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diff --git a/doc/thesis/chapters/theory.tex b/doc/thesis/chapters/theory.tex index 3262751..a9bea7d 100644 --- a/doc/thesis/chapters/theory.tex +++ b/doc/thesis/chapters/theory.tex @@ -282,7 +282,6 @@ is different from \eqref{eqn:multipath-impulse-response} consider again the plot From a signal processing perspective \eqref{eqn:discrete-multipath-impulse-response} can be interpreted as a simple tapped delay line, schematically drawn in \figref{fig:tapped-delay-line}, which confirms that the presented mathematical model is indeed a FIR filter. Simple multipath channels can be simulated with just a few lines of code, for example the data for the static fading channel in \figref{fig:multipath-frequency-response-plots} is generated in just four lines of Python. The difficulty of fading channels in practice lies in the estimation of the constantly changing parameters \(c_k(t)\) and \(\tau_k(t)\). - \subsection{Simulating multipath CIR with FIR filters} \label{sec:fractional-delay} % TODO quelle: http://users.spa.aalto.fi/vpv/publications/vesan_vaitos/ch3_pt1_fir.pdf @@ -324,17 +323,10 @@ where the odd order of the filter \(N\) should satisfy the condition for a minimal error in the approximation. It is worth mentioning that it is also possible to build FIR filters of even length with a different condition, or that do not satisfy \eqref{eqn:fractional-fir-length}, in which cases more consideration is required. An example of a fractional delay FIR filter is shown in \figref{fig:fractional-delay-sinc-plot}. \subsection{Statistical model} \label{sec:statistical-model} -%TODO: Quelle? -Because as mentioned earlier it is difficult to estimate the time-dependent parameters of \(h_l(m)\) in many cases it is easier to model the components of the CIR as stochastic processes, thus greatly reducing the number of parameters. This is especially effective for channels that are constantly changing, because by the central limit theorem the cumulative effect of many small changes tends to a normal distribution. - -Recall that \(h_l(m)\) is a function of time because \(c_k\) and \(\tau_k\) change over time. The idea of the statistical model is to replace the cumulative change caused by \(c_k\) and \(\tau_k\) (which are difficult to estimate) by picking the next CIR sample \(h_l(m +1)\) from a \emph{circularly symmetric complex Gaussian distribution}, or concisely written as -\begin{equation} - h_l \sim \mathcal{CN}(0, \sigma^2) -\end{equation} -for some parameter \(\sigma\). Loosely speaking, the distribution needs to be ``circular'' because \(h_l\) is a complex number, which is a ``two dimensional number'', it can however be understood as \(\Re{h_l} \sim \mathcal{N}(0, \sigma^2)\) and \(\Im{h_l} \sim \mathcal{N}(\mu, \sigma^2)\), i.e. having each component being normally distributed. +Because as mentioned earlier it is difficult to estimate the time-dependent parameters of \(h_l(m)\) in many cases it is easier to model the components of the CIR as stochastic processes, thus greatly reducing the number of parameters \cite{Messier,Mathis}. This is especially effective for channels that are constantly changing, because by the central limit theorem the cumulative effect of many small changes tends to a normal distribution. -%TODO : Picture of gaussian distribution +\skelpar[3]{Assumptions of the model} \begin{subequations} \begin{align} R_{l} (k) &= \E{h_l(m) h_l^*(m+k)}, \\ @@ -342,6 +334,27 @@ for some parameter \(\sigma\). Loosely speaking, the distribution needs to be `` \end{align} \end{subequations} +\paragraph{NLOS case} + +Recall that \(h(\tau, t)\) is a function of time because \(c_k\) and \(\tau_k\) change over time. The idea of the statistical model is to replace the cumulative change caused by \(c_k\) and \(\tau_k\) (which are difficult to estimate) with a single random variable \(f\). This is done as follows. + +Multipath fading is a form of multiplicative noise, as mathematically confirmed by the fact that convolving a complex baseband signal \(e^{j\omega_c t}\) with the fading CIR \(h(\tau, t)\) gives +\begin{equation} + e^{j\omega_c \tau} * h(t, \tau) = \sum_k c_k(t) e^{j\omega_c(\tau - \tau_k(t))} + = e^{j\omega_c \tau} \sum_k c_k(t) e^{-j\omega_c \tau_k(t)} + = e^{j\omega_c \tau} \cdot f(t). +\end{equation} +If there is no line of sight (NLOS), it is reasonable to assume that all path have more or less the same attenuation, i.e. all \(c_k\) are the same. Another reasonable assumption in this case is that all paths are equally likely to be taken, or in other words the delays \(-\omega_c \tau_k\) can be replaced with random variables \(\vartheta_k\) that are uniformly distributed on \([0,2\pi)\) \cite{Hoher2013,Mathis}. Finally, assuming that there are infinitely many paths the random variable for the multiplicative fading noise becomes +\begin{equation} + f = \lim_{N\rightarrow\infty} \frac{1}{\sqrt{N}} + \sum_{k=1}^{N} e^{j \vartheta_k }, +\end{equation} +where the \(c_k\) where omitted, since they are assumed to be all equal. The factor \(1/\sqrt{N}\) is introduced such that \(\expectation \{|f|^2\} = 1\). It then can be shown that the probability density function of \(|f|\) is +\begin{equation} + p(a)= 2a e^{-a^2}, \text{ or } |f| \sim \mathcal{R}, +\end{equation} +i.e. the amplitude of \(f\) is \emph{Raileigh} distributed. + \begin{figure} \centering \begin{subfigure}{.45\linewidth} @@ -360,25 +373,6 @@ for some parameter \(\sigma\). Loosely speaking, the distribution needs to be `` \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} - - In the case of the Rayleight distribution the signal has no line of sight. It can be looked at the Fading process \cite{Hoher2013} with the help of the amplitude \(a(t)\) in time and the associated phase \(\Theta(t) \in[\,0,2\pi)\,\). To find the probability function of the amplitutes it can be looked at the fading possess as a superimposition of those infinity distribute signals: -\begin{equation} \label{eqn:rayleight fading} - f(t) = \lim_{N\rightarrow\infty} \frac{1}{\sqrt{N}}\sum_{n=1}^{N} e^{j(\Theta +2\pi jf t)}. - \end{equation} -whish are nominate with the factor \(\frac{1}{\sqrt{N}}\) so that the power probability is \(\E{|f(t)|²}=1\) as mentioned earlier. From the fact that we are looking at the complex basiband and this processes is an independent one with a gaussian distributions it can be said that \(\E{f(t)}=0\) and so the propabiliti of the amplitude is: - -\begin{equation} \label{eqn:rayleight_fading_probabilety_dencety} - p(a)= 2a\exp{{-a}^2} -\end{equation} - -It can also explaint with I- and Q-contribution : -\begin{equation} - \Re{h_l(n)}, \Im{h_l(n)} - \sim \mathcal{N} \left(0, \frac{1}{2} \E{|h_l(n)|^2} \right) -\end{equation} -\skelpar[4] \paragraph{LOS case} |