Doppler effect on Baud Rate
I'm sure that you are familiar with the idea of how Doppler affects a sine wave.  When the transmitter and/or receiver move towards the other the wave arrives more quickly, which makes the wave appear to be a higher frequency.  In regards to the baud rate, it also arrives more quickly.  If you think about it you will see that this must be so.  After all, the data stream cannot be separated from the carrier wave that modulates it!

So, you have a signal that is exactly identical to the "non-Doppler" signal except it arrives faster or slower than the non-Doppler signal.  This can be modeled exactly, in the discrete time domain, through resampling.

So why is Doppler usually modeled as a simple frequency shift when it should really be modeled as resampling?  Two reasons- first, the frequency shift effect is almost always vastly larger than the baud rate change.  The reason for this is that the Doppler effect is proportional to the frequency of the original wave ($\Delta f = \frac{\Delta v}{c}f_0$).  The frequency of the carrier wave is vastly greater than the baud rate, so the effect is much larger on the carrier wave.

The second reason most people don't model it as resampling is that it is difficult to do very small arbitrary sample rate changes, like 3013/3014.  If either the numerator or denominator is a prime number or has really large factors then it can become impractical to do traditional fractional resampling.  In that case polynomial interpolation is probably the best way to go.