I have been reading about doppler ambiguity and I have two explanations for its occurrence that I can't relate to each other.

1st Explanation: In the stationary case (radar and target are stationary) we would have a peak frequency that define the range to a target $f_r$. If the target is moving a doppler shift $f_d$ will be added/subtracted to $f_r$. In the case where the target is near ($f_r$ is close to zero) and fast ($f_d$ is large) having $f_d>f_r$, we would have the peak being shifted to the negative side and by FFT, we would have mirroring for real signals and so we won't be able to know the coming from the going targets. So we would be ambiguous about the direction of the target.

$f_d$: doppler frequency

$f_r$: peak frequency at stationary case referring to the range to a target

$f_d \pm f_r$: peak frequency at motion case

2nd Explanation: In FMCW modulation we have chirps where after having 1D FFT, we should make sure that Shannon Nyquist theorem is applied to avoid anti-aliasing $f_s > 2 f_0$. Now to do a 2-D FFT we have another Nyquist rule to be applied to avoid ambiguity $1/T_r > 2 f_{d_{max}}$ where $T_r$ is the Repetition time between chirps. if $f_d$ is too high then we have an aliasing receiving wrong $f_d$ than it actually is. So we would be ambiguous about the velocity value itself.

$f_0$: frequency of the signal sent

$f_s$: sampling frequency

$T_r$: Repetition time between chirps

$f_{d_{max}}$: maximum possible $f_d$ that could take place depending on the relative velocity between radar and target.

So now, how can I relate a

  • $f_d$ that exceeds $f_r$ and
  • $f_d$ that exceeds the doppler signal band $1/2T_r$

to each other?

This Paragraph is an illustrations with a drawing for the answer by Dan Boschen

![![enter image description here

solid line peak: original peak after being shifted at $f_r+f_d$

dotted line peak: what we see after FT mirroring peaks into the band $[0,f_s]$

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    $\begingroup$ the scale of your bottom drawing is off- the distance from f_s to 2f_s is half the distance from 0 to f_s as drawn. You almost have the frequency periodicity shown correct: everything that you draw in the range from 0 to f_s will also appear in the range from f_s to 2f_s and every range of Nf_s to (N+1)f_s. Also complex signals are assumed since what is in 0 to f_s/2 is independent of f_s/2 to fs. $\endgroup$ Apr 22, 2022 at 10:32
  • $\begingroup$ I cleared it out by removing $2f_s$ and also I was a bit confused with $f_s/2$ but now I cleared my confusion. I did the shifts as per your description and now the same question that I asked in my last comment. I can't separate the pink peak from the dotted blue peak or the dotted light blue peak from the solid orange peak except by complex conjugate multiplier. What I mean is If I have seen a peak, I am not sure if it was shifted or was originally there by only applying Nyquist theorem on $f_{d_{max}}$. Is that correct? $\endgroup$
    – Mour_Ka
    Apr 22, 2022 at 10:49
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    $\begingroup$ Right you translate frequency only in one direction with a complex conjugate multiplier (which would model a Doppler shift as well). Any shift is periodic over period f_s, so a tiny shift negative is the same as a large shift positive. But you can always distinguish a small shift negative from a small shift positive. Typically you would use a ramp range that far exceeds your Doppler so there would be no confusion with this $\endgroup$ Apr 22, 2022 at 10:54

1 Answer 1


There is no fundamental reason to limit the processing to real signals and incur the first ambiguity on Doppler direction.

Starting with the following conceptual block diagram, we can see both ambiguity effects due to range and FFT processing:

The multiplier shown is a full complex conjugate multiplier, with the wider arrows representing complex data paths (implemented with quadrature splitters in actual hardware). Optionally with additional filtering the quadrature splitter on the receive path could be eliminated. The point is this can and is implemented and avoids one of the ambiguities mentioned.


The signals introduced are the instantaneous frequency waveforms versus time. $f_T(t)$ is the locally generated chirp that is both transmitted and multiplied with the received chirp waveform $f_R(t)$. The time domain complex conjugate product results in the frequency difference between the two waveforms, which we can then process with a DFT to observe the frequency content directly.

The plot below shows the transmit and receive waveforms with a chirp duration of $T$ seconds, and a round-trip delay to target and back of $\tau$ seconds. Here we see the frequency difference that is created by the delay when comparing the transmit and receive waveforms. Also introduced is the static (in this case) frequency offset of $f_D$ due to Doppler:

Transmit and Receive Waveforms

Now before we even consider the effects of sampling and the FFT, we can immediately see the range ambiguity introduced by $T$ and $\tau$: The difference in frequency will be given by $\tau$ modulo $T$, such that the first range bin is for $0<\tau \le T$ and then repeating after that. For any frequency difference result (which we would without considering Doppler associate directly with a range in that value of $0<\tau \le T$), once Doppler is introduced, this will simply shift the resulting frequency difference positive or negative as given by the actual Doppler offset.

