1. Chapter 3: Data and Signals
Data Communications Taught by Mr. Kim No, M.Sc. In Telecommunications Engineering
Chapter 3:
Data and Signals
Data Communications and Networking by Behrouz A. Forouzan-4th Edition

2. Data and Signals
One of the major functions of the physical layer is to move data in the form of electromagnetic signals across a transmission medium.
Generally, the data usable to a person or application are not in a form that can be transmitted over a network. To be transmitted, data must be transformed to electromagnetic signals.
1. Analog and Digital
Both data and the signals that represent them can be either analog or digital in form.

3. Analog and Digital
1.1. Analog and Digital Data
Data can be analog or digital. The term analog data refers to information that is continuous; digital data refers to information that has discrete states.
For example, an analog clock, that has hour, minute, and second hands gives information in a continuous form; the movements of the hands are continuous.
On the other hand, a digital clock will change suddenly from 8:05 to 8:06.

4. Analog and Digital
Analog data, such as the sounds made by a human voice. When someone speaks, an analog wave is created in the air. This can be captured by a microphone and converted to an analog signal and converted to a digital signal.
Data are stored in computer memory in the form of Os and 1s. They can be converted to a digital signal or modulated into an analog signal for transmission across a medium.

5. Analog and Digital
1.2. Analog and Digital Signals
Like the data they represent, signals can be either analog or digital.
An analog signal has many levels of intensity over a period of time. As the wave moves from value A to value B, it passes through many number of values along its path.
A digital signal, on the other hand, can have only a limited number of defined values. Although each value can be any number, it is often as simple as 1 and O.

6. Analog and Digital

7. Analog and Digital 2. Analog Signals
In data communications, we use analog signals (periodic) and digital signals (non-periodic).
2. Analog Signals
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. Periodic Analog Signals
2.1. Sine Wave
The sine wave is the most fundamental form of a periodic analog signal. When we visualize it as a simple oscillating curve, its change over the course of a cycle is smooth and consistent, a continuous, rolling flow.
A sine wave can be represented by three parameters: the peak amplitude, the frequency, and the phase.

9. Periodic Analog Signals

10. Periodic Analog Signals
Peak Amplitude
The peak amplitude of a signal is the absolute value of its highest intensity, proportional to the energy it carries. For electric signals, peak amplitude is normally measured in volts.
Example: The power in U.S. home can be represented by a sine wave with a peak amplitude of 110 to 120 V.

11. Periodic Analog Signals

12. Periodic Analog Signals
Period and Frequency
Period refers to the amount of time, in seconds, a signal needs to complete 1 cycle. Frequency refers to the number of periods in 1 s.
Note that period and frequency are just one characteristic defined in two ways. Period is the inverse of frequency, and frequency is the inverse of period, as the following formulas show.

13. Periodic Analog Signals

14. Periodic Analog Signals
Period is formally expressed in seconds. Frequency is formally expressed in Hertz (Hz), which is cycle per second. Units of period and frequency are shown in below table:
Units of period and frequency

15. Periodic Analog Signals
Example 1: The power we use at home has a frequency of 60 Hz. The period of this sine wave can be determined as follows:
This means that the period of the power for our lights at home is s or 16.6 ms.

16. Periodic Analog Signals
Example 2: Express a period of 100 ms in microseconds.
Solution
From the table 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:.

17. Periodic Analog Signals
Example 3: 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).

18. Periodic Analog Signals
Phase
The term phase describes the position of the waveform relative to time 0. If we think of the wave as something that can be shifted backward or forward along the time axis, phase describes the amount of that shift. It indicates the status of the first cycle.
Phase is measured in degrees or radians [360° is 2π rad; 1° is 2π/360 rad, and 1 rad is 360/(2π)].
A phase shift of 360° corresponds to a shift of a complete period; a phase shift of 180° corresponds to a shift of one-half of a period; and a phase shift of 90° corresponds to a shift of one-quarter of a period.

19. Periodic Analog Signals

20. Periodic Analog Signals
Example: A sine wave is offset 1/6 cycle with respect to time 0. What is its phase in degrees and radians?
Solution
We know that 1 complete cycle is 360°. Therefore, 1/6 cycle is:

21. Periodic Analog Signals
Wavelength
Wavelength is another characteristic of a signal traveling through a transmission medium.
The wavelength is the distance a simple signal can travel in one period.

