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I have been reading a paper that mentions Doppler model that may be included in the fading characteristics of impulse response. One example of bell shaped Doppler Spectrum is the following

$$S(f) = \frac{1}{1+A\frac{f}{f_d}}\,\,\,\,\,\,\,\, |f|\leq f_{\max}$$

where $f_d$ is the Doppler Spread, $A$ is a constant, $f_{max}$ is maximum frequency component of the Doppler power spectrum. Does anyone know how the following function affects the channel model?

Thanks

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This is not a full answer, but I'll try to provide some intuition. First, imagine a static channel -- neither the antennas nor the reflectors around them move. In such a channel, the impulse response is always the same -- it's time invariant. The channel characteristics are fixed.

Now, introduce movement of the antennas and/or the reflectors. The channel impulse response now varies with time. The channel coherence time indicates how fast the channel impulse response changes, and for how long can we consider the channel to remain relatively static.

If there is movement of the antennas and/or the reflectors, we also get a Doppler shift in the transmitted signal. The more movement there is, the higher the shift.

The channel coherence time and the Doppler shift are, then, two manifestations of the same phenomenon -- movement. If one increases, the other decreases. The easiest way to introduce this into the channel model is to assume that the channel is fixed for a certain amount of time, after which it changes to a completely different, independent channel.

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  • $\begingroup$ thank you for the answer and the details therin. I am just wondering given $S (f) $ what can I learn from it? so if you give me $S (f) $ what can I tell about the fading model? $\endgroup$ – Tyrone May 14 '15 at 2:55
  • $\begingroup$ @Tyrone The Doppler spread is approximately equal to $1/T_c$, where $T_c$ is the channel coherence time. For example, if the Doppler spread is 80 Hz, then the channel coherence time is approximately 1/80=12.5 ms. $\endgroup$ – MBaz May 14 '15 at 14:37
  • $\begingroup$ thank you very much for your answers, I am wondering if you have an opinion on my question here dsp.stackexchange.com/questions/23394/… $\endgroup$ – Tyrone May 14 '15 at 20:58
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I agree with the above response, i.e, " The easiest way to introduce this into the channel model is to assume that the channel is fixed for a certain amount of time, after which it changes to a completely different, independent channel." But you need to make the changing according to the Doppler Spectra that you have: if it is a Gaussian Doppler Spread, Log normal, ...

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