Exercise 3.5: GMSK Modulation

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 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:

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

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

${\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)| \ = \ $

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

$${\rm Max} \hspace{0.12cm}[q_{\rm G}(t)] = g(t = 0) - 2 \cdot g(t = T) = 0.737 - 2 \cdot 0.131 \hspace{0.15cm} \underline {= 0.475 }\hspace{0.05cm}.$$