Exercise 3.4: GMSK Modulation

Various signals of the GMSK modulation

The modulation method used for GSM is known as Gaussian Minimum Shift Keying (GMSK). This is a type of Frequency Shift Keying (FSK) with continuous phase adjustment $(\rm CP-FSK)$, in which

the modulation index is smallest possible to still satisfy the orthogonality condition
$h = 0.5$ ⇒ Minimum Shift Keying,
a Gaussian low-pass filter with impulse response $h_{\rm G}(t)$ is introduced before the FSK modulator,
to further save bandwidth.

The graphic illustrates the facts:

The digital message is represented by the amplitude coefficients $a_{\nu} ∈ ±1$ to which a Dirac pulse is applied. Note that the plotted (red) sequence is assumed for the subtask (3).

Let the rectangular pulse be dimensionless, symmetric and have the GSM bit duration $T_{\rm B} = T$:

$$g_{\rm R}(t) = \left\{ \begin{array}{c} 1 \\ 0 \\ \end{array} \right.\quad \begin{array}{*{5}c} {\rm{{\rm{for}}}} \\ {\rm{{\rm{for}}}} \\ \end{array}\begin{array}{*{5}c} |\hspace{0.05cm} t \hspace{0.05cm}| < T/2 \hspace{0.05cm}, \\ |\hspace{0.05cm} t \hspace{0.05cm}| > T/2 \hspace{0.05cm}. \\ \end{array}$$

This gives the following for the rectangular wave 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:

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

where the system theoretic cutoff frequency $f_{\rm G}$ is used. However, in the GSM specification, the $3 \hspace{0.05cm}\rm dB$ cutoff frequency is given as $f_{\rm 3\hspace{0.05cm} dB} = 0.3/T$ . From this $f_{\rm G}$ can be calculated directly.

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 called the frequency pulse. For this holds:

$$g(t) = q_{\rm R} (t) \star h_{\rm G}(t) \hspace{0.05cm}.$$

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:

$$f_{\rm A}(t) = f_{\rm T} + \Delta f_{\rm A} \cdot q_{\rm G} (t)\hspace{0.05cm}.$$

For your calculations, use the example values $f_{\rm T} = 900 \ \rm MHz$ and $\Delta f_{\rm A} = 68 \ \rm kHz$.

Hints:

Table of the Gaussian error function

This exercise belongs to the chapter "Radio Interface".
Reference is made in particular to the page "Gaussian Minimum Shift Keying"

Use the Gaussian integral to solve this exercise (see adjacent table):

$$\phi(x) =\frac {1}{\sqrt{2 \pi}} \cdot \int^{x} _{-\infty} {\rm e}^{-u^2/2}\,{\rm d}u \hspace{0.05cm}.$$

Questions

${\rm Max} \ \big [f_{\rm A}(t) \big ] \ = \ $

$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 $a_{\nu}$ are equal to $+1$, $q_{\rm R}(t) = 1$ is a constant.

The Gaussian low-pass therefore has no effect and $q_{\rm G}(t) = 1$ is obtained.
The maximum frequency is therefore

$${\rm Max}\hspace{0.05cm}[f_{\rm A}(t)] = f_{\rm T} + \Delta f_{\rm A} \hspace{0.15cm} \underline{= 900.068\,{\rm MHz}} \hspace{0.05cm}.$$

The minimum of the instantaneous frequency results when all amplitude coefficients are negative:

$${\rm Min}\hspace{0.05cm}[f_{\rm A}(t)] = f_{\rm T} - \Delta f_{\rm A} \hspace{0.15cm} \underline { = 899.932\,{\rm MHz}} \hspace{0.05cm}$$

In this case, $q_{\rm R}(t) = q_{\rm G}(t) = -1$.

(2) The frequency at which the logarithmized power transfer function is $3 \ \rm dB$ smaller than $f = 0$ is called the $3\hspace{0.05cm}\rm dB$ cutoff frequency.

This can also be expressed as follows:

$$\frac {|H(f = f_{\rm 3\hspace{0.05cm}dB})|}{|H(f = 0)|}= \frac{1}{\sqrt{2}} \hspace{0.05cm}.$$

In particular, for the Gaussian lowpass 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 $f_{\rm G} \approx 1.5 \cdot f_{\rm 3\hspace{0.05cm}dB}$.
From $f_{\rm 3\hspace{0.05cm}dB} \cdot T = 0.3$ thus follows $f_{\rm G} \cdot T \hspace{0.15cm}\underline{\approx 0.45}$.

(3) The frequency pulse is obtained by convolution of square wave function $g_{\rm R}(t)$ and impulse response $h_{\rm G}(t)$:

$$g(t) = g_{\rm R} (t) \star h_{\rm G}(t) = 2 f_{\rm G} \cdot \int \limits^{t + T/2} _{t - T/2} {\rm e}^{-\pi\cdot (2 f_{\rm G}\cdot \tau)^2}\,{\rm d}\tau \hspace{0.05cm}.$$

With the substitution $u^{2 } = 8π \cdot f_{\rm G}^{2} \cdot \tau^{2}$ and the function $\phi (x)$ can also be written for this:

$$g(t) \ = \ \frac {1}{\sqrt{2 \pi}} \cdot \int \limits^{2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t + T/2)} _{2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t - T/2)} {\rm e}^{-u^2/2}\,{\rm d}u = \phi(2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t + T/2))- \phi(2 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot(t - T/2)) \hspace{0.05cm}.$$

For the time $t = 0$, taking into account $\phi (-x) = 1 - \phi (x)$ and $f_{\rm G} \cdot T = 0.45$:

$$g(t = 0) \ = \ \phi(\sqrt{2 \pi} \cdot f_{\rm G} \cdot T)- \phi(-\sqrt{2 \pi} \cdot f_{\rm G} \cdot T)= 2 \cdot \phi(\sqrt{2 \pi} \cdot f_{\rm G} \cdot T)-1 \approx 2 \cdot \phi(1. 12)-1 \hspace{0.15cm} \underline {= 0.737} \hspace{0.05cm}.$$

(4) With $a_{3} = +1$, $q_{\rm G}(t = 3 T) = 1$ would result. Thus, due to linearity:

$$q_{\rm G}(t = 3 T ) = 1 - 2 \cdot g(t = 0)= 1 - 2 \cdot 0.737 \hspace{0.15cm} \underline {= -0.474} \hspace{0.05cm}.$$

(5) Using the result of subtask (3) and $f_{\rm G} \cdot T = 0.45$ we obtain:

$$g(t = T) \ = \ \phi(3 \cdot \sqrt{2 \pi} \cdot f_{\rm G} \cdot T)- \phi(\sqrt{2 \pi} \cdot f_{\rm G} \cdot T) \approx \ \phi(3.36)-\phi(1.12) = 0.999 - 0.868 \hspace{0.15cm} \underline { = 0.131} \hspace{0.05cm}.$$

(6) In the alternating sequence, for symmetry reasons, the amounts $|q_{\rm G}(\nu \cdot T)|$ are all the same for all multiples of the bit duration $T$.

All intermediate values at $t \neq \nu - T$ are smaller.
Considering $g(t ≥ 2T) \approx 0$, every single pulse value $g(0)$ is reduced by the preceding pulse with $g(t = T)$, additionally by the following one with $g(t = -T)$.
Thus, pulse interference results and one obtains:

$${\rm Max} \hspace{0.08cm}q_{\rm G}(t) = g(0) - 2 \cdot g(T) = 0.737 - 2 \cdot 0.131 \hspace{0.15cm} \underline {= 0.475 }\hspace{0.05cm}.$$