Data Communications Taught by Mr. Kim No, Lecturer @RUPP M.Sc. In Telecommunications Engineering...

Data CommunicationsTaught byMr. Kim No, Lecturer @RUPPM.Sc. In Telecommunications Engineering

Chapter 3:Data and Signals

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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 DigitalBoth data and the signals that represent them

can be either analog or digital in form. 2

Analog and Digital1.1. 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.

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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.

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Analog and Digital1.2. 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. 5

Analog and Digital

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Analog and DigitalIn data communications, we use analog

signals (periodic) and digital signals (non-periodic).

2. Analog SignalsAnalog 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.

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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.

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2.1.1. 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.

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2.1.2. Period and FrequencyPeriod 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.

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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

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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 0.0166 s or 16.6 ms.

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Example 2: Express a period of 100 ms in microseconds.

SolutionFrom 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:.

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Example 3: The period of a signal is 100 ms. What is its frequency in kilohertz?

SolutionFirst we change 100 ms to seconds, and then

we calculate the frequency from the period (1 Hz = 10−3 kHz).

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2.1.3. PhaseThe 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. 18


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Example: A sine wave is offset 1/6 cycle with respect to time 0. What is its phase in degrees and radians?

SolutionWe know that 1 complete cycle is 360°.

Therefore, 1/6 cycle is:

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2.1.4. WavelengthWavelength is another characteristic of a

signal traveling through a transmission medium.

The wavelength is the distance a simple signal can travel in one period.

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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/fWhere: λ expresses in m or µm; c is 3*108

m/s

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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

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2.1.5. Time and Frequency DomainThe 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.

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2.2. Composite SignalsA 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.

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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. 28


A composite periodic signal

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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.

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2.3. BandwidthThe 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 5000 - 1000, or 4000.

Note: The bandwidth of a composite signal is the difference between the highest and the lowest frequencies contained in that signal.

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B = fh - fl


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Periodic Analog SignalsExample1: 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.

SolutionLet fh be the highest frequency, fl the lowest

frequency, and B the bandwidth. Then

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The spectrum has only five spikes, at 100, 300, 500, 700, and 900 Hz


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Periodic Analog SignalsExample2: 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.

SolutionLet fh be the highest frequency, fl the lowest

frequency, and B the bandwidth. Then

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The spectrum contains all integer frequencies. We show this by a series of spikes


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Digital Signals

3. Digital SignalsIn 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.

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Digital Signals3.1. Bit Rate and Bit Interval Bit RateMost 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).

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Digital SignalsExample: 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

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Digital Signals Bit IntervalBit 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 = 0.0005s = 500 µs 41

1 s = 8 bit intervals

Bit Interval

Digital Signals3.2. 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.

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Digital Signals

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Bandwidth requirements

B = n / 2 (Harmonic 1)B = 3n / 2 (Harmonic 1, 3)B = 5n / 2 (Harmonic 1, 3, 5)

Where B is the required bandwidth (Hz); n is bit rate (bps)

Digital SignalsExample1:If we need to send 1 Mbps by using digital

transmission, What is the required bandwidth?Solution:The answer depends on the accuracy desired.a. The minimum bandwidth, is B = n / 2 = 500 kHz

b. A better solution is to use the first and the third

harmonics with B = 3n / 2 = 1.5 MHzc. Still a better solution is to use the first, third,

and fifth harmonics with B = 5n / 2 = 2.5 MHz

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Transmission Impairment

4. Transmission ImpairmentSignals 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.

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4.1 Attenuation (ការថយចុះ��ថាមពល)

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.

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DecibelTo 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.

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dB = 10Log10 P2/P1

Transmission ImpairmentExample1: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.

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Transmission ImpairmentExample2: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

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Transmission ImpairmentOne 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.

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4.2. Distortion (ការប្រែ � ��លរ ��រាង)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. 53


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4.3. Noise (ការរ�ខាន)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.

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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.

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Data Rate Limits

5. Data Rate LimitsA 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 available2. The level of the signals we use3. 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.

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Data Rate Limits5.1. 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

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Bit Rate = 2 x Bandwidth x Log2L

Data Rate LimitsExample: 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:

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Data Rate Limits5.2. 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

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Capacity = Bandwidth x Log2(1 + SNR)

Data Rate LimitsSNR 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. 61

Data Rate LimitsExample: 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 3162. For this channel the capacity is calculated as

This means that the highest bit rate for a telephone line is 34.860 kbps.

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Data Rate LimitsUsing 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.

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Data Rate LimitsThe 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.

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Performance

6. PerformanceOne important issue in networking is the

performance of the network-how good is it?6.1. ThroughputThe 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.

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Performance6.2. 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

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Data Communications Taught by Mr. Kim No, Lecturer @RUPP M.Sc. In Telecommunications Engineering...

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