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

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[[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  ''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  "Gaussian Minimum Shift Keying",  short  $\rm GMSK$.  This is a special type of  $\rm  FSK$  ("Frequency Shift Keying")  with  $\rm  CP-FSK$  ("Continuous Phase Matching"), where:
*the modulation index has the smallest value that just satisfies the orthogonality condition:   $h = 0.5$   ⇒   ''Minimum Shift Keying'',
+
*The modulation index has the smallest value that just satisfies the orthogonality condition:   $h = 0.5$   ⇒   "Minimum Shift Keying"  $\rm  (MSK)$,
*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.
+
*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 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 digital message is represented by the amplitude coefficients  $a_{\mu} ∈ \{±1\}$  which are applied to a "Dirac comb"  (Dirac delta train).  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:
+
*The symmetrical rectangular pulse with duration  $T = T_{\rm B}$  (GSM bit duration)  is dimensionless:
 
:$$g_{\rm R}(t) = \left\{ \begin{array}{c} 1 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{5}c} {\rm{{\rm{f\ddot{u}r}}}} \\ {\rm{{\rm{f\ddot{u}r}}}} \\ \end{array}\begin{array}{*{5}c} |\hspace{0.05cm} t \hspace{0.05cm}| < T/2 \hspace{0.05cm}, \\ |\hspace{0.05cm} t \hspace{0.05cm}| > T/2 \hspace{0.05cm}. \\ \end{array}$$
 
:$$g_{\rm R}(t) = \left\{ \begin{array}{c} 1 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{5}c} {\rm{{\rm{f\ddot{u}r}}}} \\ {\rm{{\rm{f\ddot{u}r}}}} \\ \end{array}\begin{array}{*{5}c} |\hspace{0.05cm} t \hspace{0.05cm}| < T/2 \hspace{0.05cm}, \\ |\hspace{0.05cm} t \hspace{0.05cm}| > T/2 \hspace{0.05cm}. \\ \end{array}$$
 
*This results for the rectangular signal
 
*This results for the rectangular signal
 
:$$q_{\rm R} (t) = q_{\rm \delta} (t) \star g_{\rm R}(t) = \sum_{\nu} a_{\nu}\cdot g_{\rm R}(t - \nu \cdot T)\hspace{0.05cm}.$$
 
:$$q_{\rm R} (t) = q_{\rm \delta} (t) \star g_{\rm R}(t) = \sum_{\nu} a_{\nu}\cdot g_{\rm R}(t - \nu \cdot T)\hspace{0.05cm}.$$
 
*The Gaussian low-pass is given by its frequency response or impulse response:
 
*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
 
:$$q_{\rm G} (t) = q_{\rm R} (t) \star h_{\rm G}(t) = \sum_{\nu} a_{\nu}\cdot g(t - \nu \cdot T)\hspace{0.05cm}.$$
 
:$$q_{\rm G} (t) = q_{\rm R} (t) \star h_{\rm G}(t) = \sum_{\nu} a_{\nu}\cdot g(t - \nu \cdot T)\hspace{0.05cm}.$$
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":
 
:$$g(t) = q_{\rm R} (t) \star h_{\rm G}(t) \hspace{0.05cm}.$$
 
:$$g(t) = q_{\rm R} (t) \star h_{\rm G}(t) \hspace{0.05cm}.$$
  
*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:
 
+
:$$f_{\rm A}(t) = f_{\rm T} + \Delta f_{\rm A} \cdot q_{\rm G} (t)\hspace{0.05cm}.$$
 
 
 
 
  
  
  
 +
[[File:P_ID2226__Bei_A_3_4b.png|right|frame|Some Gaussian integral values]]
  
 
''Notes:''  
 
''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”.  
+
*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; $\Delta f_{\rm A} = 68 \ \rm kHz$.
 
