From 743056f7631e1af196f040ca2099eb95bb44b83c Mon Sep 17 00:00:00 2001 From: Nguyễn Gia Phong Date: Sat, 18 Apr 2020 16:20:12 +0700 Subject: [usth/ICT2.10] Communicate mobile networks --- usth/ICT2.10/homework/solutions.pdf | Bin 0 -> 190401 bytes usth/ICT2.10/homework/solutions.tex | 405 +++++++++++++++++++++++++++ usth/ICT2.10/midterm/A-GPS.png | Bin 0 -> 185117 bytes usth/ICT2.10/midterm/BY.png | Bin 0 -> 34691 bytes usth/ICT2.10/midterm/CC.png | Bin 0 -> 32442 bytes usth/ICT2.10/midterm/SA.png | Bin 0 -> 33152 bytes usth/ICT2.10/midterm/USTH.jpg | Bin 0 -> 375099 bytes usth/ICT2.10/midterm/lobes.png | Bin 0 -> 320391 bytes usth/ICT2.10/midterm/main.pdf | Bin 0 -> 1628832 bytes usth/ICT2.10/midterm/main.tex | 439 ++++++++++++++++++++++++++++++ usth/ICT2.10/midterm/one-way-to-earth.png | Bin 0 -> 149364 bytes usth/ICT2.10/midterm/two-way.png | Bin 0 -> 292038 bytes 12 files changed, 844 insertions(+) create mode 100644 usth/ICT2.10/homework/solutions.pdf create mode 100644 usth/ICT2.10/homework/solutions.tex create mode 100644 usth/ICT2.10/midterm/A-GPS.png create mode 100644 usth/ICT2.10/midterm/BY.png create mode 100644 usth/ICT2.10/midterm/CC.png create mode 100644 usth/ICT2.10/midterm/SA.png create mode 100644 usth/ICT2.10/midterm/USTH.jpg create mode 100644 usth/ICT2.10/midterm/lobes.png create mode 100644 usth/ICT2.10/midterm/main.pdf create mode 100644 usth/ICT2.10/midterm/main.tex create mode 100644 usth/ICT2.10/midterm/one-way-to-earth.png create mode 100644 usth/ICT2.10/midterm/two-way.png diff --git a/usth/ICT2.10/homework/solutions.pdf b/usth/ICT2.10/homework/solutions.pdf new file mode 100644 index 0000000..2cde2c5 Binary files /dev/null and b/usth/ICT2.10/homework/solutions.pdf differ diff --git a/usth/ICT2.10/homework/solutions.tex b/usth/ICT2.10/homework/solutions.tex new file mode 100644 index 0000000..1dbdf4f --- /dev/null +++ b/usth/ICT2.10/homework/solutions.tex @@ -0,0 +1,405 @@ +\documentclass[a4paper,12pt]{article} +\usepackage[english,vietnamese]{babel} +\usepackage{amsmath} +\usepackage{booktabs} +\usepackage{circuitikz} +\usepackage{enumerate} +\usepackage{lmodern} +\usepackage{mathtools} +\usepackage{pgfplots} +\usepackage{siunitx} +\usepackage{textcomp} +\usepackage{tikz} +\usetikzlibrary{arrows,automata} + +\newcommand{\N}{\mathcal N} +\newcommand{\ud}{\,\mathrm{d}} +\newcommand{\SIR}{\mathrm{SIR}} +\newcommand{\baud}{\mathrm{Bd}} +\newcommand{\bit}{\mathrm{b}} +\newcommand{\chip}{\mathrm{c}} +\newcommand{\problem}[1]{\noindent\textbf{#1.}} + +\title{Mobile Wireless Communication} +\author{Nguyễn Gia Phong} +\date{Spring 2020} + +\begin{document} +\maketitle +\setcounter{section}{1} +\section{Characteristics of Radio Environment} +\subsection{Propagation Models} +\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*} + &\bar P_R = d^{-n_1}\left(1 + \frac{d}{d_b}\right)^{-n_2} P_T\\ + \iff &P_T = d^{n_1}\left(1 + \frac{d}{d_b}\right)^{n_2}\bar P_R +\end{align*} + +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 +\begin{align*} + &P_{T,\si{\deci\bel}} - P_{R,\si{\deci\bel}} + = 13\lg d + 35\lg\frac{d + 90}{90}\\ + \iff &P_{R,\si{\deci\bel}} - P_{T,\si{\deci\bel}} + = 35\lg\frac{90}{d + 90} - 13\lg d +\end{align*} +\pagebreak + +This is plotted in the figure below +\begin{center} + \begin{tikzpicture} + \begin{axis}[ + xlabel={$d$ (\si{\metre})}, + ylabel={$P_{R,\si{\deci\bel}}-P_{T,\si{\deci\bel}}$ (\si{\deci\bel})}] + \addplot[domain=0:200]{35*ln(90/(x+90))/ln(10) - 13*ln(x)/ln(10)}; + \end{axis} + \end{tikzpicture} +\end{center} + +\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 $\sigma = \SI{6}{\deci\bel}$. +\begin{enumerate}[(a)] + \item Given $\bar P_R = \SI{1}{\milli\watt}$ at $d = \SI{100}{\metre}$. + \[P_R > \bar P_R \iff \sqrt[10]{10^X} > 1 \iff X > 0\] + Since $X$ is zero-mean and normally distributed, the probability + the received power at a mobile at that distance from the base station + will exceed \SI{1}{\milli\watt} is \SI{50}{\percent}, and so is the + probability it is less than \SI{1}{\milli\watt}. + + \item Let $Y = X/\sigma$, $Y \sim \N(0, 1)$ + and $F_X(x) = \Phi(X/\sigma) = \Phi(X/6)$. + + The probability a mobile has an acceptable received signal at + \SI{10}{\milli\watt} or higher is + \begin{align*} + P\left(P_R \ge 10\bar P_R\right) + &= 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} + \end{align*} + + \item For $\sigma = \SI{10}{\deci\bel}$, $F_X(x) = \Phi(X/10)$. + The probability a mobile has an acceptable received signal at + \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*} + 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} + +\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: +\[\Pi(t) = \begin{dcases} + 0, &\text{if }|t| > \frac{1}{2}\\ + \frac{1}{2}, &\text{if }|t| = \frac{1}{2}\\ + 1, &\text{if }|t| < \frac{1}{2}\\ +\end{dcases}\] + +The convolution sum of $x$ and $h$ is +\begin{align*} + y(t) &= x(t - 1)\\ + &= \int_{-\infty}^\infty e^{t-z-1}(\Pi(z-2) - \Pi(z-4))\ud z\\ + &= \int_{-\infty}^\infty e^{t-z-1}\Pi(z-2)\ud z + - \int_{-\infty}^\infty e^{t-z-1}\Pi(z-4)\ud z\\ + &= \int_{1.5}^{2.5} e^{t-z-1}\ud z + - \int_{3.5}^{4.5} e^{t-z-1}\ud z\\ + &= e^{t-1}\left(e^{2.5} - e^{1.5} - e^{4.5} + e^{3.5}\right) +\end{align*} + +\subsection{Fading} +\problem 1 Consider several delay spreeds $D$ of \SI{0.5}{\micro\second}, +\SI{1}{\micro\second} and \SI{6}{\micro\second}. + +\begin{itemize} + \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}. + 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}. + +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 +radio environments would result in flat fading for each of these data rates. + +\section{Cellular Concept} +\subsection{Channel Allocation} +\problem 1 Assume the simplest path-loss model of $g(d) = d^{-3}$, calculate +down-link SIR at point P at the corner of a hexagonal cell in a 3-reuse case. + +Using to path-loss model, the signal-to-interference ratio +can be approximated from the six first-tier interferers as follows +\[\SIR \approx \frac{1}{\left(\frac{R}{D-R}\right)^3 + + \left(\frac{R}{D+R}\right)^3 + + 4\left(\frac{R}{D}\right)^3}\] + +In a 3-reuse case, $D = \sqrt{3C}R = 3R$, and thus +\[\SIR \approx \frac{1}{\left(\frac{R}{2R}\right)^3 + + \left(\frac{R}{4R}\right)^3 + + 4\left(\frac{R}{3R}\right)^3} = \frac{1728}{499}\] + +\problem 2 Calculate the worst-case uplink SIR assuming the co-channel +interference is caused only by the closest interfering mobiles in radio cells +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 +\[\SIR \approx \frac{P_t/R^4}{6P_t/\left(\frac{3D}{4}\right)^4} += \frac{(3D/4)^4}{6R^4}\] + +With $D = 3.46R$ (4-reuse), this becomes +\[\SIR \approx \frac{(3\cdot3.46/4)^4}{6} = 7.56\] + +\subsection{Erlang-B Formula and Sizing a Cell} +\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. + +\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. +Given