I am struggling to find or derive the relationship between Doppler frequency and the associated phase shift. Let's say I have a signal transmitted at a given frequency $f_c$. I then down convert from $f_c$ to baseband at the receiver, but due to speed between the transmitter and receiver, I receive a frequency shifted by a Doppler frequency $\Delta f$.

If I wish to generate a matching signal to that being received, I can use

$$ \cos(2 \pi \Delta f t + \phi) + i \sin(2 \pi \Delta f t + \phi) $$

The part I'm struggling to get to grips with is how does $\phi$ relate to $\Delta f$? Is it simply the integral of $\Delta f$ with respect to time?

$$ \phi = \int_0^t \Delta f dt $$


1 Answer 1


You are correct that the total accumulated phase offset due to Doppler could be calculated by integrating the Doppler shift $\Delta f(t)$ over time. In addition to Doppler shift, however, you would also need to take into account any frequency difference between the transmitter and receiver's reference oscillators.

However, I would question the value of trying to do so. In most communications systems, there will be a random, unknown phase offset between the transmitter and receiver oscillators. This makes absolute phase measurements meaningless unless you have some reference. In real systems, phase synchronization is either achieved by exploiting some known characteristic of the signal (e.g. the known constellation of a PSK signal), or in some rare cases, via the explicit synchronization of the transmitter and receiver local oscillators.

  • $\begingroup$ Thanks, Jason. It is somewhat of an academic question trying to get my head around something. What I'm actually trying to do is simulate the received signal without having a transmitter in real life to then test the code that tracks the signal with a PLL. $\endgroup$
    – Darran
    Jan 17, 2013 at 16:05
  • $\begingroup$ Can you introduce phase change in you're simulated signal? Add a variable that varies with time to the function that generates the simulation signal. $\endgroup$
    – user2718
    Jan 17, 2013 at 17:36

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