AWR1642: What happens if the velocity of an object exceeds the maximum detectable velocity of radar?

Hi,

    Yes there is a periodic pattern. Take for instance if the objects are moving away from radar (velocity is positive). And if objects are moving towards the radar, velocity is negative. So in your example, if 8m/s is the maximum velocity, then for a object moving at 10m/s, it would be measured as -6m/s I guess. But if you see the object is moving away from the radar and you see -6m/s, then you know the velocity should have been positive. So in order to correct the velocity, you need to add 2*Vmax to it.

     So if -6m/s is measured velocity for a object moving away, then to get the actual velocity add 2*Vmax, i.e, -6+2*8 = 10m/s to get the actual velocity.  Usually, in FMCW radar we do correct velocity only up to 2*Vmax. 

     But for velocity > 2*Vmax, you would need to find how may times phasor completes rotation and all. So I would request you to please have Vmax in such a way that you would need to detect velocity up to 2*Vmax only.

     Let me know if you have any questions.

Thanks and Regards,

Akshay.

Hi Akshay,

  First, if the Vmax is calculated as 8m/s, it means radar can measure the speed of plus or minus 4m/s.

  Second, if the closing object is -5m/s, the radar  calculate velocity is 1m/s

  Is that right?

Thank you

Hi Charles,

                 No I think there is a small confusion in your understanding. If Vmax is 8m/s, then the objects travelling with velocity in the range of +16m/s to -16m/s can be detected easily. The rule is:

1. If the object is moving away from the radar (Velocity is positive). Till Vmax the velocity is taken as such. If it travels with a velocity greater than Vmax and less than two times Vmax, it would occur as negative velocity (the measured velocity). Therefore to get the actual velocity, add the measured negative velocity with two times Vmax.                                                                                                                                                                              For ex, if Vmax = 8m/s, an object travelling at 10m/s away from radar will be measured as -6m/s, so we add it with two times Vmax. Therefore, -6m/s+2 x 8m/s = 10m/s (the actual velocity).                                                                                                                                                                         
2. If the object is moving towards the radar (Velocity is negative). Till -Vmax the velocity is taken as such. If it travels with a velocity less than -Vmax and greater than two times -Vmax, it would occur as positive velocity (the measured velocity). Therefore to get the actual velocity, add the measured positive velocity with two times -Vmax.                                                                                                                                                             For ex, if Vmax = 8m/s, an object travelling at -14m/s towards the radar will be measured as 2m/s, so we add it with two times -Vmax. Therefore, 2m/s+2 x -8m/s = -14m/s (the actual velocity).

           Hopefully with above two examples and explanation, this topic is clear to you. Let me know if you have any further questions.

Thanks and Regards,

Akshay.

Hi Akshay,

  Thank you for your answer, but I still have a confusion. If I don't konw the object towards the radar or away from radar, how do I add the measured value with two times -Vmax or Vmax?

  And as you said that if a object is measured  -6m/s, I can't confirm with the object is away from radar by 10m/s, or towards the radar by -6m/s.

Thank you