406 lines
16 KiB
TeX
406 lines
16 KiB
TeX
\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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\newcommand{\N}{\mathcal N}
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\newcommand{\ud}{\,\mathrm{d}}
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\newcommand{\SIR}{\mathrm{SIR}}
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\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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\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
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distance under \SI{200}{\metre} in Orlando. The average receive power is given by
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\[\bar P_R = g(d) P_T G_T G_R\]
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Assume the attenna gains are both 1 and apply the inverse variation of power
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with distance for two-slope model we get
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\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
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\end{align*}
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Substituting $d_b = \SI{90}{\metre}$, $n_1 = 1.3$ and $n_2 = 3.5$ gives us
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\[P_T = d^{1.3}\left(1 + \frac{d}{90}\right)^{3.5}\bar P_R\]
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With average power effect experienced, $P_{R,\si{\deci\bel}} = 10\lg P_R
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= 10\lg \bar P_R$ 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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This is plotted in the figure below
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\begin{center}
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\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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\problem 2 Consider a log-normal shadow fading propagation, the receive power
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is given by \[P_R = \sqrt[10]{10^X} g(d) P_T G_T G_R\] where $X$ is a zero-mean
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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)\\
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&= 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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\item For $\sigma = \SI{10}{\deci\bel}$, $F_X(x) = \Phi(X/10)$.
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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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\[P\left(P_R \ge 10\bar P_R\right) = 1 - F_X(10)
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= 1 - \Phi(1) = \SI{15.87}{\percent}\]
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\item If the lower limit for an acceptable received signal is
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\SI{6}{\milli\watt}, with $\sigma = 6$, the probability a received signal
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is acceptable is
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\begin{align*}
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P\left(P_R \ge 6\bar P_R\right)
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&= P\left(\sqrt[10]{10^X} \ge 6\right)
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= P\left(X \ge \lg{6^{10}}\right)\\
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&= 1 - F_X\left(\lg{6^{10}}\right)
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= 1 - \Phi\left(\frac{\lg{6^{10}}}{6}\right) = \SI{9.73}{\percent}
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\end{align*}
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With $\sigma = 10$, the probability a received signal is acceptable is
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\[P\left(P_R \ge 6\bar P_R\right) = 1 - F_X\left(\lg{6^{10}}\right)
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= 1 - \Phi\left(\frac{\lg{6^{10}}}{10}\right) = \SI{21.82}{\percent}\]
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\end{enumerate}
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\subsection{Random Channel Characterization}
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Given $x(t) = e^t * (\Pi(t - 1) - \Pi(t - 3))$ and $h(t) = \delta(t - 1)$,
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where $\Pi$ is the rectangular function:
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\[\Pi(t) = \begin{dcases}
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0, &\text{if }|t| > \frac{1}{2}\\
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\frac{1}{2}, &\text{if }|t| = \frac{1}{2}\\
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1, &\text{if }|t| < \frac{1}{2}\\
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\end{dcases}\]
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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)
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\end{align*}
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\subsection{Fading}
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\problem 1 Consider several delay spreeds $D$ of \SI{0.5}{\micro\second},
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\SI{1}{\micro\second} and \SI{6}{\micro\second}.
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\begin{itemize}
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\item For IS-95 and cdma2000 which uses the transmission bandwidth of
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\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
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greater than this (\SI{1}{\micro\second} and \SI{6}{\micro\second}).
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\item For WCDMA which uses the bandwidth of \SI{5}{\mega\hertz},
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the symbol interval is \SI{0.2}{\micro\second}, thus symbols
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are resolvable in all cases.
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\end{itemize}
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\problem 2 Indicate the condition for flat fading for each of the following
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data rates with transmission in binary form: \SI{8}{kbps}, \SI{40}{kbps},
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\SI{100}{kbps}, \SI{6}{Mbps}.
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Assume information is transmitted in rectangular waves, the symbol interval
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are \SI{125}{\micro\second}, \SI{25}{\micro\second}, \SI{10}{\micro\second}
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and \SI{1/6}{\micro\second} respectively. For flat fading to occur,
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the delay spread must be significantly less than the symbol interval.
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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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\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
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can be approximated from the six first-tier interferers as follows
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\[\SIR \approx \frac{1}{\left(\frac{R}{D-R}\right)^3
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+ \left(\frac{R}{D+R}\right)^3
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+ 4\left(\frac{R}{D}\right)^3}\]
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In a 3-reuse case, $D = \sqrt{3C}R = 3R$, and thus
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\[\SIR \approx \frac{1}{\left(\frac{R}{2R}\right)^3
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+ \left(\frac{R}{4R}\right)^3
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+ 4\left(\frac{R}{3R}\right)^3} = \frac{1728}{499}\]
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\problem 2 Calculate the worst-case uplink SIR assuming the co-channel
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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
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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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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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\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
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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
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and 7-reuse, there are over 118 channels per cell. With the probility of
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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 $\lambda = \SI{0.505}{\mathrm{calls}\per\second}$.
