Data Communications & Computer Networks Chapter 3 Data ...staff.fit.ac.cy/com.cl/03-DataTransmission.pdf · ACOE312 Data Transmission 1 Data Communications & Computer Networks Chapter

ACOE312 Data Transmission 1

Data Communications &

Computer Networks

Chapter 3

Data Transmission

Fall 2008

Agenda

• Terminology and basic concepts

• Analog and Digital Data Transmission

• Transmission impairments

• Channel capacity

• Home Exercises


Terminology and basic concepts

1. Terminology (1)

• Transmitter

• Receiver

• Medium

—Guided medium

• e.g. twisted pair, optical fiber

—Unguided medium

• e.g. air, water, vacuum


Terminology (2)

• Direct link

—No intermediate devices

• Point-to-point

—Direct link

—Only 2 devices share link

• Multi-point

—More than two devices share the link

Terminology (3)

• Simplex

—One direction

• e.g. Television

• Half duplex

—Either direction, but only one way at a time

• e.g. police radio

• Full duplex

—Both directions at the same time

• e.g. telephone


Time-Domain Concepts

• Analog signal

—Varies in a smooth way over time

• Digital signal

—Maintains a constant level then changes to another constant level

• Periodic signal

—Pattern of signal is repeated over time

• Aperiodic signal

—Pattern of signal is not repeated over time

Analogue & Digital Signals


Periodic

Signals

Sine Wave

Square Wave

Sine Wave characteristics

• Peak Amplitude (A)

—maximum strength of signal

—volts

• Frequency (f)

—Rate of change of signal

—Hertz (Hz) or cycles per second

—Period = time for one repetition (T)

—T = 1/f

• Phase (φ)

—Relative position in time


Varying Sine Waves

s(t) = A sin(2πft +φ)

Wavelength

• Distance occupied by one cycle

• Distance between two points of corresponding phase in two consecutive cycles

• λ=wavelength

• Assuming signal velocity v— λ = v·T

— λ·f = v

— c =2,98*108 m/s (approximately 3*108 m/s) speed of light in free space


Frequency Domain Concepts

• Signal usually made up of many frequencies

• Components are sine waves

• Can be shown (Fourier analysis) that any signal is made up of component sine waves

• Can plot frequency domain functions

Addition of

Frequency

Components

(T=1/f)

(1/3) sin(2π(3f)t)

sin(2πft)

(4/π) [sin(2πft)+(1/3)sin(2π(3f)t)]


Spectrum & Bandwidth

• Spectrum—range of frequencies contained in signal

• Bandwidth (BW)—Narrow band of frequencies containing most of the signal energy

—Absolute bandwidth: Width of the spectrum

—Effective bandwidth (or bandwidth): energy of signal contained in a narrow band of frequencies (usually expressed as the –3 dB points)

• DC Component—Component of zero frequency

Frequency

Domain

Representations

Fundamental frequency (f)

Signal spectrum

Absolute bandwidth=

3f-1f=2f

This signal has an infinite bandwidth.

Its effective bandwidth is

limited in a relatively narrow band of frequencies where the most

energy of the signal is contained

(4/π) [sin(2πft)+(1/3)sin(2π(3f)t)]

s(t)=1, -X/2<t<X/2


Signal with DC Component

Time Domain

Bandwidth

s(t) = 1 + (4/π) [sin(2πft)+(1/3)sin(2π(3f)t)]

Frequency Domain

Square

wave

Square wave signal consists of an infinite number of odd harmonics

(4/π)Σ[sin(2πkft)]/k

for odd values of k

(4/π) [sin(2πft)+(1/3)sin(2π(3f)t)+(1/5)sin(2π(5f)t)]

(4/π) [sin(2πft)+(1/3)sin(2π(3f)t)+(1/5)sin(2π(5f)t) +(1/7)sin(2π(7f)t)]]


Data Rate and Bandwidth (1)

• Any transmission system has a limited band of frequencies

• This limits the data rate that can be carried


• Suppose a digital transmission system is capable of transmitting signals with a BW of 4MHz. Let us attempt to transmit a square wave signal (i.e. a sequence of alternating 0s and 1s. What is the achievable data rate?



Case 1:Assume that the square wave is approximated to this signal.

BW=fupper – flower = 5f – f =4f

If f=1MHz, then the BW=4MHz.Since T=1/f then signal period is 1/1MHz=1µsSince one bit occurs every 0.5T then Data rate=1/0.5T=2Mbps

So, for this particular example, for a BW of 4MHz, the Data Rateachieved is 2Mbps





If f=2MHz, then the BW=8MHz.Since T=1/f then signal period is 1/2MHz=0.5µsSince one bit occurs every 0.5T then Data rate=1/0.25T=4Mbps







If f=2MHz, then the BW=4MHz.Since T=1/f then signal period is 1/2MHz=0.5µsSince one bit occurs every 0.5T then Data rate=1/0.25T=4Mbps


(4/π) [sin(2πft)+(1/3)sin(2π(3f)t)]


