Aufgaben:Exercise 3.5: GMSK Modulation

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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

1

In what range of values can the instantaneous frequency  $f_{\rm A}(t)$  fluctuate? Which requirements must be met?

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

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

2

Which (normalized) system-theoretical cut-off frequency of the Gaussian low pass results from the requirement  $f_{\rm 3dB} \cdot T = 0.3$?

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

3

Calculate the frequency pulse  $g(t)$  using the function  $\phi (x)$. How large is the pulse value  $g(t = 0)$?

$g(t = 0) \ = \ $

4

Which signal value results for  $q_{\rm G}(t = 3T)$  with  $a_{3} = -1$  and  $a_{\mu \ne 3} = +1$? What is the instantaneous frequency  $f_{\rm A}(t = 3T)$?

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

5

Calculate the pulse values  $g(t = ±T)$  of the frequency pulse

$g(t = ±T) \ = \ $

6

The amplitude coefficients are alternating. What is the maximum amount of  $q_{G}(t)$ ? Consider  $g(t ≥ 2 T) \approx 0$.

${\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$f_{\rm G} \approx 1.5 \cdot f_{\rm 3dB}$.