[usth/ICT2.10] Communicate mobile networks
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\documentclass[a4paper,12pt]{article}
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\usepackage[english,vietnamese]{babel}
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\usepackage{amsmath}
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\usepackage{booktabs}
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\usepackage{circuitikz}
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\usepackage{enumerate}
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\usepackage{lmodern}
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\usepackage{mathtools}
|
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\usepackage{pgfplots}
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\usepackage{siunitx}
|
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\usepackage{textcomp}
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\usepackage{tikz}
|
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\usetikzlibrary{arrows,automata}
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|
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\newcommand{\N}{\mathcal N}
|
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\newcommand{\ud}{\,\mathrm{d}}
|
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\newcommand{\SIR}{\mathrm{SIR}}
|
||||
\newcommand{\baud}{\mathrm{Bd}}
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\newcommand{\bit}{\mathrm{b}}
|
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\newcommand{\chip}{\mathrm{c}}
|
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\newcommand{\problem}[1]{\noindent\textbf{#1.}}
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|
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\title{Mobile Wireless Communication}
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\author{Nguyễn Gia Phong}
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\date{Spring 2020}
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\begin{document}
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\maketitle
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\setcounter{section}{1}
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\section{Characteristics of Radio Environment}
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\subsection{Propagation Models}
|
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\problem 1 Consider radio waves propagating by two-slope model over the
|
||||
distance under \SI{200}{\metre} in Orlando. The average receive power is given by
|
||||
\[\bar P_R = g(d) P_T G_T G_R\]
|
||||
|
||||
Assume the attenna gains are both 1 and apply the inverse variation of power
|
||||
with distance for two-slope model we get
|
||||
\begin{align*}
|
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&\bar P_R = d^{-n_1}\left(1 + \frac{d}{d_b}\right)^{-n_2} P_T\\
|
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\iff &P_T = d^{n_1}\left(1 + \frac{d}{d_b}\right)^{n_2}\bar P_R
|
||||
\end{align*}
|
||||
|
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Substituting $d_b = \SI{90}{\metre}$, $n_1 = 1.3$ and $n_2 = 3.5$ gives us
|
||||
\[P_T = d^{1.3}\left(1 + \frac{d}{90}\right)^{3.5}\bar P_R\]
|
||||
|
||||
With average power effect experienced, $P_{R,\si{\deci\bel}} = 10\lg P_R
|
||||
= 10\lg \bar P_R$ and $P_{T,\si{\deci\bel}} = 10\lg P_T$ and thus
|
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\begin{align*}
|
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&P_{T,\si{\deci\bel}} - P_{R,\si{\deci\bel}}
|
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= 13\lg d + 35\lg\frac{d + 90}{90}\\
|
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\iff &P_{R,\si{\deci\bel}} - P_{T,\si{\deci\bel}}
|
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= 35\lg\frac{90}{d + 90} - 13\lg d
|
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\end{align*}
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\pagebreak
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|
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This is plotted in the figure below
|
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\begin{center}
|
||||
\begin{tikzpicture}
|
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\begin{axis}[
|
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xlabel={$d$ (\si{\metre})},
|
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ylabel={$P_{R,\si{\deci\bel}}-P_{T,\si{\deci\bel}}$ (\si{\deci\bel})}]
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\addplot[domain=0:200]{35*ln(90/(x+90))/ln(10) - 13*ln(x)/ln(10)};
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\end{axis}
|
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\end{tikzpicture}
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\end{center}
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|
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\problem 2 Consider a log-normal shadow fading propagation, the receive power
|
||||
is given by \[P_R = \sqrt[10]{10^X} g(d) P_T G_T G_R\] where $X$ is a zero-mean
|
||||
normal random variable with STD $\sigma = \SI{6}{\deci\bel}$.
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\begin{enumerate}[(a)]
|
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\item Given $\bar P_R = \SI{1}{\milli\watt}$ at $d = \SI{100}{\metre}$.
|
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\[P_R > \bar P_R \iff \sqrt[10]{10^X} > 1 \iff X > 0\]
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Since $X$ is zero-mean and normally distributed, the probability
|
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the received power at a mobile at that distance from the base station
|
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will exceed \SI{1}{\milli\watt} is \SI{50}{\percent}, and so is the
|
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probability it is less than \SI{1}{\milli\watt}.
|
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|
||||
\item Let $Y = X/\sigma$, $Y \sim \N(0, 1)$
|
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and $F_X(x) = \Phi(X/\sigma) = \Phi(X/6)$.
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|
||||
The probability a mobile has an acceptable received signal at
|
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\SI{10}{\milli\watt} or higher is
|
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\begin{align*}
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P\left(P_R \ge 10\bar P_R\right)
|
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&= P\left(\sqrt[10]{10^X} \ge 10\right) = P(X \ge 10)\\
|
||||
&= 1 - F_X(10) = 1 - \Phi\left(\frac{10}{6}\right) = \SI{4.78}{\percent}
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\end{align*}
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|
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\item For $\sigma = \SI{10}{\deci\bel}$, $F_X(x) = \Phi(X/10)$.
