data and computer communications chapter 3 – data transmission
TRANSCRIPT
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Data and Computer Data and Computer CommunicationsCommunications
Chapter 3 – Data Transmission Chapter 3 – Data Transmission
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TransmissionTransmission TerminologyTerminology
data transmission occurs between a data transmission occurs between a transmitter & receiver via some mediumtransmitter & receiver via some medium
guidedguided medium medium e.g. twisted pair, coaxial cable, optical fibere.g. twisted pair, coaxial cable, optical fiber
unguidedunguided / wireless medium / wireless medium e.g. air, water, vacuume.g. air, water, vacuum
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TransmissionTransmission TerminologyTerminology
direct link (guided & unguided)direct link (guided & unguided) no intermediate devicesno intermediate devices
point-to-point (guided)point-to-point (guided) direct link direct link only 2 devices share linkonly 2 devices share link
multi-pointmulti-point more than two devices share the linkmore than two devices share the link
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TransmissionTransmission TerminologyTerminology
simplexsimplex one directionone direction
• eg. televisioneg. television
half duplexhalf duplex either direction, but only one way at a timeeither direction, but only one way at a time
• eg. police radioeg. police radio
full duplexfull duplex both directions at the same timeboth directions at the same time
• eg. telephoneeg. telephone
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Frequency, Spectrum and Frequency, Spectrum and BandwidthBandwidth
time domain conceptstime domain concepts analog signalanalog signal
• varies in a smooth way over timevaries in a smooth way over time digital signal (discrete)digital signal (discrete)
• maintains a constant level then changes to another maintains a constant level then changes to another constant levelconstant level
periodic signalperiodic signal• pattern repeated over timepattern repeated over time
aperiodic signalaperiodic signal• pattern not repeated over timepattern not repeated over time
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Analogue & Digital SignalsAnalogue & Digital Signals
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PeriodicPeriodicSignalsSignals
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Sine WaveSine Wave
peak amplitude (A)peak amplitude (A) maximum strength of signalmaximum strength of signal voltsvolts
frequency (f)frequency (f) rate of change of signalrate of change of signal Hertz (Hz) or cycles per secondHertz (Hz) or cycles per second period = time for one repetition (T)period = time for one repetition (T) T = 1/fT = 1/f
phase (phase ()) relative position in timerelative position in time
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Varying Sine WavesVarying Sine Wavess(t) = A sin(2s(t) = A sin(2ft +ft +))
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Wavelength (Wavelength ())
is is distancedistance occupied by one cycle occupied by one cycle between two points of corresponding between two points of corresponding
phase in two consecutive cyclesphase in two consecutive cycles assuming signal velocity assuming signal velocity v, we v, we have have = vT = vT or equivalently or equivalently f = vf = v especially when especially when v=cv=c
c = 3*10c = 3*108 8 msms-1 -1 (speed of light in free space)(speed of light in free space)
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ProblemProblem
## In a multipoint configuration, a central In a multipoint configuration, a central control may be used that enables only one control may be used that enables only one device to transmit. What is the merit and device to transmit. What is the merit and demerit of such control as compared to demerit of such control as compared to distributed control?distributed control?
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3.3.1212
According to Fourier analysis, any composite signal is a combination of
simple sine waves with different frequencies, amplitudes, and phases.
Note
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Frequency Domain ConceptsFrequency Domain Concepts
Signal is made up of Signal is made up of manymany frequencies frequencies components are sine wavescomponents are sine waves Fourier analysis can show that any signal Fourier analysis can show that any signal
is made up of component sine wavesis made up of component sine waves can plot frequency domain functionscan plot frequency domain functions
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Addition of Addition of FrequencyFrequency
ComponentsComponents(T=1/f)(T=1/f)
c is sum of c is sum of f & 3f f & 3f (with different (with different amplitudes)amplitudes)
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FrequencyFrequencyDomainDomain
Representations Representations
freq domain function freq domain function of Fig 3.4cof Fig 3.4c
freq domain function freq domain function of single square pulseof single square pulse
-ve -ve amplitude signify? amplitude signify?
