Derivation of expression for a Gaussian Filter with 3 dB bandwidth

[ratings]
In GMSK modulation (used in GSM and DECT standard), a GMSK signal is generated by shaping the information bits in NRZ format through a Gaussian Filter. The filtered pulses are then frequency modulated to yield the GMSK signal. GMSK modulation is quite insensitive to non-linearities of power amplifier and is robust to fading effects. But it has a moderate spectral efficiency.

An expression for the Gaussian Filter with 3dB Bandwidth is derived here.

The requirements for a gaussian filter used for GMSK modulation in GSM/DECT standard are as follows,

$latex T = \text{ bit duration}&s=2&fg=0000A0$

$latex B =3\text{ dB Bandwidth of the filter}&s=2&fg=0000A0$

$latex BT =0.3 \text{ for GSM}&s=2&fg=0000A0$

$latex BT =0.5 \text{ for DECT}&s=2&fg=0000A0$

Now the challenge is to design a Gaussian Filter f_G(t) that satifies the 3dB bandwidth requirement i.e. in the frequency domain at some frequency f=B, the filter should posses -3dB gain ( in otherwords => half power point located at f=B)

The probability density function for a Gaussian Distribution with mean=0 and standard deviation=σ is given by

$latex f(t) = \frac{1}{\sqrt{2 \pi \sigma}} e^{ -\frac{t^{2}}{2 \sigma^{2}} }&s=2&fg=0000A0$

The expression for the required Gaussian Filter can be obtained by choosing the variance of the above mentioned distribution so that the Fourier Transform of the above mentioned expression has a -3dB power gain at f=B.

The fourier transform of the above mentioned expression is

$latex F[f(t)]=e^{-2 \sigma^{2} ( \pi f) }&s=2&fg=0000A0$

Setting f=B,
$latex e^{-2 \sigma^{2} ( \pi B) } = \frac{1}{\sqrt{2}} \Rightarrow \sigma = \frac{\sqrt{ln 2}}{2 \pi B}&s=2&fg=0000A0$

$latex \therefore f_{G}(t) = \sqrt{ \frac{2 \pi }{ ln 2}} B e^{ – \frac{2}{ln2}(\pi B t)^{2}} &s=2&fg=0000A0$