• Conclusions

—In general, any digital waveform has infinite BW

—If a digital waveform is transmitted over any medium, the transmission system will limit the BW that can be transmitted

—For any given medium, the greater the BW transmitted, the greater the cost

—Limiting the BW creates distortions, which makes the task of interpreting the received signal more difficult

—The more limited the BW, the greater the distortion, and the greater the potential for error by the receiver


Analog and Digital Data Transmission

2. Analog and Digital Data

Transmission

• Data

—Entities that convey information

• Signals

—Electric or electromagnetic representations of data

—Signaling is the physical propagation of the signal along a suitable medium

• Transmission

—Communication of data by propagation and processing of signals


Analog and Digital Data

• Analog

—Continuous values within some interval

—e.g. sound, video

• Digital

—Discrete values

—e.g. text, integers

Acoustic Spectrum (Analog)

(log scale)


Analog and Digital Signals

• Means by which data are propagated

• Analog signals

—Continuously variable

—Various media

• wire, fiber optic, space

—Speech bandwidth 100Hz to 7kHz

—Telephone bandwidth 300Hz to 3400Hz

—Video bandwidth 4MHz

• Digital signals

—Use two DC components (binary 0 and 1)

Advantages & Disadvantages

of Digital signals

• Advantages

—Cheaper

—Less susceptible to noise

• Disadvantages

—Greater attenuation

• Pulses become rounded and smaller

• Leads to loss of information


Attenuation of Digital Signals

Components of Speech

• Frequency range (of hearing) 20Hz-20kHz

—Speech 100Hz-7kHz

• Easily converted into electromagnetic signal for transmission

• Sound frequencies with varying volume converted into electromagnetic frequencies with varying voltage

• Limit frequency range for voice channel

—300-3400Hz


Conversion of Voice Input into

Analogue Signal

Binary Digital Data

• From computer terminals etc.

• Two dc components

• Bandwidth depends on data rate


Conversion of PC Input to

Digital Signal

Data and Signals

• Usually use digital signals for digital data and analog signals for analog data

• Can use analog signal to carry digital data

—Modem

• Can use digital signal to carry analog data

—Compact Disc audio


Analog Signals Carrying Analog

and Digital Data

Digital Signals Carrying Analog

and Digital Data

Voice


Analog Transmission

• Analog signal transmitted without regard to content

• May be analog or digital data

• Attenuated over distance

• Use amplifiers to boost signal

• However, amplifiers or signal boosters also amplify noise

Digital Transmission

• Concerned with content

• Integrity endangered by noise, attenuation etc.

• Repeaters are used

—A repeater receives digital signal, recovers the bit pattern (0 or 1) and retransmits new signal. Thus, attenuation is overcome

• Noise is not amplified


Advantages of Digital

Transmission

• Digital technology— Low cost large-scale and very-large scale integration technology

• Data integrity— Longer distances over lower quality lines

• Capacity utilization—High bandwidth links economical

— High degree of multiplexing easier with digital techniques

• Security & Privacy— Encryption

• Integration—Can treat analog and digital data similarly

— Economies of scale and convenience can be achieved by integrating voice, video and digital data

Transmission Impairments


3. Transmission Impairments

• Signal received may differ from signal transmitted

• For Analog signals— degradation of signal quality

• For Digital signals— bit errors may occur

• Most significant transmission impairments are—Attenuation and attenuation distortion

— Delay distortion

— Noise

Attenuation

• Signal strength reduces with distance over any transmission medium

• Depends on medium

• Received signal strength:

—must be enough to be detected

—must be sufficiently higher than noise to be received without error

• Attenuation is an increasing function of frequency, i.e. the higher the frequency, the more the attenuation attenuation


Delay Distortion (DD)

• Only in guided media

• It occurs because the propagation velocity of a signal through a guided medium varies with frequency

• Received signal is distorted due to varying delays experienced at its constituent frequencies

• DD is particularly critical for digital signals—some of the signal components of one bit may spill over into other bit positions, causing intersymbolinterference, which limits the maximum data rate over a transmission channel

Noise (1)

• Additional signals inserted between transmitter and receiver

• Noise is the major limiting factor in communication system performance

• Noise can be divided into 4 main categories

—Thermal

—Intermodulation

—Crosstalk

—Impulse noise


Noise (2)

• Thermal— Due to thermal agitation of electrons in all electronic devices

— Uniformly distributed across the bandwidth

— Also referred to a white noise

• Intermodulation— Signals that are the sum and difference of original frequencies sharing

the same transmission medium

— Example: mixing of signals at f1 and f2 may produce energy at f1±f2, which could interfere with an intended signal at (f1+f2) or (f1-f2)

• Crosstalk— Unwanted coupling between signal paths

— Antennas or wires may pick up other unwanted signals, eg. phone line

• Impulse— Non continuous, consisting of irregular pulses or noise spikes of short

duration but of high amplitude

— e.g. External electromagnetic interference, such as lightning

Channel capacity


4. Channel Capacity

• As we have seen so far, there is a variety of impairments that distort or corrupt a signal. To what extent do these impairments limit the maximum achievable data rate?

