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In fmcw radar rather than sending a signal at constant freq. we are sending a chirp which goes e.g. from 6-GHz to 6.1-GHz in 5ms.

What i cannot fit in my mind is this:(Nmbers are just for example)

The rf signal is moving at speed of light so while we are changing the chrip freq. linearly e.g. we start from 6Ghz and waited like 500us then increased it to 6.01 (this goes up to 6.1 ghz with sometime between each freq. change) and we already received the 6ghz signal from receiver without completing 5ms chirp period. So up to this point we have nothing to do with other frequencies of chirp (6.02,6.03...6.1). How is this generating IF like 5-10kHz based on distance. If so then by applying the formula in the picture we can find the distance, why we are changing the tarnsmitted signals' frequency since we already had a IF for a constant 6ghz part of signal chirp signal.

Shortly should i just simultaneously change chirp from 6to 6.1 and 6.1 to 6.0 back and forth and at the same time a/d convert the upcoming signal and save it somewhere until it makes 5ms then plot the fft of the whole 5ms block.

I hope i am clear. Please show me what i am missing here:)

LATER UPDATE: Everyone was trying make me understand the theoretical background where i already know but thank you guys for your time.I had a moment of enlightment and now i understand :)

I am teling this with examples so people who have same problem in future can understand. While sending a chirp we start sending with a fix freq. as a start point like 6ghz. While it is traveling to the target and coming back a little bit of time passed and during this time our TX chirp signals freq. changed a little bit linearly like 6+X GHz so multiplexer difference is X Hz. Therefore this change depend on the distance of the target since let say target is so far far away we will maybe receive the first signal which was at 6GHz while our chirp is at its halfway of 6 to 6.1 GHz cycle. So the difference will be like 6.05-6=0.5GHz.

Another way of saying, for stationary object nothing happening on the transmitted signal(doppler effect says reflected data is radited with same structure if object is stationary) but until the transmitted signal come we changed the transmitted signal so the upcoming signal will be multiplexed with different freq. signal and based on this change we are telling this is object at this much distance. Thank you

enter image description here

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  • \$\begingroup\$ If you don't use a a chirp, but just use a fixed frequency, then the difference frequency will be zero (DC). You can do it this way if you use in-phase and quadrature mixer, and step through the TX frequencies discretely. The difference will still be DC, but by examining how the DC values vary of both in-phase and quadrature, you can get the same image data you would get from the chirp. \$\endgroup\$
    – user57037
    Commented Feb 6, 2017 at 2:25
  • \$\begingroup\$ I suppose you could use multi fixed freq and I/Q detection (I would need to carefully consider this to be sure) but I think the necessary accuracy and stability would be a tough to obtain. The beauty of the continuously-swept frequency is that we can use low peak power, compared to pulse radar, because the return gets detected for the entire duration of the chirp. We get good distance resolution without needing ridiculously narrow pulses at very high peak power. The narrower the pulse, the more peak power we need to get enough return power to process. FMCW is not without drawbacks though. \$\endgroup\$
    – Paul Elliott
    Commented Feb 6, 2017 at 3:07
  • \$\begingroup\$ @PaulElliott, agree 100%. It is much easier to put power on target using CW. Strictly speaking, range resolution is determined by chirp bandwidth only. Gating of pulses (and also gating of CW radar) may be needed to exclude clutter from the measurement, or resolve range ambiguity caused by aliasing. At least that is how I remember it. It has been a while. \$\endgroup\$
    – user57037
    Commented Feb 8, 2017 at 4:43
  • \$\begingroup\$ If you expect the difference frequency to be 0.5MHz, you will need to sample it at over 1 GHz. If you sample it at a lower frequency, the target will alias in to a different apparent range. \$\endgroup\$
    – user57037
    Commented Feb 8, 2017 at 8:36
  • \$\begingroup\$ I gave 6.05 as an extreme example. I am aware of nyquist rate. Also speed of light so fast that it will be definetly around khz band difference for a few kms or less. \$\endgroup\$
    – snrIcmn
    Commented Feb 9, 2017 at 11:31

2 Answers 2

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I'm not sure that you understand the basic operation of the FMCW radar.

