british computer society (bcs) computer networks digital communications data and signals

British Computer Society (BCS)

Computer Networks

Digital Communications Data and Signals

Lesson Aim An understanding of digital data and analog data An understanding of digital transmission and

analog transmission and various transmission impairments in communication systems

An understanding of guided and unguided transmission media

Lesson Objective To differentiate digital signals with analog one Explain advantages of digital transmission over analog one Understand the effects of transmission impairments in

communication channel Understand the techniques use to transmit data digitally Understand the mechanism for modulating digital data into

an analog signal and an analog to an analog signal Explain the characteristics and applications of transmission

medium

Digital Communications

Data and Signals 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. Analog data take on continuous values. Digital data take on discrete values.

Digital Communications Signals:

Information travels through a transmission system is in the form of electric or electromagnetic signals

Signaling is the physical propagation of the signal along a transmission media

Signals are represented in two ways: Analog and Digital Analog signals:

Analog signals can have an infinite number of values in a range. As the wave moves from value A to value B, it passes through and includes an infinite number of values along its paths

Digital Signal: Digital signals can have only a limited number of values.

Although each value can be any number, it is often as simple as 1 and 0


Figure 1.1: Comparison of analog and digital signals


Figure 1.1 above illustrate an analog signal and a digital signal.

The curve representing the analog signal passes through an infinite number of points

The vertical lines of the digital signal, however, demonstrate the sudden jump that the signal makes from value to value

Digital Communications Periodic and Nonperiodic signals:

Both analog and digital signals can be periodic or non periodic

A periodic signal completes a pattern within a measurable time frame, called a period, and repeats that pattern over subsequent identical periods

The completion of one full pattern is called a cycle A nonperiodic signal changes without exhibiting a pattern

or cycle that repeats over time In data communications, we commonly use periodic analog

signals and nonperiodic digital signals


Periodic Analog Signals: Sine Wave:

The sine wave is the most fundamental form of a periodic analog signal

A sine wave can be represented by three parameters: the peak amplitude, the frequency and the phase

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


Figure 1.2: A Sine wave


Figure 1.3: Two signals with the same phase and frequency, but different amplitude


Period and Frequency: Period is the amount of time, in seconds, a signal needs

to complete 1 cycle Frequency is the number of periods in 1 seconds Period and frequency are the inverse of each other, as

the following formulas show: Period is measured in seconds whereas frequency is

measured in Hertz (Hz)

f= 1/T and T= 1/f


Figure 1.4: Two signals with the same amplitude and phase, but different frequencies


Table 1.1 Units of period and frequency


Example 1.1:

The power we use at home has a frequency of 60 Hz. The period of this sine wave can be determined as follows:

Solution:


Example 1.2:

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).


Phase: The term phase describes the position of the waveform

relatives to time 0. Phase is measured in degrees or radians A phase shift of 3600 corresponds to a shift of a complete

period A phase shift of 1800 corresponds to a shift of one half of a

period A phase shift of 900 corresponds to a shift of one quarter of

a period


Figure 1.5: Three sine waves with the same amplitude and frequency, but different phases


Example 1.3:

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

Digital Communications Wavelength:

Wavelength is another characteristic of a signal travelling through a transmission medium

Wavelength binds the frequency of a simple sine wave to the propagation speed of the medium

While the frequency of a signal is independent of the medium, the wavelength depends on both the frequency and the medium

If wavelength is represented by‘’ propagation speed by ‘c’ and frequency by ‘f’, we get

wavelength = propagation speed* period

= propagation speed/ frequency


Figure 1.6: Wavelength and period

Digital Communications Time and Frequency Domain:

A sine wave is characterise by amplitude, frequency and phase

A sine wave can be represented by a time domain plot The time domain plot shows changes in signal

amplitude with respect to time To show the relationship between amplitude and

frequency, a frequency domain plot is used A frequency domain plot is concerned with only the

peak value and the frequency. Change of amplitude during one period are not considered.


Figure 1.7 The time-domain and frequency-domain plots of a sine wave


Note

A complete sine wave in the time domain can be represented by one single spike in

the frequency domain.