*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$)$.
[[File:P_ID2226__Bei_A_3_4b.png|right|frame|Some Gaussian integral values]]
 
 
:$$\phi(x) =\frac {1}{\sqrt{2 \pi}} \cdot \int^{x} _{-\infty} {\rm e}^{-u^2/2}\,{\rm d}u \hspace{0.05cm}.$$
 
:$$\phi(x) =\frac {1}{\sqrt{2 \pi}} \cdot \int^{x} _{-\infty} {\rm e}^{-u^2/2}\,{\rm d}u \hspace{0.05cm}.$$
 
<br clear=all>
 
<br clear=all>
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<quiz display=simple>
 
<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="{}"}
 
|type="{}"}
 
${\rm Max} \  \big[f_{\rm A}(t)\big] \ = \hspace{0.2cm} $ { 900.068 0.01% } $\ \rm MHz$
 
${\rm Max} \  \big[f_{\rm A}(t)\big] \ = \hspace{0.2cm} $ { 900.068 0.01% } $\ \rm MHz$
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$f_{\rm G} \cdot T \ = \ $ { 0.45 3% }  
 
$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="{}"}
 
|type="{}"}
 
$g(t = 0) \ = \ $ { 0.737 3% }  
 
$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="{}"}
 
|type="{}"}
 
$q_{\rm G}(t = 3T) \ = \ $ { -0.51822--0.42978 }  
 
$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="{}"}
 
|type="{}"}
 
$g(t = ±T) \ = \ $ { 0.131 3% }  
 
$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="{}"}
 
|type="{}"}
 
${\rm Max} \ |q_{\rm G}(t)| \ = \ $ { 0.475 3% }  
 
${\rm Max} \ |q_{\rm G}(t)| \ = \ $ { 0.475 3% }  
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{{ML-Kopf}}
 
{{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
 
*The maximum frequency is thus
 
:$${\rm Max}\ [f_{\rm A}(t)] = f_{\rm T} + \Delta f_{\rm A} \hspace{0.15cm} \underline {= 900.068\,{\rm MHz}} \hspace{0.05cm}.$$
 
:$${\rm Max}\ [f_{\rm A}(t)] = f_{\rm T} + \Delta f_{\rm A} \hspace{0.15cm} \underline {= 900.068\,{\rm MHz}} \hspace{0.05cm}.$$
*The minimum instantaneous frequency
+
*The minimum instantaneous frequency is
 
:$${\rm Min}\ [f_{\rm A}(t)] = f_{\rm T} - \Delta f_{\rm A} \hspace{0.15cm} \underline { = 899.932\,{\rm MHz}} \hspace{0.05cm}$$
 
:$${\rm Min}\ [f_{\rm A}(t)] = f_{\rm T} - \Delta f_{\rm A} \hspace{0.15cm} \underline { = 899.932\,{\rm MHz}} \hspace{0.05cm}$$
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:
 
*This can also be expressed as follows:
 
:$$\frac {|H(f = f_{\rm 3dB})|}{|H(f = 0)|}= \frac{1}{\sqrt{2}} \hspace{0.05cm}.$$
 
:$$\frac {|H(f = f_{\rm 3dB})|}{|H(f = 0)|}= \frac{1}{\sqrt{2}} \hspace{0.05cm}.$$
*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}
 
:$$ 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}.$$
 
\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 ${\rm g}(t)$ results from the convolution of the rectangular function $g_{\rm R}(t)$ with the pulse response $h_{\rm G}(t)$:
+
'''(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; $h_{\rm G}(t)$:
 
:$$g(t) = g_{\rm R} (t) \star h_{\rm G}(t) = 2 f_{\rm G} \cdot \int^{t + T/2} _{t - T/2} {\rm e}^{-\pi\cdot (2 f_{\rm G}\cdot \tau)^2}\,{\rm d}\tau \hspace{0.05cm}.$$
 
:$$g(t) = g_{\rm R} (t) \star h_{\rm G}(t) = 2 f_{\rm G} \cdot \int^{t + T/2} _{t - T/2} {\rm e}^{-\pi\cdot (2 f_{\rm G}\cdot \tau)^2}\,{\rm d}\tau \hspace{0.05cm}.$$
*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:
 
:$$g(t) = \ \frac {1}{\sqrt{2 \pi}} \cdot \int^{2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t + T/2)} _{2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t - T/2)} {\rm e}^{-u^2/2}\,{\rm d}u = \ \phi(2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t + T/2))- \phi(2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t - T/2)) \hspace{0.05cm}.$$
 
