%% This LaTeX-file was created by Mon Mar 13 13:14:36 2000 %% LyX 1.0 (C) 1995-1999 by Matthias Ettrich and the LyX Team %% Do not edit this file unless you know what you are doing. \documentclass[11pt]{article} \usepackage[T1]{fontenc} \setlength\parskip{\medskipamount} \setlength\parindent{0pt} \makeatletter %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% LyX specific LaTeX commands. \providecommand{\LyX}{L\kern-.1667em\lower.25em\hbox{Y}\kern-.125emX\@} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% User specified LaTeX commands. \setlength{\oddsidemargin}{0in} \setlength{\evensidemargin}{0in} \setlength{\topmargin}{0.0in} \setlength{\headheight}{0in} \setlength{\headsep}{0.5in} \setlength{\textwidth}{6.5in} \setlength{\textheight}{8.5in} \setlength{\parindent}{0in} \setlength{\parskip}{2mm} \setlength{\columnseprule}{0.4pt} \usepackage{times} \usepackage{psfig} \usepackage{fancyheadings} \makeatother \begin{document} \pagestyle{fancy} \cfoot{Channelization: \thepage} \title{Channelization of a Communications Medium} \author{Norman Matloff \\ \\ University of California at Davis\\ \copyright{}2001, N. Matloff } \date{March 12, 2000} \maketitle \section{Overview} The term \textbf{channelization} refers to the sharing of a point-to-point communications medium.\footnote{ That medium could be a physical wire, an optical fiber, a radio link, and so on. } For example, many telephone conversations (or in our context, computer-to-computer network transactions) can be submitted simultaneously on a single wire, with each conversation being on a separate \textbf{channel}. The notion of a channel is very closely related to the household concept of radio and TV channels. The frequency spectrum for television, for instance, is divided into subranges called channels, and these correspond to our everyday concept of TV channels. Each channel is used to transmit different information, all simultaneously. There are three main ways of doing this: \begin{itemize} \item In \textbf{time-division} multiplexing (TDMA),\footnotemark{}% different sources transmit on the line at different times, each taking (very short) turns. This is used in long-distance phone lines, for example.\footnotemark{}% \item In \textbf{frequency-division} multiplexing (FDMA), the different sources attached to the line send on different frequencies (e.g. different radio frequencies, or different light frequencies, i.e. different colors). This is used for radio and television transmission, and increasingly for computer-to-computer network transactions. \item In \textbf{code-division multiplexing} (CDMA), all nodes on the network send at the same time, on the same frequency, but using different codes. (Think of one node using a 4B/5B code, another using a second kind of code, and so on.) This is used in some cellular telephone systems. \end{itemize}\addtocounter{footnote}{-1}\footnotetext{ The full formal title is Time-Division Multiple Access. }% \stepcounter{footnote}\footnotetext{ Technically, this would seem to violate our definition at the outset of this unit, in which said that the various channels operate simultaneously. However, since the turns are so short, each channel's turn is followed very quickly by its next turn, so we have the illusion that it is simultaneous. }% \section{TDMA} \subsection{Synchronous TDMA} At one end of the point-to-point line we have n source nodes. Messages from these sources are \textbf{multiplexed} onto the line: The n sources continually send in \textbf{Round Robin} fashion, meaning that they take turns: Source node 0 sends, then source node 1, then source node 2,..., then source node n-1, then source node 0, then source node 1, and so on. Each cycle of n transmissions (one from each source, starting with source node 0) is called a \textbf{frame}. The hardware that does the multiplexing is called, not surprisingly, a \textbf{multiplexer}. At the other end of the line there may be a host computer, in which case the n sources are typically terminals. Or there may be a \textbf{demultiplexer} at the destination end, distributing the messages from the n sources to n destination entities. Each source sends the same amount of data---typically set at one byte or one bit---per turn. This way the destination end of the line knows whose data is arriving at any given time. If the unit of data is bytes, for instance, then you know that any byte which arrives at time k, where k mod n = 5, is from source node 5. A specific example is the T1 lines which are common for long-distance communication in North America. Here n = 24 and the unit of data is bytes. Thus there should be \( 24\times 8=192 \) bits sent on the line per frame.