0
$\begingroup$

I have started doing some reading into the effects of the doppler shift on signals. In particular, radar signals from ships moving relative to a receiver. This might seem weird but I understand why the shift in frequency happens what I don't understand is how that changes the actual signal. If we receive the signal and perform an FFT on it do we get the original frequency plus another component that corresponds to the doppler effect? Is the doppler shift similar to some sort of distortion on the signal? To be honest it is hard for me to ask the right question. I think I am missing an intuitive picture of what happens to the signal as a result of a doppler shift. If anyone would help i would really appreciate it.

$\endgroup$

3 Answers 3

2
$\begingroup$

It is well-known that Doppler effect shifts the received frequencies compared with the emitting frequencies. On the other hand, the time period of the sound signal is often ignored.

Say a sound source is approaching you. For the sound source itself, it emits a sound signal for $T$ seconds, but you will hear a sound less than $T$ seconds, causing an increase in the frequency. Let's take an extreme example, if the sound source is running towards you at the speed of sound, all the wavefronts and the sound source itself reach you at the same moment, and you will hear a huge amount of sound energy, which is called sonic boom.

It's not a distortion, only a stretch on the time scale. For discrete signal, all you have to do is to change the playback speed, or to interpret the sampling rate and do a resample.

Due to the connection between the time scaling and the frequency shifting, audio time stretching and pitch scaling have become research interests, which mean changing the speed without affecting the pitch, and the opposite, changing the pitch without affecting the speed.

$\endgroup$
1
$\begingroup$

So, let's take this from the top down:

  1. Thanks to Fourier, we know we can describe all signals as combination of harmonic oscillations.
  2. For a harmonic oscillation sent at frequency $f_t$, you know the received frequency is $f_r=\left(1+\frac{\Delta v}c\right)f_t$.
  3. So $\left(1+\frac{\Delta v}c\right)=:d$ is a constant (for any fixed speed) that you multiply with all frequencies, and you get the new frequencies

So, that technically answers your question: you "stretch" all frequencies out by a factor $>1$ if approaching, $<1$ if leaving.

The trick is that usually, radars use high carrier frequencies, and bandwidths that are significantly smaller. So, for all frequencies "close to" $f_t$, it looks like they're subject to the same frequency "shift" (altough it's not actually a shifting, it's a scaling; just an approximation!).

So, what you often find is that people say "we use the narrowband approximation", and then, simply, say all frequencies get shifted by the same amount – the carrier frequency times our Doppler shift constant. That's actually why it's called Doppler shift: it shifts spectra (locally. Globally, it scales them.).

$\endgroup$
0
$\begingroup$

Observing a ship via radar waves is exactly analogous to observing a fire truck via sound waves. If a moving sound source (it doesn't matter if the source or the observer is moving, the relative velocity is what counts) passes your ear; do you hear more tones than when it's at rest? No. The Doppler shift changes the frequency of the signal, it does not add frequencies.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.