1. Chapter # 3 Data and Signals

2. Introduction
One of the major functions of physical layer is to move data in the form of electromagnetic signals across a transmission medium.
Thus, the data must be transformed to electromagnetic signals to be transmitted.

3. 1. Analog and Digital Data
Data can be analog or digital.
The term analog refers to information that is continuous
e.g. analog clock hh:mm:ss
The term digital data refers to information that has discrete states.
e.g. digital clock hh:mm
Analog data take on continuous values.
Digital data take on discrete values.

4. Analog and Digital Signals

5. 2. Periodic and Nonperiodic Signals
Both analog and digital signals can take one of two forms: periodic or nonperiodic
A periodic signal
Completes a pattern within a measureable time frame.
Repeats that pattern over subsequent identical period
A nonperiodic signal
Changes without exhibiting a pattern or cycle that repeats over time.

6. Periodic and Nonperiodic Signals
In data communications, we commonly use:
Periodic analog signals ( because they need less bandwidth).
and nonperiodic digital signals ( because they can represent variation in data)

7. A. Periodic Analog Signals
Periodic analog signals can be classified as simple or composite.
A simple periodic analog signal, a sine wave, cannot be decomposed into simpler signals.
A composite periodic analog signal is composed of multiple sine waves.

8. 1) Sine Waves
The sine wave is the most fundamental form of a periodic analog signal.
A sine wave is represented by three parameters: Peak amplitude, Frequency, and Phase.
Peak amplitude: it is the absolute value of the highest intensity. It is normally measured in volts.

9. Sine Waves
Frequency: it refers to the number of periods in 1 s. It is formally expressed in Hertz (Hz).
Period is the amount of time, in seconds, a signal needs to complete one cycle (the completion of one full pattern).
Therefore , frequency and period are the inverse of each other.
Note : Frequency is the rate of change with respect to time.
Change in a short span of time means high frequency.
Change over a long span of time means low frequency.
If a signal does not change at all, its frequency is zero

10. Units of period and frequency
TABLE 3.1

11. Examples
Q1 . The power we use at home has a frequency of 60 Hz. The period of this sine wave can be determined as follows:
Q2. Express a period of 100 ms in microseconds.
Solution
From Table 3.1 we find the equivalents of 1 ms (1 ms is 10−3 s) and 1 s (1 s is 106 μs). We make the following substitutions:

12. Examples
Q3. The period of a signal is 100 ms. What is its frequency in kilohertz?
Solution
First we change 100 ms to seconds, and then we calculate the frequency from the period (1 Hz = 10−3 kHz).

13. Sine Waves – cont’ A sine wave with a phase 0° is not shifted.
It describes the position of the waveform relative to time 0.
It is measured in degree or radian
To look to the phase is in term of shift or offsit:
A sine wave with a phase 0° is not shifted.
A sine wave with a phase 90° is shifted to the left by ¼ cycle.
A sine wave with a phase 180° is shifted to the left by ½ cycle.

14. Three sine waves with the same amplitude and frequency, but different phases

15. The figure below Two signals with the same amplitude and phase, but different frequencies
The figure below show Two signals with the same phase and frequency, but different amplitudes

16. Wavelength
Wavelength binds the period or frequency of the simple sine wave to the propagation speed of the medium.
Wavelength depends on both the frequency and the medium.
Wavlength = propgation speed X period = progation speed/ frequency

17. Example# 1
In a vacuum, light is propagated with a speed of 3 X m/s. (that speed is lower in air and cable.) . The frequency of red light is 4 X 1014
Wavelength is normally measured in micrometers.
Solution
Wavelength= c/f = (3 X 108 ) / (4 X 1014)
= 0.75 X m= 0.75 μm

18. Time and Frequency Domain
A complete sine wave in the time domain can be represented by one single spike in the frequency domain.

19. Example
This example Shows three sine waves, each with different amplitude and frequency. All can be represented by three spikes in the frequency domain.

20. 2) Composite Signals
A single-frequency sine wave is not useful in data communications; we need to send a composite signal, a signal made of many simple sine waves.
e.g. if we use single sine wave to convey a conversation over the telephone. It would just hear a buzz.
According to Fourier analysis, any composite signal is a combination of simple sine waves with different frequencies, amplitudes, and phases.