|
||||
The probability a mobile has an acceptable received signal at
|
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\SI{10}{\milli\watt} or higher is
|
||||
\[P\left(P_R \ge 10\bar P_R\right) = 1 - F_X(10)
|
||||
= 1 - \Phi(1) = \SI{15.87}{\percent}\]
|
||||
|
||||
\item If the lower limit for an acceptable received signal is
|
||||
\SI{6}{\milli\watt}, with $\sigma = 6$, the probability a received signal
|
||||
is acceptable is
|
||||
\begin{align*}
|
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P\left(P_R \ge 6\bar P_R\right)
|
||||
&= P\left(\sqrt[10]{10^X} \ge 6\right)
|
||||
= P\left(X \ge \lg{6^{10}}\right)\\
|
||||
&= 1 - F_X\left(\lg{6^{10}}\right)
|
||||
= 1 - \Phi\left(\frac{\lg{6^{10}}}{6}\right) = \SI{9.73}{\percent}
|
||||
\end{align*}
|
||||
With $\sigma = 10$, the probability a received signal is acceptable is
|
||||
\[P\left(P_R \ge 6\bar P_R\right) = 1 - F_X\left(\lg{6^{10}}\right)
|
||||
= 1 - \Phi\left(\frac{\lg{6^{10}}}{10}\right) = \SI{21.82}{\percent}\]
|
||||
\end{enumerate}
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|
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\subsection{Random Channel Characterization}
|
||||
Given $x(t) = e^t * (\Pi(t - 1) - \Pi(t - 3))$ and $h(t) = \delta(t - 1)$,
|
||||
where $\Pi$ is the rectangular function:
|
||||
\[\Pi(t) = \begin{dcases}
|
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0, &\text{if }|t| > \frac{1}{2}\\
|
||||
\frac{1}{2}, &\text{if }|t| = \frac{1}{2}\\
|
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1, &\text{if }|t| < \frac{1}{2}\\
|
||||
\end{dcases}\]
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||||
|
||||
The convolution sum of $x$ and $h$ is
|
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\begin{align*}
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y(t) &= x(t - 1)\\
|
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&= \int_{-\infty}^\infty e^{t-z-1}(\Pi(z-2) - \Pi(z-4))\ud z\\
|
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&= \int_{-\infty}^\infty e^{t-z-1}\Pi(z-2)\ud z
|
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- \int_{-\infty}^\infty e^{t-z-1}\Pi(z-4)\ud z\\
|
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&= \int_{1.5}^{2.5} e^{t-z-1}\ud z
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- \int_{3.5}^{4.5} e^{t-z-1}\ud z\\
|
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&= e^{t-1}\left(e^{2.5} - e^{1.5} - e^{4.5} + e^{3.5}\right)
|
||||
\end{align*}
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|
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\subsection{Fading}
|
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\problem 1 Consider several delay spreeds $D$ of \SI{0.5}{\micro\second},
|
||||
\SI{1}{\micro\second} and \SI{6}{\micro\second}.
|
||||
|
||||
\begin{itemize}
|
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\item For IS-95 and cdma2000 which uses the transmission bandwidth of
|
||||
\SI{1.25}{\mega\hertz}, their symbol interval is \SI{0.8}{\micro\second}.
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For the multipath rays to be resolvable, the delay spread must be
|
||||
greater than this (\SI{1}{\micro\second} and \SI{6}{\micro\second}).
|
||||
\item For WCDMA which uses the bandwidth of \SI{5}{\mega\hertz},
|
||||
the symbol interval is \SI{0.2}{\micro\second}, thus symbols
|
||||
are resolvable in all cases.
|
||||
\end{itemize}
|
||||
|
||||
\problem 2 Indicate the condition for flat fading for each of the following
|
||||
data rates with transmission in binary form: \SI{8}{kbps}, \SI{40}{kbps},
|
||||
\SI{100}{kbps}, \SI{6}{Mbps}.
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||||
|
||||
Assume information is transmitted in rectangular waves, the symbol interval
|
||||
are \SI{125}{\micro\second}, \SI{25}{\micro\second}, \SI{10}{\micro\second}
|
||||
and \SI{1/6}{\micro\second} respectively. For flat fading to occur,
|
||||
the delay spread must be significantly less than the symbol interval.
|
||||
Since no data is provided or found, no conclusion is drawn on which
|
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radio environments would result in flat fading for each of these data rates.
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|
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\section{Cellular Concept}
|
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\subsection{Channel Allocation}
|
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\problem 1 Assume the simplest path-loss model of $g(d) = d^{-3}$, calculate
|
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down-link SIR at point P at the corner of a hexagonal cell in a 3-reuse case.
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|
||||
Using to path-loss model, the signal-to-interference ratio
|
||||
can be approximated from the six first-tier interferers as follows
|
||||
\[\SIR \approx \frac{1}{\left(\frac{R}{D-R}\right)^3
|
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+ \left(\frac{R}{D+R}\right)^3
|
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+ 4\left(\frac{R}{D}\right)^3}\]
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||||
|
||||
In a 3-reuse case, $D = \sqrt{3C}R = 3R$, and thus
|
||||
\[\SIR \approx \frac{1}{\left(\frac{R}{2R}\right)^3
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||||
+ \left(\frac{R}{4R}\right)^3
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+ 4\left(\frac{R}{3R}\right)^3} = \frac{1728}{499}\]
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|
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\problem 2 Calculate the worst-case uplink SIR assuming the co-channel
|
||||
interference is caused only by the closest interfering mobiles in radio cells
|
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a distance $D = 3.46R$ away from the cell. Assume the simplest path-loss model
|
||||
of $g(d) = d^{-4}$, the signal-to-interference ratio is approximated by
|
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\[\SIR \approx \frac{P_t/R^4}{6P_t/\left(\frac{3D}{4}\right)^4}
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= \frac{(3D/4)^4}{6R^4}\]
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|
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With $D = 3.46R$ (4-reuse), this becomes
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\[\SIR \approx \frac{(3\cdot3.46/4)^4}{6} = 7.56\]
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|
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\subsection{Erlang-B Formula and Sizing a Cell}
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\problem 1 An user who makes a call attempt every 15 minutes, with each call
|
||||
lasts an average of 2 minutes, generate the load of 2/15 erlangs.