22. Periodic Analog Signals
If the propagation speed (the speed of light) and the period of the signal are given, Wavelength can be calculated via below formula:
Wavelength = Propagation Speed x Period
= Propagation Speed x (1/Frequency)
= Propagation Speed / Frequency
We represent wavelength by λ, propagation speed by c (speed of light), and frequency by f.
λ = c/f
Where: λ expresses in m or µm; c is 3*108 m/s

23. Periodic Analog Signals
For example 1: The wavelength of yellow light (frequency = 5 x 1014 Hz) in air is:
λ = c/f = (3 x 108)/(5 x 1014) = 0.6 x 10-6 m = 0.6 µm
For example 2: The wavelength of red light (frequency = 4 x 1014 Hz) in air is:
λ = c/f = (3 x 108)/(4 x 1014) = 0.75 x 10-6 m = 0.75 µm

24. Periodic Analog Signals
Time and Frequency Domain
The time-domain plot shows changes in signal amplitude with respect to time.
A frequency-domain plot shows changes in signal amplitude with respect to frequency.
Note: A complete sine wave in the time domain can be represented by one single spike in the frequency domain.

25. Periodic Analog Signals

26. Periodic Analog Signals

27. Periodic Analog Signals
2.2. Composite Signals
A single-frequency sine wave is not useful in data communications; we need to send a composite signal which is a signal made of many simple sine waves.
In the early 1900s, the French mathematician Jean-Baptiste Fourier showed that any composite signal is actually a combination of simple sine waves with different frequencies, amplitudes, and phases.

28. Periodic Analog Signals
According to Fourier analysis, any composite signal is a combination of simple sine waves with different frequencies, amplitudes, and phases.
A composite signal can be periodic or non-periodic.
If the composite signal is periodic, the decomposition gives a series of signals with discrete frequencies.
If the composite signal is non-periodic, the decomposition gives a combination of sine waves with continuous frequencies.

29. Periodic Analog Signals
A composite periodic signal

30. Periodic Analog Signals

31. Periodic Analog Signals
The amplitude of the sine wave with frequency f is almost the same as the peak amplitude of the composite signal.
The amplitude of the sine wave with frequency 3f is one-third of that of the first.
The amplitude of the sine wave with frequency 9f is one-ninth of the first.
Frequency of first sine wave is 3Hz, therefore the frequency of second sine wave is 9Hz and the third is 18Hz.

32. Periodic Analog Signals
2.3. Bandwidth
The range of frequencies contained in a composite signal is its bandwidth. The bandwidth is normally a difference between two numbers. For example, if a composite signal contains frequencies between 1000 and 5000, its bandwidth is , or 4000.
Note: The bandwidth of a composite signal is the difference between the highest and the lowest frequencies contained in that signal.
B = fh - fl

33. Periodic Analog Signals

34. Periodic Analog Signals
Example1: 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

35. Periodic Analog Signals
The spectrum has only five spikes, at 100, 300, 500, 700, and 900 Hz

36. Periodic Analog Signals
Example2: 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

37. Periodic Analog Signals
The spectrum contains all integer frequencies. We show this by a series of spikes

38. Digital Signals 3. 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.

39. Digital Signals 3.1. Bit Rate and Bit Interval Bit Rate
Most digital signals are non-periodic, and thus period and frequency are not appropriate characteristics. Another term-bit rate (instead of frequency) is used to describe digital signals. The bit rate is the number of bits sent in 1 second, expressed in bits per second (bps).

40. Digital Signals
Example: Assume we need to download text documents at the rate of 100 pages per second. What is the required bit rate of the channel?
Assume that a page is an average of 24 lines with 80 characters in each line and one character requires 8 bits.
Solution

41. Digital Signals Bit Interval
Bit Interval is a duration (expressed in second) of one bit. It is the inverse of bit rate. Bit Interval = 1/Bit rate
Example: A digital signal has bit rate 2000 bps. What is its bit interval?
Solution: bit interval = 1/bit rate = 1/2000 s = s = 500 µs
1 s = 8 bit intervals
Bit Interval

42. Digital Signals 3.2. Transmission of Digital Signals
In digital transmission, the required bandwidth is proportional to the bit rate; if we need to send bits faster, we need more bandwidth.

43. Digital Signals Bandwidth requirements B = n / 2 (Harmonic 1)
Where B is the required bandwidth (Hz); n is bit rate (bps)

44. Digital Signals Example1:
If we need to send 1 Mbps by using digital transmission, What is the required bandwidth?
Solution:
The answer depends on the accuracy desired.
The minimum bandwidth, is B = n / 2 = 500 kHz
A better solution is to use the first and the third harmonics with B = 3n / 2 = 1.5 MHz
Still a better solution is to use the first, third, and fifth harmonics with B = 5n / 2 = 2.5 MHz

45. 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.

46. Transmission Impairment

47. Transmission Impairment
4.1 Attenuation​ (ការថយចុះថាមពល)
Attenuation means a loss of energy. When a signal, simple or composite, travels through a medium, it loses some of its energy in overcoming the resistance of the medium.
That is why a wire carrying electric signals gets warm after a while. Some of the electrical energy in the signal is converted to heat. To compensate for this loss, amplifiers are used to amplify the signal.

