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The last couple of days I read a lot about cw-radars, I/Q modulation and quadrature amplitude modulation (QAM). But one things, which is never mentioned, really confuses me. Which signal is really transmitted out of the antenna? Do these radars use QAM to send a signal like this: \begin{align} v(t)=m_1(t)\cos(2\pi f_ct)-m_2(t)\sin(2\pi f_ct), \end{align} with \begin{align} m_1(t) = A_m\cos\phi_m \quad \text{and} \quad m_2(t) = \underbrace{A_m\sin\phi_m}_{\mathrm{Quadrature}}, \end{align} to send I and Q signal at the same time? This would lead to confusion on about the doppler shift can be calculated afterwards, since $m_1(t), m_2(t)$ would have a second frequency next to $f_c$. So you would have a pretty complex doppler effect calculation (or am I missing something here)?

Or ist just a signal \begin{align} v(t) = I\cos(2\pi f_ct) + Q \sin(2\pi f_ct) = \sqrt{I^2+Q^2}\cos(2\pi f_ct - \arctan(Q/I)) \end{align} transmitted at the end? But wouldn't this lead to an ineffective bandwidth usage? And how could you calculate I and Q from the received signal?

I am probably missing something here, but I would appreciate any kind of help. Just for your information, I want to calculate small distance changes using a CW-radar and am struggling to REALLY understand what is going on here. I also haven't found a good book explaining this, so any suggestions are welcome.

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  • $\begingroup$ OK, you're throwing around a lot of terms, and I think that's gotta be relatively confusing for you. I'd recommend understanding IQ/the quadrature mixer first; the rest is just applications of the same math. I wonder where the difference between the two formulas you show actually are; set $m_1(t) = I(t)$ and $m_2(t) = -Q(t)$ and they are identical. $\endgroup$
    – Marcus Müller
    Commented Sep 5 at 12:37
  • $\begingroup$ There's quite a bit here. For starters, to measure range you can't use a simple CW tone, it will either need to be modulated somehow are use a series of CW tones at different frequencies. $\endgroup$
    – Envidia
    Commented Sep 5 at 14:45
  • $\begingroup$ @Envidia CW does not mean a single unmodulated tone, but "Constant Wave". it just means the transmit signal is always active (on) in contrast to pulse radar where the signal actually turns on and off. $\endgroup$
    – Dan Boschen
    Commented Sep 6 at 7:29
  • $\begingroup$ @DanBoschen In the radar community, a "CW radar" without mentioning any qualifiers, is assumed to be a radar that transmits a single tone continuously. That is why qualifiers to "CW" exist for things like FMCW, where the transmitter is always on (for the time of interest, which really doesn't make it true), but the signal itself is modulated. Over time, a lot of radar terms aren't truly what they used to mean, given advances in technology on how signals are processed. $\endgroup$
    – Envidia
    Commented Sep 6 at 14:20
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    $\begingroup$ Thanks @Envidia I’ve seen CW refer to both modulated and unmodulated (and that appeals to me since “CW” is constant wave not constant tone) but good to get your perspective on that. $\endgroup$
    – Dan Boschen
    Commented Sep 6 at 17:46

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The use of IQ is fundamentally to create complex baseband waveforms which can then be subsequently translated to an RF signal as a real passband signal and through a single antenna. This can be used to send QAM modulation as the OP is inferring, but IQ does not necessitate QAM.

"CW" for radar refers to "Continuous Wave" where the transmit signal is always on, in contrast to pulse radar where the transmit signal is turned on and off during the radar operation. “CW” can refer to modulated or unmodulated radars (see https://en.m.wikipedia.org/wiki/Continuous-wave_radar ), where the modulated case is required to resolve range. A very common modulated CW radar implementation is "FMCW" where the transmit frequency is ramped (chirped). Qasim Chaudhari wrote a nice series of blog articles recently with more details about FMCW specifically. Qasim is also lined up to make an interesting presentation on FMCW at the 2024 DSP Online Conference coming up at the end of October. Also see my related posts here regarding resolving range and Doppler with FMCW radar:

Range-Doppler Coupling in FMCW

https://dsp.stackexchange.com/a/82571/21048

By using "I" and "Q" referring to the in-phase and quadrature components of a complex baseband signal, we're able to create waveforms where the upper and lower sidebands of a passband transmitted signal are completely independent of each other. With a complex baseband signal, the spectrum that is centered on DC (f=0) will exactly represent the spectrum as it will be centered on any other carrier frequency. With that we can implement FMCW at baseband by starting with a negative frequency and ramping that through DC to a positive frequency.

To get an intuition for this, see the following spectrums below for the Fourier transform of a real sinusoid, vs the transforms for complex exponentials representing a single negative frequency tone or a single positive frequency tone:

Fourier Transforms

If we were to upconvert the real sinusoid to a higher RF carrier frequency, we would see each of the tones shown as two sidebands on that carrier. (Double Sideband Suppressed Carrier specifically). However upconverting either of the other two cases (using a complex baseband signal) would result in a single tone at RF either slightly higher or slightly lower than the carrier frequency. We therefore would have full control of that single tone, including ramping it for the FMCW application.

We note how Euler's formula represents this desired complex signal into it's I and Q components for implementation:

$$e^{j\omega t} = \cos(\omega t) + j\sin(\omega t) = I + jQ$$

That is the significance of "I" and "Q" and nothing to do here with QAM.

The "upconvert to a higher RF carrier frequency" functionally proceeds as follows in many different variants that all have this same result:

Multiply the complex baseband IQ with a complex Local Oscillator and select the real part:

$$(I_b + jQ_b)(I_{LO} - jQ_{LO}) = I_bI_{LO} + Q_bQ_{LO} + j(I_{LO}Q_b - I_bQ_{LO})$$

If we select the real part of this we get the desired RF output as:

$$I_bI_{LO} + Q_bQ_{LO}$$

Why and how this works is further detailed in DSP.SE# 31355.

Please see this related post on how to generate FM signals from IQ:

https://dsp.stackexchange.com/a/73460/21048

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  • $\begingroup$ +1, but CW == 'Continuous Wave'. $\endgroup$
    – AnonSubmitter85
    Commented Sep 11 at 18:43
  • $\begingroup$ @AnonSubmitter85 Yes! Much better, thank you $\endgroup$
    – Dan Boschen
    Commented Sep 11 at 23:40

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