\footnote{ Plus one more bit, to be described below. } We say that a T1 line consists of 24 \textbf{channels} of one byte each. In TDMA, we are dividing time, instead of frequency range as in the radio/TV case. For instance, the totality of all the third bytes from all frames---that is, all the bytes sent by the third source---would be called channel 2 (the first one being numbered 0). Suppose a channel is used for voice transmission, sampled at regular intervals and digitized. The sample taken at any given time consists of the numerical value of the loudness of the sound at that time, which we represent as an 8-bit number, i.e. one byte. Recall that the bandwidth of human voice is about 4 KHz. One can prove that this means we need to sample at \( 2\times 4,000=8,000 \) times per second to accurately capture the voice signal. So, on a T1 line we must send 8,000 frames per second, thus \( 8,000\times 192 \) or 1.54 million bits per second. What if instead we use some or all of the channels to transmit data? T1 lines are set up to send 7-bit sets instead of bytes, with the eighth bit used for encoding control information. Thus we are sending 56,000 bits per second on each channel. However, since the clock at the source-end multiplexer will not exactly match that of the destination end, some drift will occur between them. Eventually the destination end will drift ahead of or behind the sources by one entire bit's duration of time, resulting in reception of wrong data. Thus each frame has an extra bit for \textbf{synchronization} purposes, so that the two ends can synchronize their clocks at regular intervals. In T1 lines, then, this will be the 193rd bit in each frame. The totality of these bits is then called the \textbf{control channel}. In it the transmitter will send the repeating pattern 101010101... If the receiver receives a 0 on that channel when it is expecting a 1, or vice versa, the receiver will know that it has, literally, ``gone out of sync''; it will then have to look at several subsequent frames for the 1010101... sequence before getting back on track. Each channel will operate its own HDLC process, independent of the others, to set up sliding-window operations and so on, though of course not for clock synchronization. \subsection{Statistical TDMA} The problem with synchronous TDMA is that some source nodes may sometimes have nothing to send; they simply are idle during turns when they have nothing to send. This of course wastes the line, which may be leased at a very expensive monthly fee. \textbf{Statistical} TDMA remedies this by allowing a source node to skip its turn in such a situation. Each time a node has something to send, it places the item on a first-in, first-out (FIFO) queue into the multiplexer. This approach makes better use of the line's bandwidth. Of course we do pay a price, in the sense that each source now must sent its source number in addition to the data, so that the receiver on the other end of the line can determine who sent the data. Thus part of the line's capacity will be used for sending these source numbers, which we did not have in the synchronous TDMA case. \section{FDMA} If we send data in their original form, this is called \textbf{baseband} transmission. A much more flexible alternative is \textbf{broadband} transmission, in which we \textbf{modulate}, i.e. transform the data. The ``AM'' in ``AM radio'' stands for \textbf{amplitude modulation}. Here analog data, say voice, is still sent in an analog signal, but in the form of a product of the orginal data and a sine wave (or a cosine wave, which is after all just a sine wave which has been ``moved over''). Say x(t) is the voice waveform, and consider a radio station at a \textbf{carrier frequency} \( g_{c} \) (this is the frequency at which you tune your radio in order to receive that station's broadcast). Then the radio station will transmit the product \begin{equation} p(t)=x(t)cos(2\pi g_{c}t) \end{equation} What is the effect of this? Well, x(t) has a Fourier series, as we saw in our unit on the Physical Layer: \begin{equation} x(t)={\sum _{n=0}^{\infty }{a_{n}}\cos (2\pi nf_{0}t)}+{\sum _{n=1}^{\infty }{b_{n}}\sin (2\pi nf_{0}t)} \end{equation} Consider one cosine term in that series, \begin{equation} {a_{n}}\cos (2\pi nf_{0}t) \end{equation} and its associated term(s) in the Fourier series for p(t). In other words, what we are asking is, what will happen when this \( {a_{n}}\cos (2\pi nf_{0}t) \) term gets multiplied by \( cos(2\pi g_{c}t) \)? Using high-school trig identities, we can show that \begin{equation} cos(a)cos(b)=\frac{1}{2}[cos(a+b)+cos(a-b)] \end{equation} so \begin{equation} {a_{n}}\cos (2\pi nf_{0}t)\cdot cos(2\pi g_{c}t)=\frac{a_{n}}{2}[cos(2\pi [g_{c}+nf_{0}]t)+cos(2\pi [g_{c}-nf_{0}]t)] \end{equation} In other words, the component which was of frequency \( nf_{0} \) in the original voice waveform x(t) has now been transformed into two frequencies for p(t), centered around the radio station's carrier frequency \( g_{c} \): \( g_{c}+nf_{0} \) and \( g_{c}-nf_{0} \). Now the Federal Communications Commission, which regulates radio stations, will make sure that the \( g_{c} \) values for various radio stations are spaced well apart. So the bandwidth interval for a given station will be a \textbf{channel} \begin{equation} [g_{c}-d,g_{c}+d] \end{equation} for some value of d. You can see that since the bandwidth interval for human voice is about {[}300,3400{]}, setting \begin{equation} d\geq 3400 \end{equation} will ensure that radio stations at a given geographical location and having adjacent carrier frequencies will not interfere with each other.