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|
||||
\problem 2 Consider a mobile system supporting 832 frequency channels
|
||||
and 7-reuse, there are over 118 channels per cell. With the probility of
|
||||
call blocking of $P_B \le \SI{1}{\percent}$, the traffic is around 101 erlangs.
|
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Given the average call-holding time $h = \SI{200}{\second}$, the arrival rate
|
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can be calculated to be $\lambda = \SI{0.505}{\mathrm{calls}\per\second}$.
|
||||
Since an user makes a call every \SI{900}{\second} on average, there are
|
||||
approximately 454.5 users. As the density of mobile terminals is
|
||||
\SI{2}{\mathrm{terminals}\per\square{\kilo\metre}}, the area is
|
||||
\SI{227.25}{\square{\kilo\metre}}, which indicates a cell radius
|
||||
of $R = \SI{9.35}{\kilo\metre}$, assuming a hexagonal topology.
|
||||
|
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\section{Modulation Techniques}
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\problem 1 Consider communication system operating at the transmission
|
||||
bandwidth of \SI{1}{\mega\hertz} with the rolloff factor of 0.25.
|
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\begin{itemize}
|
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\item Achievable data traffic rate is
|
||||
\[R_s = \frac{B}{1 + \beta} = \frac{10^6}{1 + 0.25}
|
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= \SI{800}{\kilo\baud\per\second}\]
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||||
\item Delay spread that no ISI occurs is much less than the symbol interval,
|
||||
which is $T = B^{-1} = \SI{1}{\micro\second}$.
|
||||
\item Using OFDM with $N = 16$ equally spead carriers, for each subcarrier,
|
||||
$\Delta f = \SI{62.5}{\kilo\hertz}$, $R_s = \SI{50}{\kilo\baud\per\second}$
|
||||
and $T = \SI{16}{\micro\second}$.
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\item Additionally use 16-QAM, the bit rate is
|
||||
$R_\bit = \SI{800}{\kilo\bit\per\second}$.
|
||||
\end{itemize}
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|
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\problem 2 Given $B = \SI{1}{\mega\hertz}$, $\beta = 0.25$,
|
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$R_\bit = \SI{4.8}{\mega\bit\per\second}$ and $T = \SI{25}{\micro\second}$.
|
||||
|
||||
$R_s = B/(1+\beta) = \SI{0.8}{\mega\baud\per\second}$, thus 64-QAM is used.
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|
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For OFDM, $N = R_s/\Delta f = R_\bit T \approx 128$.
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|
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\problem 3 Consider a transmission of bandwidth $B = \SI{2}{\mega\hertz}$,
|
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where phase-shift keying and Nyquist rolloff shaping is used.
|
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|
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For rolloff factors of 0.2, 0.25, 0.5, the traffic rates are respectively
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\SI{1.67}{\mega\baud\per\second}, \SI{1.6}{\mega\baud\per\second} and
|
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\SI{1.33}{\mega\baud\per\second}.
|
||||
|
||||
In order to transmit at a rate of $R_\bit = \SI{6.4}{\mega\bit\per\second}$
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||||
when $\beta = 0.25$, 16-QAM should be used.
