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I have been reading about "all-digital" or "software-defined radars". Lately, I was trying to simulate an FMCW short range radar in Matlab. One thing that struck me during the simulation is that to achieve a range resolution of around a 100 cm or less, the signal bandwidth is quite large -- in multiple GHz. This means, if one wishes to realise a digital implementation, one needs a really high speed ADC - multiple GigaSamples/second. Such ADCs are either not available and if you find something towards the top end of the high speed ADCs out there, the processing requirements are going to be ridiculously high. And of course, if you are clocking your baseband processor in GigaHertz, the power requirements of the device (typically an FPGA) are again going to be impractical.

My questions:

  1. How, if at all, high-resolution, high-bandwidth digital radars are realised in practice?

  2. what would make sense in case of FMCW radar is to mix the transmitted signal and the target return signal and digitize only the beat frequency for Inverse FFT and further analysis. In that case, could you still call it a digital or software defined radar?

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  • $\begingroup$ This is a problem - maybe consider stepped frequency continous frequency radar if you can support it. The IF bandwidth is ultra low $\endgroup$
    – johnnymopo
    Commented Jan 13, 2016 at 20:45
  • $\begingroup$ That is SFCW - stepped frequency continous WAVE, not frequency $\endgroup$
    – johnnymopo
    Commented Jan 13, 2016 at 20:47
  • $\begingroup$ You are actually doing direct conversion to baseband (in 2). You need to maintain the proper bandwidth for the expected signal. For 1 - you have a band limited signal, and assuming that the out of band noise is low, you may under sample in rates that are 10 or more times lower. $\endgroup$
    – Moti
    Commented Jan 13, 2016 at 23:23

2 Answers 2

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A quick scribble on paper shows that FMCW radar has a resolution of

$$ \Delta d = \frac{c}{2b} $$ with $c$ being the speed of wave propagation (aka speed of light, essentially), $b$ being the bandwidth, and $2$ coming into play because the wave has to trave both ways. For your $\Delta d\overset != 1\mathrm m$,

$$ 1\mathrm{m} = \frac{3 \cdot 10^8 \frac{\mathrm{m}}{\mathrm{s}}}{2b} $$

which means you'd only need $b\ge 150\, \mathrm{MHz}$ in my calculation.

That's something that software defined radio devices can do nowadays.

Hence, to answer your questions: My questions:

How, if at all, high-resolution, high-bandwidth digital radars are realised in practice?

Simply be mixing the signal down to baseband (or a low IF), and then sampling it with an ADC.

what would make sense in case of FMCW radar is to mix the transmitted signal and the target return signal and digitize only the beat frequency for Inverse FFT and further analysis.

Yes, that's the way it's done in most "small" FMCW radars, for example automotive radars.

In that case, could you still call it a digital or software defined radar?

Puh. Digital: yes. Software defined: well, if everyone asked me, no, but paper doesn't blush, so there's probably someone out there who calls this kind of radar "software defined". Please don't do that.

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  • $\begingroup$ thanks. I am satisfied with your explanations except that instead of specifying 10cm range resolution (typical in short range radars) I stated 100 cm :( using 10 cm in your calculation, the bandwidth requirements jumps to 1.5GHz which is already pushing today's techology $\endgroup$
    – user4673
    Commented Jan 14, 2016 at 23:08
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Your 2. is (as far as I know) the standard way to implement a FMCW radar. The major advantage of both the FMCW and the SFCW (which was mentioned in the comments), is the sample rate of the ADC is greatly reduced. (This is sometimes called down-conversion, or pulse-compression).

After the mixer, you have a mixer sum and mixer difference, filtering out so that you only have the difference, there is a linear relation between the frequency and range of the target:

$f_\text{beat} = \frac{B\tau}{T_m}$

where $\tau=2r/c$ is the two way-travel time for a target at range $r$ and $c$ is the speed-of-light, $B$ is the bandwidth and $T_m$ the chirp length. The really neat thing here, is that you do not need to sample the entire bandwidth $B$, but only the $f_\text{beat}$ spectrum.

To take an (extreme) numerical example, lets say we want a resolution of 1.5cm, this requires a bandwidth

$B=\frac{c}{2\cdot 1.5cm}=10\,GHz$

which is difficult to deal with on the transmitter and receiver part (generator, PA, LNA and mixer), but after the mixer, with a sweep time of say $T_m=1/3\,s\approx0.33\,s$ and a target at 100 meters, the beat frequency is only

$f_\text{beat} = \frac{10\,GHz\cdot(\frac{2\cdot 100\,m}{3e8\,m/s})}{(1/3)\,s} = 20\,kHz$

which could be sampled by an audio card.

Whether this system is "all-digital" or "software-defined radars", depends on the implementation. If you can build your chirp generator, amplifiers and mixers using only digital components, then your system is "all-digital". If your components are sufficiently adaptable/programmable, then if would be (in my mind) "software-defined".

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  • $\begingroup$ +1. This is the most complete answer because you included the chirp length, which affects the slope of the frequency ramp. This makes intuitive sense. A shallower ramp will naturally produce a lower beat. $\endgroup$
    – AaronDanielson
    Commented Apr 6, 2017 at 14:37

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