Phase noise influence on ADC in bistatic FMCW Radar

Question

I am currently studying the topic of oscillator phase noise in frequency modulated continuous wave (FMCW) radar. In my specific case, the radar is implemented on an FPGA (AMD RFSoC). Thus, the echo is directly digitized by the ADC after the antenna (and some filter and amp. stages, but no down mixing). Deramping will be done in the digital domain, followed by a digital filter and downsampling to reduce the memory demand.

Now I see a lot of studies on phase noise and radar, but usually they assume the deramping in the analog domain. I have been wondering whether I will get different effects with direct RF-sampling.

For example: The local oscillator used to synthesize the ADC and DAC sampling clocks will have some phase noise and maybe frequency drifts. This would lead to jitter and a sampling frequency drift as I understand. I have not encountered this in literature when the deramped signal is sampled by the ADC.

Now, assuming I receive the echo with an ADC that is driven by a different LO, the phase noise will not cancel out. My idea of an echo model (from point target) looks like this: $$ s_{rx}[T'_s n] = e^{j2\pi\left( f_0[T'_sn - \tau +\Delta t]+ \frac{\alpha}{2}[T'_sn - \tau +\Delta t]^2 \right)} $$ where $\tau$ is the travel delay, $f_0$ the start frequency (assumed perfect for simplicity), $\alpha$ the ramp rate, $T'_s=T_s(1+\epsilon)$ the sampling interval of the ADC that potentially deviates from the assumed ideal sampling interval $T_s$, $\Delta t$ the clock jitter and $n=0, 1, \dots, N$ the sample index.

The deramping would be done with an ideal stored chirp in memory (assuming the timing is synchronised) $$ s_{ref}[T_sn]=e^{j2\pi(f_0[T_sn]+\frac{\alpha}{2}[T_sn]^2)} $$

I have written out the outcoming beat signal ($s*_{rx}\cdot s_{ref}$), which gets pretty blown up. Especially when the time delay $\tau$ becomes time dependent due to movement.

EDIT: Here is the deramped signal (without movement) $$ \begin{align} s_{deramp}[n] = &s^*_{rx}\times s_{ref}\\ &e^{j2\pi\left( f_0(T_s-T'_s)n + \frac{\alpha}{2}(T_s-T'_s)^2n^2 \right)} \times\\ &e^{j2\pi\left(f_0(\tau-\Delta t) - \frac{\alpha}{2}(\tau^2+\Delta t^2) + \alpha (\tau-\Delta t)T'_s n + \alpha\tau\Delta t \right)} \end{align} $$ My question is:
Is this model correct or somewhat in the correct direction? Am I misunderstanding something? I am lacking the experience to judge if some of these effects are potentially negligible. If anyone knows resources I could read up about this, a tip would be greatly appreciated!

Hope the question is clear, otherwsie I clarify!

Answer 1

I think you are on the right track, but it's hard to know for sure as you haven't written out what you get for the deramp process. The idea behind dechirp-on-receive (or deramp or stretch processing) is that at any instant in time, you are relatively narrowband, just with a changing center frequency.

Assuming no jitter, if you have a transmit signal \begin{equation} s_{tx}(t,n) = Ae^{j2\pi\left(f_{0}(t) + \frac{\alpha}{2}(t)^{2}\right)} \end{equation} your received signal will be a time delayed version of itself, delayed by $\tau$. The reason why dechirp-on-receive is so popular for SAR is that you usually have some idea of how far away you are from the scene being imaged. So, you know that $\tau = \frac{2R}{c}$ under the assumption you know $R$.

If we have a reference point $ref$ for pulse $n$, we know that the range to this reference point is \begin{equation} r_{ref}(n) = \sqrt{x_{ref}(n)^{2} + y_{ref}(n)^{2} + z_{ref}(n)^{2}} \end{equation} Assuming we discretize the scene into a set of $k$ points, the range relative to the reference point is \begin{equation} r_{k}(n) = \sqrt{(x_{k}(n) - x_{ref}(n))^{2} + (y_{k}(n) - y_{ref}(n))^{2} + (z_{k}(n) - z_{ref}(n))^{2}} \end{equation} We can then define the differential range as $\Delta R_{k}(n) = r_{k}(n) - r_{ref}(n)$ with relative round trip time to each point as $\tau_{rel} = \frac{2\Delta R_{k}(n)}{c}$.

Our receive signal model is \begin{equation} s_{rx}(t,n) = Ae^{j2\pi\left(f_{0}(t-\tau) + \frac{\alpha}{2}(t-\tau)^{2}\right)} \end{equation} Performing the dechirp-on-receive we get \begin{align} s_{stretch}(t,n) &= A^{2}e^{j2\pi\left(f_{0}t - f_{0}(t-\tau) + \frac{\alpha}{2}t^{2} - \frac{\alpha}{2}(t-\tau)^{2}\right)} \\ &= A^{2}e^{j2\pi\left(f_{0}\tau -\frac{\alpha}{2}(\tau^{2}-2t\tau)\right)} \\ &= A^{2}e^{j2\pi\left(f_{0}\tau + \alpha t\tau - \frac{\alpha}{2}\tau^{2}\right)} \\ &= \sum_{k}A^{2}e^{j\frac{4\pi}{c}\left(f_{0}+\alpha t\right) \Delta R_{k}(n) }e^{j\frac{4\pi\alpha}{c^{2}}\Delta R_{k}(n)^{2}} \end{align} Notice, the quadratic phase term has no dependence on $t$, so we can say $A_{k} = A^{2}e^{j\frac{4\pi\alpha}{c^{2}}\Delta R_{k}(n)^{2}}$. Also notice that $f_{0} + \alpha t$ represents the range bins. So, the sampled dechirp-on-receive signal can be represented by \begin{equation} s_{stretch}(f_{m},n) = \sum_{k}A_{k}e^{j\frac{4\pi f_{m}}{c}\Delta R_{k}(n)} \end{equation}

Adding in a jitter $\Delta t$ adds an additional phase shift of \begin{equation} \phi_{j}(t) = -2\pi(f_{0}\Delta t + \alpha t \Delta t + \frac{\alpha}{2}\Delta t^{2}) \end{equation} So, it will introduce a range shift as well as alter the amplitude slightly, from what it looks like.

Also, it's important to note that stretch processing only saves you bandwidth if your range swath is less than range resolution of an uncompressed pulse. Hope this helps!