\documentclass[11pt]{article} \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}{0.1in} \usepackage{times} \usepackage{psfig} \usepackage{fancyheadings} \begin{document} \pagestyle{fancy} \cfoot{Physical Layer: \thepage} \title{Computer Networks: the Physical Layer} \author{Norman Matloff \\ Dept. of Computer Science \\ University of California at Davis\\ \copyright{}2001, N. Matloff} \date{April 17, 2001} \maketitle \section{Goals of This Unit} In this unit we will look at the physical properties of media which are used in computer networks. Our main emphasis is on the capacity of a network, meaning the maximum number of bits per second which can be loaded onto it. We will use the mathematical technique of {\bf Fourier series} to illustrate the role of {\bf bandwidth} in our network and in our data. This will help not only to understand the physical limitations of networks, but also to lay the groundwork for our later unit on {\bf multiplexing}. \section{Media} Network signals might be propagated through {\bf guided} media such as metal wires (twisted pairs, coaxial cable, etc.) or optical fibers (flashing light, off for 0, on for 1), or in free space, such as with radio waves or microwaves. \section{Signal Types} A signal may be in {\bf digital} form, e.g. 0-1 bits with a low voltage meaning 0 and a high voltage meaning 1. The graph of a digital signal over time has a ``square wave'' shape. On the other hand, {\bf analog} forms look ``wavy.'' A modem, for instance, sends a low-pitched sound for a 0 and a high pitch for a 1; the sounds themselves look like sine waves when graphed against time. \section{Data Types} Just as the signals of interest can be either digital or analog, the data encoded by those signals could be either digital, such as numbers or characters, or analog, such as voices. Either type of data can be encoded within either type of signal. We noted above, for instance, that a modem encodes digital data onto analog signals. Voice data, on the other hand, though inherently analog, can be {\bf digitized}: A person's voice graphed against time is a continuous curve, but we can sample it at regular intervals, typically 8,000 times per second. At each sample point, we record the numerical height of the curve (i.e. the numerical value of the loudness). This gives us numbers, which comprise digital data. These can be stored in a disk file as with a game program, sent through the Internet as with Real Audio transmission of radio programs, and so on. \section{Spectral Analysis} Recall from calculus that we can write a function x(t) as a Taylor series, which is an ``infinite polynomial'': \begin{equation} x(t) = \sum_{n=0}^{\infty} c_n t^n \end{equation} For instance, for $e^t$, \begin{equation} e^t = \sum_{n=0}^{\infty} \frac{1}{n!} t^n \end{equation} Recall also that we say a function x(t) (where t is time) is {\bf periodic} with period T if if x(u+T) = x(u) for all u. The {\bf fundamental frequency} of x is then defined to be the number of periods per unit time, \begin{equation} f_0 = {1 \over T } \end{equation} Suppose x(t) is a digital or analog periodic signal transmitted in some medium. We can write x as an ``infinite trig polynomial,'' i.e. a {\bf Fourier series}: \begin{equation} x(t) = {\sum_{n=0}^{\infty } {a_n} \cos ( 2 \pi n f_0 t)} + {\sum_{n=1}^{\infty} {b_n} \sin ( 2 \pi n f_0 t)} \end{equation} Here, instead of having a series of terms \begin{equation} 1 ,\ t ,\ t^2 ,\ t^3 ,\ ... \end{equation} as in a Taylor series, we have a series of terms \begin{equation} 1 ,\ \cos (2 \pi f_0 t) ,\ \cos (4 \pi f_0 t) ,\ \cos (6 \pi f_0 t),\ ... \end{equation} and similar sine terms. Note that the frequences in those cosines are integer multiples, called {\bf harmonics}, of the fundamental frequency of x, $f_0$. The set of coefficients \{${a_n}$\}, \{$ {b_n}$\} is called the {\bf frequency spectrum} of x.\footnote{We can also write x as an \underline{integral} of trig functions, rather than a sum of such functions. Then the spectrum is a continuous range of numbers, rather than the discrete points $a_n$ and $b_n$. We will not pursue this point here.} The coefficients are calculated as follows: \begin{equation} a_0 = \frac{1}{T} \int_0^T x(t) dt \end{equation} \begin{equation} \label{eight} a_n = \frac{2}{T} \int_0^T x(t) cos(2 \pi n f_0 t) dt \end{equation} \begin{equation} b_n = \frac{2}{T} \int_0^T x(t) sin(2 \pi n f_0 t) dt \end{equation} In other words, suppose x(t) is the loudness of your voice at time t. Your voice is a sum of components at various frequencies, and $a_n$ and $b_n$ represent the loudness of the n-th frequency. Because of the periodic nature of the functions involved, we can shift the range of integration by equal amounts on the lower and upper bounds, and it is often convenient to do so. Any lower and upper bounds which differ by the amount T will give the same answer. For example, consider cases in which the function x(t) is symmetric around 0. Then it