Example 1.4:The frequency domain is more compact and useful when we are dealing with more than one sine wave. For example, Figure 1.8 shows three sine waves, each with different amplitude and frequency. All can be represented by three spikes in the frequency domain.


Figure 1.8 The time domain and frequency domain of three sine waves

Digital Communications Composite Signals:

A composite signal is a combination of simple sine waves with different frequencies, amplitude, 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


Example 1.5:Figure 1.9 shows a periodic composite signal with frequency f. This type of signal is not typical of those found in data communications. We can consider it to be three alarm systems, each with a different frequency. The analysis of this signal can give us a good understanding of how to decompose signals


Figure 1.9 A composite periodic signal

Digital Communications Figure 1.10 shows the result of decomposing the above signal

in both the time and frequency domain The amplitude of the sine wave with frequency f is almost the

same as the peak amplitude of the composite signal. This is also called the fundamental frequency or 1st harmonic

The amplitude of the sine wave with frequency 3f is one third of that of the first. This is called the third harmonic

The amplitude of the sine wave with frequency 9f is one ninth of the first. This is called the ninth harmonic

Note that the frequency decomposition of the signal is discrete; it has frequencies of f, 3f, and 9f . The frequency domain of a periodic composite signal is always made of discrete spikes


Figure 1.10 Decomposition of a composite periodic signal in the time and frequency domains


Example 1.6:Figure 1.11 shows a non periodic composite signal. It can be the signal created by a microphone or a telephone set when a word or two is pronounced. In this case, the composite signal cannot be periodic, because that implies that we are repeating the same word or words with exactly the same tone.


Figure 1.11 The time and frequency domains of a nonperiodic signal


In a time domain representation of this composite signal, there are an infinite number of simple sine frequencies

Although the number of frequencies in a human voice is infinite, the range is limited. A normal human being can create a continuous range of frequencies between 0 and 4 KHz

Note that the frequency decomposition of the signal yields a continuous curve. There are an infinite number of frequencies between 0.0 and 4000.0 (real values). To find the amplitude related to frequency f, we draw a vertical line at f to intersect the envelope curve. The height of the vertical line is the amplitude of the corresponding frequency

Digital Communications Bandwidth:

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

Figure 1.12 below shows the concept of bandwidth. The figure depicts two composite signals, one periodic and

the other non periodic The bandwidth of the periodic signal contains all integer

frequencies between 1000 and 5000 (1000,1001, 1002,….) The bandwidth of the non periodic signals has the same

range, but the frequencies are continuous


Figure 1.12: The bandwidth of periodic and nonperiodic composite signals


Digital Signals: An information can be represent by a digital signal A digital information ‘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 Figure 1.13 shows two signals, one with two levels

and the other with four In general, if a signal has ‘L’ levels, each level needs

log2 L bits

Digital CommunicationsFigure 1.13 Two digital signals: one with two signal levels and the other with four signal levels


Example 1.7A 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


Bit Rate: Bit rate is the number of bits sent in one second It is expressed in bits per second (bps)

Bit Length: The bit length is the distance one bit occupies on the

transmission medium

Bit length = propagation speed * bit duration


Example 1.8:Assume we need to download text documents at the rate of 100 pages per minute. What is the required bit rate of the channel?

SolutionA page is an average of 24 lines with 80 characters in each line. If we assume that one character requires 8 bits, the bit rate is

Digital Communications Digital Signal as a Composite Analog Signal:

Base on Fourier analysis, a digital signal is a composite analog signal

If the digital signal is periodic, which is rare in data communications, the decomposed signal has a frequency domain representation with an infinite bandwidth and discrete frequencies

If the digital signal is non periodic, the decomposed signal still has an infinite bandwidth, but the frequencies are continuous

Figure 1.14 shows a periodic and a nonperiodic digital signal and their bandwidths


Figure 1.14 The time and frequency domains of periodic and non periodic digital signals


Transmission of Digital signal: To transmit a digital signal from point ‘A’ to point

‘B’, one of two different approaches can be used Baseband Transmission Broadband Transmission

Digital Communications Baseband Transmission:

Baseband transmission means sending a digital signal over a channel without changing the digital signal to an analog signal.

