`"' *The minimum instantaneous frequency :'"`UNIQ-MathJax69-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-MathJax70-QINU`"' *In particular the Gauss low pass because of $H(f = 0) = 1$: :'"`UNIQ-MathJax71-QINU`"' *The numerical evaluation leads to $f_{\rm G} \approx 1.5 \cdot f_{\rm 3dB}$.