21. 2) Composite Signals
A composite signals can be periodic or non periodic.
A periodic composite signal can be decomposed into a series of simple sine waves with discrete frequencies ( with integer values [1,2,3 ,….] ).
A nonperiodic composite signal can be decomposed into a combination of an infinite number of simple sine waves with continuous frequencies. ( with real value)

22. Decomposition of a composite periodic signal in the time and frequency domains

23. The time and frequency domains of a nonperiodic signal
In this case, the composite signal cannot be periodic, because it implies that we are repeating the same word or words with exactly the same tone.
This can be the signal created by a microphone or a telephone set when a word or two is pronounced.

24. Bandwidth
The bandwidth of a composite signal is the difference between the highest and the lowest frequencies contained in that signal.
e.g. if a composite signal contain frequencies between and 5000, its bandwidth is = 4000

25. Example#1
If a periodic signal is decomposed into five sine waves with frequencies of 100, 300, 500, 700, and 900 Hz, what is its bandwidth? Draw the spectrum, assuming all components have a maximum amplitude of 10 V.
Solution
Let fh be the highest frequency, fl the lowest frequency, and B the bandwidth. Then

26. Example#2
A periodic signal has a bandwidth of 20 Hz. The highest frequency is 60 Hz. What is the lowest frequency? Draw the spectrum if the signal contains all frequencies of the same amplitude.
Solution
Let fh be the highest frequency, fl the lowest frequency, and B the bandwidth. Then

27. Example#3
A nonperiodic composite signal has a bandwidth of 200 kHz, with a middle frequency of 140 kHz and peak amplitude of 20 V. The two extreme frequencies have an amplitude of 0. Draw the frequency domain of the signal.
Solution
The lowest frequency must be at 40 kHz and the highest at 240 kHz. Figure 3.15 shows the frequency domain and the bandwidth.

28. B- DIGITAL SIGNALS
In addition to being represented by an analog signal, information can also be represented by a digital signal. For example, a 1 can be encoded as a positive voltage and a 0 as zero voltage.
A digital signal can have more than two levels. In this case, we can send more than 1 bit for each level.

29. Figure 3.16 Two digital signals: one with two signal levels and the other with four signal levels

30. Example # 1
A digital signal has eight levels. How many bits are needed per level? We calculate the number of bits from the formula
Each signal level is represented by 3 bits.

31. Example #2
A digital signal has nine levels. How many bits are needed per level?
We calculate the number of bits by using the formula:
Log2 L= number of bits in each level
Log2(9)=3.17bits.
However, this answer is not realistic. The number of bits sent per level needs to be an integer as well as a power of 2.
For this example, 4bits can represent one level.

32. Bit rate and bit interval
Most digital signals are nonperiodic, frequency and period are not appropriate. Another terms instead of frequency is bit rate and instead of period: bit interval(bit duration)
Bit rate: number of bits per second bps
Bit interval=1/bit rate
Bit length =propagation speed x Bit interval

33. Example
Assume we need to download text documents at the rate of 100 pages per minute. What is the required bit rate of the channel?
Solution
A page is an average of 24 lines with 80 characters in each line. If we assume that one character requires 8bits
The bit rate is:
=100x24x80x8/60
=25.6Kbps

34. Digital Signal as a composite Analog Signal
Note
A digital signal is a composite analog signal with an infinite bandwidth.
Fourier analysis can be used to decompose a digital signal
If the digital signal is periodic (rare in data communications), the decomposed signal has a frequency domain representation with an infinite Bandwidth and discrete frequencies.
If it is nonperiodic, the decomposed signal still has infinite Bandwidth, but the frequencies are continuous.

35. Figure 3.17 The time and frequency domains of periodic and nonperiodic digital signals

36. Transmission of Digital Signals
How can we send a digital signal from point A to point B?
We can transmit a digital signal by using one of two different approach :
Baseband transmission
Broadband transmission

37. 1. Baseband transmission
Means sending a digital signal over a channel without changing the digital signal to an analog signal
Base band transmission required a low-pass channel (channel with a B-W that starts from zero)

38. Figure 3.20 Baseband transmission using a dedicated medium
Note
Baseband transmission of a digital signal that preserves the shape of the digital signal is possible only if we have a low-pass channel with an infinite or very wide bandwidth.

39. 2. Broadband Transmission (modulation)
Means changing the digital signal to an analog signal for transmission. Modulation use a band-pass channel (a channel with a B-W that doesn't start from Zero). This type of channel is more available than a low-pass channel.

40. Note
If the available channel is a bandpass channel, we cannot send the digital signal directly to the channel; we need to convert the digital signal to an analog signal before transmission.

