Aufgaben:Exercise 3.5: GMSK Modulation - Revision history

Hwang at 06:08, 18 September 2022

2022-09-18T06:08:34Z

Javier: Text replacement - "Dirac impulse" to "Dirac delta"

2021-05-26T07:04:55Z

Text replacement - "Dirac impulse" to "Dirac delta"

Javier: Text replacement - "Category:Exercises for Mobile Communications" to "Category:Mobile Communications: Exercises"

2021-03-23T12:37:32Z

Text replacement - "Category:Exercises for Mobile Communications" to "Category:Mobile Communications: Exercises"

Guenter at 14:31, 22 February 2021

2021-02-22T14:31:25Z

Guenter at 12:56, 23 January 2021

2021-01-23T12:56:30Z

Guenter at 17:00, 22 January 2021

2021-01-22T17:00:52Z

Guenter at 16:49, 22 January 2021

2021-01-22T16:49:05Z

Javier: Text replacement - "===Sample solution===" to "===Solution==="

2020-12-16T13:16:08Z

Text replacement - "===Sample solution===" to "===Solution==="

Javier: Text replacement - "low pass" to "low-pass"

2020-12-15T14:19:21Z

Text replacement - "low pass" to "low-pass"

Javier: Text replacement - "Mobile Kommunikation/Die Charakteristika von GSM" to "Mobile_Communications/Characteristics_of_GSM"

2020-12-15T14:03:16Z

Text replacement - "Mobile Kommunikation/Die Charakteristika von GSM" to "Mobile_Communications/Characteristics_of_GSM"

@@ Line 68: / Line 68: @@
 $g(t = ±T) \ = \ $ { 0.131 3% }
-{The amplitude coefficients are alternating.&nbsp; What is the maximum amount of&nbsp; $q_{\rm G}(t)$&nbsp;?&nbsp; Consider&nbsp; $g(t ≥ 2 T) \approx 0$.
+{The amplitude coefficients are alternating.&nbsp; What is the maximum magnitude of&nbsp; $q_{\rm G}(t)$&nbsp;?&nbsp; Consider&nbsp; $g(t ≥ 2 T) \approx 0$.
 |type="{}"}
 ${\rm Max} \ |q_{\rm G}(t)| \ = \ $ { 0.475 3% }

@@ Line 12: / Line 12: @@
 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 comb"&nbsp; (Dirac impulse train).&nbsp; It should be noted that the sequence drawn in is assumed for the subtask&nbsp; '''(3)'''.
+*The digital message is represented by the amplitude coefficients&nbsp; $a_{\mu} ∈ \{±1\}$&nbsp; which are applied to a "Dirac comb"&nbsp; (Dirac delta train).&nbsp; It should be noted that the sequence drawn in is assumed for the subtask&nbsp; '''(3)'''.
 *The symmetrical rectangular pulse with duration&nbsp; $T = T_{\rm B}$&nbsp; (GSM bit duration)&nbsp; is dimensionless:

← Older revision		Revision as of 12:37, 23 March 2021
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−	[[Category:~~Exercises for~~ Mobile Communications\|^3.3 Characteristics of GSM^]]	+	[[Category:Mobile Communications: Exercises\|^3.3 Characteristics of GSM^]]

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Mob_A_3_5.png|right|frame|different signals of GMSK-Modulation]]
+[[File:EN_Mob_A_3_5.png|right|frame|Different signals of GMSK-Modulation]]
 The modulation method used for GSM is&nbsp; "Gaussian Minimum Shift Keying",&nbsp; short&nbsp; $\rm GMSK$.&nbsp; This is a special type of&nbsp; $\rm  FSK$&nbsp; ("Frequency Shift Keying")&nbsp; with&nbsp; $\rm  CP-FSK$&nbsp; ("Continuous Phase Matching"), where:
@@ Line 12: / Line 12: @@
 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.&nbsp; It should be noted that the sequence drawn in is assumed for the subtask&nbsp; '''(3)'''.
+*The digital message is represented by the amplitude coefficients&nbsp; $a_{\mu} ∈ \{±1\}$&nbsp; which are applied to a "Dirac comb"&nbsp; (Dirac impulse train).&nbsp; It should be noted that the sequence drawn in is assumed for the subtask&nbsp; '''(3)'''.
 *The symmetrical rectangular pulse with duration&nbsp; $T = T_{\rm B}$&nbsp; (GSM bit duration)&nbsp; is dimensionless:

