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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?

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    $\begingroup$ 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, 2019 at 14:34
  • $\begingroup$ @anpar That is an answer, not a comment! $\endgroup$ Feb 28, 2019 at 15:26
  • $\begingroup$ OK, let me post that then. $\endgroup$
    – anpar
    Feb 28, 2019 at 15:29

1 Answer 1

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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.

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  • $\begingroup$ 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, 2019 at 7:40

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