the average call-holding time $h = \SI{200}{\second}$, the arrival rate +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. + +\section{Modulation Techniques} +\problem 1 Consider communication system operating at the transmission +bandwidth of \SI{1}{\mega\hertz} with the rolloff factor of 0.25. +\begin{itemize} + \item Achievable data traffic rate is + \[R_s = \frac{B}{1 + \beta} = \frac{10^6}{1 + 0.25} + = \SI{800}{\kilo\baud\per\second}\] + \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 $T = \SI{16}{\micro\second}$. + \item Additionally use 16-QAM, the bit rate is + $R_\bit = \SI{800}{\kilo\bit\per\second}$. +\end{itemize} + +\problem 2 Given $B = \SI{1}{\mega\hertz}$, $\beta = 0.25$, +$R_\bit = \SI{4.8}{\mega\bit\per\second}$ and $T = \SI{25}{\micro\second}$. + +$R_s = B/(1+\beta) = \SI{0.8}{\mega\baud\per\second}$, thus 64-QAM is used. + +For OFDM, $N = R_s/\Delta f = R_\bit T \approx 128$. + +\problem 3 Consider a transmission of bandwidth $B = \SI{2}{\mega\hertz}$, +where phase-shift keying and Nyquist rolloff shaping is used. + +For rolloff factors of 0.2, 0.25, 0.5, the traffic rates are respectively +\SI{1.67}{\mega\baud\per\second}, \SI{1.6}{\mega\baud\per\second} and +\SI{1.33}{\mega\baud\per\second}. + +In order to transmit at a rate of $R_\bit = \SI{6.4}{\mega\bit\per\second}$ +when $\beta = 0.25$, 16-QAM should be used. + +\problem 4 Given the input sequence 1001111010 +and the following QPSK signal pairs +\begin{center} + \begin{tabular}{c c c} + \toprule + Successive Signal & $a_i$ & $b_i$\\ + \midrule + 0 0 & $-1$ & $-1$\\ + 0 1 & $-1$ & $+1$\\ + 1 0 & $+1$ & $-1$\\ + 1 1 & $+1$ & $+1$\\ + \bottomrule + \end{tabular} +\end{center} + +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] + \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} + +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} diff --git a/usth/ICT2.10/midterm/A-GPS.png b/usth/ICT2.10/midterm/A-GPS.png new file mode 100644 index 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--git a/usth/ICT2.10/midterm/main.pdf b/usth/ICT2.10/midterm/main.pdf new file mode 100644 index 0000000..837437b Binary files /dev/null and b/usth/ICT2.10/midterm/main.pdf differ diff --git a/usth/ICT2.10/midterm/main.tex b/usth/ICT2.10/midterm/main.tex new file mode 100644 index 0000000..0a33ca1 --- /dev/null +++ b/usth/ICT2.10/midterm/main.tex @@ -0,0 +1,439 @@ +\documentclass[pdf]{beamer} +\usepackage[english,vietnamese]{babel} +\usepackage{amsmath} +\usepackage{booktabs} +\usepackage{graphicx} +\usepackage{hyperref} +\usepackage{lmodern} +\usepackage{siunitx} + +\mode{} +\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}. + \end{thebibliography} +\end{frame} + +\begin{frame}{Copying}\Large + \begin{center} + \includegraphics[width=0.2\textwidth]{CC.png} + \includegraphics[width=0.2\textwidth]{BY.png} + \includegraphics[width=0.2\textwidth]{SA.png} + \end{center} + + This work is licensed under a + \href{https://creativecommons.org/licenses/by-sa/4.0/}{Creative Commons + Attribution-ShareAlike 4.0 International License}. +\end{frame} +\end{document} diff --git a/usth/ICT2.10/midterm/one-way-to-earth.png b/usth/ICT2.10/midterm/one-way-to-earth.png new file mode 100644 index 0000000..965acc4 Binary files /dev/null and b/usth/ICT2.10/midterm/one-way-to-earth.png differ diff --git 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