# Channel model: LOS Component and Rician K-factor

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 ?

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?

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

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.