:$$g(t) = \ \frac {1}{\sqrt{2 \pi}} \cdot \int^{2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t + T/2)} _{2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t - T/2)} {\rm e}^{-u^2/2}\,{\rm d}u = \ \phi(2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t + T/2))- \phi(2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t - T/2)) \hspace{0.05cm}.$$
*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; $t = 0$:
 
:$$g(t = 0) = \ \phi(\sqrt{2 \pi} \cdot f_{\rm G} \cdot T)- \phi(-\sqrt{2 \pi} \cdot f_{\rm G} \cdot T)= \ 2 \cdot \phi(\sqrt{2 \pi} \cdot f_{\rm G} \cdot T)-1 \approx 2 \cdot \phi(1.12)-1 \hspace{0.15cm} \underline {= 0.737} \hspace{0.05cm}.$$
 
:$$g(t = 0) = \ \phi(\sqrt{2 \pi} \cdot f_{\rm G} \cdot T)- \phi(-\sqrt{2 \pi} \cdot f_{\rm G} \cdot T)= \ 2 \cdot \phi(\sqrt{2 \pi} \cdot f_{\rm G} \cdot T)-1 \approx 2 \cdot \phi(1.12)-1 \hspace{0.15cm} \underline {= 0.737} \hspace{0.05cm}.$$
  
  
  
'''(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:
 
:$$q_{\rm G}(t = 3 T ) = 1 - 2 \cdot g(t = 0)= 1 - 2 \cdot 0.737 \hspace{0.15cm} \underline {= -0.474} \hspace{0.05cm}.$$
 
:$$q_{\rm G}(t = 3 T ) = 1 - 2 \cdot g(t = 0)= 1 - 2 \cdot 0.737 \hspace{0.15cm} \underline {= -0.474} \hspace{0.05cm}.$$
  
  
  
'''(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
 
:$$g(t = T) = \ \phi(3 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot T)- \phi(\sqrt{2 \pi} \cdot f_{\rm G} \cdot T) \approx  \phi(3.36)-\phi(1.12) = 0.999 - 0.868 \hspace{0.15cm} \underline { = 0.131} \hspace{0.05cm}.$$
 
:$$g(t = T) = \ \phi(3 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot T)- \phi(\sqrt{2 \pi} \cdot f_{\rm G} \cdot T) \approx  \phi(3.36)-\phi(1.12) = 0.999 - 0.868 \hspace{0.15cm} \underline { = 0.131} \hspace{0.05cm}.$$
*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
 
:$${\rm Max} \hspace{0.12cm}[q_{\rm G}(t)] = g(t = 0) - 2 \cdot g(t = T) = 0.737 - 2 \cdot 0.131 \hspace{0.15cm} \underline {= 0.475 }\hspace{0.05cm}.$$
 
:$${\rm Max} \hspace{0.12cm}[q_{\rm G}(t)] = g(t = 0) - 2 \cdot g(t = T) = 0.737 - 2 \cdot 0.131 \hspace{0.15cm} \underline {= 0.475 }\hspace{0.05cm}.$$
  
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[[Category:Exercises for Mobile Communications|^3.3 Characteristics of GSM^]]
+
[[Category:Mobile Communications: Exercises|^3.3 Characteristics of GSM^]]