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3.3.1616
A composite periodic signal
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3.3.1717
Decomposition of a composite periodic signal in the time and frequency domains
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Spectrum & BandwidthSpectrum & Bandwidth
spectrumspectrum range of frequencies contained in signalrange of frequencies contained in signal
absolute bandwidthabsolute bandwidth width of spectrumwidth of spectrum
effective bandwidtheffective bandwidth often just often just bandwidthbandwidth narrow band of frequencies containing most energynarrow band of frequencies containing most energy
DC ComponentDC Component component of zero frequencycomponent of zero frequency
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Data Rate and BandwidthData Rate and Bandwidth
any transmission system can accommodate a any transmission system can accommodate a limitedlimited band of frequencies band of frequencies
this limits the data rate that can be carriedthis limits the data rate that can be carried Square wave: Square wave: infiniteinfinite components and hence components and hence
infinite bandwidthinfinite bandwidth but but most energy most energy is in first is in first few componentsfew components limited bandwidth increases distortionlimited bandwidth increases distortion has a direct relationship between data rate & has a direct relationship between data rate &
bandwidthbandwidth
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Data Rate-Bandwidth RelationData Rate-Bandwidth Relation
Square Wave transmissionSquare Wave transmission
Case 1:Case 1: Three sinusoidal frequency Three sinusoidal frequency components – f, 3f, 5f => Bandwidth = (5-components – f, 3f, 5f => Bandwidth = (5-1) f = 4f. Let f = 1 MHz, Bandwidth = 4 1) f = 4f. Let f = 1 MHz, Bandwidth = 4 MHz,MHz,
T = 1µs => 1 bit needs 0.5 µsT = 1µs => 1 bit needs 0.5 µs Data rate = 2 MBPSData rate = 2 MBPS
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Data Rate-Bandwidth RelationData Rate-Bandwidth Relation
Square Wave transmissionSquare Wave transmission
Case 2: Case 2: Three sinusoidal frequency Three sinusoidal frequency components – f, 3f, 5f => Bandwidth = (5-components – f, 3f, 5f => Bandwidth = (5-1) f = 4f. Let f = 1) f = 4f. Let f = 22 MHz, Bandwidth = MHz, Bandwidth = 8 8 MHz,MHz,
T = 0.5µs => 1 bit needs 0.25 µsT = 0.5µs => 1 bit needs 0.25 µs Data rate = Data rate = 44 MBPS MBPS
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Data Rate-Bandwidth RelationData Rate-Bandwidth Relation
Square Wave transmissionSquare Wave transmission
Case 3:Case 3: Two sinusoidal frequency Two sinusoidal frequency components – f, 3f only => Bandwidth = (3-components – f, 3f only => Bandwidth = (3-1) f = 2f. Let f = 1) f = 2f. Let f = 22 MHz, Bandwidth = MHz, Bandwidth = 44 MHz,MHz,
T = 0.5µs => 1 bit needs 0.25 µsT = 0.5µs => 1 bit needs 0.25 µs Data rate = 4 MBPSData rate = 4 MBPS Shape of signal?Shape of signal?