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Since an user makes a call every \SI{900}{\second} on average, there are
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approximately 454.5 users. As the density of mobile terminals is
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\SI{2}{\mathrm{terminals}\per\square{\kilo\metre}}, the area is
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\SI{227.25}{\square{\kilo\metre}}, which indicates a cell radius
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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
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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
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\[R_s = \frac{B}{1 + \beta} = \frac{10^6}{1 + 0.25}
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= \SI{800}{\kilo\baud\per\second}\]
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\item Delay spread that no ISI occurs is much less than the symbol interval,
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which is $T = B^{-1} = \SI{1}{\micro\second}$.
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\item Using OFDM with $N = 16$ equally spead carriers, for each subcarrier,
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$\Delta f = \SI{62.5}{\kilo\hertz}$, $R_s = \SI{50}{\kilo\baud\per\second}$
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and $T = \SI{16}{\micro\second}$.
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\item Additionally use 16-QAM, the bit rate is
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$R_\bit = \SI{800}{\kilo\bit\per\second}$.
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\end{itemize}
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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 $T = \SI{25}{\micro\second}$.
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$R_s = B/(1+\beta) = \SI{0.8}{\mega\baud\per\second}$, thus 64-QAM is used.
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For OFDM, $N = R_s/\Delta f = R_\bit T \approx 128$.
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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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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}.
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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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\problem 4 Given the input sequence 1001111010
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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 & $+1$ & $+1$\\
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\bottomrule
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\end{tabular}
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\end{center}
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Let the carrier frequency be some multiple of 1/T
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\begin{tikzpicture}
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\begin{axis}[scale only axis, width=0.8\textwidth, height=0.16\textwidth,
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xlabel=In-phase Carrier, xtick={0,1,2,3,4,5},
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xticklabels={0,T,2T,3T,4T,5T}, ymin=-2, ymax=2, samples=420]
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\addplot[domain=0:5]{cos(x*720)};
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\end{axis}
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\end{tikzpicture}
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\begin{tikzpicture}
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\begin{axis}[scale only axis, width=0.8\textwidth, height=0.16\textwidth,
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xlabel=Quadrature-phase Carrier, xtick={0,1,2,3,4,5},
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xticklabels={0,T,2T,3T,4T,5T}, ymin=-2, ymax=2, samples=420]
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\addplot[domain=0:5]{sin(x*720)};
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\end{axis}
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\end{tikzpicture}
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The output QPSK signal would then be
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\begin{tikzpicture}
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\begin{axis}[scale only axis, width=0.8\textwidth, height=0.16\textwidth,
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xlabel=In-phase Component, xtick={0,1,2,3,4,5},
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xticklabels={0,T,2T,3T,4T,5T}, ymin=-2, ymax=2, samples=69]
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\addplot[domain=0:1]{+cos(x*720)};
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\addplot[domain=1:2]{-cos(x*720)};
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\addplot[domain=2:3]{+cos(x*720)};
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\addplot[domain=3:4]{+cos(x*720)};
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\addplot[domain=4:5]{+cos(x*720)};
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\end{axis}
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\end{tikzpicture}
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\begin{tikzpicture}
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\begin{axis}[scale only axis, width=0.8\textwidth, height=0.16\textwidth,
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xlabel=Quadrature-phase Component, xtick={0,1,2,3,4,5},
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xticklabels={0,T,2T,3T,4T,5T}, ymin=-2, ymax=2, samples=69]
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\addplot[domain=0:1]{-sin(x*720)};
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\addplot[domain=1:2]{+sin(x*720)};
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\addplot[domain=2:3]{+sin(x*720)};
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\addplot[domain=3:4]{-sin(x*720)};
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\addplot[domain=4:5]{-sin(x*720)};
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\end{axis}
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\end{tikzpicture}
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\begin{tikzpicture}
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\begin{axis}[scale only axis, width=0.8\textwidth, height=0.16\textwidth,
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xlabel=Output Signal, xtick={0,1,2,3,4,5},
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xticklabels={0,T,2T,3T,4T,5T}, ymin=-2, ymax=2, samples=69]
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\addplot[domain=0:1]{+cos(x*720)-sin(x*720)};
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\addplot[domain=1:2]{-cos(x*720)+sin(x*720)};
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\addplot[domain=2:3]{+cos(x*720)+sin(x*720)};
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\addplot[domain=3:4]{+cos(x*720)-sin(x*720)};
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\addplot[domain=4:5]{+cos(x*720)-sin(x*720)};
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\end{axis}
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\end{tikzpicture}
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\section{Multiple Access Techniques}
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\subsection{Time-Division Multiple Access}
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Transmission bit rate is the rate at which the bits are transmitted, while
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the user information bit rate is the rate at which per data are transmitted.
