0
$\begingroup$

I am asking about Rician K- Factor in this model:

http://web.stanford.edu/group/introstwc/Course%20Notes/lect3.pdf

I know that if $K=0$ then it is a pure Rayleigh fading, if $K= \infty$, then non fading Channel.

What is it, if $0<K<\infty$ ?

If i represet the NLOS component using the Kronecker model and take expected value from $\mathbf{H}$, then I get only loscomponent. Will be it right?

$\endgroup$
  • 2
    $\begingroup$ For $0 < K < \infty$, the channel is a combination of both a deterministic component (i.e., LOS) and a fading component. As the $K$-factor is the ratio of the energy in the deterministic Line-of-Sight (LOS) component to the energy in the aggregation of the random scattered paths (i.e., the fading component), higher $K$ means that the channel is more deterministic. As for the 2nd part of your question, I'm not sure to understand what you meant. $\endgroup$ – anpar Feb 28 at 14:34
  • $\begingroup$ @anpar That is an answer, not a comment! $\endgroup$ – Dilip Sarwate Feb 28 at 15:26
  • $\begingroup$ OK, let me post that then. $\endgroup$ – anpar Feb 28 at 15:29
1
$\begingroup$

For $0 < K < \infty$, the channel is a combination of both a deterministic component (i.e., LOS) and a fading component.

As the $K$-factor is the ratio of the energy in the deterministic Line-of-Sight (LOS) component to the energy in the aggregation of the random scattered paths (i.e., the fading component), higher 𝐾 means that the channel is more deterministic.

$\endgroup$
  • $\begingroup$ thanks. the second question: If i represet the NLOS component using the Kronecker model and take expected value from H, Did i get then only LOS-component? $\endgroup$ – nani Mar 4 at 7:40

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.