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|
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\problem 4 Given the input sequence 1001111010
|
||||
and the following QPSK signal pairs
|
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\begin{center}
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\begin{tabular}{c c c}
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\toprule
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Successive Signal & $a_i$ & $b_i$\\
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\midrule
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0 0 & $-1$ & $-1$\\
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0 1 & $-1$ & $+1$\\
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1 0 & $+1$ & $-1$\\
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1 1 & $+1$ & $+1$\\
|
||||
\bottomrule
|
||||
\end{tabular}
|
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\end{center}
|
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|
||||
Let the carrier frequency be some multiple of 1/T
|
||||
|
||||
\begin{tikzpicture}
|
||||
\begin{axis}[scale only axis, width=0.8\textwidth, height=0.16\textwidth,
|
||||
xlabel=In-phase Carrier, xtick={0,1,2,3,4,5},
|
||||
xticklabels={0,T,2T,3T,4T,5T}, ymin=-2, ymax=2, samples=420]
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||||
\addplot[domain=0:5]{cos(x*720)};
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
|
||||
\begin{tikzpicture}
|
||||
\begin{axis}[scale only axis, width=0.8\textwidth, height=0.16\textwidth,
|
||||
xlabel=Quadrature-phase Carrier, xtick={0,1,2,3,4,5},
|
||||
xticklabels={0,T,2T,3T,4T,5T}, ymin=-2, ymax=2, samples=420]
|
||||
\addplot[domain=0:5]{sin(x*720)};
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
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|
||||
The output QPSK signal would then be
|
||||
|
||||
\begin{tikzpicture}
|
||||
\begin{axis}[scale only axis, width=0.8\textwidth, height=0.16\textwidth,
|
||||
xlabel=In-phase Component, xtick={0,1,2,3,4,5},
|
||||
xticklabels={0,T,2T,3T,4T,5T}, ymin=-2, ymax=2, samples=69]
|
||||
\addplot[domain=0:1]{+cos(x*720)};
|
||||
\addplot[domain=1:2]{-cos(x*720)};
|
||||
\addplot[domain=2:3]{+cos(x*720)};
|
||||
\addplot[domain=3:4]{+cos(x*720)};
|
||||
\addplot[domain=4:5]{+cos(x*720)};
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
|
||||
\begin{tikzpicture}
|
||||
\begin{axis}[scale only axis, width=0.8\textwidth, height=0.16\textwidth,
|
||||
xlabel=Quadrature-phase Component, xtick={0,1,2,3,4,5},
|
||||
xticklabels={0,T,2T,3T,4T,5T}, ymin=-2, ymax=2, samples=69]
|
||||
\addplot[domain=0:1]{-sin(x*720)};
|
||||
\addplot[domain=1:2]{+sin(x*720)};
|
||||
\addplot[domain=2:3]{+sin(x*720)};
|
||||
\addplot[domain=3:4]{-sin(x*720)};
|
||||
\addplot[domain=4:5]{-sin(x*720)};
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
|
||||
\begin{tikzpicture}
|
||||
\begin{axis}[scale only axis, width=0.8\textwidth, height=0.16\textwidth,
|
||||
xlabel=Output Signal, xtick={0,1,2,3,4,5},
|
||||
xticklabels={0,T,2T,3T,4T,5T}, ymin=-2, ymax=2, samples=69]
|
||||
\addplot[domain=0:1]{+cos(x*720)-sin(x*720)};
|
||||
\addplot[domain=1:2]{-cos(x*720)+sin(x*720)};
|
||||
\addplot[domain=2:3]{+cos(x*720)+sin(x*720)};
|
||||
\addplot[domain=3:4]{+cos(x*720)-sin(x*720)};
|
||||
\addplot[domain=4:5]{+cos(x*720)-sin(x*720)};
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
|
||||
\section{Multiple Access Techniques}
|
||||
\subsection{Time-Division Multiple Access}
|
||||
Transmission bit rate is the rate at which the bits are transmitted, while
|
||||
the user information bit rate is the rate at which per data are transmitted.
|
||||
|
||||
In particular, GSM gives each time slot \SI{576.92}{\micro\second},
|
||||
minus \SI{30.46}{\micro\second} guard time. During this duration,
|
||||
\SI{148}{\bit} are tramsmitted, thus the transmission bit rate is
|
||||
$\SI{148}{\bit}/(\SI{576.92}{\micro\second}-\SI{30.46}{\micro\second})
|
||||
= \SI{270.834}{\kilo\bit\per\second}$. Of the \SI{148}{\bit},
|
||||
\SI{114}{\bit} are data bits. Furthermore, only one slot per GSM eight-slot
|
||||
frame and 24 out of 26 frames are used to carry information. Therefore,
|
||||
the user bit rate is $\SI{114}{\bit}/\SI{4.615}{\milli\second}\cdot 24/26
|
||||
= \SI{22.8}{\kilo\bit\per\second}$.
|
||||
|
||||
Similarly, IS-136 has the transmission bit rate of $\SI{1944}{\bit}
|
||||
/ \SI{40}{\milli\second} = \SI{48.6}{\kilo\bit\per\second}$ and $\SI{520}{\bit}
|
||||
/ \SI{40}{\milli\second} = \SI{13}{\kilo\bit\per\second}$.
|
||||
|
||||
\subsection{Code-Division Multiple Access}
|
||||
Consider IS-95 with the bit rate of \SI{9.6}{\kilo\bit\per\second}
|
||||
and the chip rate of \SI{1.2288}{\mega\chip\per\second}, the speading gain
|
||||
is 128 chips per bit.
|
||||
|
||||
\section{Channel Coding Techniques}
|
||||
|
||||
\subsection{Block Coding}
|
||||
Consider the generator matrix
|
||||
\[\mathbf G = [\mathbf I_k \mathbf P] = \begin{pmatrix}
|
||||
1&0&0&0&1&0&1\\
|
||||
0&1&0&0&1&1&1\\
|
||||
0&0&1&0&1&1&0\\
|
||||
0&0&0&1&0&1&1
|
||||
\end{pmatrix}\]
|
||||
it is trivial that $n = 7$, $k = 4$ and
|
||||
\[\mathbf P = \begin{pmatrix}
|
||||
1&0&1\\
|
||||
1&1&1\\
|
||||
1&1&0\\
|
||||
0&1&1
|
||||
\end{pmatrix}\]
|
||||
|
||||
The parity check matrix is then given by
|
||||
\[\mathbf H = [\mathbf P^T \mathbf I_{n-k}] = \begin{pmatrix}
|
||||
1&0&0&1&1&1&0\\
|
||||
0&1&0&0&1&1&1\\
|
||||
0&0&1&1&1&0&1
|
||||
\end{pmatrix}\]
|
||||
|
||||
\subsection{Convolutional Coding}
|
||||
Consider a $K = 3$, rate \textonehalf{} convolution encoder with generators
|
||||
$g_1 = [101]$ and $g_2 = [011]$.
