Chapter 3: Physical Layer-Overview
Physical layer is the bottom layer in OSI model and is responsible for sending and receiving data. The data exchanged at this layer are bits. Bits are transmitted either in an analog signal or in a digital signal.
An analog signal can be represented in a continuous, smooth waveform over time. An example is a voice signal in its original form before it is digitized.
A digital signal is in a pulse shape where 0’s and 1’s are represented at difference voltage levels. An example of a digital signal is the 0’s and 1’s used inside a computer where 0’s have 0 volts and 1’s have 5 volts.
Digital signal Analog signal
Data bits are transmitted through a transmission medium. There are several kinds of media: twisted pair, coaxial cable, optical fiber and the space(the medium for microwave, satellite, and wireless communications).
We need to grasp some of the basic concepts such as signal bandwidth, transmission line(channel) bandwidth, data rate, transmission impairments.
In this chapter, we will look at how data bits can be transmitted through a medium. We will discuss basic concepts such as analog/digital transmission, FDM(Frequency Division Multiplexing)/TDM(Time Division Multiplexing), Signal Bandwidth, Channel Bandwidth. Ultimately, we would like to understand how we can achieve a data rate(bits/sec) on a given transmission medium. Let’s first explore some of the basic concepts and terms.
Analog signal and Digital signal
An analog signal is a continuous signal in which the signal intensity (voltage) changes in a smooth path over time. There is no break in the signal.
A digital signal is a pulse shaped signal in which the signal intensity (voltage) changes abruptly (e.g. voltage changes from 0 volt to 5 volts instantaneously).
In absolute sense, all signals are analog signals; a perfect digital signal which changes voltages infinitely fast can not be achieved. If we look at the digital pulse closely, we would see that the signal changes quickly but not infinitely fast as in an ideal digital pulse. In other words, since both analog signals and digital signals have the same physical, electrical properties, we can analyze both in equal terms.
3.1 Bandwidths of signals and channels
* Bandwidth of a signal
A signal, either analog or digital, has many frequency components that range between certain frequencies. This range(e.g. between f1 and f2 ) is called a bandwidth or signal bandwidth. For example, a human voice signal is said to have frequency components between 20Hz and 20KHz. This is the voice signal’s bandwidth.
* Bandwidth of a channel
A given channel has a limited range of frequencies that the channel is capable of transporting. This range is called as the bandwidth of the channel. The bandwidth of a channel determines its capacity(channel capacity or data rate of a channel). An analogy for bandwidth is a “thickness of a water pipe”. The thicker the pipe, the more water it can carry. For example, a telephone channel has a channel bandwidth of 3100Hz (ranges from 300Hz to 3400Hz).
The following figure shows the bandwidth of various transmission media. You may easily notice that the scale of the frequency axis is in an exponential scale. Each block is 10 times the size of the previous block.
* Relationship between Signal bandwidth and channel bandwidth
A signal is transmitted through a channel. For example, a voice signal is transmitted through a telephone channel. Obviously, a channel’s bandwidth must be equal or larger than the signal’s bandwidth. An analogy is; the thickness of a pipe(channel bandwidth) must be big enough to carry the water flowing in(signal bandwidth).
Therefore, we can say channel bandwidth >= Signal bandwidth.
From the above discussions about a voice signal’s bandwidth and a telephone channel’s bandwidth , you may have noticed that there is discrepancy between them; A voice signal’s bandwidth(about 20KHz=(20Hz – 20KHz)) is not matching to a telephone channel bandwidth (3100Hz=(300Hz – 3400Hz)). This discrepancy is due to the design of telephone systems. An extensive analysis of voice signals resulted in the conclusion that the most of the signal’s energy and the intelligence of voice are concentrated between 300Hz and 3400Hz. Therefore, a telephone channel is limited to the 3400Hz without losing much of the quality of a voice. By limiting a telephone channel, a telephone system can multiplex(carry multiple voice channels) many channels in a given system. An analogy is: by limiting the size of packages, post office can handle many packages in a truck.
One of the most important results from an extensive analysis of human voices is the fact that the most of the energy of human voice is concentrated in a rather small range of frequencies (between 300 Hz to 3400Hz). Also the intelligence of human voice shows similar frequency distribution. Therefore, it is safe to conclude that the human voice can be restricted(by cutting off the frequencies lower than 300 Hz and also cutting off the frequencies higher than 3400 Hz) to the frequencies between 300 Hz and 3400 Hz.
This range is adopted as a telephone channel’s bandwidth. Since most of the energy and intelligence are concentrated in this range, the quality of a voice would not be greatly degraded by this limitation.
