2
$\begingroup$

I am performing some simulations for FMCW radar for target detection. I am working with a dechirp on receive system. Therefore, the target distance can be determined from the Intermediate Frequency or beat frequency.

$$F_{beat} = \frac{\text{rate}_{\text{chirp}} \cdot 2 \text{dist}_{\text{target}}}{c}$$

I realised that the maximum theoretical distance that can be detected by the radar is determined by $F_s$ (sampling frequency). What is the relationship between the two? Is there a particular formula or concept that links both parameters?

$\endgroup$

2 Answers 2

3
$\begingroup$

They are related, and their relationship is so subtle that you may not realize that your equation already shows the relationship. Let's make the relationship explicit between the generated beat frequencies $f_b$ and the chirp with pulse width $\tau$ and bandwidth $\beta$:

$$f_b = \frac{2R\beta}{c\tau}$$

If everything remained the same, then processing out to a larger range $R$ requires a larger $f_b$, and thus you must be able to sample it. Then, this means that you need to increase the sample rate $f_s$.

So given a fixed waveform, if you want to see out farther in range, you need a higher sample rate, and thus they are related. The answer Marcus has is technically correct, but it misses the question-answer to how you determine the bandwidth you need to sample.

$\endgroup$
1
$\begingroup$

They are not inherently related. You need to design your system with a sampling rate high enough to cover the bandwidths you need, that's all.

$\endgroup$
1
  • $\begingroup$ For a given range, increasing the waveform bandwidth pushes the mapped beat frequency to be higher, and you must sample at a higher rate. So in this way they are related, and you will cause range aliasing if you don't sample fast enough. $\endgroup$
    – Envidia
    Jun 8, 2023 at 18:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.