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author | sara <sara.halter@gmx.ch> | 2021-12-16 16:43:31 +0100 |
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committer | sara <sara.halter@gmx.ch> | 2021-12-16 16:43:31 +0100 |
commit | 2c0214f3e8d7ff15cacb2acd5f9c04308d5c86ec (patch) | |
tree | 64f3404258bf42d826c74c1debad0c5bd7693e84 /doc/thesis/chapters/theory.tex | |
parent | Correct parameters for every flowgraph (diff) | |
download | Fading-2c0214f3e8d7ff15cacb2acd5f9c04308d5c86ec.tar.gz Fading-2c0214f3e8d7ff15cacb2acd5f9c04308d5c86ec.zip |
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-rw-r--r-- | doc/thesis/chapters/theory.tex | 12 |
1 files changed, 6 insertions, 6 deletions
diff --git a/doc/thesis/chapters/theory.tex b/doc/thesis/chapters/theory.tex index 6cf3520..d8dd696 100644 --- a/doc/thesis/chapters/theory.tex +++ b/doc/thesis/chapters/theory.tex @@ -378,7 +378,7 @@ i.e. the amplitude of \(f\) is \emph{Raileigh} 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)}. + 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 \vartheta_k }. \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. @@ -389,13 +389,13 @@ For this distribution model the expectation value for the real part is \(\E{\Re{ 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)}) + p(a)= 2a(1+K)e^{(-K-{a}^2(K+1))}\cdot I_0(2a\sqrt{K(1+K)}) \end{equation} Where \(I_0\) the zero ordered modified besselfunction represent. -\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) -\end{equation} +%\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) +%\end{equation} \skelpar[4] |