48. Transmission Impairment

49. Transmission Impairment
Decibel
To show that a signal has lost or gained strength, engineers use the unit of the decibel.
The decibel (dB) measures the relative strengths of two signals or one signal at two different points. Note that the decibel is negative if a signal is attenuated and positive if a signal is amplified.
Variables P1 is the power of a signal at points 1 and P2 is the power of a signal at point 2.
dB = 10Log10 P2/P1

50. Transmission Impairment
Example1:
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.

51. Transmission Impairment
Example2:
A signal travels through an amplifier, and its power is increased 10 times. This means that P2 = 10P1 . In this case, the amplification (gain of power) can be calculated as

52. Transmission Impairment
One reason that engineers use the decibel to measure the changes in the strength of a signal is that decibel numbers can be added (or subtracted) when we are measuring several points instead of just two.

53. Transmission Impairment
4.2. Distortion (ការប្រែប្រួលរូបរាង)
Distortion means that the signal changes its form or shape. Distortion can occur in a​composite signal made of different frequencies.
Each signal component has its own​ propagation speed​​​​ through a medium and, therefore, its own delay in arriving at the final destination. Differences in delay may create a difference in phase if the delay is not exactly the same as the period duration.
In other words, signal components at the receiver have phases different from what they had at the sender. The shape of the composite signal is therefore not the same.

54. Transmission Impairment

55. Transmission Impairment
4.3. Noise (ការរំខាន)
Noise is another cause of impairment. Several types of noise, such as thermal noise,​ induced noise, crosstalk, and impulse noise, may corrupt the signal.
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. These devices act as a sending antenna, and the transmission medium acts as the receiving antenna.

56. Transmission Impairment
Crosstalk is the effect of one wire on the other. One wire acts as a sending antenna and the other as the receiving antenna.
Impulse noise is a signal with high energy in a very short time that comes from power lines, lightning, and so on.

57. Data Rate Limits 5. 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:
1. The bandwidth available
2. The level of the signals we use
3. The quality of the channel (the level of noise)
Two formulas were developed to calculate the data rate: one by Nyquist for a noiseless channel. another by Shannon for a noisy channel.

58. Data Rate Limits Bit Rate = 2 x Bandwidth x Log2L
5.1. Noiseless Channel: Nyquist Bit Rate
For a noiseless channel, the Nyquist bit rate formula defines the maximum bit rate.
Where:
- Bit Rate expresses in bps
- Bandwidth expresses in Hz
- L is the number of signal levels used to represent data
Bit Rate = 2 x Bandwidth x Log2L

59. Data Rate Limits Example:
Consider a noiseless channel with a bandwidth of 3000 Hz transmitting a signal with two signal levels. The maximum bit rate can be calculated as:
Consider the same noiseless channel transmitting a signal with four signal levels (for each level, we send 2 bits). The maximum bit rate can be calculated as:

60. Data Rate Limits Capacity = Bandwidth x Log2(1 + SNR)
5.2. Noisy Channel: Shannon Capacity
In reality, we cannot have a noiseless channel; the channel is always noisy. In 1944, Claude Shannon introduced a formula, called the Shannon capacity, to determine highest data rate for a noisy channel:
Where:
- Capacity is the capacity of the channel in bps
- Bandwidth expresses in Hz
- SNR is the signal-to-noise ratio
Capacity = Bandwidth x Log2(1 + SNR)

61. Data Rate Limits
SNR is actually the ratio of what is wanted (signal) to what is not wanted (noise). A high SNR means the signal is less corrupted by noise; a low SNR means the signal is more corrupted by noise.
Example: Consider an extremely noisy channel in which the value of the signal-to-noise ratio is almost zero. In other words, the noise is so strong that the signal is faint. For this channel the capacity C is calculated as:
This means that the capacity of this channel is zero regardless of the bandwidth. In other words, we cannot receive any data through this channel.

62. Data Rate Limits
Example: We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000 Hz (300 to 3300 Hz) assigned for data communications.
The signal-to-noise ratio is usually For this channel the capacity is calculated as
This means that the highest bit rate for a telephone line is kbps.

63. Data Rate Limits Using Both Limits
In practice, we need to use both methods to find the limits and signal levels.
Example:
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.

64. Data Rate Limits
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.
Note: The Shannon capacity gives us the upper limit; the Nyquist formula tells us how many signal levels we need.

65. Performance 6. Performance
One important issue in networking is the performance of the network-how good is it?
6.1. Throughput
The throughput is a measure of how fast we can actually send data through a network. Although, at first glance, bandwidth in bits per second and throughput seem the same, they are different. A link may have a bandwidth of B bps, but we can only send T bps through this link with T always less than B.
In other words, the bandwidth is a potential measurement of a link; the throughput is an actual measurement of how fast we can send data.

66. Performance 6.2. Latency (Delay)
The latency or delay 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. We can say that latency is made of four components: propagation time, transmission time, queuing time and processing delay.
Latency = propagation time + transmission time + queuing time + processing delay