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Diffstat (limited to 'doc/thesis/chapters/theory.tex')
| -rw-r--r-- | doc/thesis/chapters/theory.tex | 4 |
1 files changed, 3 insertions, 1 deletions
diff --git a/doc/thesis/chapters/theory.tex b/doc/thesis/chapters/theory.tex index 50759bb..54668d0 100644 --- a/doc/thesis/chapters/theory.tex +++ b/doc/thesis/chapters/theory.tex @@ -300,6 +300,8 @@ where the odd order of the filter \(N\) should satisfy the condition N = 2 \lfloor \tau \rfloor + 1 \end{equation} for a minimal error in the approximation \cite{Valimaki1995}. It is worth mentioning that it is also possible to build FIR filters of even length with a different condition, or ones 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}. +Where the sinc function does not select an integer sample, which in turn means that the other sampled values do not add up to zero. +Thus, they will be distributed among the other whole numbers. \begin{figure} \centering @@ -396,7 +398,7 @@ Extending the previous NLOS case, if there is a line of sight (LOS) path (red si \frac{1}{\sqrt{N}}\sum_{n=1}^{N} e^{j \vartheta_k} \right). \end{equation} -Notice that by letting \(K = 0\), that is, no power in the LOS path, \eqref{eqn:mult-fading-los} becomes \eqref{eqn:mult-fading-nlos} or Rayleigh distributed (as expected). Conversely when \(K \to \infty\), i.e. no power in the NLOS paths, then \(f \to 1\) and the fading disappears. The new amplitude density in this case is: +Notice that by letting \(K = 0\), that is, no power in the LOS path, \eqref{eqn:mult-fading-los} becomes \eqref{eqn:mult-fading-nlos} or Rayleigh distributed (as expected). For a factor \(K = 5.1\) the probability function is gaussian distributed. Conversely when \(K \to \infty\), i.e. no power in the NLOS paths, then \(f \to 1\) and the fading disappears. The new amplitude density in this case is: \begin{equation} p(a)= 2a(1+K) \exp{\left(-K -a^2 (K+1) \right)} I_0 \left(2a\sqrt{K(1+K)} \right), \end{equation} |