\footnote{ Actually, we can do even better than this, using a \textbf{single sideband} technique; this means that the radio station would only transmit the frequency range \( [g_{c},g_{c}+d] \) (which is no loss of information, since the Fourier components in this interval mirror those in \( [g_{c}-d,g_{c}] \)). In this way, we save half the bandwidth of the radio spectrum, and can fit twice as many stations in. } Yes, the human voice does have higher-frequency components and these will cause some overlap between stations, but these components are so weak that the listeners of one station will not hear those of the neighboring station. Again, there are many refinements to this which are used in actual practice, but the basic principles are clear: If we use broadband trasmission (on any medium, not just radio or microwave), we can send many signals at different frequencies, with the spacings between adjacent frequencies being determined by the the highest frequency (in the effective bandwidth) in the original signal, such as \( 4\times 10^{6} \) in the ``101010...'' example in the unit on the Physical Layer. This is called \textbf{frequency division multiplexing}. When done in an optical-fiber medium, FDMA is called Wave-Division Multiple Access (WDMA). Each channel uses a different color of light. \section{CDMA} Here each data bit will be sent as G transmission bits. (Recall that in Manchester coding each data bit is sent as two transmission bits, and in 4B/5B coding each group of four data bits is sent as five transmission bits.) All stations which have data to send will send simultaneously, using the entire bandwidth of the line. (CDMA is called a \textbf{spread spectrum} method, alluding to the fact that the bandwidth of a given signal, i.e. its ``spectrum,'' will be much broader in its transmitted form than in its data form.) Each node j will have its own code, defined as \( c_{j0},...,c_{jg} \) where g = G-1. The code values will be generated at random (actually in pseudorandom fashion, in hardware), as a sequence of coin tosses, with the \( k^{th} \) toss determining whether \( c_{jk} \) is +1 (for heads) or -1 (for tails). Denoting the data bit sent by node j as \( d_{j} \), then node j will send G transmission bits \( d_{j}c_{j0},...,d_{j}c_{jg} \). (For simplicity, we will assume that all nodes are now sending, though it may be the case that only some of them are currently sending.) Consider a node which is currently receiving from node i. The receiving node knows node i's code (which it has been notified of previously, say on a special channel). Keep in mind that this receiving node will receive a signal which is the sum of all the signals sent by the various nodes, not just node i. The receiving node will see the signals \( s_{k}= \)\( \sum _{j}d_{j}c_{jk} \) , k = 0,...,g. Since this node is receiving from node i, it will take the ``dot product'' of this \( s_{k} \) sequence with node i's code sequence \( c_{ik} \): \( D_{i}=\sum ^{g}_{k=0}s_{k}c_{ik} \) Now, here is the neat part: Even though this value \( D_{i} \) comes from a mixture of the signals from all the nodes, not just the signal from node i, the signals from the nodes other than i will actually ``cancel out,'' as seen in the following: \( D_{i}=\sum ^{g}_{k=0}s_{k}c_{ik}=\sum ^{g}_{k=0}\sum _{j}d_{j}c_{jk}c_{ik}=\sum _{j}d_{j}\sum ^{g}_{k=0}c_{ik}c_{jk} \) Recall that the code values c are random +1s and -1s. Thus \( \sum ^{g}_{k=0}c_{ik}c_{jk}\approx \left\{ \begin{array}{rl} G, & \textrm{if j}=\textrm{i}\\ 0 & \textrm{if j}\neq \textrm{i} \end{array}\right. \) because terms approximately cancel out in the case j \( \neq \) i. In other words, \( D_{i}\approx \left\{ \begin{array}{rl} +G, & \textrm{if }d_{i}=+1\\ -G, & \textrm{if }d_{i}=-1 \end{array}\right. \) So, even though the receiving node sees a signal which is a mixture of all the nodes' signals, the receiving node can tell specifically what node i's data bit was, and thus can recover node i's data from the mixture. Of course, this method is only approximate, and thus errors can occur (just as with any method, though maybe more so), so error detection/correction techniques must be used. The reader might at first think that CDMA is inefficient, since it takes G transmission bits just to send out one data bit. But this is no different from having G channels in TDMA or FDMA. One advantage of CDMA is security. Since each +1 is changed to +G and each -1 is changed to -G, we can set the +1 and -1 to low signal levels, i.e. low power. That would make it harder for eavesdroppers to pick up i's data. \end{document}