By narrowing the bandwidth of a voice signal, we can multiplex(mix) many voice signals into a given medium so that multiple voices travel together thus saving the cost of long distance carriers.
Physical layer is responsible for transmitting and receiving bits. Therefore, the Protocol Data Unit (PDU) of the physical layer is a bit. When a physical layer device sends bits(in analog or digital signals) through a physical medium, the success of the transmission depends upon the strength and quality of the signal(source signal) and the characteristics of the transmission medium such as channel bandwidth, distance, interference, error rate. Ultimately, we would like to know the maximum data rate of a given channel. Before discussing the basic theory, let us first look at a generic model of communications. The model is adopted by ITU-T.
DTE --- DCE -------------------------------------- DCE --- DTE
Transmission system
DTE: Data Terminal Equipment—a device which is a source and a destination of data
DCE: Data Circuit-terminating Equipment—a device which transmits and receives bits
This model can be applied to most of the communication systems.
Example #1:
DTE --- DCE -------------------------------------- DCE --- DTE
Desktop --- Modem ---------------------------- Modem --- Server
Telephone system
Example #2:
DTE --- DCE -------------------------------------- DCE --- DTE
Desktop --- Network ----------------------------- Network --- Server
Card LAN/WAN card
Connection
When we analyze a data signal transmitted through a transmission medium, we need to look at “signal’s bandwidth”, “transmission channel’s bandwidth”, “data rate of a channel”, etc.
We can analyze a signal in time domain (signal as a function of time) or in frequency domain (signal as a function of frequency).
The analysis of signals in frequency domain gives us a good insight about signals. French mathematician Fourier established solid ground work and we will borrow essential part of his work.
The main idea of Fourier’s theorem can be summarized as:
"A given signal (either analog or digital) can be proved to be made up of components at various frequencies, where each component is a sinusoid"
It would be out of scope of this book to prove his theorem mathematically. But there is a way to (sort of) prove his theorem graphically.
Here is a diagram which shows the summarization of several sinusoids.
S(t) = sin(2pf)t + 1/3 sin3(2pf)t + 1/5 sind5(2pf)t + …..
As we can visualize from the diagram, when we add several signals of different frequencies the resulting signal looks close to a digital signal (bits of ones and zeroes). In other words, a digital signal is proven to have many components at various frequencies. It is the essential point in Fourier’s theory. Now, some of the remaining questions are; “Which components are needed to make up a signal?”, “How many components are needed to make up the signal accurately”, “How about the strength of each component?”.
From the above diagram, we can easily see that the summed signal will look closer to a digital bit as you keep on adding more components. In fact, to get a pulse shape closer to a perfect one, we need to add more and more components (components are called as 1st harmonic, 2nd harmonic, 3rd ….). To get an absolutely perfect pulse shape, we need to add infinite number of components. From this fact, we can state the following.
The spectrum of a signal is the range of frequencies that the signal is composed of no matter how wide the range might be. This width of the spectrum is the absolute bandwidth of the signal. As an example, for a perfect pulse, the absolute bandwidth is infinitely wide (from 0Hz to “infinity” Hz). However, most of the energy in a signal is concentrated in a relatively narrow width. This width is referred to as the “effective bandwidth” or simply the “bandwidth”.
We have already seen an example for the effective bandwidth in the case of a voice signal’s bandwidth. A voice signal’s absolute bandwidth is commonly known as between 20Hz and 20KHz. Since most of energy of voice signals is concentrated in certain range, we can set the effective bandwidth. The range happens to be between 300Hz and 3400Hz.
Now, let’s turn our attention to channel bandwidth and look at an example to draw a conclusion from the preceding discussion about bandwidth. Let’s assume that we need to send perfect pulse-shaped signals (as digital data) through a transmission channel which as a limited bandwidth. We know that a perfect pulse is comprised of infinitely many harmonics (components), therefore requires a channel with an infinite bandwidth. Since we have a channel with a limited bandwidth, it is rather obvious that the result of the transmission will not be ideal. In other words, when we send a perfect pulse, we will get a not-so-perfect pulse at the receiving end. Let’s look at this in a diagram.
As we increase the channel’s bandwidth, the signal becomes more accurately represented. However, if we increase a channel’s bandwidth, the cost will also increase. Therefore, we need to find a compromise between the channel bandwidth (directly related to cost) and the quality of transmission. The actual calculation of proper bandwidth is out of scope for this book. It is left to communication engineers.
Let’s clarify certain terms before we proceed further. Often the terms, “channel” and “line”.