Fig 1.15 shows the baseband transmission Baseband transmission is carried out on a low pass channel, a channel

with a bandwidth that starts from zero. This is the case if we have a dedicated medium with a bandwidth constituting only one channel.

For example, the entire bandwidth of a cable connecting two computers is one single channel

Figure 1.16 shows two low pass channel: one with a narrow bandwidth and the other with a wide bandwidth

As a low pass channel with infinite bandwidth is ideal, but cannot achieve such a channel in real life. However, we can get close to it


Figure 1.15 Baseband transmission


Figure 1.16 Bandwidths of two low-pass channels


Broadband Transmission: Broadband transmission means changing the digital signal

to an analog signal for transmission. Broad band transmission use a band pass channel, a channel

with a bandwidth that does not start from zero Figure 1.17 shows a band pass channel Figure 1.18 shows the modulation of a digital signal

In the figure , a digital signal is converted to a composite analog signal. This composite analog signal is transmit through band pass channel . At the receiver the received analog signal is converted to digital and the result is a replica of what has been sent


Figure 1.17 Bandwidth of a band pass channel


Figure 1.18 Modulation of a digital signal for transmission on a bandpass channel

Digital Communications Transmission Impairments:

Signals that are received may differ from the signals that are transmitted due to various transmission impairments

For analog signals these impairments can degrade the signal quality

For digital signals, bit errors may be introduced

The most significant impairments are Attenuation Distortion Noise

Digital Communications Attenuation:

Attenuation means a loss of energy When a signal travels through a medium, it loses some

of its energy in overcoming the resistance of the medium. In other word, the signal strength falls off with distance over any transmission medium

To compensate for this loss, amplifiers are used to amplify the signals

Figure 1.19 below shows the effect of attenuation and amplification


Figure 1.19: Attenuation

Digital Communications Decibel:

Decibel (dB) measures the relatives strength of two signals or one signal at two different points.

Decibel (dB) is the unit use to show that a signal has lost or gained strength

Note that the decibel (dB) is negative if a signal is attenuated and positive if a signal is amplified

dB = 10 log10 P2/P1

Where P1 and P2 are the powers of a signal at point 1 and 2 respectively


Example 1.9:

Suppose a signal travels through a transmission medium and its power is reduced to one-half. This means that P2 is (1/2)P1. Calculate attenuation.

Solution:

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.


Example 1.20:

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

Solution:


Example 1.21:

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 (cascading) instead of just two. In Figure 1.20 a signal travels from point 1 to point 4. In this case, the decibel value can be calculated as

Solution: In this case, the decibel value can be calculate as


Figure 1.20: Decibels for example 3.28


Example 1.22:

Sometimes the decibel is used to measure signal power in milliwatts. In this case, it is referred to as dBm and is calculated as dBm = 10 log10 Pm , where Pm is the power in milliwatts. Calculate the power of a signal with dBm = −30.

Solution:

We can calculate the power in the signal as

Digital CommunicationsExample 1.23:

The loss in a cable is usually defined in decibels per kilometer (dB/km). If the signal at the beginning of a cable with −0.3 dB/km has a power of 2 mW, what is the power of the signal at 5 km?

Solution:

The loss in the cable in decibels is 5 × (−0.3) = −1.5 dB. We can calculate the power as

Digital Communications Distortion:

Distortion means that the signal changes it form or shape Distortion occurs in a composite signal due to 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. The figure 1.21 shows the effect of distortion on a composite signal


Figure 1.21: Distortion

Digital Communications Noise:

An unwanted signals that are inserted somewhere between transmission and reception on the transmission medium due to various distortion is known as noise

Noise may be divided into four categories Thermal Noise Intermodulation Noise Crosstalk Noise Impulse Noise

Digital Communications Thermal Noise:

Thermal noise is due to the random motion of electrons in a conductor which create an extra signal not originally sent by the transmitter. It is present in all electronic devices and transmission media and is a function of temperature

Thermal noise is uniformly distributed across the bandwidths typically used in communications system.

