I'm trying to implement this GMSK demodulator. GMSK Demodulation here

  1. Is the input to deviator in degrees? do I have to change it to radians?

  2. Does the output of the deviator has to be NRZ or Have to convert it to NRZ?

  3. I'm doing quadrature demodulation so I've to multiply the quadrature components of the incoming signal with $\cos(n\omega_0)$ and $\sin(n\omega_0)$. right?

    In this diagram, the incoming signal is $cos(n\omega+\phi)$ which was the resultant output of adding the quadrature components of the GMSK modulation part. I am unable to understand this.

I'm done with the modulator part. GMSK Modulation

  • 1
    $\begingroup$ why would 1) matter ? Derivation is linear, and the conversion between radian and degree is linear, so the only thing that'd change would be the scaling of the signal coming out. 2) think about what part does what. 3) no, not right. what do the crossed circles in your diagram do, exactly?! $\endgroup$ Commented Apr 28, 2017 at 7:42
  • $\begingroup$ clear about (1). About (2), output of the tangent inverse will be the phase and derivation of phase gives frequency. how can I use this frequency to get NRZ? About(3), crossed circles are for modulation? $\endgroup$
    – Sumbul
    Commented Apr 28, 2017 at 8:49
  • $\begingroup$ @S.G.K These are not exactly modulators on their own, but other devices (mixer or multiplier) that can help building a modulator. $\endgroup$
    – Jürgen
    Commented Apr 28, 2017 at 9:01
  • $\begingroup$ I'm multiplying quadrature components with cos and sine of bandwidth of the signal..isn't it modulation? After this step, I modulate the quadrature components to the RF frequency. why do I have to do the first modulation? $\endgroup$
    – Sumbul
    Commented Apr 28, 2017 at 10:20
  • $\begingroup$ Yes, the mixers/multipliers are crucial parts of your modulator, but indeed only part of it (this is meant just as a clarification). --- When you write "to do the modulation first" you refer to the Gaussian filter? This is more or less a lowpass filter with a special characteristic. It would be possible (at least theoretically) to implement it in the RF region, but tedious and costly. $\endgroup$
    – Jürgen
    Commented Apr 28, 2017 at 11:14


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.