|
||||
\begin{center}
|
||||
\begin{circuitikz}
|
||||
\draw (0,3) node (input) {input}
|
||||
(1,3) node[inputarrow] (in) {}
|
||||
(2,3) node[circ] (m1) {}
|
||||
(3.5,3) node[twoportshape] (port1) {}
|
||||
(5,3) node[circ] (m2) {}
|
||||
(6.5,3) node[twoportshape] (port2) {}
|
||||
(8,3) node[circ] (m3) {}
|
||||
(input) -- (in) -- (m1) -- (port1) -- (m2) -- (port2) -- (m3)
|
||||
|
||||
(5,5.5) node[xor port, rotate=90] (xor1) {}
|
||||
(9,6) node[flowarrow] (out1) {}
|
||||
(10,6) node{$n_1$}
|
||||
(m1) |- (xor1.in 1)
|
||||
(m3) |- (xor1.in 2)
|
||||
(xor1.out) |- (out1)
|
||||
|
||||
(6.5,0.5) node[xor port, rotate=270] (xor2) {}
|
||||
(9,0) node[flowarrow] (out2) {}
|
||||
(10,0) node{$n_2$}
|
||||
(m3) |- (xor2.in 1)
|
||||
(m2) |- (xor2.in 2)
|
||||
(xor2.out) |- (out2);
|
||||
\end{circuitikz}
|
||||
\end{center}
|
||||
|
||||
Initialize the encoder with 01, we get the following state diagram
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[->,>=latex,shorten >=1pt,auto,node distance=42mm]
|
||||
\node[initial,state] (01) {01};
|
||||
\node[state] (00) [above right of=01] {00};
|
||||
\node[state] (11) [below right of=01] {11};
|
||||
\node[state] (10) [below right of=00] {10};
|
||||
|
||||
\path (00) edge [dashed, loop above] node {00} (00)
|
||||
edge node {10} (10)
|
||||
(01) edge [dashed] node {11} (00)
|
||||
edge [bend left] node {01} (10)
|
||||
(10) edge [dashed, bend left] node {01} (01)
|
||||
edge node {11} (11)
|
||||
(11) edge [dashed] node {10} (01)
|
||||
edge [loop below] node {00} (11);
|
||||
|
||||
\node [below of=10] {%
|
||||
\begin{tabular}{c c}
|
||||
\raisebox{2pt}{\tikz{\draw[dashed] (0,0) -- (10mm,0);}} & 0\\
|
||||
\raisebox{2pt}{\tikz{\draw (0,0) -- (10mm,0);}} & 1
|
||||
\end{tabular}};
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
Given the input bit sequence of 10011011, the output would be
|
||||
0101111011100111.
|
||||
\end{document}
|
After Width: | Height: | Size: 181 KiB |
After Width: | Height: | Size: 34 KiB |
After Width: | Height: | Size: 32 KiB |
After Width: | Height: | Size: 32 KiB |
After Width: | Height: | Size: 366 KiB |
After Width: | Height: | Size: 313 KiB |
|
@ -0,0 +1,439 @@
|
|||
\documentclass[pdf]{beamer}
|
||||
\usepackage[english,vietnamese]{babel}
|
||||
\usepackage{amsmath}
|
||||
\usepackage{booktabs}
|
||||
\usepackage{graphicx}
|
||||
\usepackage{hyperref}
|
||||
\usepackage{lmodern}
|
||||
\usepackage{siunitx}
|
||||
|
||||
\mode<presentation>{}
|
||||
\usetheme[hideothersubsections]{Hannover}
|
||||
\usecolortheme{crane}
|
||||
\usefonttheme[onlymath]{serif}
|
||||
\usebackgroundtemplate{
|
||||
\includegraphics[width=\paperwidth,height=\paperheight]{USTH.jpg}}
|
||||
\renewcommand{\thefootnote}{\fnsymbol{footnote}}
|
||||
\setcounter{tocdepth}{2}
|
||||
|
||||
\title{Satellite Internet}
|
||||
\author[Group 1]{Nguyễn Như Hiếu---BI9-103\\
|
||||
Ngô Ngọc Đức Huy---BI9-119\\
|
||||
Ngô Xuân Minh---BI9-167\\
|
||||
Nguyễn Gia Phong---BI9-184\\
|
||||
Nguyễn Hồng Quang---BI9-194\\
|
||||
Trần Minh Vương---BI9-239}
|
||||
\institute{University of Science and Technology of Hà Nội}
|
||||
\date{\selectlanguage{english}\today}
|
||||
|
||||
\begin{document}
|
||||
\frame{\titlepage}
|
||||
\selectlanguage{english}
|
||||
\begin{frame}{Contents}
|
||||
\tableofcontents
|
||||
\end{frame}
|
||||
|
||||
\section{Introduction}
|
||||
\frame{\tableofcontents[currentsection]}
|
||||
\begin{frame}{Usage}\Large
|
||||
\begin{block}{Popular Use}
|
||||
\begin{itemize}
|
||||
\item Airplane
|
||||
\item Cruise Ship
|
||||
\item Rural Area
|
||||
\end{itemize}
|
||||
\end{block}
|
||||
\begin{block}{Similarity}
|
||||
All three are either in or travel through area \\
|
||||
with little to no ground station.
