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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,
Now the challenge is to design a Gaussian Filter fG(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
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
![F[f(t)]=e^{-2 \sigma^{2} ( \pi f) }](https://s0.wp.com/latex.php?latex=F%5Bf%28t%29%5D%3De%5E%7B-2+%5Csigma%5E%7B2%7D+%28+%5Cpi+f%29+%7D&bg=ffffff&fg=0000A0&s=2&c=20201002)
Setting f=B,
See also :
[1] Correlative Coding – Modified Duobinary Signaling
[2] Correlative Coding – Duobinary signaling
[3] Nyquist and Shannon Theorem
[4] Correlative coding – Duobinary Signaling
[5] Square Root Raised Cosine Filter (Matched/split filter implementation)
[6] Introduction to Inter Symbol Interference
External Resources:
[1] The care and feeding of digital, pulse-shaping filters – By Ken Gentile
[2] Inter Symbol Interference and Root Raised Cosine Filtering – Complex2real
Recommended Books
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f(t) should be 1/sqrt(2pi) * 1/𝜎 -not- 1/(sqrt(2pi*𝜎)
The Fourier transform result should have the (pi*f) term squared as Chris stated
Wolfram-alpha yields the F[f(t)] as:
e^-½(2π𝜎f)² = e^(-2π²𝜎²f²)
I think you’re missing a square on (pi*f) in the Fourier transform and on the sigma under the square root in the PDF.