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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} |