I have some clarifications required in this topic So initially I know that to shift a signal $X(f)$ by $f_0$ we need to take its inverse fourier transform and multiply it with a factor of $\exp(−j2\pi f_0)$.

  • Now as I understand it, in the frequency domain the spectrum is shifted by the value of $f_0$ from its original frequency, correct?

  • So therefore, is the frequency of my original signal $x(t)$ actually increasing/decreasing?(depending again on the value specified by $f_0$)

Now coming to my problem at hand, I'm trying to design a receiver which compensates for doppler effect.

  • And as I've read up a doppler shift is nothing but an increase in frequency so am I on the right track in considering a frequency shift or should I actually consider a frequency scaling approach?

  • Or can someone tell me what exactly does a Doppler shift in frequency have an effect in the time domain?


1 Answer 1


The term Doppler Shift is actually a bit of a misnomer. The frequencies are not actually shifted but they are scaled (see http://fourier.eng.hmc.edu/e101/lectures/handout3/node2.html for definition of shifting vs. scaling). It's a relative change not an absolute one.

Both time and frequency domains are scaled: when the source is moving towards you, the time axis compresses and the frequency axis expands. If the source is moving a away from you the time axis expands and the frequency axis compresses.

Let's look at a simple example. A car is 100m away from you and moves at a speed of 20m/s towards you. We assume speed of sound to 100 m/s (it's actually 344 m/s but I'm too lazy to do the math that way). Let's now assume that the car honks its horn for 1 second at a frequency of 1 kHz. So the honk consists of 1000 wiggles of air molecules. Let's call the start of the honk time t = 0.

At the beginning of the honk, the car is 100m away so the beginning of the honk arrives at your ear at time t = 1s. It takes one second for the sound to travel 100m. By the end of the honk the car has traveled 20m so it's only 80m away. The end of the honk arrives at time t = 1.8s. The sound only takes 0.8s to travel 80m.

The honk at the listening location is only 0.8s long, so the time has been compressed. It's still 1000 wiggles but they happen in less time. So the frequency is now 1000 wiggles/0.8s = 1250 Hz. The frequency is higher so the frequency axis has expanded.

It's important to note the frequency doesn't go up by a constant shift frequency but by a multiplicative factor. If the honk would have been 2 kHz the resulting Doppler frequency would be 2500 Hz and NOT 2250 Hz.

  • $\begingroup$ I guess I was looking at it the wrong way,I'll use the frequency scaling theorem and compensate for the new frequency If I run into any other problems I will post it on here Thank you so much for your help :) $\endgroup$ Oct 19, 2015 at 14:30
  • $\begingroup$ That is the prettiest explanation of Doppler shifting I have ever seen in my life and I am a physicists ! Congratulations and thanks you ! $\endgroup$
    – Dim Galan
    Dec 12, 2017 at 21:33
  • $\begingroup$ @DimGalan Please do not post answers that are just comments. $\endgroup$
    – Peter K.
    Dec 12, 2017 at 22:13

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