41. Figure 3.24 Modulation of a digital signal for transmission on a bandpass channel

42. Example
An example of broadband transmission using modulation is the sending of computer data through a telephone subscriber line, the line connecting a resident to the central telephone office. These lines are designed to carry voice with a limited bandwidth. The channel is considered a bandpass channel. We convert the digital signal from the computer to an analog signal, and send the analog signal. We can install two converters to change the digital signal to analog and vice versa at the receiving end. The converter, in this case, is called a modem.

43. Digital modulation and multiplexing
Baseband transmission: the signal occupies frequencies from zero up to a maximum that depends on the signaling rate.
It is common for wires.
Passband transmission: the signal occupies a band of frequencies around the frequency of the carrier signal.
It is common for wireless and optical channels.
Multiplexing: Channels are often shared by multiple signals.

44. 1. Baseband transmission
Line codes: (a) Bits, (b) NRZ, (c) NRZI, (d) Manchester, (e) Bipolar or AMI.
Balanced Signals: Signals that have as much positive voltage as negative voltage even over short periods of time are called balanced signals.

45. 2- Passband Transmission
We can take a baseband signal that occupies 0 to B Hz and shift it up to occupy a passband of S to S +B Hz.
Digital modulation is accomplished with passband transmission by regulating or modulating a carrier signal that sits in the passband.
ASK (Amplitude Shift Keying), two different amplitudes are used to represent 0 and 1.
FSK (Frequency Shift Keying), two or more different tones are used.
PSK (Phase Shift Keying): the carrier wave is systematically shifted 0 or 180 degrees at each symbol period.

46. Passband Transmission
(a) A binary signal. (b) Amplitude shift keying. (c) Frequency shift keying. (d) Phase shift keying.

47. Multiplexing
1- Frequency Division Multiplexing
FDM: divides the spectrum into frequency bands, with each user having exclusive possession of some band.

48. 2- Time Division Multiplexing
TDM : The users take turns (in a round-robin fashion), each one periodically getting the entire bandwidth for a little burst of time.
Bits from each input stream are taken in a fixed time slot and output to the aggregate stream.

49. 3 - Code Division Multiplexing
CDM : is a form of spread spectrum communication.
Chip sequences for four stations.
Signals the sequences represent

50. 3-4 TRANSMISSION IMPAIRMENT
Signals travel through transmission media, which are not perfect. The imperfection causes signal impairment.
This means that the signal at the beginning of the medium is not the same as the signal at the end of the medium. What is sent is not what is received.
Three causes of impairment are attenuation, distortion, and noise.

51. 1. Attenuation – a loss of energy
when Signal travels through a medium, it losses some of its energy in overcoming the resistance of the medium. To compensate for this loss, amplifiers are used to amplify the signal.
Decibel: Measure the relative power(attenuation)
dB=10 log10 P2 / P1

52. See the book for more Examples
Suppose a signal travels through a transmission medium and its power is reduced to one-half. This means that P2 is (1/2)P1. In this case, the attenuation (loss of power) can be calculated as
A loss of 3 dB (–3 dB) is equivalent to losing one-half the power.
See the book for more Examples

53. 2. Distortion
Distortion : means that signal changes its form or shape.
Each signal component has its own propagation speed through the medium and therefore ,its own delay in arriving final destination

54. 3. Noises
Thermal noise: is the random motion of electrons in a wire which creates an extra signal not originally sent by the transmitter
Induced noise: Comes from sources such as motors and appliances.
Crosstalk noise: Is the effect of one wire on the other.
Impulse Noise: is a spike ( a signal with high energy in a very short time) that comes from power lines, lighting and so on.

55. Signal-to-Noise Ratio
SNR: ratio between signal power to the noise power
A high SNR: means the signal is less corrupted by noise
A low SNR: means the signal is more corrupted by noise.
SNR can be described in db units: SNR db=10 log10 SNR

56. Example
The power of a signal is 10 mW and the power of the noise is 1 μW; what are the values of SNR and SNRdB ?
Solution
The values of SNR and SNRdB can be calculated as follows:
, SNRdb= 10 log10 10,000 = 40

57. Data Rate Limits
A very important consideration in data communications is how fast we can send data, in bits per second, over a channel. Data rate depends on three factors:
The bandwidth available
The level of the signals we use
The quality of the channel (the level of noise)
Two theoretical formulas were developed to calculate the data rete :
By Nyquist for a noiseless channel
By Shannon for a noisy channel

58. Nyquist Theorem (noiseless channel)
The relation between bandwidth and data rate in a noiseless channel (throughput)
Maximum data rate = 2 x bandwidth x log2 L
where L: No of signal levels used to represent data
Example :
Consider a noiseless channel with a bandwidth of 3 KHz transmitting a signal with two signal levels.
The maximum bit rate/ troughput can be calculated as:

59. Example
We need to send 265 kbps over a noiseless channel with a bandwidth of 20 kHz. How many signal levels do we need?
Solution
We can use the Nyquist formula as shown:
Since this result is not a power of 2, we need to either increase the number of levels or reduce the bit rate. If we have 128 levels, the bit rate is 280 kbps. If we have 64 levels, the bit rate is 240 kbps.