A “line”, “transmission line”, “circuit”, or sometimes called as “transmission system” refers to a hardware system including the actual copper or fiber cables and the associated equipments such as transceivers (a device with both the transmitter and the receiver), connection mechanisms, and any other supporting systems.
A “channel”, or “transmission channel” refers to a specific bandwidth assigned to a given application.
Usually, a “transmission line” can carry one or more channels depending upon its configuration. So, in a precise sense, a channel is not always the same as a line.
Let’s look at an example. A cable TV system has a line coming into a house and this one cable carries many TV channels. In some cases, a line may carry only one channel, for example, a telephone line between a house and a switching office (central office) carries only one voice channel. This is why we said that a line can carry one or more channels depending upon its configuration. Actually, a UTP between Central Offices carries 24 digitized voices in a Time Division Multiplex (TDM) method.
Unfortunately, the terms, line and channel, have been used interchangeably. Therefore, we should look at the context of their usage.
3.2 Transmission impairment
Transmission lines can not be made perfect. They have impairments which affect their bandwidths, therefore, their data rates. Transmission impairments can be classified into two categories: (a) Non-transient impairments, (b) Transient impairments.
(a) Non-transient impairments
These impairments are static. In other words, they do not change with time. Therefore, engineers can do something to improve the quality of the signal. There are several kinds in this category:
· Attenuation—Signals loose power as they travel along the transmission lines. We need to enhance the signals by amplifiers and repeaters. Another factor that needs attention is the fact that the attenuation is not uniform across frequencies of the signals. In other words, different frequency signals attenuate at different amount. This is called as “attenuation distortion” and is cured by electronic circuits called as “equalizer”.
Attenuation
· Dispersion
Signals tend to spread as they travel, with the amount of spreading dependent on the frequency.
· Delay distortion
Since the velocities of propagation vary with frequency, various frequency components of a signal arrive at the receiver at different times.
· Echoes
Due to imperfect circuits and connection points, echoes do occur. There are circuits that can detect and reduce echoes. They are called as “echo cancellation” circuits.
· Others: Amplitude jitter, Phase jitter, etc.
(b) Transient impairments
These impairments change with time and unpredictable. In other words, they occur randomly. They are collectively called as “noise”. There are several kinds of noises.
· Thermal noise
This noise occurs due to the thermal agitation of electrons in conductors, and is a function of the temperature. It is often called as “white noise” because it affects all frequencies.
· Intermodulation noise
This noise is from interference of different frequencies in a medium.
For example, the mixing of signals at frequencies f1 and f2 might produce energy at the frequency f1 + f2 . This signal could interfere with the signal at frequency f1 + f2 .
· Crosstalk
This occurs when more than several conductors run together which typically occurs to wires in a building, telephone wiring, and LAN wiring, etc. Crosstalk is due to the electromagnetic coupling between adjacent wires. When a electric current flows on one pair of wires, the magnetic field formed along the wires affect the electric field of the adjacent wires. Crosstalk can be reduced by twisting wires. Wires with larger bandwidth have more twist per foot.
· Impulse noise
This type of noise is due to electrical surges by lightening, switching of high powered equipments nearby, etc. It usually arrives as a short, high amplitude spikes. It is one of the major causes of error in communications.
A transmission line with all of the above impairments obviously has a limit in its ability in carrying signals. This ability is measured by data rate, in modern digital communications, bits per second. The data rate is directly related to the channel’s bandwidth.
Data rate of a channel can not be determined solely by mathematical calculations but rather determined by the combination of mathematics and engineering experiments. In the following section, we will discuss some of the ways to calculate the data rate of a channel mathematically even though these calculations are far from the actual data rate.
3.2 Date rate of a channel
Before we look at a more elaborate analysis of a channel’s data rate, let’s start with an idealistic view of a channel and its data rate. An ideal channel is the one which does not have any impairment. Therefore, the channel is error-free and also we assume that we have infinitely accurate transmitters and receivers. An American scientist Nyquist studied this ideal case scenario and came up with the Nyquist’s Law.
* Nyquist Law
Nyquist(1924) proved that “The maximum signaling rate achievable by a “noiseless channel” equals to two times its bandwidth(2*H)”. In a simpler words, “the number of signals(or signal changes per second) that can be sent over a channel is two times its bandwidth”. Therefore, the bandwidth of a channel determines the maximum number of signal changes. Now if each signal carries one bit of data, then the maximum data rate is 2*H. If each signal carries two bits of data, then the maximum data rate is 2*(2*H).
