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IWR6843: Maximum velocity and velocity resolution calculation in mixed chirp frames

Part Number: IWR6843

Hi,

I watched the TI tutorials on velocity calculations and following are the two key formulas I use:

1 - Maximum velocity ( V(m) ) = L / (4Tc)  where L is the wavelength and Tc is the chirp period in a given frame

2 - Velocity resolution ( V(r) ) = L / (2Tf) where L is the wavelength and Tf is the frame period.

I am trying to calculate V(m) and V(r) for the "People Tracking and Counting" configuration file provided for IWR6843ISK as shown below (i.e. see underlined parameters I am interested in)

I studied the frame structure for the above example and following figure summarizes my understanding. There are two different chirps defined for each frame (i.e. indicated with black and red below) These chirps are based on the same profile ID and the only difference is each chirp type is transmitted using different antennas. According to the chirp period, Tc, (which I highlighted in the TI chirp reference figure further below) each red and black chirps have a duration of 92 us as shown. The total active time (Ta) where chirps are transmitted can be calculated as 256 x 92 us = 23.552 ms. Given that Tf = 50 ms, the gap at the end of each frame (Tg) is automatically calculated (by the mmWave Studio) as 26.448 ms (i.e. 50 ms - 23.552 ms = 26.448 ms) Therefore, the duty cycle of each frame is Ta/Tf = 47.1% (i.e. just below 50%)

Next, I will walk you through my calculations for V(r) and V(m) here and request your help on where I seem to be making a mistake.

The starting frequency is specified as 60.6 GHz and therefore the wavelength (L) is 4.95 mm in this example.

1 - My V(r) calculation:

 According to the reference formula V(r) = L / (2Tf), I plug in L=4.95e-3 and the Tf = 50 ms (from the above frame structure).

Therefore, V(r) = 0.0495 m/sec = 0.178 km/hr

According to the table I shared above, V(r) is supposed to be 0.36 km/hr, which is almost twice as what I am calculating. The first thing that comes to mind is the duty cycle (which is almost 50%), which is not factored in my calculation. Therefore, it seems I should be factoring that in as:

V(r) = 0.178 / (47.1%) = 0.378 km/hr , which is close to the listed result above.

I am not sure if the duty cycle is the reason why my first calculation is off, and also whether it should be factored in as highlighted above.

2 - My V(m) calculation:

According to the reference formula V(m) = L / (4Tc), I plug in L=4.95e-3 and the Tc = 92 us (as per the above frame structure).

V(m) = 13.451 m/sec (i.e. 48.42 km/hr)

My V(m) seems to be almost twice as large as the listed result in the table above, and I think this is attributed to the duty cycle again. (As my calculation implicitly assumes transmitting chirps for the entire duration of the frame, which is actually not the case!)

Therefore, my intuition tells me I should factor in the duty cycle at the end as follows to get the correct answer:

V(m) = 48.42 * (47.1%) = 22.8 km/hr, which is closer to the listed result in the table above.

Could you please shed light onto the duty cycle aspect and any other term I might be missing in my V(r) and V(m) calculations I listed above? I presume in more complicated frame and chirp configurations (with different Tc values for example), calculation of V(r) and V(m) are also going to have to vary. However, if I get an understanding of this particular case, I think I can generalize it to other cases.

Thank you for your help.

Regards,

Cagri

  • HI, Cagri:

    You can find the velocity calculation from the following document at section 2.2. https://www.ti.com/lit/an/swra553a/swra553a.pdf

    1) Maximum velocity ( V(m) ) = L / (4Tc) 

    The chirp repetition time Tc should be the duration from one chirp to the next chirp, it is not only the ramp time, it should include the idle time, ADC start time, ADC sample time and the excess time.  In addition, if you have 3 chirp to cover TDM-MIMO, then Tc should be the time between the two chirp of the same chirp ID, which is 3 times of the one chirp duration.   The velocity estimation is based on the phase change from one chirp to the next chirp of the same antenna pair.  Therefore, if you increase any part of the repetition time, the neighboring two chirps will separated longer in time domain.  The same movement in speed will cause bigger phase change from one chirp to the next chirp.   Note that the maximum velocity is actually the maximum velocity without ambiguity, if the velocity is beyond this speed range (-Vm, Vm), the phase change between the two chirp will be beyond the phase range of (-PI, PI).

    2) - Velocity resolution should be: ( V(r) ) = V(m) / numLoop 

    Best,

    Zigang

  • Hi Zigang,

    Thank you for your detailed description and prompt response. Much appreciated. I wanted to recap my calculations based on your guidance to make sure we are on the same page. Please have a look and let me know if it all makes sense.

    By the way, I have also read the document you cited in terms of velocity calculations and they concur with my descriptions earlier.

    One thing I seem to have missed is that duty cycle of a frame is completely irrelevant to velocity max and resolution calculations. Therefore, Tg I indicated earlier can be ignored.

    The Tc period I mentioned earlier covers all of the delays you mentioned (i.e.  idle time, ADC start time, ADC sample time and the excess time) since it is the sum of Idle Time and the Ramp End Time as I referred to in the TI chirp reference figure in my previous message. (Please let me know if there is any other time delay I should include except the Tc = Idle Time + Ramp End Time)

    1) Maximum velocity ( V(m) ) = L / (4Tc) 

    I updated my previous chirp frame reference for calculating the V(m) below based on your description. Basically, one needs to use either the red or the black chirps to calculate the V(m) but not both. Therefore, as you said, the delay between the end of one black chirp and the start of the next black chirp increases by one red chirp duration as shown below. Therefore, the Tc I should be using is 184 us rather than 92 us as I used in my previous message.

    So using the V(m) = L / (4Tc) formula again:

    V(m) = 4.95e-3 / (4 x 184e-6) = 6.726 m/sec = 24.21 km/hr (which is close to the 23.6 km/hr above so this works)

    2 - Velocity resolution calculation using V(r) = L / (2Tf)

    I think in your last message related to the calculation of V(r) you meant to say:

    V(r) = V(m) / (0.5 x numLoop) 

    I revisited sections 2.2.1 and 2.2.2 in  https://www.ti.com/lit/an/swra553a/swra553a.pdf and the 0.5 factor is needed.

  • HI, Cagri:

    1) You are right, that "duty cycle of a frame is completely irrelevant to velocity max and resolution calculations"

    2) Yes, you are right,  Tc = idle time + Ramp End time

    3) Regarding the calculation, it probably the best to use center frequency instead of starting frequency when calculate lambda.

    V(m) = 4.95e-3 / (4 x 184e-6) = 6.726 m/sec = 24.21 km/hr (which is close to the 23.6 km/hr above so this works)

    2 - Velocity resolution calculation using V(r) = L / (2Tf)

    I think in your last message related to the calculation of V(r) you meant to say:

    V(r) = V(m) / (0.5 x numLoop) 

    I revisited sections 2.2.1 and 2.2.2 in  https://www.ti.com/lit/an/swra553a/swra553a.pdf and the 0.5 factor is needed.

    [ZY], yes, you are right.  because the total velocity range is (-Vm, Vm), so the resolution is 2*Vm/numLoop. 

    Best,

    Zigang

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