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In particular, GSM gives each time slot \SI{576.92}{\micro\second},
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minus \SI{30.46}{\micro\second} guard time. During this duration,
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\SI{148}{\bit} are tramsmitted, thus the transmission bit rate is
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$\SI{148}{\bit}/(\SI{576.92}{\micro\second}-\SI{30.46}{\micro\second})
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= \SI{270.834}{\kilo\bit\per\second}$. Of the \SI{148}{\bit},
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\SI{114}{\bit} are data bits. Furthermore, only one slot per GSM eight-slot
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frame and 24 out of 26 frames are used to carry information. Therefore,
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the user bit rate is $\SI{114}{\bit}/\SI{4.615}{\milli\second}\cdot 24/26
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= \SI{22.8}{\kilo\bit\per\second}$.
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Similarly, IS-136 has the transmission bit rate of $\SI{1944}{\bit}
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/ \SI{40}{\milli\second} = \SI{48.6}{\kilo\bit\per\second}$ and $\SI{520}{\bit}
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/ \SI{40}{\milli\second} = \SI{13}{\kilo\bit\per\second}$.
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\subsection{Code-Division Multiple Access}
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Consider IS-95 with the bit rate of \SI{9.6}{\kilo\bit\per\second}
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and the chip rate of \SI{1.2288}{\mega\chip\per\second}, the speading gain
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is 128 chips per bit.
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\section{Channel Coding Techniques}
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\subsection{Block Coding}
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Consider the generator matrix
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\[\mathbf G = [\mathbf I_k \mathbf P] = \begin{pmatrix}
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1&0&0&0&1&0&1\\
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0&1&0&0&1&1&1\\
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0&0&1&0&1&1&0\\
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0&0&0&1&0&1&1
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\end{pmatrix}\]
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it is trivial that $n = 7$, $k = 4$ and
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\[\mathbf P = \begin{pmatrix}
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1&0&1\\
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1&1&1\\
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1&1&0\\
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0&1&1
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\end{pmatrix}\]
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The parity check matrix is then given by
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\[\mathbf H = [\mathbf P^T \mathbf I_{n-k}] = \begin{pmatrix}
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1&0&0&1&1&1&0\\
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0&1&0&0&1&1&1\\
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0&0&1&1&1&0&1
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\end{pmatrix}\]
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\subsection{Convolutional Coding}
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Consider a $K = 3$, rate \textonehalf{} convolution encoder with generators
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$g_1 = [101]$ and $g_2 = [011]$.
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\begin{center}
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\begin{circuitikz}
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\draw (0,3) node (input) {input}
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(1,3) node[inputarrow] (in) {}
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(2,3) node[circ] (m1) {}
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(3.5,3) node[twoportshape] (port1) {}
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(5,3) node[circ] (m2) {}
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(6.5,3) node[twoportshape] (port2) {}
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(8,3) node[circ] (m3) {}
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(input) -- (in) -- (m1) -- (port1) -- (m2) -- (port2) -- (m3)
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(5,5.5) node[xor port, rotate=90] (xor1) {}
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(9,6) node[flowarrow] (out1) {}
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(10,6) node{$n_1$}
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(m1) |- (xor1.in 1)
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(m3) |- (xor1.in 2)
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(xor1.out) |- (out1)
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(6.5,0.5) node[xor port, rotate=270] (xor2) {}
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(9,0) node[flowarrow] (out2) {}
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(10,0) node{$n_2$}
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(m3) |- (xor2.in 1)
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(m2) |- (xor2.in 2)
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(xor2.out) |- (out2);
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\end{circuitikz}
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\end{center}
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Initialize the encoder with 01, we get the following state diagram
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\begin{center}
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\begin{tikzpicture}[->,>=latex,shorten >=1pt,auto,node distance=42mm]
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\node[initial,state] (01) {01};
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\node[state] (00) [above right of=01] {00};
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\node[state] (11) [below right of=01] {11};
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\node[state] (10) [below right of=00] {10};
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\path (00) edge [dashed, loop above] node {00} (00)
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edge node {10} (10)
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(01) edge [dashed] node {11} (00)
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edge [bend left] node {01} (10)
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(10) edge [dashed, bend left] node {01} (01)
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edge node {11} (11)
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(11) edge [dashed] node {10} (01)
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edge [loop below] node {00} (11);
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\node [below of=10] {%
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\begin{tabular}{c c}
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\raisebox{2pt}{\tikz{\draw[dashed] (0,0) -- (10mm,0);}} & 0\\
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\raisebox{2pt}{\tikz{\draw (0,0) -- (10mm,0);}} & 1
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\end{tabular}};
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\end{tikzpicture}
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\end{center}
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Given the input bit sequence of 10011011, the output would be
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0101111011100111.
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\end{document}
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