|
||||
\end{block}
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}{Future Usage}\Large
|
||||
Provide Internet for the whole world
|
||||
|
||||
\begin{block}{Fact}
|
||||
Over 3.7 Billion people are living without being \\
|
||||
connected to the internet.
|
||||
\end{block}
|
||||
\end{frame}
|
||||
|
||||
\section{How It Works}
|
||||
\frame{\tableofcontents[currentsection]}
|
||||
\begin{frame}{Components}
|
||||
\begin{itemize}\Large
|
||||
\item Geostationary satellite (GEO)
|
||||
\item Gateway
|
||||
\item Antenna
|
||||
\item Others:
|
||||
\begin{itemize}\Large
|
||||
\item Modem
|
||||
\item Centralized NOC
|
||||
\end{itemize}
|
||||
\end{itemize}
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}{Components Interaction}
|
||||
\begin{figure}
|
||||
\includegraphics[width=0.9\textwidth]{A-GPS.png}
|
||||
\caption{GPS using A-GPS and GSM network}
|
||||
\end{figure}
|
||||
\end{frame}
|
||||
|
||||
|
||||
\begin{frame}{One-way Satellite Network}
|
||||
\Large
|
||||
\begin{center}
|
||||
\includegraphics[width=\textwidth]{one-way-to-earth.png}
|
||||
\end{center}
|
||||
\end{frame}
|
||||
|
||||
\subsection{1-way from Earth}
|
||||
\begin{frame}{Components}\large
|
||||
\begin{itemize}
|
||||
\item Upstream: Data travelling through telephone modem
|
||||
\item Downstream: Download through satellite
|
||||
\end{itemize}
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}{Characteristics}
|
||||
\begin{itemize}
|
||||
\item Upload speed: Same as that of the dial-up internet
|
||||
\item Download speed: Much faster than dial-up internet
|
||||
\item Latency: Still high,much lower than two way satellite internet
|
||||
\item You have to tie up the telephone lie when you use the Internet
|
||||
\end{itemize}
|
||||
\end{frame}
|
||||
|
||||
\subsection{1-way to Earth}
|
||||
\begin{frame}{One-way to Earth}\Large
|
||||
\begin{block}{Components}
|
||||
\begin{itemize}
|
||||
\item 1 transmitting hub station (usually very large)
|
||||
\item Multiple receive-only Earth stations
|
||||
\end{itemize}
|
||||
\end{block}
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}{Characteristics}\Large
|
||||
\begin{itemize}
|
||||
\item Usage: IP multicast-based data,\\
|
||||
audio and video distribution
|
||||
\item Interactivity: Little user interface,\\
|
||||
similar to TV or radio content
|
||||
\end{itemize}
|
||||
\end{frame}
|
||||
|
||||
\subsection{2-way}
|
||||
\begin{frame}{Two-way Satellite Network}
|
||||
\begin{center}
|
||||
\includegraphics[width=0.8\textwidth]{two-way.png}
|
||||
\end{center}
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}{Components}\LARGE
|
||||
\begin{itemize}
|
||||
\item VSAT: Send and receive data
|
||||
\item Telecommunication port:\\ Relay data through Internet
|
||||
\end{itemize}
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}{Condition}\LARGE
|
||||
Satellite dish must be precisely pointed\\
|
||||
to avoid interference.
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}{Characteristics}\large
|
||||
\begin{itemize}
|
||||
\item Both TDMA and single channel per carrier
|
||||
\item Mostly Ku-band, but also C-band and Ka-band
|
||||
\item May utilize telephone modem to reduce latency
|
||||
\item Home-user's bandwidth based on payment
|
||||
\item Difficult on moving vehicles
|
||||
\end{itemize}
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}{Portable Satellite Internet}
|
||||
\begin{block}{Portable}
|
||||
\begin{itemize}
|
||||
\item Use self-contained box pointed in general direction of Satellite
|
||||
\item Expensive
|
||||
\end{itemize}
|
||||
\end{block}
|
||||
\begin{block}{Satellite phone}
|
||||
\begin{itemize}
|
||||
\item Omnidirectional antenna so no alignment needed
|
||||
\item Low bandwidth so slow to browse net,useful for sending email
|
||||
\end{itemize}
|
||||
\end{block}
|
||||
\end{frame}
|
||||
|
||||
\section{Limits and Challenges}
|
||||
\frame{\tableofcontents[currentsection]}
|
||||
|
||||
\subsection{Weather}
|
||||
\begin{frame}{Heavy rain or Blizzard}\LARGE
|
||||
\begin{itemize}
|
||||
\item Fading
|
||||
\item Accumulating raindrop or snow
|
||||
\item Wind
|
||||
\end{itemize}
|
||||
\end{frame}
|
||||
|
||||
\subsection{Latency}
|
||||
\begin{frame}{Latency}\large
|
||||
\begin{block}{Satellite altitude}
|
||||
\begin{itemize}
|
||||
\item LEO: $<$ \SI{2000}{\kilo\meter}
|
||||
\item MEO: 2000--\SI{35786}{\kilo\meter}
|
||||
\item GEO: $>$ \SI{35786}{\kilo\meter}
|
||||
\end{itemize}
|
||||
\end{block}
|
||||
\begin{block}{Result}
|
||||
GEO has 12 times higher latency than terrestrial base networks.
|
||||
LEO and MEO have a bit lower delay.
