Antennas coupling

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

Oct 4, 2015
93
6
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:


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

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

Thank you.

Oct 4, 2015
93
6
Perhaps I understand how the example comes to this solution ...

Given:

image=https://latex.codecogs.com/gif.latex?\dpi{120}&space;\begin{aligned}&space;f(\varphi)&space;=&space;1&space;-&space;0.77\,\mathrm{e}^{\mathrm{i}\,(0.62\cos(\varphi)&space;+&space;0.67)}&space;=&space;1&space;-&space;0.77\,\mathrm{e}^{\mathrm{i}\,(0.62\cos(\varphi))}\mathrm{e}^{\mathrm{i}\,0.67}&space;\qquad&space;\theta&space;=&space;(\pi/2)&space;\end{aligned}&hash=1e6791df0a7498d267243bf999869ca6

and its derivative is:

$image=https://latex.codecogs.com/gif.latex?\frac{\mathrm{d}f(\varphi)}{\mathrm{d}\varphi}&space;=&space;\mathrm{i}\,0.77(0.628)\sin(\varphi)\,\mathrm{e}^{\mathrm{i}\,(0.62\cos(\varphi))}\mathrm{e}^{\mathrm{i}\,0.67}&hash=f1cc15a6a363faa7241fe325b42df20a$

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