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AWR1843BOOST: Azimuth angle estimation

Mario
Prodigy 85 points
Part Number: AWR1843BOOST
Other Parts Discussed in Thread: AWR1642

Hello,

i am trying to figure out how the azimuth-angle estimation in the demo example is being calculated. I am using the AWR1843Boost.

I understand, that the angle of a single reflector can be calculated using the formula phi = arcsin(omage / pi), where omega is the phase shift between the signal of two adjacent receive antennas. This works fine as long as I have only a single reflector in a given range/velocity bin. However, if I have multiple reflectors "located" in the same range/velocity bin I first have to perform the  "angle-FFT" to separate the different phase-shift components. For the AWR1843EVM I have a maximum of 8-virtual antennas that I can use. This allows me to separate objects with a phase shift larger than delta_omega = (2pi / 8). Due to the nature of the fft I now only have 8 different (fixed) values for the possible phase shifts that I am able to observe. Using the above mentioned formula (phi = arcsin(omage / pi)) I can now only distinguish if an object is located around -49deg, -30deg, -14deg, 0deg, 14deg, 30deg or 49deg.

However the accuracy in the demo example seems to be way higher than the resolution. Even when multiple object fall into the same range/velocity - bin. Can you tell me how this i being done? 

Kind regards

Mario 

  • 0
    TI__Guru*** 149044 points

    Hi

    Please let us know where this information is documented

    Thank you

    cesar

    However the accuracy in the demo example seems to be way higher than the resolution. 

  • 0 in reply to Cesar
    Prodigy 85 points

    Hi Cesar,

    thanks for your response. As far as I know, this information is not documented anywhere. This is more an observation from several experiments i have made.  The following picture may help to clarify what i mean.

    In the picture you see three different snapshots, recorded with the "mmWave Demo Visualizer". In each snapshot two objects are placed in the same range bin. The range resolution was set to approx. 20cm. The reference object is placed at 0 degrees. The other object, which I marked with the red circle, is placed at approx. 15, 22 and 30 degrees. As you can see, the object is clearly located at three different locations within the scatter plot. However, if the accuracy would be equal to the resolution, the object would only be able to be located at two different location, i.e. 15 and 30 degrees. 

    To me it seems like an additional step must be taken after calculating the angle-fft, to locate the object with a higher accuracy. Maybe through beam-forming or general interpolation of the corresponding phase shift. Can you tell me how this is done for the demo?

    Kind regards 

    Mario 

  • 0 in reply to Mario
    TI__Guru*** 149044 points

    Thank you for the additional details.

    Please give me a few days to discuss with the systems team

    Thank you

    Cesar

  • 0 in reply to Cesar
    Prodigy 85 points

    Ok great, thank you very much.

  • 0 in reply to Mario
    TI__Guru*** 149044 points

    Hi,

    I have reviewed your question with systems team and I think there is some confusion.

    The angular resolution is the capability two separate 2 objects in the same range bin.

    The angular accuracy is the difference between the physical position (angle) of an object and the computed position (angle) of the object.

    So, in the pictures you shared there is no problem detecting the object at different locations. The only challenge would be detecting two objects in the same range bin at an angle smaller than approx 15 deg

    thank you

    Cesar

  • 0 in reply to Cesar
    Prodigy 85 points

    Hi Cesar,

    thanks very much for your response. 

    No, i dont think there is any confusion. I totally understand the difference between accuracy and resolution. I try to rephrase the question:

    Say you have two objects at that are located at the same range with the same radial velocity. Say the objects are placed at the (azimuth) angles of -20 degrees and +40degrees. A possible output of the angle-FFT may look as follows:

    My question is, how does the OOB Demo compute the angles from here? 

    Thanks in advance.

    Kind regards 

    Mario

  • +1 in reply to Mario
    TI__Guru*** 149044 points

    Hi,

    Please see this snapshot from an older version of the SDK that shows how the angle (x,Y) was computed.

    Kmax is the index of the peak in the angle FFT. If there are two objects in the same range bin there will be two peaks in the angle FFT.

    This is the extent of support our team can provide for this question. For more information please refer to the source code and SDK documentation.

    Tu understand the basics of angle computation you may want to download the mmwave SDK 2.1. This older SDK supports AWR1642. The source code is simpler to understand because it was implemented only on the DSP and also the documentation may be more detailed.

    C:\ti\mmwave_sdk_02_01_00_04\docs

    thank you

    Cesar

    SDK_2_1_X_Y_estimation.pdf

  • 0 in reply to Cesar
    Prodigy 85 points

    Hi Cesar,

    thanks very much for your reply. The pdf does not quite answer my question, but i will work through the code as you suggested. Thanks for your advice.

    Kind regards

    Mario 

  • 0 in reply to Mario
    TI__Guru*** 149044 points

    Thank you!

    Cesar