This maximum signaling rate is called as “Nyquist rate”. Its unit is called as “Baud”. For example, if H= 1000Hz, then the Nyquist rate is 2*1000= 2000 Baud. Baud can be interpreted as “Signals per second”. Now, if each signal carries one bit, then it has 2000BPS(Bits Per Second) and Baud rate equals BPS. If each signal carries two bits, then it still has 2000 Baud but 4000 BPS. Therefore, in this case, BPS/Baud= 2.
Nyquist's Law (C = 2H log 2 M) Baud
Assuming a noiseless channel, the channel’s data rate(C bits/sec) is determined by its bandwidth (H) and the number of encoding levels (M) in the signal.
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Four-levels |
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Encoding of signal using two levels and four levels
(note that the number of signals are the same for both)
In easier terms, Nyquist’s Law can be stated as follows:
Given a channel with bandwidth H, the maximum number of signal changes per second is 2H (also called as a “signaling rate” of the channel).
Each signal can be used to carry one bit of information when we use two level signaling.
The data rate doubles if we use four level signaling (notice that we do not change the signaling rate).
Assuming a noiseless channel, in theory, we can increase the data rate as high as we want by increasing the number of levels in each signal.
Nyquist Law provides us with some insight into the data rate but unfortunately it is not applicable to a real world since it assumes a “noiseless channel” but we know that noise is present in every circuit.
* Shannon’s Law
A more in depth and more practical study was done by Claude Shannon of Bell Labs in late 1940s. His theory laid a firm foundation for information theory. The major contribution of his theory is that it gives us a guideline on the maximum data rate that we can squeeze out of a given channel.
Bell Labs' Claude Shannon, creator of
modern information theory.
From http://www.lucent.com/minds/infotheory/what.html
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Data rate of a given channel (e.g. a voice channel of 4Khz bandwidth) is a function of:
· Bandwidth: The limitations due to the channel’s physical properties
· Error rate: Due to the impairments of the channel
Shannon's Law: C = 2W log 2 (1 + S/N)
§ S = Signal strength
§ N = Noise strength (but assumes thermal noise only)
§ Example (telephone channel):
Bandwidth(H) = 3100Hz (300-3400Hz)
Typical S/N = 1000/1 (30 dB)
Channel Capacity = C = 2H log 2 (1 + S/N)
Where S/N is the power ratio of Signal(S) and Noise(N)
If we plug in the above values to the Channel Capacity formula;
C = 2*3100* log 2 (1 + 1000/1) = 30894 BPS
Note: Decibel(dB or db) is a measure often used in communications theory to represent the power ratio of signals. It(dB) is defined as:
dB = 10 log 10 P1/P2 where P1/P2 is the ratio of power of the signals
In the above case where we need to compare the ratio of Signal(S) and Noise(N), the formula is dB = 10 log 10 S/N and if we plug in the S/N=1000/1 into the formula we have dB = 10 log 10 1000/1 which calculates to 30dB.
Another example of dB is:
Category 5 UTP, as specified in ANSI/TIA/EIA-568, at 16MHz, must have a loss less than 10 dB for 100 meters. Let’s see what this means. Using the formula above; 10 dB = 10 log 10 P1/P2 and from this formula we can derive that the power ratio P1/P2 is 10/1. In other words, a Cat 5 UTP cable must not lose its input signal power (P1) more than 10 to 1 ratio at the end of 100 meters.
Here are some values of dB to know:
3 dB = 2/1 ratio
10 dB = 10/1
20 dB= 100/1
30 dB = 1000/1
Unfortunately, Shannon’s formula was derived considering only the “Thermal noise”. No other types of noise could be put into the formula. It is due to the fact that only the “thermal noise” can be mathematically formulated since it (thermal noise) is present in every frequency in a given channel.
Actual data rate of a channel is usually lower than the Shannon’s capacity due to other types of noises. The significance of Shannon’s work is that his formula gives us a guideline or a goal for the maximum achievable data rate for a channel.
* Application of Shannon’s formula
For a telephone channel with bandwidth of 3100 Hz, the formula gives us: C = 2*3100* log 2 (1 + 1000/1) = 30894 BPS.
This data rate was an ultimate goal of modem designers for a long time. Here, you might raise a question: “A modern modem can achieve up to 56000 BPS and how can it be higher than the Shannon’s rate which is supposed to be the theoretical maximum data rate?”.
Answer: Still, the maximum data rate is limited by Shannon’s formula but with current technique of “data compression”. So, the actual data rate is lower than “30894 BPS” but with compression we can achieve up to 56000 BPS.