The plot below in bold blue represents the frequency versus time of the difference signal that we would expect for the example above, regardless of subsequent sampling and FFT processing used. Here we can introduce sampling and its aliasing and considerations of the FFT. If we were to make the sampling rate of the signal match the difference in frequency range of the chirp ($F2-F1$) or equivalently the maximum possible frequency in our complex difference signal without considering Doppler, we would introduce the horizontal dashed lines in the plot indicating the sampling rate boundaries ($f_s$) at positive and negative the sampling rate ($f_s$) and positive and negative half the sampling rate ($f_s/2$).

Difference Signal

Once sampled, the frequency domain in periodic, so everything that is shown in the range from $f=0$ to $f=f_s$ would repeat exactly in every other range $f=Nf_s$ to $f=N(f_s+1)$ for all integers $N$. Thus the single blue trace which we now completely understand from the perspective of a continuous time signal, becomes the repeating red traces, appearing in each of these frequency ranges of the sampled system. Finally we note that the DFT presents the unique samples starting from $f=Nf_s$ to $f=N(f_s+1)$ (so we would typically associate the first DFT bin with $f=0$ and the last DFT bin as being one sample less than $f= f_s$. This is periodicity in frequency associated with sampled waveforms such that for any tone with a frequency $f$ in the primary Nyquist zone of $-f_s/2$ to $+f_s/2$, we could also have other tones at frequencies $N f_s + f$ that would all map to this same location.

From this we can understand clearly the ambiguity introduced by sampling and observed to us from the DFT result-- any frequency difference will be modulo $f_s$, and the effect of Doppler just shifts this result up and down without any additional direction ambiguity (a small shift up is not ambiguous from a small shift down, but a small shift up is identical to a very large shift down similar to the range ambiguity already introduced).

  • $\begingroup$ That is great and I am currently reading about IQ receivers. From your answer you are confirming that the first ambiguity explanation has nothing to do with the second ambiguity? The reason that I am asking this question is because some publications are actually using the first ambiguity explanation problem but it is mentioning that decreasing $T_r$ (2nd ambiguity explanation solution) increases $f_r$ and so it is shifting the problem but yet not solving it. That is exactly what confused me. $\endgroup$
    – Mour_Ka
    Apr 14, 2022 at 12:08
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    $\begingroup$ @Mour_Ka Stay tuned, the answer is in process, I just save the iterims for fear of losing my progress.... I think it is helpful to consider the whole thing first in continuous time w/o the effect of sampling and the FFT, and then as long as you understand the sampling process, the FFT and how aliasing occurs in those domains it should all be very clear. If you don't really understand those things, then that would be the first step to dig into and then come back to my complete answer here. $\endgroup$ Apr 14, 2022 at 12:10
  • $\begingroup$ After spending some time reading your answer and more about aliasing. I have 2 questions. I still didn't get the formula of $f=fsN+N$ as how come we are adding count to frequency to get frequency. The nearest Interpretation is that $f = fs.N + \Delta f.N$ where $\Delta f$ is the frequency gab between two bins. Second I still don't see the point of without any direction ambiguity at the end. If the distance is so low and the shift of $f_d$ is large in the negative direction then we can still have the peak shifted into the negative side still we can have direction ambiguity. $\endgroup$
    – Mour_Ka
    Apr 20, 2022 at 5:17
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    $\begingroup$ @Mour_Ka I don't think I wrote that formula correctly. What I was trying to say is given a sampling rate of $f_s$ and a tone at $f$, any other tone at $N_f_s +f$ would also appear as $f$. I will fix that. For your second case that would make make a small shift negative appear like a very big shift positive, which for that amount of shift would occur wherever we are (even if distance is large or small). That particular ambiguity is covered by the first one mentioned on Doppler bin range. I thought you considered small shifts in Doppler would have direction ambiguity. I will clarify the range. $\endgroup$ Apr 20, 2022 at 11:24
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    $\begingroup$ See if the edits in the last paragraph make that clearer. There wasn't anything wrong with the formulas as I was indicating the range, but I added the formula you were trying to resolve, and clarified what I meant by Doppler ambiguity. $\endgroup$ Apr 20, 2022 at 11:33

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