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Analog and Digital Data Analog and Digital Data TransmissionTransmission
datadata entities that convey meaning / informationentities that convey meaning / information
signals signals & signaling& signaling electric or electromagnetic representations of electric or electromagnetic representations of
data, physically propagates along mediumdata, physically propagates along medium TransmissionTransmission
propagation and processing of signalspropagation and processing of signals
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Acoustic Spectrum (Analog)Acoustic Spectrum (Analog)
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Audio SignalsAudio Signals
freq range 20Hz-20kHz (speech 100Hz-7kHz)freq range 20Hz-20kHz (speech 100Hz-7kHz) easily converted into electromagnetic signalseasily converted into electromagnetic signals varying volume converted to varying voltagevarying volume converted to varying voltage can limit frequency range for can limit frequency range for voice channel voice channel to to
300-3400Hz with acceptable reproduction 300-3400Hz with acceptable reproduction
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Video Signals - BandwidthVideo Signals - Bandwidth
525 lines x 30 scans = 15750 lines per sec525 lines x 30 scans = 15750 lines per sec =>=> 63.5 63.5s per lines per line 1111s for s for horizontalhorizontal retrace, so 52.5 retrace, so 52.5 s per video lines per video line
max frequency if line alternates black and whitemax frequency if line alternates black and white 483 lines per frame 483 lines per frame
USA has 525 lines but 42 lost during USA has 525 lines but 42 lost during verticalvertical retrace retrace
horizontal resolution is about 450 lines giving horizontal resolution is about 450 lines giving 225 cycles of wave in 52.5 225 cycles of wave in 52.5 ss
max frequency of 4.2MHzmax frequency of 4.2MHz
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Digital DataDigital Data as generated by computers etc.as generated by computers etc. has two dc componentshas two dc components bandwidth depends on sequence of 1s & 0sbandwidth depends on sequence of 1s & 0s
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Analog SignalsAnalog Signals
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Digital SignalsDigital Signals
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Advantages & Disadvantages Advantages & Disadvantages of Digital Signalsof Digital Signals
cheapercheaper less susceptible to noiseless susceptible to noise but greater attenuationbut greater attenuation digital now preferred choicedigital now preferred choice
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Preferred MethodPreferred Method
Digital, because:Digital, because:
- Technology support of VLSI- Technology support of VLSI - Security (Encryption)- Security (Encryption) - Integration (data, audio, video)- Integration (data, audio, video)
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Transmission ImpairmentsTransmission Impairments
signal received may differ from signal signal received may differ from signal transmitted causing:transmitted causing: analog - degradation of signal qualityanalog - degradation of signal quality digital - bit errors (‘1’ as ‘0’ or vice-versa)digital - bit errors (‘1’ as ‘0’ or vice-versa)
most significant impairments aremost significant impairments are attenuationattenuation delay distortiondelay distortion noisenoise
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AttenuationAttenuation where signal where signal strengthstrength falls off with distance falls off with distance depends on mediumdepends on medium received signal strength must be:received signal strength must be:
strong enough to be detectedstrong enough to be detected sufficiently higher than noise to receive without errorsufficiently higher than noise to receive without error
so so increase strength increase strength using amplifiers/repeatersusing amplifiers/repeaters is also an increasing function of frequencyis also an increasing function of frequency so so equalizeequalize attenuation across band of attenuation across band of
frequencies usedfrequencies used e.g. using loading coils (voice grade) or amplifierse.g. using loading coils (voice grade) or amplifiers
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ProblemProblem
## A signal has passed through three A signal has passed through three cascaded amplifiers, each with a 4 dB cascaded amplifiers, each with a 4 dB gain. What is the total gain in dB? How gain. What is the total gain in dB? How much is the signal amplified? What does a much is the signal amplified? What does a negative dB value signify?negative dB value signify?
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Delay DistortionDelay Distortion
only occurs in guided mediaonly occurs in guided media propagation propagation velocity varies velocity varies with frequencywith frequency hence various frequency components hence various frequency components
arrive at different timesarrive at different times particularly critical for digital dataparticularly critical for digital data since parts of one bit spill over into otherssince parts of one bit spill over into others causing causing inter-symbolinter-symbol interference interference
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NoiseNoise
additional additional signals inserted signals inserted between between transmitter and receivertransmitter and receiver
thermalthermal due to thermal agitation of electronsdue to thermal agitation of electrons uniformly distributed uniformly distributed across typical bandwidthacross typical bandwidth white noisewhite noise
Inter-modulationInter-modulation signals that are the sum and difference (or signals that are the sum and difference (or