@@ Line 79: / Line 79: @@
 {{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.
+'''(1)'''&nbsp; If all amplitude coefficients&nbsp; $a_{\mu}$&nbsp; are equal to&nbsp; $+1$, then&nbsp; $q_{\rm R}(t) = 1$&nbsp; is a constant.&nbsp; Thus, the Gaussian low-pass has no influence and&nbsp; $q_{\rm G}(t) = 1$&nbsp; 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
+*The minimum instantaneous frequency is
 :$${\rm Min}\ [f_{\rm A}(t)] = f_{\rm T} - \Delta f_{\rm A} \hspace{0.15cm} \underline { = 899.932\,{\rm MHz}} \hspace{0.05cm}$$
-is obtained when all amplitude coefficients are negative. In this case $q_{\rm R}(t) = q_{\rm G}(t) = -1$.
+:is obtained when all amplitude coefficients are negative.&nbsp; 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.
+'''(2)'''&nbsp; The frequency at which the logarithmic power transfer function is&nbsp; $3 \ \rm dB$&nbsp; less than for&nbsp; $f = 0$&nbsp; 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 $H(f = 0) = 1$:
+*In particular the Gauss low-pass because of&nbsp; $H(f = 0) = 1$:
 :$$ H(f = f_{\rm 3dB})= {\rm e}^{-\pi\cdot ({f_{\rm 3dB}}/{2 f_{\rm G}})^2} = \frac{1}{\sqrt{2}}\hspace{0.3cm}
 \Rightarrow \hspace{0.3cm}(\frac{f_{\rm 3dB}}{2 f_{\rm G}})^2 = \frac{{\rm ln}\hspace{0.1cm}\sqrt{2}}{\pi} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm G} = \sqrt{\frac{\pi}{4 \cdot {\rm ln}\hspace{0.1cm}\sqrt{2}}}\cdot f_{\rm 3dB}\hspace{0.05cm}.$$
-*The numerical evaluation leads to  $f_{\rm G} \approx 1.5 \cdot f_{\rm 3dB}$.
+*The numerical evaluation leads to&nbsp;  $f_{\rm G} \approx 1.5 \cdot f_{\rm 3dB}$.
-*From $f_{\rm 3dB} \cdot T = 0.3$ follows $f_{\rm G} \cdot T \underline{\approx 0.45}$.
+*From&nbsp; $f_{\rm 3dB} \cdot T = 0.3$&nbsp; follows&nbsp; $f_{\rm G} \cdot T \underline{\approx 0.45}$.
+'''(3)'''&nbsp; The frequency pulse&nbsp; ${\rm g}(t)$&nbsp; results from the convolution of the rectangular function&nbsp; $g_{\rm R}(t)$&nbsp; with the impulse response&nbsp; $h_{\rm G}(t)$:
 :$$g(t) = g_{\rm R} (t) \star h_{\rm G}(t) = 2 f_{\rm G} \cdot \int^{t + T/2} _{t - T/2} {\rm e}^{-\pi\cdot (2 f_{\rm G}\cdot \tau)^2}\,{\rm d}\tau \hspace{0.05cm}.$$
-*With the substitution $u^{2} = 8π \cdot {f_{G}}^{2} \cdot \tau^{2}$ and the function $\phi (x)$ you can also write for this:
+*With the substitution&nbsp; $u^{2} = 8π \cdot {f_{G}}^{2} \cdot \tau^{2}$&nbsp; and the function&nbsp; $\phi (x)$&nbsp; you can also write for this:
 :$$g(t) = \ \frac {1}{\sqrt{2 \pi}} \cdot \int^{2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t + T/2)} _{2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t - T/2)} {\rm e}^{-u^2/2}\,{\rm d}u = \ \phi(2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t + T/2))- \phi(2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t - T/2)) \hspace{0.05cm}.$$
-*For the time $t = 0$ is valid considering $\phi (-x) = 1 - \phi (x)$ and $f_{\rm G} \cdot T = $0.45
+*Considering&nbsp; $\phi (-x) = 1 - \phi (x)$&nbsp; and&nbsp; $f_{\rm G} \cdot T = 0.45$,&nbsp; you get for the time&nbsp; $t = 0$:
 :$$g(t = 0) = \ \phi(\sqrt{2 \pi} \cdot f_{\rm G} \cdot T)- \phi(-\sqrt{2 \pi} \cdot f_{\rm G} \cdot T)= \ 2 \cdot \phi(\sqrt{2 \pi} \cdot f_{\rm G} \cdot T)-1 \approx 2 \cdot \phi(1.12)-1 \hspace{0.15cm} \underline {= 0.737} \hspace{0.05cm}.$$
-'''(4)'''&nbsp; With $a_{3} = +1$ the result would be $q_{\rm G}(t = 3 T) = 1$. Due to the linearity, the following therefore applies:
+'''(4)'''&nbsp; With&nbsp; $a_{3} = +1$&nbsp; the result would be&nbsp; $q_{\rm G}(t = 3 T) = 1$.&nbsp; Due to the linearity therefore applies:
 :$$q_{\rm G}(t = 3 T ) = 1 - 2 \cdot g(t = 0)= 1 - 2 \cdot 0.737 \hspace{0.15cm} \underline {= -0.474} \hspace{0.05cm}.$$
-'''(5)'''&nbsp; With the result of (3) and $f_{\rm G} \cdot T = $0.45 you get
+'''(5)'''&nbsp; With the result of&nbsp; '''(3)'''&nbsp; and&nbsp; $f_{\rm G} \cdot T = 0.45$&nbsp; you get
 :$$g(t = T) = \ \phi(3 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot T)- \phi(\sqrt{2 \pi} \cdot f_{\rm G} \cdot T) \approx  \phi(3.36)-\phi(1.12) = 0.999 - 0.868 \hspace{0.15cm} \underline { = 0.131} \hspace{0.05cm}.$$
-*The pulse value $g(t = -T)$ is exactly the same due to the symmetry of the Gaussian low-pass.
+*The pulse value&nbsp; $g(t = -T)$&nbsp; is exactly the same due to the symmetry of the Gaussian low-pass.
-'''(6)'''&nbsp; With alternating sequence, the absolute values $|q_{\rm G}(\mu \cdot T)|$ are all the same for all multiples of the bit duration $T$ for reasons of symmetry.
+'''(6)'''&nbsp; With alternating sequence, the absolute values&nbsp; $|q_{\rm G}(\mu \cdot T)|$&nbsp; are all the same for all multiples of the bit duration&nbsp; $T$&nbsp; for reasons of symmetry.
-*All intermediate values at $t \approx \mu \cdot T$ are smaller.
+*All intermediate values at&nbsp; $t \approx \mu \cdot T$&nbsp; are smaller.
-* Taking $g(t ≥ 2T) \approx 0$ into account, each individual pulse value $g(0)$ is reduced by the preceding pulse with $g(t = T)$, and also by the following pulse with $g(t = -T)$.
+* Taking&nbsp; $g(t ≥ 2T) \approx 0$&nbsp; into account, each individual pulse value&nbsp; $g(0)$&nbsp; is reduced by the preceding pulse with&nbsp; $g(t = T)$, and by the following pulse with&nbsp; $g(t = -T)$.
-*So there will be impulse interference and you get
+*So there will be "intersymbol interference" and you get
 :$${\rm Max} \hspace{0.12cm}[q_{\rm G}(t)] = g(t = 0) - 2 \cdot g(t = T) = 0.737 - 2 \cdot 0.131 \hspace{0.15cm} \underline {= 0.475 }\hspace{0.05cm}.$$