Thermal noise is also referred to as white noise Thermal noise cannot be eliminated and therefore places an

upper bound on communication system performance

Digital Communications The amount of thermal noise to be found in a bandwidth of

1Hz in any device or conductor is given by:

No= KT (W/Hz)

Where,

No= noise power density in watts per 1Hz of bandwidth

K= Boltzmann’s constant=1.38* 10-23 J/K

T= temperature in kelvins (absolute temperature)


Room temperature is usually specified as T = 170 c. At this temperature , the thermal noise power density is

Solution:

N = (1.38 * 10-23 ) * 290 = 4* 10-21 W/Hz = - 204 dB/Hz

Where dBW is the decibel watt


The noise is assumed to be independent of frequency. Thus the thermal noise in watts present in a bandwidth of B Hertz can be expressed as

N= K TB

Thermal noise in decibal-watts is expressed as

N= 10 log K +10 log T + 10 log B


Intermodulation Noise: when signals at different frequencies share the same

transmission medium, the result may be intermodulation noise

The effect of intermodulation noise is to produce signals at a frequency that is the sum or difference of the two original frequencies or multiples of frequencies

For example, the mixing of signals at frequencies f1 and f2 might produce energy at the frequency f1 + f2 . This derived signal could interfere with an intended signal at frequency f1 + f2

Digital Communications Crosstalk Noise:

A signal from one line is picked up by another Crosstalk noise is occurred by poor electrical coupling

between nearby twisted pairs or coaxial cable lines carrying multiple signals

Impulse noise: A irregular pulses or noise spikes of short duration and of

relatively high amplitude. It is generated from a variety of causes, including external

electromagnetic disturbances such as lightning and fault and flaws in the communications system


Signal to Noise Ratio (SNR): The signal to noise ratio is defined as

SNR = avg. signal power/avg. noise power

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


SNR is the ratio of two powers, it is often described in decibel units. SNR(dB), defined as

SNR(dB) = 10 log10 (SNR)


Example 1.25:

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:


Example 1.26:

The values of SNR and SNRdB for a noiseless channel are

Solution:

We can never achieve this ratio in real life; it is an ideal

Digital Communications Channel Capacity:

The maximum rate at which data can be transmitted over a given communication path or channel under given condition is known a channel capacity

Channel capacity is measured in bits per second Channel capacity depends on three factors:

The bandwidth available The level of the signal use The quality of the channel ( level of noise present)

Digital Communications Nyquist Channel Capacity (Noiseless Channel):

Nyquist has assumed noiseless channel, In this environment, the limitation on data rate is simply the bandwidth

of the signal Nyquist states that if the rate of signal transmission is 2B, then a signal

with frequencies no greater than B is sufficient to carry the signal rate. The converse is also true.

C = 2 B log2 L

Where, C= channel capacity in bit/sec

B= Bandwidth of physical channel

L= no. of signal level or voltage level used to represent data Note: Increasing the levels of a signal may reduce the reliability of the

system


Example 1.27:

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

Solution:


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.


Shannon Channel Capacity (Noisy Channel): 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 determined the theoretical highest data rate for a noisy channel:

This is

C= Blog2 (1+SNR)

Where,

C= capacity of channel in bits per second

B= bandwidth of the channel in Hertz


Example 1.29:

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

Solution:

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.


We can calculate the theoretical highest bit rate of a regular telephone line. A telephone line normally has a bandwidth of 3000. The signal-to-noise ratio is usually 3162. For this channel the capacity is calculated as

Solution:

This means that the highest bit rate for a telephone line is 34.860 kbps. If we want to send data faster than this, we can either increase the bandwidth of the line or improve the signal-to-noise ratio.


Example 1.31:

The signal-to-noise ratio is often given in decibels. Assume that SNRdB = 36 and the channel bandwidth is 2 MHz. The theoretical channel capacity can be calculated as

Solution:


Example 1.32:

For practical purposes, when the SNR is very high, we can assume that SNR + 1 is almost the same as SNR. In these cases, the theoretical channel capacity can be simplified to

Solution:

For example, we can calculate the theoretical capacity of the previous example as


Network Performance: Throughput:

The throughput is a measure of how fast data travels through a network.