|
||||
\end{block}
|
||||
\end{frame}
|
||||
|
||||
\subsection{Others}
|
||||
\begin{frame}{Other Limitations}\Large
|
||||
\begin{block}{Economically}
|
||||
Costly: \SI{2}{\mega b\per\second} costs around \$100 a month.
|
||||
\end{block}
|
||||
\begin{block}{Environmentally}
|
||||
Space junk: Only 2000 out of 5000 launched satellites are still in function.
|
||||
\end{block}
|
||||
\end{frame}
|
||||
|
||||
\section{Mitigations}
|
||||
\frame{\tableofcontents[currentsection]}
|
||||
\subsection{Techniques}
|
||||
\begin{frame}{Fade Mitigation Techniques}\Large
|
||||
Common functions:
|
||||
\begin{itemize}
|
||||
\item \emph{Monitor} link quality by continuous measurements
|
||||
\item \emph{Predict} short-term behavior and duration\\
|
||||
of satellite channel's next state
|
||||
\item \emph{Set} parameters based on previous estimation
|
||||
\end{itemize}
|
||||
\end{frame}
|
||||
|
||||
\subsubsection{EIRP Control Techniques}
|
||||
\begin{frame}{Effective Isotropic Radiated Power}\Large
|
||||
\begin{itemize}
|
||||
\item EIRP = tranmitted power $\times$ antenna gain
|
||||
\item EIRP control = adjusting carrier power
|
||||
or antenna gain to compensate for power losses
|
||||
\end{itemize}
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}{Power Control System}
|
||||
\begin{enumerate}\large
|
||||
\item Open loop: Based on recently received power.
|
||||
\begin{itemize}\large
|
||||
\item Non-reliable
|
||||
\item Responsive
|
||||
\end{itemize}
|
||||
\item Closed loop: Based on channel power measurements.
|
||||
\begin{itemize}\large
|
||||
\item More comprehensive
|
||||
\item Large propagation delay
|
||||
\end{itemize}
|
||||
\end{enumerate}
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}{Uplink Power Control}
|
||||
\begin{itemize}
|
||||
\item Vary carrier power at the earth station
|
||||
\item Restoration of side lobes might lead to\\
|
||||
adjacent \emph{channel} interference\\
|
||||
\includegraphics[width=0.54\textwidth]{lobes.png}
|
||||
\item Increase of earth station transmit power may cause
|
||||
adjacent \emph{satellite} interference\footnote{Satellites
|
||||
are separated by 2--3 degrees on the geostationary orbit.\\}
|
||||
\item Effective and preferred by many satellite operators
|
||||
\end{itemize}
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}{Downlink Power Control}
|
||||
\begin{itemize}\Large
|
||||
\item Vary carrier power on-board the satellite
|
||||
\item Difficult to implement due to\\ satellite size and weight limitations
|
||||
\item Subject to
|
||||
\begin{enumerate}\large
|
||||
\item Adjacent \emph{channel} interference
|
||||
\item Inter\emph{modulation} interference
|
||||
\item Inter\emph{system} interference (with terrestrial networks)
|
||||
\end{enumerate}
|
||||
\end{itemize}
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}{Spot Beam Shaping}\Large
|
||||
\begin{itemize}
|
||||
\item Adjust antenna gain on-board the satellite\\
|
||||
for a certain geographical region
|
||||
\item Shape satellite antenna for nearly constant\\
|
||||
ground receive power, even under rainfall
|
||||
\item Does \textbf{not} need expensive calculations\\
|
||||
for attenuation estimation\footnote{SBS compensates
|
||||
the entire coverage area instead of a single site.\\}
|
||||
\item Technology and research are WIP
|
||||
\end{itemize}
|
||||
\end{frame}
|
||||
|
||||
\subsubsection{Adaptive Transmission Techniques}
|
||||
\begin{frame}{Adaptive Transmission Techniques}
|
||||
\begin{itemize}\Large
|
||||
\item Modify processing/transmission\\ manner of signals
|
||||
\item Resource-shared techniques
|
||||
\item Categories:
|
||||
\begin{enumerate}\large
|
||||
\item Hierarchical coding
|
||||
\item Hierarchical modulation
|
||||
\item Data rate reduction
|
||||
\end{enumerate}
|
||||
\end{itemize}
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}{Hierarchical Coding}\large
|
||||
\begin{itemize}
|
||||
\item Add redundancy to the information signal
|
||||
\item Trade-off between bandwidth and error probability
|
||||
\item Different conditions require different coding schemes
|
||||
\item Prioritize users with less efficient coding schemes, i.e.\\
|
||||
longer bursts (TDMA) or larger bandwidth (FDMA)
|
||||
\end{itemize}
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}{Hierarchical Modulation}\large
|
||||
\begin{itemize}
|
||||
\item Provide lower quality fallback in case of weak signals
|
||||
\item Exchange bandwidth efficiency for power requirements
|
||||
\item Suitable for localized satellite systems, e.g. VSAT
|
||||
\item Users with lower-order modulation get more resources
|
||||
\end{itemize}
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}{Data Rate Reduction}\large
|
||||
\begin{itemize}
|
||||
\item Reduce information data rate for power gain
|
||||
\item Distribute satellite resources equally to every user