60. Shannon Theorem (Noisy Channel)
The maximum throughput of a noisy channel of bandwidth B with a signal to noisy ratio of S/N is:
Shannon Capacity
Capacity = bandwidth x log2(1+SNR)
is the capacity of the channel in bps
(Max data rate/ Max throughput)
Example :
Telephone line Bandwidth=3kHz; S/N=30 dB ?
Max throughput = 3000 * log2(1+1000) =~ bps = kbps

61. Example (using both limits)
We have a channel with a 1-MHz bandwidth. The SNR for this channel is 63. What are the appropriate bit rate and signal level?
Solution
First, we use the Shannon formula to find the upper limit.
The Shannon formula gives us 6 Mbps, the upper limit. For better performance we choose something lower, 4 Mbps, for example. Then we use the Nyquist formula to find the number of signal levels.

62. Note
The Shannon capacity gives us the upper limit; the Nyquist formula tells us how many signal levels we need.

63. 3-6 PERFORMANCE
One important issue in networking is the performance of the network—how good is it?
Bandwidth
Throughput
Latency (Delay)

64. 1. Bandwidth In networking, we use the term bandwidth in two contexts.
❏The first, bandwidth in hertz, refers to the range of frequencies in a composite signal or the range of frequencies that a channel can pass.
❏The second, bandwidth in bits per second, refers to the speed of bit transmission in a channel or link.

65. 2. Throughput
Is a measure of how fast we can actually send data through a network.
The bandwidth is potential measurement of a link; the throughput is an actual measurement of how fast we can send data.
Throughput less than Bandwidth

66. Example
A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network?
Solution
We can calculate the throughput as
The throughput is almost one-fifth of the bandwidth in this case.

67. 3. Latency ( Delay)
Latency defines how long it takes for an entire message to completely arrive at the destination from the time the first bit is sent out from the source
Latency (Delay) =
propagation time + transmission time
+queuing time + processing time

68. 1) Propagation time
It is the time required for a bit to travel from the source to the destination
Propagation speed depend on the medium and on the frequency of the signal
Example :
light propagate by 3x108m/s in vacuum. It is lower in air ; it is much lower in cable.

69. 2. Transmission time
It is the time required for transmission of a message .
It depends on the size of the message and the bandwidth of the channel.

70. Example #1 :
What are the propagation time and the transmission time for a 2.5-kbyte message (an ) if the bandwidth of the network is 1 Gbps? Assume that the distance between the sender and the receiver is 12,000 km and that light travels at 2.4 × 108 m/s.
Solution
Note that in this case, because the message is short and the bandwidth is high, the dominant factor is the propagation time, not the transmission time. The transmission time can be ignored.

71. Example # 2
What are the propagation time and the transmission time for a 5-Mbyte message (an image) if the bandwidth of the network is 1 Mbps? Assume that the distance between the sender and the receiver is 12,000 km and that light travels at 2.4 × 108 m/s.
Solution
Note that in this case, because the message is very long and the bandwidth is not very high, the dominant factor is the transmission time, not the propagation time. The propagation time can be ignored.

72. 3. Queuing time
It is the time needed for each end device to hold the message before it can be processed.
It changes with the load imposed on the network, if there is heavy traffic on the network , the queuing time increases.

73. Chapter # 7 Transmission Media

74. Transmission medium and physical layer
A transmission media defined as anything that carry information between a source to a destination
Located below the physical layer and are directly controlled by the physical layer

75. Classes of Transmission Media

76. 7-1 GUIDED MEDIA
Guided media, which are those that provide a conduit from one device to another, include twisted-pair cable, coaxial cable, and fiber-optic cable.
Twisted –pair cables and coaxial cable:
use metallic (copper) conductors that transport signals in the form of electric current
Optical fiber :
transport signals in the form of the light

77. 1. Twisted-pair cable
One of the wire used to carry signal and the other as a ground. The receiver uses the difference between the two.
If the two wires are parallel, the effect of interference noise and crosstalk is big
Twisting the pair of wire balance the effect of unwanted signal and reduce it.

