# Why CW radar I-Q signals do not contain Doppler frequency component?

I have a question about I/Q signals in Doppler radar, but I couldn't find a solution for this.

As far as I understand, the received signal at receiver side will be down-converted by mixing it with the transmit signal (direct conversion). And in order to generate I/Q signals, the received signal is also mixed with a 90degree phase shift of the transmit signal.

I have read several papers, and all of them say that

$$I(t) = \cos(\phi(t))$$ $$Q(t) = \sin(\phi(t))$$

My question is: where is the Doppler frequency component (assume having a moving target, for simplicity I ignore their amplitudes). I expected to see something like:

$$I(t) = \cos(f_D 2\pi t + \phi(t))$$ $$Q(t) = \sin(f_D 2\pi t + \phi(t))$$

Why the Doppler frequency does no more remain in the IQ signals?

• I'm guessing those are the I/Q signals from the source, before they were mixed with the reflected signal... generating sum and difference frequencies that you are seeking... A link to the source of your confusion would help. – George Herold Sep 30 '15 at 12:58
• However as far as i know, the I- and Q-signals are generated by mixing the received (reflected) signal with the transmit (source) signal and with a 90degree phase-shifted version of the transmit signal respectively. Therefore I think I-Q signals are not from the source but the reflected signal. About the link to the source of my confusion, unfortunately I don't have a link to give you, however I'm working on a topic which using Doppler radar for detect vital sign. I think there are plenty of paper about this and you can find the general structure of the radar system – bienle Oct 1 '15 at 12:27

I've been reading your comments and I see your understanding has grown a little.

Your transmit signal is set to some transmit frequency and reference phase: $$f_t\ and\ \phi$$ the phase could be anything but let's say it's zero for our purposes. A simple pulse-Doppler radar will transmit a pulse of the form: $$\cos(2{\pi}f_t{t})$$

Your return signal from a moving target will contain a frequency shifted version with an added phase: $$\cos(2{\pi}(f_t + f_D)t + \phi(t))$$

The signal will come to the mixer that will split and then mix the split signals individually with: $$\cos(2{\pi}f_t{t}) \ \ \sin(2{\pi}f_t{t})$$

Which produces your I and Q channels (cosine and sine, respectively). The resulting signals after low-pass filtering will remove the transmit frequency and you'll end up with: $$\cos(2{\pi}f_D{t} + \phi(t)) \ and\ \sin(2{\pi}f_D{t} + \phi(t))$$

At this point, the information you want is available. You can pass either I or Q through a Doppler filter bank to determine the Doppler frequency of the signal. Doing a Fast Fourier Transform (FFT) will give you the spectrum of the received signal which will peak at the Doppler frequency received. The phase of the signal at this point is not as important.

Finding the magnitude from I and Q is for just that, finding the magnitude of the returned signal mainly for detection and display purposes.

This is a very simple overview and Doppler radar demodulation contains more steps that are not shown here, but this should give you a good idea.

Set $\phi(t)$ from the papers equal to $f_D 2\pi t + \phi(t)$? It's just different notation and they are calling the entire thing $\phi(t)$ for a phase, and not explicitly breaking out each separate term perhaps. Again, citing one of those "several papers" where you are seeing this would help us answer your question.

• Hi, thanks for the reply. Although I'm no longer working on this topic, I really like to discuss this question. i don't think ϕ(t) = fD2πt+ϕ(t), since for extracting the phase of the signal, an arctangent approach can be used (i.e: ϕ(t) = atan(Q/I)). Therefore if ϕ(t) = fD2πt+ϕ(t), then the phase of the signal cannot be extracted using arctangent. For citing the papers, unfornately I lost track of them, so it would be hard for me to find them again. – bienle Mar 13 '16 at 10:09