|
||||
\item Utilizable where significant information rate reduction\\
|
||||
is tolerable, e.g.\ video or data but voice transmission
|
||||
\end{itemize}
|
||||
\end{frame}
|
||||
|
||||
\subsubsection{Diversity Protection Schemes}
|
||||
\begin{frame}{Diversity Protection Schemes}\large
|
||||
\begin{itemize}
|
||||
\item Use multiple channels with different characteristics
|
||||
\item Oriented against rain fades and highly efficient
|
||||
\item Performance criteria
|
||||
\begin{itemize}
|
||||
\item Diversity gain: difference between site attenuation\\
|
||||
and joint attenuation, for the same probability level
|
||||
\item Diversity improvement: ratio of site exceedence probability
|
||||
to the joint one, for the same attenuation value
|
||||
\end{itemize}
|
||||
\end{itemize}
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}{Diversity Techniques}
|
||||
\begin{tabular}{l p{0.39\textwidth} l p{0.18\textwidth}}
|
||||
\toprule
|
||||
\textbf{Diversity} & \textbf{Setup} & \textbf{Efficiency} & \textbf{Cost}\\
|
||||
\midrule
|
||||
Site & Connected earth stations & High & High\\
|
||||
Orbital & Earth station may choose\newline between satellites & Low & Low\\
|
||||
Frequency & Use lower frequency\newline on higher attenuation
|
||||
& Adaptive & Terrestrial equipments\\
|
||||
Time & Repeat faded data & Selective\footnote{\ldots of fade duration} & N/A\\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\end{frame}
|
||||
|
||||
\subsection{Comparison}
|
||||
\begin{frame}{EIRP Control Techniques}
|
||||
\begin{table}
|
||||
\begin{tabular}{l l p{0.27\textwidth} p{0.2\textwidth}}
|
||||
\toprule
|
||||
\textbf{Tech} & \textbf{Availability}
|
||||
& \textbf{Max gain} (dB) & \textbf{Cons} \\
|
||||
\midrule
|
||||
ULPC & 0.01--10 \% & 5 (VSAT)\newline 15 (hubs)& power range \\
|
||||
DLPC & 0.01--10 \% & 3 (sat.~TWTA) & power range \\
|
||||
SBS & 0.01--1 \% & 5 (sat.~antenna) & immature research \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\caption{Comparisons between EIRP control techniques}
|
||||
\end{table}
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}{Adaptive Transmission Techniques}
|
||||
\begin{table}
|
||||
\begin{tabular}{l l p{0.25\textwidth} p{0.23\textwidth}}
|
||||
\toprule
|
||||
\textbf{Tech} & \textbf{Availability}
|
||||
& \textbf{Max gain} (dB) & \textbf{Cons} \\
|
||||
\midrule
|
||||
HC/HM & 0.01--10 \% & 10--15\newline ($E_b/N_0$ range)
|
||||
& fading in \newline many stations \\
|
||||
DDR & 0.01--10 \% & 3--9 & low rate\newline intolerant \\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\caption{Comparisons between adaptive transmission techniques}
|
||||
\end{table}
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}{Diversity Protection Schemes}
|
||||
\begin{table}
|
||||
\begin{tabular}{l l p{0.25\textwidth} p{0.24\textwidth}}
|
||||
\toprule
|
||||
\textbf{Tech} & \textbf{Availability}
|
||||
& \textbf{Max gain} (dB) & \textbf{Cons} \\
|
||||
\midrule
|
||||
SD & 0.001--0.1 \% & 10--30\newline (conv.~rain) & cost \\
|
||||
OD & 0.001--1 \% & 3--10 & satellite switch \\
|
||||
FD & 0.01--10 \% & 30 (Ka--Ku) & cost\\
|
||||
\bottomrule
|
||||
\end{tabular}
|
||||
\caption{Comparisons between diversity protection schemes}
|
||||
\end{table}
|
||||
\end{frame}
|
||||
|
||||
\section{Conclusion}
|
||||
\frame{\tableofcontents[currentsection]}
|
||||
\begin{frame}{Conclusion}\LARGE
|
||||
\begin{itemize}
|
||||
\item Have many potentials
|
||||
\item Challenging
|
||||
\item Need more research
|
||||
\end{itemize}
|
||||
\end{frame}
|
||||
|
||||
\begin{frame}{References}
|
||||
\begin{thebibliography}{69}
|
||||
\setbeamertemplate{bibliography item}[article]
|
||||
\bibitem{KuKaV} Athanasios D.~Panagopoulos,\\
|
||||
Pantelis-Daniel M.~Arapoglou and Panayotis G.~Cottis.\\
|
||||
``Satellite communications at Ku, Ka, and V bands:
|
||||
Propagation impairments and mitigation techniques''.\\
|
||||
\emph{Communications Surveys \& Tutorials}, vol.~6, p.~2--14.\\
|
||||
IEEE, 2004. doi:10.1109/COMST.2004.5342290.
|
||||
\setbeamertemplate{bibliography item}[online]
|
||||
\bibitem{wiki} Satellite Internet access. \emph{Wikipedia}.
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\end{thebibliography}
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\includegraphics[width=0.2\textwidth]{CC.png}
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\includegraphics[width=0.2\textwidth]{BY.png}
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\includegraphics[width=0.2\textwidth]{SA.png}
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\end{center}
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This work is licensed under a
|
||||
\href{https://creativecommons.org/licenses/by-sa/4.0/}{Creative Commons
|
||||
Attribution-ShareAlike 4.0 International License}.
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