78. Applications of Twisted pair
Coaxial (Coax) cable
Coax cable carries signals of higher frequency ranges than those in Twisted pair cablebecause the two media are constructed quite differently
The outer conductor serves both as a shield against noise and as second conductor, which complete the circuit
Applications of Twisted pair
Used in
1. Telephone lines to provide voice and data channels
2. The DSL lines that are used by the telephone companies to provide high-data-rate connections
3. Local area networks, such as 10-base-Tand 100base-T

79. 2. Coaxial cable
Coax cable carries signals of higher frequency ranges than those in Twisted pair cable because the two media are constructed quite differently.
The outer conductor serves both as a shield against noise and as second conductor, which complete the circuit

80. Applications of coaxial cable
1.Analog telephone network where a single cable could carry 10,000 voice signals. Later it was used in Digital telephone networks where cable can carry 600Mbps
2.Cable TV network: hybrid network use coaxial cable only at the network boundaries , near the consumer.
3.Traditional Ethernet LANs.

81. Is made of glass or plastic and transmit signals in the form of light.
3. Fiber Optic Cable
Is made of glass or plastic and transmit signals in the form of light.
Light travels in a straight line as long as it is moving through a single uniform substance. If a ray of light traveling through one substance enters another substance of different density , the ray change direction as shown:
As the figure shows, if the angle of incidence I (the arIgle the ray makes with the
line perpendicular to the interface between the two substances) is less than the critical
angle, the ray refracts and moves closer to the surface. If the angle of incidence is
equal to the critical angle, the light bends along the interface. If the angle is greater than
the critical angle, the ray reflects (makes a turn) and travels again in the denser substance.
Note that the critical angle is a property of the substance, and its value differs
from one substance to another.

82. Optical fiber
Fiber Optical : uses reflection to guide light through a channel. A glass or plastic core is surrounded by a cladding of less dense glass or plastic
Back to the book for advantages and disadvantages

83. Applications for Fiber Optic cable
Used in :
1.Cable TV network: hybrid network use a combination of optical fiber and coax cable. Optical provides the backbone while coaxial cable provide the connation to the user.
2.Local area networks such as ( fast Ethernet)
3.Backbone networks because its wide bandwidth

84. Propagation modes using fiber optics
Multimode Fiber: any light ray incident on the boundary above the critical angle will be reflected internally, many different rays will be bouncing around at different angles.
Single-mode Fiber: light can propagate only in a straight line, without bouncing.

85. Fiber cable composition
(a) Side view of a single fiber.
(b) End view of a sheath with three fibers.
Core: 50 microns for multi-mode, 8-10 microns for single mode
Cladding: glass with a lower refraction index, to keep the light in the core

86. 7- 2 UNGUIDED MEDIA - wireless
Unguided media transport electromagnetic waves without using a physical conductor.
Signals are normally broadcast through free space and thus are available to anyone who has a device capable of receiving them
Unguided signals can travel from the source to destination in several ways:
Ground propagation
Sky propagation
Line – of – sight propagation

88. The Electromagnetic Spectrum
The electromagnetic spectrum and its uses for communication.

89. 1. Radio Transmission
(a) In the VLF, LF, and MF bands, radio waves follow the curvature of the earth.
(b) In the HF band, they bounce off the ionosphere.

90. 1. Radio Transmission Frequency ranges: 3 KHz to 1 GHz Omnidirectional
Susceptible to interference by other antennas using same frequency or band
Ideal for long-distance broadcasting
May penetrate walls
Apps: AM and FM radio, TV, maritime radio, cordless phones, paging

91. 2. Microwaves Frequencies between 1 and 300 GHz Unidirectional.
Narrow focus requires sending and receiving antennas to be aligned.
Issues:
Line-of-sight (curvature of the Earth; obstacles)
Cannot penetrate walls

92. 2. Satellite Microwaves
Similar to terrestrial microwave except the signal travels from a ground station on earth to a satellite and back to another ground station.
Satellite receives on one frequency, amplifies or repeats signal and transmits on another frequency
Satellite is relay station
Applications
Television
Long distance telephone
Private business networks

93. Communication Satellites
The principal satellite bands.

94. 3. Infrared Frequencies between 300 GHz and 400 THz.
Short-range communication in a closed area.
High frequencies cannot penetrate walls.
Requires line-of-sight propagation.
Advantage: prevents interference between systems in adjacent rooms.
Disadvantage: cannot use for long-range communication or outside a building due to sun’s rays.

95. The End