As soon as the transmitter sends out its signal, this signal is shifting in frequency at a particular rate, in Hz per Second. The radar receiver will see no returns until the transmitter signal bounces off a target and returns to the antenna. There will be a speed-of-light time delay, so the received signal will be at a different frequency than the signal currently being sent (the difference being Hz per Second times the round-trip delay). The receiver mixes this return signal with the currently-transmitted signal to get the difference signal. The reflected signal will be continuously received for the duration of the transmission, and for a stationary target the difference between the TX and RX frequencies will be constant. You can perform your FFT of this difference over the full duration of the transmit signal. Different targets at different distances will give other difference signals, and the FFT can resolve all these simultaneous returns.

These transmissions aren't exactly continuous, but they are much longer than traditional pulsed-radar. Multiple chirps are sent as the antenna rotates, so you have angle as well as distance information. The amplitude of the return will indicate the "radar cross section" (size) of the target.

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  • \$\begingroup\$ Thanks for comment. You are saying that we send signal it is reflected from target object and freq shift occurs so after receiver we put it to mult. with the original signal we send and difference is giving the IF. If this is how it works then why do we need to modulate the frequency and send a chirp rather than constant freq. signal. \$\endgroup\$
    – snrIcmn
    Commented Feb 6, 2017 at 1:19
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    \$\begingroup\$ See mkeith's and my comments above. If the transmit frequency isn't changing then the receive signal will be at the same frequency as the transmit, and there will be no frequency difference to measure. The only freq shift is at the transmitter -- this is the chirp.. The target reflects back exactly what hits it. A moving target or transmitter will cause a Doppler shift in the returned signal, but that is not what we're talking about here. \$\endgroup\$
    – Paul Elliott
    Commented Feb 6, 2017 at 3:11
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\$X= c\frac{|\delta f|}{2df/dt}\$

It is easier to measure this for low resolution with high carrier f and X >c/f for multiple wavelengths or very fast echo times.

Options are ramp rate, f range , carrier f, power, path loss and modulation.

For modulation, you may choose; sawtooth, triangle, staircase, Square wave, butterfly square wave of decreasing deviation. Each has advantages and disadvantages for resolution , range , reducing ambiguity and ADC sampling rate & BW.

250MHz deviation with 50MHz/ms might be a reasonable starting point with triangular modulation for mobile doppler shift (velocity) and range separation.

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  • \$\begingroup\$ To be sure: Are you suggesting that i should take fft of each freq. part of the chirp and compare fb change between consecutive frequencies like take fft of 6gzh part of chirp then take fft of 6.05(50mhz per ms) and there will be 2 fb values then make distance calculation. And keep doing this \$\endgroup\$
    – snrIcmn
    Commented Feb 6, 2017 at 0:17
  • \$\begingroup\$ Not sure on the FFT tradeoffs for resolution , data size, SNR but I would start with 1 or more full cycles then do FFT. These are tradeoffs. \$\endgroup\$
    – D.A.S.
    Commented Feb 6, 2017 at 0:23
  • \$\begingroup\$ parabolic dish and CNR are key for range. \$\endgroup\$
    – D.A.S.
    Commented Feb 6, 2017 at 0:31
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    \$\begingroup\$ Sir thanks alot. The problem in my mind occured because people said this system only works because of chirp etc. Technically constant freq. works. When we change freqeuncy the IF is always fix for a target at constant point so even we modulate the frequency the target at fix point will always give us the same amount of shift. Then the reason of the fm is because we will have better performance like SNR etc. Now my mind is clear and thank you. \$\endgroup\$
    – snrIcmn
    Commented Feb 6, 2017 at 1:51
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    \$\begingroup\$ Constant frequency does not work. FMCW requires a changing transmit frequency. This is the "chirp". The change can be sinusoidal, triangular, or other waveforms, but the typical chirp is a linear ramp. There are other reasons for modulating the transmit frequency that have to do with improving SNR and resolution, but that's a different subject. FMCW requires the chirp in order to function at all. \$\endgroup\$
    – Paul Elliott
    Commented Feb 6, 2017 at 17:05

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