Suppose a medium has a bandwidth of ‘B’ bps, but we can send ‘T’ bps through this medium with ‘T’ always less than ‘B’. That means, the bandwidth is a potential measurement of a medium, the throughput is an actual measurement of how fast we can send data.


Example 1.33:

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.


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.

Latency = propagation time + transmission time + Queuing time + Processing delay


Propagation Time: Propagation time measures the time required for a bit

to travel from the source to the destination. The propagation time is calculated by dividing the distance by the propagation speed

Propagation time= Distance/ Propagation speed


Example 1.34:

What is the propagation time if the distance between the two points is 12,000 km? Assume the propagation speed to be 2.4 × 108 m/s in cable.

Solution: We can calculate the propagation time as

The example shows that a bit can go over the Atlantic Ocean in only 50 ms if there is a direct cable between the source and the destination


Transmission Time: Transmission time is defined as the time required for

transmission of a message size over the bandwidth of the channel

Transmission time = Message size/Bandwidth


Example 1.35:

What are the propagation time and the transmission time for a 2.5-kbyte message (an e-mail) 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:

We can calculate the propagation and transmission time as shown on the next slide:


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.


Queuing Time: The time needed for each intermediate or end device to hold the

message before it can be processed The queuing time is not a fixed factor; it changes with the load imposed

on the network When there is a heavy traffic on the network, the queuing time

increases

Jitter: Jitter is a delay variation for different data packets received by the

receiver in the network For example, if the delay for the 1st packet is 20ms, for the 2nd packet is

45 ms and for the 3rd is 40ms, then the real application that uses the packets endures jitter

Digital Communications1.1 Exam Question (April 2009)

A digital transmission system uses a coding scheme that defines a ‘symbol’ as a voltage that can have one of eight possible values. If the system operates at a transmission rate of 200 symbols per second, determine the data rate measured in:

i) Baud

ii) Bits per second

Solution:

A symbol is a voltage level that can have one of 8 possible values. Eight levels are represented by 3 binary bits. Therefore one symbol represents three bits of data. Baud is defined as a rate of one symbol per second. Therefore if the system transmits at 200 symbols per second then the data rate is also 200 Baud.

If each symbol represents three bits then the transmission rate in bits per second will be 200 x 3 = 600 bits per second.

Digital Communications1.2 Exam Question (April 2006)

Define the terms bandwidth and capacity as applied to data communication systems and briefly explain their significance.

In a data communication system, if the signal power is 10mW, the noise power is 1000 nW and the bandwidth of the system is 100kHz, calculate the maximum capacity of the system.

Solution:

Bandwidth: The range of frequencies contained in a signal.

Capacity: Directly proportional to the bandwidth of the signals that the system carries.

Bandwidth indicates the amount of information that can be sent through the system. An important parameter to determine the capacity of the system


Capacity = B Log 2[ 1 +S/N]

where B= bandwidth, S= signal power, N= Noise Power

Also,

S/N = 10mW/1000nW = 104

Capacity = 100* 103 Log2 [1+ 104]

= 105 * Log10 [104] / Log10 [2] bits/sec = 13.3 *103 bits/sec

Digital Communications1.3 Exam Question (October 2005)

Discuss the sources of noise in data communication systems. Why it is important to consider the effect of noise on data communication systems?

Solution:

Sources of noise:

- Thermal noise

- Cross-talk

- Impulse noise 5 marks

Explanation- corruption of data bits, bandwidth implications – signal-to-noise ratio

Digital Communications References:

Forouzan A. Behrouz, Data Communications and Networking (Fourth Edition), McGraw Hill

William Stallings, Data and Computer Communication ( 7th Edition), Prentice Hall International Edition

Andrew Tanenbaum, Computer Networks (4TH Edition), Prentice Hall

Website: www.pearsoned.co.uk Website: http://en.wikipedia.org/wiki/

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