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Angular resolution when mounted over head

Krishna Kumar4
Expert 1565 points

Part Number: IWR6843ISK-ODS
Other Parts Discussed in Thread: IWR6843ISK, AWR1642

I refer to this link  for my understanding of the term "Angular resolution" http://dev.ti.com/tirex/content/mmwave_industrial_toolbox_3_1_1/.metadata/.html/antennas.html

Scene setup

IWR6843ISK-ODS is mounted on the ceiling at a height of 3m.

Since this board has 120 degree field of view in both azimuth and elevation, it simplifies my question.

2 Objects of equal size, for ex: 2 stationary cube shaped boxes of same size and reflective property are placed on the floor. From the field of fiew of the radar, these two boxes are placed in a straight line along one of the axis (X).

Now as per the link above the azimuth angular resolution of IWR6843ISK in Azimuth plane is 15 degrees. And let us assume the azimuth plane is covering the X axis. If so to identify these two boxes as two separate

items.

the centroid of these two boxes have to be separated by (tan 15)*3m = 0.87m (3m is the distance from ceiling to the floor) => is my theory correct?

If the above is correct then if I apply the same concept the distance of separation to identify two objects along the elevation plane is (tan 58) * 3m = 4.8m => as angular resolution is 58 degrees on the elevation

plane

These appear to be large values, considering the fov of the radar from that height may not even extend to these distances. Can some one please clarify what is the correct equation that should be used to calculate the separation

distance between two stationary objects for a MIMO radar when angular resolution is given in degrees.

  • 0
    TI__Mastermind 45825 points
    HI,

    First IWR6843ISK and IWR6843ISK-ODS are different in antenna pattern, and therefore they are different in FOV and angle resolution.
    1) IWR6843ISK has about 120 degree FOV in azimuth, but 30 degree FOV in elevation. And it has 15 degree angle resolution in azimuth, and 58 degree of angle resolution in elevation.
    1) IWR6843ISK has about 120 degree FOV in both azimuth and elevation. But it has about 30 degree angle resolution in azimuth, and 30 degree of angle resolution in elevation.

    Radar is more sensitive to separate targets in range and Doppler. Only two targets that are the same in range and same in Doppler , needs to separate using angle resolution. So, very often, we should be able to separate two static boxes if they are slightly different in range (check the range resolution). If they are absolutely the same in range, then we need to consider separate them in angle domain.

    Now back to your calculation, if the angle resolution is 15 degree, and the two boxes are placed symmetrically to x = zero. Then theoretically, they can be separated if they are away in x by 2*tand(15/2)*3 = 0.7899m. For a angle resolution of 58 degree, you will get 2*tand(58/2)*3 = 3.3m.

    And that is theoretical analysis, practically, if one reflect much more energy than the other, it will be harder to detect the weaker signal. In addition, the DOA algorithm you develop has to be able to support multiple target detection in the same range bin. And sometimes, it is a trade-off between false detection from side-lobe vs. miss detection from a weaker target.

    Hope it helps,
    Zigang
  • 0 in reply to zigang Yang
    Expert 1565 points
    I am getting what you are implying. Could you please provide me with angular resolution for the awr1642-ODS antenna pattern in elevation and azimuth planes, I couldn't find it in the spruik1.pdf

    For static object identification with similar range, let me look in to how it has been implemented on ODS
  • 0 in reply to Krishna Kumar4
    TI__Mastermind 45825 points
    IN general, the angle resolution can be estimated as 2/M*180/pi
    So, if you have 4 antenna element in azimuth angle, it will be 2/4*180/pi = 28.65

    For AWR1642 ODS antenna, we have 4 element in azimuth and 3 element in the elevation, so you get about 28.65 degree angle resolution in azimuth, and 38 degree in elevation.


    Best,
    Zigang
  • 0 in reply to zigang Yang
    Expert 1565 points
    I am aware of 2/(N or M) = resolution angle in radians, but in case of IWR6843ISK - the resolution angle in elevation is around 58 degrees that means N =2. Even though it has 3 Tx and 4Rx in total, for elevation only N=2 and for azimuth N=8 (resulting in 14.32 deg =15deg). I refer to Page 5 of swra554a.pdf

    However in ODS there are only 2 Tx and 4 RX and they are placed in some angle, I don't understand how the matrix is formed to get the 4 elements in Azimuth and 3 elements in elevation. Could you kindly draw the array for my understanding.
  • 0 in reply to Krishna Kumar4
    TI__Mastermind 45825 points

    For IWR6843ISK, the visual antenna pattern is listed below, so N = 8 for azimuth and N = 2 for elevation.

    ........o..o..o..o

    o..o..o..o..o..o..o..o

    For IWR1642-ODS, the visual antenna pattern is listed below, so N = 4 for azimuth and N = 3 for elevation.

    o..o..o..o

    o..o

    o..o

    Best,

    Zigang

  • 0 in reply to zigang Yang
    Expert 1565 points
    Dear Zigang

    Many thanks for taking your time to resolve my query.