@@ Line 32: / Line 32: @@
+[[File:P_ID2226__Bei_A_3_4b.png|right|frame|Some Gaussian integral values]]
 ''Notes:''
-*This exercise belongs to the chapter&nbsp;  [[Mobile_Communications/Characteristics_of_GSM|Characteristics_of_GSM]].
+*This exercise belongs to the chapter&nbsp;  [[Mobile_Communications/Characteristics_of_GSM|Characteristics of GSM]].
-*Reference is also made to the chapter&nbsp;  [[Examples_of_Communication_Systems/Funkschnittstelle|Funkschnittstelle]]&nbsp; in the book „Beispiele von Nachrichtensystemen”.
+*Reference is also made to the chapter&nbsp;  [[Examples_of_Communication_Systems/Radio_Interface|Radio Interface]]&nbsp; in the book "Examples of Communication Systems".
 *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)
+*Use the Gaussian integral to solve the task&nbsp; $($some numerical values are given in the table$)$.
 :$$\phi(x) =\frac {1}{\sqrt{2 \pi}} \cdot \int^{x} _{-\infty} {\rm e}^{-u^2/2}\,{\rm d}u \hspace{0.05cm}.$$
 <br clear=all>
@@ Line 51: / Line 47: @@
 <quiz display=simple>
-{In what range of values can the instantaneous frequency&nbsp; $f_{\rm A}(t)$&nbsp; fluctuate? Which requirements must be met?
+{In what range of values the instantaneous frequency&nbsp; $f_{\rm A}(t)$&nbsp; can fluctuate?&nbsp; Which requirements must be met?
 |type="{}"}
 ${\rm Max} \  \big[f_{\rm A}(t)\big] \ = \hspace{0.2cm} $ { 900.068 0.01% } $\ \rm MHz$
@@ Line 60: / Line 56: @@
 $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; $g(t = 0)$?
+{Calculate the frequency pulse&nbsp; $g(t)$&nbsp; using the function&nbsp; $\phi (x)$.&nbsp; What is the result for&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; $f_{\rm A}(t = 3T)$?
+{Which signal value results for&nbsp; $q_{\rm G}(t = 3T)$&nbsp; with&nbsp; $a_{3} = -1$&nbsp; and&nbsp; $a_{\mu \ne 3} = +1$?&nbsp; 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
+{Calculate the 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; $g(t ≥ 2 T) \approx 0$.
+{The amplitude coefficients are alternating.&nbsp; What is the maximum amount of&nbsp; $q_{\rm G}(t)$&nbsp;?&nbsp; Consider&nbsp; $g(t ≥ 2 T) \approx 0$.
 |type="{}"}
 ${\rm Max} \ |q_{\rm G}(t)| \ = \ $ { 0.475 3% }

