0
\$\begingroup\$

In an article on Wiki, there is a plot that shows a positive reflection coefficient. How is this possible? I can't seem to see how you can get a positive one from this equation alone:

$$ \Gamma_= \frac{sin(\theta)-X}{sin(\theta) + X} $$ $$ X_{horiz}= \sqrt{\epsilon_c - cos(\theta)^2} $$ $$ X_{vert}= \frac{\sqrt{\epsilon_c - cos(\theta)^2}}{\epsilon_c} $$

Image in question: plot

http://en.wikipedia.org/wiki/File:Reflection_co-efficient_ground_reflection_of_radio_waves.jpg

Article with the image: http://en.wikipedia.org/wiki/2-Ray_Ground_Reflection_Model

I tried plotting it in MATLAB w.r.t theta, and I don't see the plateau, nor do I see the positive reflection coefficients. I'd appreciate some clarification.

\$\endgroup\$
  • \$\begingroup\$ I think you want to look at is as the magnitude of the reflection coefficient being less than or equal to 1. A coefficient of 1 means everything is reflected back if I remember right. -1 means the reflected wave is inverted? This is covered in books on time varying EM fields. \$\endgroup\$ – HL-SDK Nov 30 '13 at 2:52
1
\$\begingroup\$

Reflection coefficients are limited to the range -1 to +1 for passive systems, so it is certainly possible to get a positive one.

In this case, when theta is near 90 degrees, you get (1 - X) / (0 + x). If X is less than 1, this will be positive.

Over what range did you plot theta? The horizontal axis of that graph is the log of the sepatation distance. The left side corresponds to a theta of about 90 degrees and the right side corresponds to a theta of around 0 degrees.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.