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Original question:

IWR6843ISK: How can the vertical angular resolution be greater than the FoV?

IWR6843ISK: Calculating the antenna separation using the elevation FoV

Cagri Tanriover
Intellectual 885 points
Part Number: IWR6843ISK

Hi,

I watched the video on angular resolution and FoV a few times and with the understanding I have, I decided to work backward from the elevation FoV (i.e. 30 degrees) for IWR6843ISK.

I have included the diagrams I used and my calculations below. My understanding is a FoV = 30 degrees means +/-15 degrees which also means theta = 15 degrees is the maximum I should use in the formula below. This indicates an antenna separation d = 9.65 mm in the elevation direction is what I should see in the virtual array. d = 9.65 mm means roughly twice the wavelength, however, I would have expected to obtain a 'd' of half a wavelength as indicated on the virtual array diagram I included below.

Can you please provide me with guidance on where my calculation is going wrong?

Thank you for your help.

Regards,

Cagri

P.S. I performed the same calculation for the azimuth FoV as well (i.e. theta(max) = 60 degrees). My calculation suggested the antenna separation 'd' should be half a wavelength while the virtual array azimuth separation is one wavelength.

(Please note the azimuth virtual array separation has been corrected as half lambda (see blue annotation) after Sami's comment. Thanks Sami !)

  • 0
    TI__Expert 4010 points

    Hi Cagri,

    The azimuth separation for the virtual antenna array is lambda/2 and not lambda.  The same is for the elevation plane.

    Regards,

    Sami

  • 0 in reply to Sami Mardini
    Intellectual 885 points

    Hi Sami,

    Thank you for the note on the virtual azimuth antenna separation. I agree it needs to be half a wavelength. This means my calculation for azimuth gives the correct answer.

    However, the difference I highlighted in the elevation separation (i.e. 2 lambda versus half lambda) is still not clear. Do you have an explanation for that difference? The same calculation with the half lambda separation you have mentioned does not work in the elevation direction.

    Regards,

    Cagri

  • 0 in reply to Cagri Tanriover
    TI__Expert 4010 points

    Hi Cagri,

    Can you share the video you are referencing?

    Regards,

    Sami

  • 0 in reply to Sami Mardini
    Intellectual 885 points

    Sure Sami:

    https://training.ti.com/intro-mmwave-sensing-fmcw-radars-module-5-angle-estimation?context=1128486-1139153-1128546

    take a look at time stamp 8:28 for the information I was referring to.

    Regards,
    Cagri

  • 0 in reply to Cagri Tanriover
    TI__Expert 4010 points

    Hi Cagri,

    The antenna FOV that you are using in your calculation is the actual individual antenna FOV.  Every antenna has a FOV based on its geometry.  Typical it is the 6 dB drop in antenna gain from boresight.

    Regards,

    Sami

  • 0 in reply to Sami Mardini
    Intellectual 885 points

    Hi Sami.

    It sounds like there are multiple FoV definitions not covered in the technical video. I am confused a bit with your last comment.

    In the video, the FoV is computed using two antenna. Therefore, my understanding was in order to calculate the 30 deg of elevation FoV (i.e. +/- 15 deg in each direction), I would have to use any of the following virtual antenna pairs (as indicated on my original figure above) 3 and 5 OR 4 and 6 OR 7 and 9 OR 8 and 10. Given that the separation between any of these pairs of antenna are half a lambda, my calculation above should have given half a lambda rather than 2 lambda, which is what I cannot understand.

    When I repeat the same calculation in the azimuth direction (using the +/- 60 degrees FoV in each direction), the antenna separation is calculated as half lambda as expected.

    I still cannot understand why the calculation does not work for the elevation dimension. Which dimensions should I be using so that the calculation works in the elevation direction as well? In other words, which physical antennas are used to calculate the elevation FoV of 30 degrees? For clarification, can you please provide a step by step calculation for this as I have?

    Thanks for your help.

    Regards,

    Cagri

  • 0 in reply to Cagri Tanriover
    TI__Expert 4010 points

    Hi Cargi,

    The IWR6843ISK antenna has FOV of +/-60 deg in azimuth plane and +/- 15 in the elevation plane.  This FOV is for each series patch antenna (RX and TX).  in the video they talk about theta max which is the highest when d = lambda/2.  That is why we choose lambda/2 antenna separation.

    Regards,

    Sami

  • +1 in reply to Sami Mardini
    Intellectual 885 points

    Hi Sami,

    Let me summarize what I have understood.

    The FoV (azimuth -/+60 deg and elevation -/+ 15 deg) have nothing to do with the calculation explained in the video. These FoV values are purely a function of the antenna geometry and are defined as the angles from boresight where the 6 dB drop in the antenna occur.

    The FoV explained in the video is a theoretical calculation that tells us what the antenna separations physically need to be in order to maximize the FoV (potentially all the way up to -/+90 degrees either in azimuth or elevation). However, this maximum FoV cannot be achieved as the antenna geometries are the limiting factors in the first place. We can only measure the actual FoV (i.e. azimuth -/+60 deg and elevation -/+ 15 deg) based on the 6 dB drop and cannot calculate what it really is (based on the guidance in the video)

    Is the above an accurate summary? Please let me know in case I missed anything.

    Regards,

    Cagri

  • +1 in reply to Cagri Tanriover
    TI__Expert 4010 points

    Hi Cagri,

    I like your summary and it is all good.

    Regards,

    Sami