I'm having a hard time wrapping my head around doppler and how it affects a digital communications signal's baud rate. The examples that I've seen explain that the center frequency of a signal is shifted when either the receiver and/or transmitter are moving relative to each other. What I don't understand is the why doppler also affects baud rate? Can someone point me to some tutorials on this topic or explain? What I'd like to understand clearly is:

  1. Intuitively understand why this happens?
  2. Mathematically understand why this happens?
  3. I'd like to be able to calculate when this is something that a communications receiver has to take into account? My guess is that based on the doppler shift that will be experienced by the receiver, in combination with the signal's baud rate, but I guess I need some hints or help to get me along with this.

This is a follow-up to this question already posted: How to estimate and compensate for doppler shift in wireless signals?

  • 3
    $\begingroup$ Doppler shift causes the baud durations to expand or contract depending on the relative motion. If you express the Doppler shift as a PPM value, you can apply it to either the carrier or the baud rate. The baud-rate Doppler shift will typically be much smaller than the carrier Doppler shift -- the ratio is baud rate / carrier frequency. $\endgroup$
    – John
    Jan 13, 2014 at 17:01
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    $\begingroup$ The Doppler effect is actually most accurately modeled by resampling the carrier-modulated signal by a factor close to 1. This resampling causes an expansion or contraction in the baud rate. It also causes an apparent shift in the carrier frequency. Strictly speaking, this shift isn't constant over your signal bandwidth (instead, it is proportional to $f$), but as long as the bandwidth is small relative to the carrier frequency (which is usually true), it is commonly assumed to be constant for ease of analysis. $\endgroup$
    – Jason R
    Jan 13, 2014 at 17:51
  • $\begingroup$ The intuitive explanation is that the tx-rx distance is different at the beginning of the baud compared to the end. The communication receiver usually contains a timing recovery subsystem. This subsystem must be designed to work in the presence of Doppler shift as well as other sources of frequency error such as different tx-rx frequency references. $\endgroup$
    – John
    Jan 13, 2014 at 19:17
  • $\begingroup$ Lots of information can be found in the responses to the same question of the newsgroup comp.dsp $\endgroup$ Jan 14, 2014 at 20:58

3 Answers 3


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.

  • $\begingroup$ The resampling that you are referring to, how do you calculate what that resampling ratio should be? Would it be the same $1 + \frac{\Delta v}{c}$ (i.e. the same ratio as the shift in frequency?) $\endgroup$
    – Kiran K.
    Jan 13, 2015 at 15:27
  • $\begingroup$ Yes, that is exactly how you calculate the resampling ratio. $\endgroup$
    – Jim Clay
    Jan 13, 2015 at 17:39

I think a very intuitive, simple way to look at it would be to consider the modulation envelope generated by the baud rate. That envelope will expand and compress from the Doppler in the same manner that the carrier wave does. Hence, it exhibits the same Doppler effect scaled down to the baud rate.


The relationship between baud rate and the doppler effect is through the so called "coherence time" or doppler spread. The coherence time is a measure of the time interval over which the signal's amplitude is for the most part statistically correlated. Therefore it can be considered the maximum interval for a single baud. If the pulse is smaller than the coherence time than the amplitude is relatively constant over the period of one baud and equalization is easier (time-flat fading). If the pulse/buad was to stretch out longer than the coherence time than equalizing the signal would be more difficult because he amplitude could change during the pulse interval.

I've seen values for coherence time of approximately $\dfrac{0.43}{f_d}$ where $f_d$ is proportional to the carrier frequency $f_d=\dfrac{velocity}{\lambda}$, $\lambda=\dfrac{c}{f_0}$.

Why it happens? I can explain but you are better off just searching for "Jakes fading model". You should get all the info you need.


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