diff options
Diffstat (limited to '')
| -rw-r--r-- | doc/thesis/chapters/theory.tex | 23 |
1 files changed, 19 insertions, 4 deletions
diff --git a/doc/thesis/chapters/theory.tex b/doc/thesis/chapters/theory.tex index 376e799..3262751 100644 --- a/doc/thesis/chapters/theory.tex +++ b/doc/thesis/chapters/theory.tex @@ -358,6 +358,7 @@ for some parameter \(\sigma\). Loosely speaking, the distribution needs to be `` \label{fig:multipath-statistical-models} } \end{figure} +\skelpar[4]{Explain This formulars} %TODO :Maby some correction on the descreption, because mentionet earlyer Is f(t) =(c k (t) parameters)? \paragraph{NLOS case} @@ -380,13 +381,27 @@ It can also explaint with I- and Q-contribution : \skelpar[4] \paragraph{LOS case} -In the case of the Rician distribution model. The line of side exist, which means that one of the paths have a straight communication line from the transmitter to the reviser. -It can be said that a Rayleight distribution is the same as a Rician distribution with a factor K =0. -For a faktor K= 5.1 the probability function is gaussien distributed. +In the case of the Ricean distribution model the line of side exist, which means that one of the paths have a straight communication line from the transmitter to the reviser. So there are in addition to the Rayleight model direct components, whish are also gaussian distributed. + +\begin{equation} \label{eqn:rician fading} + f(t) = \sqrt{\frac{K}{K+1}}+\lim_{N\rightarrow\infty}\frac{1}{\sqrt{K+1}} \frac{1}{\sqrt{N}}\sum_{n=1}^{N} e^{j(\Theta +2\pi jf t)}. +\end{equation} + +The factor \(K\) named Ricean factor it is the ratio of the line of side power to the average power of the distributed components. +The Phase for the strait line component has no influences for the Random process therefore there set to zero. In the case when \(K = 0 \) +the Rician distribution becomes a Rayleight distribution on the other hand when \(K\rightarrow \infty \) the distribution becomes an AWGN-channel model (additive white Gaussian noise). When \(K > 0 \) is the phase not equally distributed. + +For this distribution model the expectation value for the real part is \(\E{\Re{f(t)}}=\sqrt{\frac{K}{K+1}} \) and for the imaginary part \(\E{\Im{f(t)}}=0\) + +So the probability function of the amplitude in this case is: +\begin{equation} \label{eqn:rician_fading_probabilety_dencety} + p(a)= 2a(1+K)\exp{(-K-{a}^2(K+1))}\cdot I_0(2a\sqrt{K(1+K)}) +\end{equation} + +Where \(I_0\) the zero ordered modified besselfunction represent. -\skelpar[4]{Explain statistical model with Rician distribution.} \begin{equation} \Re{h_l(n)}, \Im{h_l(n)} \sim \mathcal{N} \left( \frac{A_l}{\sqrt{2}}, \frac{1}{2} \sigma_l^2 \right) |