• Channel Capacity is the maximum rate at which data can be transmitted over a communication channel.

• Data rate

—In bits per second (bps)

—Rate at which data can be communicated

• Bandwidth

—In cycles per second, or Hertz

—Constrained by transmitter and medium

Nyquist Bandwidth• Assume a noise-free channel• If rate of signal transmission is 2B, then a signal with

frequencies no greater than B is sufficient to carry signal rate

• or, given bandwidth B, highest signal rate is 2B• Given a binary signal, the maximum data rate supported

by a channel of bandwidth B Hz is 2B bps• Maximum data rate, C, can be increased by using M

signal levels• Nyquist formula: C= 2·B·log2M in bps (bits per

second)• However, receiver must be able to distinguish one of M

possible signal elements. Noise and other transmission impairments limit the practical value of M.


Shannon Capacity Formula

• Nyquist’s formula indicates that doubling BW, doubles the data rate in a noise-free channel.

• In practice, noise is always present. So, let us consider the relationship between data rate, noise and error rate.

• Faster data rate shortens each bit duration so a burst of noise affects more bits— So, at a given noise level, the higher the data rate, the higher

the error rate

• Signal-to-Noise ratio (SNR or S/N) expressed in decibels

• SNRdB=10 log10 (Signal power/Noise power)

• Max channel Capacity is C=B·log2(1+SNR) in bps (bits per second)

• This formula is for error-free capacity and assumes white noise. In practice, data rate is lower than C.

A few things about Decibels (1)

• It is customary to express gains, losses and relative levels in decibels because

— Signal strength often falls off exponentially, so loss is easilyexpressed in terms of the decibel, which is a logarithmic unit

— The net gain or loss in a cascaded transmission path can be calculated with simple addition and subtraction

• The decibel (dB) is a measure of the ratio between two signal levels. The decibel gain is given by

GdB=10·log10 (Output power / Input power)

GdB=10·log10 (Pout/Pin)



• Gain is expressed in positive dB values (GdB)

• Loss is expressed in negative dB values (LdB)

• E.g. A gain of –3dB means that the power has halved and this is a loss of power. (Why?)

-6010-660106

-5010-550105

-4010-440104

-3010-330103

-2010-220102

-1010-110101

dBPower RatiodBPower Ratio


• Note that dB is a measure of relative, not absolute difference.

• The dB is also used to measure the difference in Voltage

• Since P = V2/R

Where, P=Power dissipated across resistance R

v = Voltage across resistance R

Then GdB = 10 log10 (Pout/Pin)

= 10 log10 [(V2out/R) /(V

2in/R)]

= 20 log10 (Vout/Vin)

Similarly LdB = 20 log10 (Vin/Vout)


Example on channel capacity

• Suppose that the spectrum of a noise-free channel is between 3 MHz and 4 MHz and SNRdB=24 dB.

—What is the maximum achievable data rate?

—How many signal levels are required to achieve this rate?

Solution of example

• Bandwidth, B=4 MHz – 3 MHz = 1 MHz = 106 Hz.

• SNRdB=24 dB = 10log10(SNR)

• Therefore, SNR=10(24/10) = 102.4 = 251.2

• Using Shannon’s formula, C=B log2(1+SNR),

C=106 log2 (1+251.2) = 7.98 x 106 ~ 8 Mbps

• Based on Nyquist’s formula, C=2B log2M in order to achieve a data rate of 8MBps in a channel bandwidth of 1MHz, then we need M signal levels, where M is equal to:

8x106 = 2x106 log2M => 4 = log2M => M=24=16


Home Exercises

Exercises (1)

Q1. What is the theoretical maximum channel capacity for the following PSTN channel of a signal-to-noise ratio of 13dB? Assume white thermal noise is only present on the channel.

Q2. Consider a signal f(t)=3sin(3000πt)+sin(9000πt) injected through a noisy channel of a signal-to-noise ratio of 20dB. What is the maximum data rate achieved?

0-3

S(f) in

dB

f (Hz)300 3400


Exercises (2)

Q3. A modem to be used with a PSTN network uses a modulation scheme with eight levels per signalling element. Assuming the same channel bandwidth as in Q1, but a noiseless channel, find the maximum possible data rate.

Q4. Given a channel with an intended capacity of 20 Mbps, the bandwidth of the channel is 3 MHz. Assuming white thermal noise, what signal to noise ratio in decibels is it required to achieve this capacity?

Q5. Fill in the missing elements in the following table

102Gains

0.10.5Losses

10987654321Decibels

Useful log identities

• logaB=X => aX=B

• logaB = (log10 B)/(log10 a)


Required Reading

• Stallings Chapter 3

• Tanenbaum Chapter 2.1

Documents

Data Communications & Computer Networks Chapter 3 Data ...staff.fit.ac.cy/com.cl/03-DataTransmission.pdf · ACOE312 Data Transmission 1 Data Communications & Computer Networks Chapter