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Q on antenna imaging effect

  1. Oct 24, 2008 #1
    Hi everyone

    Currently Im reading through antenna imaging effect. Im aware of positive and negative imagaing (essentially the polarity of the antenna is constant/reversed depending on the antenna orientation)

    However the illustrative example my lecturer has given me has caused quite a bit of confusion. For the curious, I have attached it as 2 separate doucments, however I will now try to describe it.

    1) He has a vertically oriented dipole situated [tex]\lambda/2[/tex] above the ground plane. Due to the imaging effect, a positive image of this antenna is produced [tex]\lambda/2[/tex] below the ground plane.

    2) The dipoles are replaced by point sources (shown as dots in the attachments)

    3) From the dots he draws rays being emitted at an angle of [tex]\theta[/tex] to the horizontal. He also draws an imaginary line from the origin parallel to both rays. This creates a scenario whereby the ray from the topmost point leads the origin by a distance of [tex]\(lambda/2)*sin(theta) [/tex]and the ray from the bottmomost point lags the origin by the same distance

    4) Now at the far-field point, he converts the lagging and leading distances to angles by multiplying them by [tex](2*pi)/lambda[/tex] and subsequently adding the field vectors. From this he gets his result as [tex](2*cos(pi*sin(\theta))[/tex], however I have no idea how this arises, due to the orthogonal nature of the vectors (as seen in the attachment)

    Once again, the attachments show the problem in better detail. Any help on how he gets the result would be highly appreciated


    Attached Files:

  2. jcsd
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