Revision as of 08:04, 26 May 2021

Different signals of GMSK-Modulation

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

  • The modulation index has the smallest value that just satisfies the orthogonality condition:   $h = 0.5$   ⇒   "Minimum Shift Keying"  $\rm (MSK)$,
  • 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 comb"  (Dirac delta train).  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:
$$H_{\rm G}(f) = {\rm e}^{-\pi\hspace{0.05cm}\cdot \hspace{0.05cm}[{f}/ ({2 f_{\rm G})} ]^2} \hspace{0.2cm}\bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \hspace{0.2cm} h_{\rm G}(t) = 2 \cdot f_{\rm G} \cdot {\rm e}^{-\pi\hspace{0.05cm}\cdot \hspace{0.05cm}(2 f_{\rm G}\hspace{0.05cm}\cdot \hspace{0.05cm}t)^2}\hspace{0.05cm},$$
where the system theoretical cut-off frequency  $f_{\rm G}$  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 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  $g(t)$  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  $q_{\rm G}(t)$, the carrier frequency  $f_{\rm T}$  and the frequency deviation  $\Delta f_{\rm A}$  for the instantaneous frequency at the output of the FSK modulator can thus be written:
$$f_{\rm A}(t) = f_{\rm T} + \Delta f_{\rm A} \cdot q_{\rm G} (t)\hspace{0.05cm}.$$


Some Gaussian integral values

Notes:

  • This exercise belongs to the chapter  Characteristics of GSM.
  • Reference is also made to the chapter  Radio Interface  in the book "Examples of Communication Systems".
  • 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$)$.
$$\phi(x) =\frac {1}{\sqrt{2 \pi}} \cdot \int^{x} _{-\infty} {\rm e}^{-u^2/2}\,{\rm d}u \hspace{0.05cm}.$$


Questionnaire

1

In what range of values the instantaneous frequency  $f_{\rm A}(t)$  can 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  $\phi (x)$.  What is the result for  $g(t = 0)$?

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

$g(t = ±T) \ = \ $

6

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

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


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 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 $q_{\rm R}(t) = q_{\rm G}(t) = -1$.


(2)  The frequency at which the logarithmic power transfer function is  $3 \ \rm dB$  less than for  $f = 0$  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  $f_{\rm G} \approx 1.5 \cdot f_{\rm 3dB}$.
  • From  $f_{\rm 3dB} \cdot T = 0.3$  follows  $f_{\rm G} \cdot T \underline{\approx 0.45}$.


(3)  The frequency pulse  ${\rm g}(t)$  results from the convolution of the rectangular function  $g_{\rm R}(t)$  with the impulse response  $h_{\rm G}(t)$:

$$g(t) = g_{\rm R} (t) \star h_{\rm G}(t) = 2 f_{\rm G} \cdot \int^{t + T/2} _{t - T/2} {\rm e}^{-\pi\cdot (2 f_{\rm G}\cdot \tau)^2}\,{\rm d}\tau \hspace{0.05cm}.$$
  • With the substitution  $u^{2} = 8π \cdot {f_{G}}^{2} \cdot \tau^{2}$  and the function  $\phi (x)$  you can also write for this:
$$g(t) = \ \frac {1}{\sqrt{2 \pi}} \cdot \int^{2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t + T/2)} _{2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t - T/2)} {\rm e}^{-u^2/2}\,{\rm d}u = \ \phi(2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t + T/2))- \phi(2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t - T/2)) \hspace{0.05cm}.$$
  • Considering  $\phi (-x) = 1 - \phi (x)$  and  $f_{\rm G} \cdot T = 0.45$,  you get for the time  $t = 0$:
$$g(t = 0) = \ \phi(\sqrt{2 \pi} \cdot f_{\rm G} \cdot T)- \phi(-\sqrt{2 \pi} \cdot f_{\rm G} \cdot T)= \ 2 \cdot \phi(\sqrt{2 \pi} \cdot f_{\rm G} \cdot T)-1 \approx 2 \cdot \phi(1.12)-1 \hspace{0.15cm} \underline {= 0.737} \hspace{0.05cm}.$$


(4)  With  $a_{3} = +1$  the result would be  $q_{\rm G}(t = 3 T) = 1$.  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)  With the result of  (3)  and  $f_{\rm G} \cdot T = 0.45$  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.


(6)  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.

  • All intermediate values at  $t \approx \mu \cdot T$  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 by the following pulse with  $g(t = -T)$.
  • So there will be "intersymbol interference" and you get
$${\rm Max} \hspace{0.12cm}[q_{\rm G}(t)] = g(t = 0) - 2 \cdot g(t = T) = 0.737 - 2 \cdot 0.131 \hspace{0.15cm} \underline {= 0.475 }\hspace{0.05cm}.$$