is natural to take our lower and upper bounds to be -T/2 and T/2, for the following reason. Note that the sine function is \underline{anti}-symmetric, i.e. sin(-t) = -sin(t). This implies that $x(t) sin(2 \pi n f_0 t)$, the integrand in \begin{equation} b_n = \frac{2}{T} \int_{- \frac{T}{2}}^{\frac{T}{2}} x(t) sin(2 \pi n f_0 t) dt \end{equation} will also be anti-symmetric. That implies that we know without even performing the calculation that $b_n = 0$, so we have saved ourselves some computation. Note that the $a_n$ and $b_n$ are independent of our bit rate. Here is how to see this: Suppose for example that we were to send the same signal but at half the bit rate, say because we changed modems. Call the new signal y(t) and the new period T'. Then y(t) = x(0.5t), and T' = 2T. The period in terms of numbers of bits has not changed, but the time to send those bits has doubled. Then for example $$ a_n = \frac{2}{T'} \int_0^{T'} y(t) cos(2 \pi n \frac{1}{T'} t) dt $$ If one uses the facts that y(t) = x(0.5t), and T' = 2T, and then one does the change-of-variable u = 0.5t, we find we get the same integral for $a_n$ as before, in Equation (\ref{asubn}). The intuition here is that although the graph of the curve y(t) would be a ``stretched out'' version of x(t), it would still have the same shape, and the same weighting would occur for all the sine and cosine terms. All this assumes a periodic signal. But strictly speaking, the signals we send are not periodic. Consider, for example, a text file transfer, transmitted as a digital signal. The characters in the file do not follow a periodic pattern, and thus the bits we send do not follow a periodic pattern, and the graph of the signal against time will not be periodic. However, for the purpose of analyzing the speed capacity of a given physical medium to transmit the data, we can imagine the signal to be periodic. If for instance the text file includes the sentence ``The quick brown fox jumped,'' we can ask how fast the medium could transmit the corresponding signals if the sentence were to be transmitted again and again (resulting in periodic signals). And for voice data transmitted in analog form, for instance, the signals are periodic during short time intervals. \section{Bandwidth: How to Read the {\it San Francisco Chronicle} Business Section} The popular press, especially business or technical sections, will often use the term {\bf bandwidth}. What does this mean? Any transmission medium has a natural range [$f_{min}$,$ f_{max}$] of frequencies that it can handle well. For example, an ordinary voice-grade telephone line can do a good job of transmitting signals of frequencies in the range 0 Hz to 4000 Hz, where ``Hz'' means cycles per second. Signals of frequencies outside this range suffer fade in strength, i.e are {\bf attenuated}, as they pass through the phone line. We call the frequency interval [0,4000] the {\bf effective bandwidth} (or just the {\bf bandwidth}) of the phone line. In addition to the bandwidth of a {\bf medium}, we also speak of the bandwidth of a {\bf signal}. For instance, although your voice is a mixture of many different frequencies, represented in the Fourier series for your voice's waveform, the really low and really high frequency components, outside the range [340,3400], have very low power, i.e. their $a_n$ and $b_n$ coefficients are small. Most of the power of your voice signal is in that range of frequencies, which we would call the effective bandwidth of your voice waveform.\footnote{This is also the reason why digitized speech is sampled at the rate of 8,000 samples per second. A famous theorem, due to Nyquist, shows that the sampling rate should be double the maximum frequency. Here the number 3,400 is ``rounded up'' to 4,000, and after doubling we get 8,000.} Obviously, in order for your voice to be heard well on the other end of your phone connection, the bandwidth of the phone line must be at least as broad as that of your voice signal, and that is the case. However, the phone line's bandwidth is not much broader than that of your voice signal. So, some of the frequencies in your voice will fade out before they reach the other person, and thus some degree of distortion will occur. It is common, for example, for the letter `f' spoken on one end to be mis-heard as `s'on the other end. This also explains why your voice sounds a little different on the phone than in person. Still, most frequencies are reproduced well and phone conversations work well. On the other hand, dogs have a hearing range which extends about twice as high as that of humans. Since the bandwidth of a phone line was designed to only cover most of the frequency range of human voice, then there are sounds which a dog would hear clearly at the scene of the sound, but which the dog could not hear if listening to that scene over the phone. We often use the term ``bandwidth'' to literally refer to width, i.e. the width of the interval [$f_{min}$,$ f_{max}$], $f_{max} - f_{min}$. There is huge variation in bandwidth among transmission media. As we have seen, phone lines have bandwidth intervals covering values on the order of $10^3$. For optical fibers, these numbers are more on the order of $10^{15}$. Suppose that for a given medium $f_{min} = f_0$ and $f_{max}$ = N $f_0$. Then the signal \begin{equation} x(t) = {\sum_{n=0}^{ \infty } {a_n} \cos ( 2 \pi n f_0 t)} + {\sum_{n=1}^{ \infty } {b_n} \sin ( 2 \pi n f_0 t)} \end{equation} is in effect truncated by the transmission medium to \begin{equation} x_N(t) = {\sum_{n=0}^{ N } {a_n} \cos ( 2 \pi n f_0 t)} + {\sum_{n=1}^{ N } {b_n} \sin ( 2 \pi n f_0 t)} \end{equation} i.e. an error is made and x(t) is distorted. The larger N is, the less distortion. For any given signal, the higher $f_{max}$ of the transmission medium, the less distortion.\footnote{Or, if we fix a level of quality we consider acceptable for a received signal, the larger N is, the more signals we can {\bf multiplex} onto the given medium. This will be explored later.} Now from calculus, for any given signal x(t), ${a_n},{b_n} \to 0$ as\ $n \to \infty$. Thus, the $a_n$ and $b_n$ get small after a while, say, after $a_M$ and $b_M$. So, \begin{equation} x(t) \approx {\sum_{n=0}^{ M } {a_n} \cos ( 2 \pi n f_0 t)} + {\sum_{n=1}^{ M } {b_n} \sin ( 2 \pi n f_0 t)} \end{equation} As long as we send x(t) along a transmission medium for which $N \geq M$, x(t) will be received with reasonable fidelity. \section{Effect of Bandwidth on Transmission Rates} In order to get an idea of the relation of bandwidth to transmission rate, suppose we are sending the bit pattern 101010101010..., and that in our transmission medium a level (say volts, tenths of volts, or whatever), a 1 represents a 1 bit and a -1 represents a 0. Note that we are just using 101010101010... as an example; our transmission medium must be able to send all bit patterns. But we will use this one to illustrate the issues involved, so we can see the how the bandwidth of our transmission medium limits how fast we can send bits. Suppose we are sending at the rate of $2 \times 10^6$ bits per second. What will be the period for the signal 101010101010 above? In terms of bit times, the period is two bit times. How much is that in seconds? At our 2 million bits/sec rate, each bit takes 0.5 millionths of a second, i.e. 0.5 microsecond. Thus our period is T = 1 microsecond, and the fundamental frequency $f_0$ is 1 megaherz (MHz, i.e. million cycles per second). (Recall that the fundamental frequency is the number of periods per unit time.) The period consists of two bit times, specifically one 1 bit and one 0 bit, each taking T/2 time to transmit. We could describe x(t) as having the value 1 from 0 to $\frac{T}{2}$, the value -1 from $\frac{T}{2}$ to T, and so on, repeating in this manner from $- \infty$ to $\infty$. Using the formulas, we find that all $b_n$ will be 0, and all even-numbered $a_n$ will be 0. Thus we need only calculate the integrals for $a_1, a_3, a_5$ and so on, using equation (\ref{eight}), and these coefficients turn out to be $\frac{4}{\pi}, -\frac{4}{3\pi}, \frac{4}{5\pi}$, et cetera. Note that since $f_0$ is 1 MHz, these nonzero coefficients will be for cosine terms with frequencies 1 MHz, 3 MHz, 5 MHz, et cetera. Now suppose our transmission medium only transmits frequencies up to 5 MHz. Then even though we place onto our transmission medium x(t)---a squarewave with values 1, -1, 1, -1, etc.---the medium will distort x(t) to the function \begin{equation} \frac{4}{\pi} [cos(2 \pi \cdot f_0 t) - \frac{1}{3} cos(2 \pi \cdot 3 f_0 t) + \frac{1}{5} cos(2 \pi \cdot 5 f_0 t)] \end{equation} Using microseconds for our horizontal axis (since T = 1 microsecond), a plot of this series would look like this: \vspace{0.2in} \par \psfig{file=Sine5.PS} \par So, even though x(t) is distorted, the result is not a bad approximation to the original waveform! We could design circuitry that would be able to distinguish well between a 1 bit and a 0 bit in this case.\footnote{Of course, this is not accounting for other factors, such as further distortion of the waveform, such as by line noise.} For example, we could design the receiver to treat any signal which is greater than 0.8 during 80\% of a bit time as a 1, and any signal which is less than -0.8 during 80\% of a bit time as a -1. In other words, a medium which transmitted all frequencies up to 5 MHz would be able to successfully send a 2,000,000 bit rate, or at least for the bit pattern we have looked at here, 101010... {\bf The point of all this is that the bandwidth of the transmission medium determines the maximum bit rate at which one can send.