Antennas coupling

Discussion in 'Wireless & RF Design' started by thumb2, Jan 2, 2017.

  1. thumb2

    Thread Starter Member

    Oct 4, 2015
    There is something I don't understand from a solved exercise of antennas coupling and I hope someone could help me to understand.

    This is not an exercise I have to solve.

    Given two dipoles each aligned on the z axis and placed at a distance 0.1\lambda on the x axis so

    \Psi = 0.1(2\pi) \sin(\theta)\cos(\varphi)

    the array factor given by the example is:

    F_A(\Psi) = 1 - 0.77\exp(\mathrm i(\Psi + 0.67))

    and the maximum of radiation given by the exercise is

    F_A(\Psi) = 1.09 for \theta = \pi/2 and \varphi = 0.

    The exercise explicitly says that the result is given calculating F_A(\Psi) for the angles just above.
    But when I try to solve for \varphi, I fall into a math error:

    <br />
\begin{align}<br />
& 0.1(2\pi)\sin(\theta)\cos(\varphi) + 0.67 = 0\\<br />
& \cos(\varphi) = -1.066<br />
\end{align}<br />

    since \arccos(.) is not defined for -1.066.
    So what's the problem with my math ? :confused:

    Thank you.
  2. thumb2

    Thread Starter Member

    Oct 4, 2015
    Perhaps I understand how the example comes to this solution ...



    and its derivative is:


    Now, equating it to 0, for definition we obtain the local maxima an minima in the range [0, π].
    Obviously the derivative is 0 for φ = 0 and φ = π, so for φ = 0 we have the maxima, while the minima for φ = π..
    Last edited: Jan 11, 2017