@@ Line 5: / Line 5: @@
 [[File:EN_Mob_A_3_5.png|right|frame|different signals of 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 method used for GSM is&nbsp; "Gaussian Minimum Shift Keying",&nbsp; short&nbsp; $\rm GMSK$.&nbsp; This is a special type of&nbsp; $\rm  FSK$&nbsp; ("Frequency Shift Keying")&nbsp; with&nbsp; $\rm  CP-FSK$&nbsp; ("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'',
+*The modulation index has the smallest value that just satisfies the orthogonality condition: &nbsp; $h = 0.5$ &nbsp; &rArr; &nbsp; "Minimum Shift Keying"&nbsp; $\rm  (MSK)$,
-*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.
+*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 digital message is represented by the amplitude coefficients&nbsp; $a_{\mu} ∈ \{±1\}$&nbsp; which are applied to a Dirac pulse.&nbsp; It should be noted that the sequence drawn in is assumed for the subtask&nbsp; '''(3)'''.
-*The symmetrical rectangular pulse with duration&nbsp; $T = T_{\rm B}$&nbsp; (GSM bit duration) is dimensionless:
+*The symmetrical rectangular pulse with duration&nbsp; $T = T_{\rm B}$&nbsp; (GSM bit duration)&nbsp; 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:
-:$$H_{\rm G}(f) = {\rm e}^{-\pi\cdot (\frac{f}{2 f_{\rm G}})^2} \hspace{0.2cm}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \hspace{0.2cm} h_{\rm G}(t) = 2 f_{\rm G} \cdot {\rm e}^{-\pi\cdot (2 f_{\rm G}\cdot t)^2}\hspace{0.05cm},$$
+:$$H_{\rm G}(f) = {\rm e}^{-\pi\hspace{0.05cm}\cdot \hspace{0.05cm}[{f}/ ({2 f_{\rm G})} ]^2} \hspace{0.2cm}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \hspace{0.2cm} h_{\rm G}(t) = 2 \cdot f_{\rm G} \cdot {\rm e}^{-\pi\hspace{0.05cm}\cdot \hspace{0.05cm}(2 f_{\rm G}\hspace{0.05cm}\cdot \hspace{0.05cm}t)^2}\hspace{0.05cm},$$
-: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)'''.
+:where the system theoretical cut-off frequency&nbsp; $f_{\rm G}$&nbsp; is used.&nbsp; In the GSM specification, however, the 3dB cut-off frequency is specified with&nbsp; $f_{\rm 3dB} = 0.3/T$&nbsp;.&nbsp; From this,&nbsp; $f_{\rm G}$&nbsp; can be calculated directly - see subtask&nbsp; '''(2)'''.
-*The signal after the gauss low-pass is thus
+*The signal after the Gaussian 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:
+*Here&nbsp; $g(t)$&nbsp; is referred to as "frequency pulse":
 :$$g(t) = q_{\rm R} (t) \star h_{\rm G}(t) \hspace{0.05cm}.$$
-*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::$$f_{\rm A}(t) = f_{\rm T} + \Delta f_{\rm A} \cdot q_{\rm G} (t)\hspace{0.05cm}.$$
+*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; for the instantaneous frequency at the output of the FSK modulator can thus be written:
@@ Line 37: / Line 38: @@
 ''Notes:''
-*This exercise belongs to the chapter&nbsp;  [[Mobile_Communications/Die_Charakteristika_von_GSM|Die Charakteristika von GSM]].
+*This exercise belongs to the chapter&nbsp;  [[Mobile_Communications/Characteristics_of_GSM|Characteristics_of_GSM]].
 *Reference is also made to the chapter&nbsp;  [[Examples_of_Communication_Systems/Funkschnittstelle|Funkschnittstelle]]&nbsp; in the book „Beispiele von Nachrichtensystemen”.

← Older revision		Revision as of 14:03, 15 December 2020
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−	{{quiz-Header\|Buchseite=~~Mobile Kommunikation~~/~~Die Charakteristika von GSM~~	+	{{quiz-Header\|Buchseite=Mobile_Communications/Characteristics_of_GSM
	}}		}}