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GMSK modulation
Case of frequency correction burst
Solving modulation equations
One bit change
Normal duration burst



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Case of Frequency Correction Burst

Case of Frequency Correction Burst [Under GSM > GMSK modulation]

Here is format of Frequency Correction Burst (3GPP TS 45.002:5.2.4).

Tail bits
3
Fixed bits
142
Tail bits
3
Guard period
8.25


Tail bits and Fixed bits are all "0". Let us look at how the carrier waveform looks like with the help of equations illustrated in earlier article.

Carrier waveform is represented as:

____
x(t') = (
2Ec
T
) . cos(2πf0t' + φ(t') + φ0)


The phase difference is also function of t and αi (differentially encoded bits):

φ(t') = n . t'-iT .
Σ  ( αiπh . g(u) du )
i=1 . -∞ .


αi = { +1, -1 }
h = 0.5


Let us look at φ(t') with the help of waveforms. We would separate out πh and draw φ(t') for 1 bit, 2 bits, 3 bits, and 4 bits transfer (each bit "0"):

gsm-psit-751218.png

If you observe, all waveforms start at value close to (-number of bits/2), becomes almost straight line of slope (1/T) and then saturates on (+number of bits/2). If we approximate φ(t'), we could re-write it as below:

φ(t') = πh . t'/T = 2π.(1/4T).t'

Carrier for frequency correction burst can be written as:

cos(2π(f0 + 1/4T)t')

Thus FCCH would look like non-modulated wave of frequency (1/4T=)66.7 kHz higher than carrier frequency !!!

In next article, we will solve the modulation equations (from earlier article).

References: GSM book by Mouly and Pautet.

© Copyright Samir Amberkar 2013

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