multiples) of original frequencies sharing a multiples) of original frequencies sharing a mediummedium
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NoiseNoise crosstalkcrosstalk
a signal from one path / line is picked up by a signal from one path / line is picked up by anotheranother
impulseimpulse irregular pulses or spikesirregular pulses or spikes
• e.g. external electromagnetic interferencee.g. external electromagnetic interference short durationshort duration high amplitudehigh amplitude a minor annoyance for analog signalsa minor annoyance for analog signals but a but a majormajor source of source of errorerror in digital data in digital data
• a noise spike could corrupt many bitsa noise spike could corrupt many bits
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Channel CapacityChannel Capacity
maximummaximum possible data rate on a possible data rate on a communication channel communication channel data rate - in bits per seconddata rate - in bits per second bandwidth - in cycles per second or Hertzbandwidth - in cycles per second or Hertz noise - on communication linknoise - on communication link error rate - of corrupted bitserror rate - of corrupted bits
limitations due to physical propertieslimitations due to physical properties want most efficient use of capacitywant most efficient use of capacity
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Nyquist BandwidthNyquist Bandwidth considers considers noise free noise free channelschannels if rate of signal transmission is 2B then it can if rate of signal transmission is 2B then it can
carry signal with frequencies no greater than B carry signal with frequencies no greater than B i.e. given bandwidth B, highest data rate is 2Bi.e. given bandwidth B, highest data rate is 2B
for binary signals, 2B bps needs bandwidth B Hzfor binary signals, 2B bps needs bandwidth B Hz can increase rate by using M signal levelscan increase rate by using M signal levels Nyquist Formula is: C = 2B logNyquist Formula is: C = 2B log22MM
so so increase rate increase rate by by increasingincreasing signal signal levelslevels at the cost of receiver complexityat the cost of receiver complexity limited by noise & other impairmentslimited by noise & other impairments
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Shannon Capacity FormulaShannon Capacity Formula
considers relation of data rate, considers relation of data rate, noisenoise & error rate & error rate faster data rate shortens each bit so bursts of noise faster data rate shortens each bit so bursts of noise
affects more bitsaffects more bits given noise level, higher signal strength means lower given noise level, higher signal strength means lower
errorserrors
Shannon developed Shannon developed formula relating these formula relating these to to signal to noise ratio (in decibels)signal to noise ratio (in decibels)
SNRSNRdbdb==10 log10 log10 10 (signal/noise)(signal/noise)
Capacity C=B logCapacity C=B log22(1+SNR)(1+SNR) theoretical maximumtheoretical maximum capacitycapacity get lower in practiceget lower in practice
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SummarySummary
looked at data transmission issueslooked at data transmission issues frequency, spectrum & bandwidthfrequency, spectrum & bandwidth analog vs digital signalsanalog vs digital signals transmission impairmentstransmission impairments
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ProblemsProblems
Q1.Q1. For a video signal, what increase in For a video signal, what increase in horizontal resolution is possible if a horizontal resolution is possible if a bandwidth of 5 MHz is used?bandwidth of 5 MHz is used?
Q2.Q2. For a digitized TV picture matrix of For a digitized TV picture matrix of 480×500 pixels, where each pixel can take 480×500 pixels, where each pixel can take one of 48 intensity values, assume 30 one of 48 intensity values, assume 30 pictures are sent per second. Find the pictures are sent per second. Find the source rate (bps). source rate (bps).
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ProblemsProblems Q3.Q3. For the above problem, use Shannon’s For the above problem, use Shannon’s
formula to calculate the channel capacity if formula to calculate the channel capacity if BW used is 4.5 MHz and SNR is 35 dB.BW used is 4.5 MHz and SNR is 35 dB.
Q4. Q4. A multi level signaling system operates A multi level signaling system operates at 9600 bps. A signal element encodes a at 9600 bps. A signal element encodes a 4-bit word. Find the minimum BW required 4-bit word. Find the minimum BW required of the channel.of the channel.
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ProblemsProblems Q5. Q5. A telephone line has 4 KHz BW. When A telephone line has 4 KHz BW. When
signal is 10V, the noise is 5 mV. Find the signal is 10V, the noise is 5 mV. Find the maximum data rate supported by this line.maximum data rate supported by this line.
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ProblemsProblems
Q6. Q6. Consider square wave transmission that Consider square wave transmission that involves the involves the fundamentalfundamental frequency and frequency and all odd all odd frequenciesfrequencies upto the upto the 99thth harmonic. If f = 2MHz, harmonic. If f = 2MHz, then find the bandwidth required and the data then find the bandwidth required and the data rate achieved. If only the rate achieved. If only the fundamental frequency fundamental frequency and the and the 33rdrd harmonic are transmitted, find the harmonic are transmitted, find the bandwidth required and the data rate achieved bandwidth required and the data rate achieved in this case. Comment about the receiver in this case. Comment about the receiver requirements for both the cases. requirements for both the cases.