} If the transmission medium only passed frequencies up to 4 MHz, we would only get two cosine terms, and thus a much poorer approximation to x(t), maybe so poor that it would produce bit-recognition errors; we would thus probably be forced to transmit at a lower rate. Likewise if we retain the medium with bandwidth 5 MHz but double our bit rate to 4,000,000 bits per second, we will have the same $a_n$ and $b_n$ (recall our earlier discussion that the $a_n$ and $b_n$ are independent of the bit rate), but $f_0$ would double, and thus the harmonic frequencies $n f_0$ would all double, thus removing some of them from our medium's bandwidth. \section{Bit Encodings} Where does one bit start and the next begin? For instance, suppose our line has a high voltage for a 1 and a low voltage for a 0, and suppose that our sending device were to send 10 1s in a row. Then the line would be high for 10 bit times, but how would the receiver know that this is not, for example, only nine 1s? There is no ``boundary'' between bits. Theoretically speaking, this problem is solved by having a clock on the receiver end, carefully marking off bit times. If we are sending at, say, 10,000 bits per second, that would mean that each bit takes 100 millionths of a second, i.e. 100 microseconds. So if the receiver sees the line is high for 1000 microseconds, then the receiver knows that this must be 1000/100 = 10 1 bits. The problem with this is that there will be discrepancies between the sender and receiver clocks. Over time, this discrepancy will accumulate and eventually the receiver will be off a full bit time from the sender---and thus the receiver will look at the wrong bit. In order to deal with this, we design the receiver to frequently resynchronize itself with the sender. The way this is typically done is to have the receiver look for {\bf line transitions}, i.e. changes from 1 to 0 or vice versa. Since a transition occurs between two consecutive bits, the receiver can use this to resynchronize its clock with that of the sender. But if we happen to send a long string of 1s, the receiving clock won't have this chance to resynchronize. One method of dealing with this is {\bf non-return-to-zero-inverted} (NRZI) coding, which works as follows. If we wish to send a 1 bit, we place a voltage opposite to what we sent in the previous bit. In other words, if the previous line level had been low, we now put a high value on the line, and vice versa. If we wish to send a 0, we send the same voltage level as in the previous bit. The idea here is to force frequent line transitions. This scheme does this, since every 1 bit will be coded as a line transition. Thus the receiver has frequent opportunities to resynchronize its clock. That solves the problem of having a long string of 1s in our message, but it doesn't cover cases in which we happen to send long strings of 0s. Both problems are addressed by {\bf Manchester} coding, which forces a bit transition in {\it each} bit time. Under this system, if we wish to send a 0 bit, we hold the line at low voltage for one transmission bit time (which will be half a data bit time) and then bring it up to high voltage for one transmission bit time. If we wish to send a 1 bit, we have the line high for a transmission bit time and then low for a transmission bit time. This guarantees a line transition during each data bit, so the receiver has a chance to resynchronize its clock during every data bit. But Manchester is wasteful---each data bit now needs two transmission bit times to send. Thus for any given transmission line, and thus for any given maximum line bandwidth, and thus for any given maximum transmission bit rate, it cuts down the allowable data bit transmission rate by half. Note that Manchester coding also changes the pattern in which we uses our transition line's frequency bandwidth. For instance, recall our earlier example, in which we were transmitting at 2 MHz, and our message consisted of 101010... Using Manchester coding, the values on our line would be 100110011001..., which has a quite different Fourier series than would the 101010... pattern we would use without Manchester coding. So, compromise schemes have been developed, such as the famous 4B/5B system. Under this scheme, each group of four message bits is coded in five bit times. Here is the translation table: \begin{verbatim} data coding ---- ------ 0000 11110 0001 01001 0010 10100 0011 10101 0100 01010 0101 01011 0110 01110 0111 01111 1000 10010 1001 10011 1010 10110 1011 10111 1100 11010 1101 11011 1110 11100 1111 11101 \end{verbatim} The five-bit code is then sent using NRZI. For example, suppose we wish to send 1011, and that the last state of the line had been high. Then we would send 1011 as low-low-high-low-high. (Make sure you understand this point.) Note that in all the five-bit codes above, there is never more than one leading 0 and never more than two trailing 0s. So, if two data items are sent back-to-back, we cannot have more than three consecutive 0s. This solves the original NRZI problem concerning long strings of 0s discussed earlier, and since NRZI itself handles the problem of long strings of 1s, you can see that the 4B/5B scheme works pretty well. \end{document}