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Data Communications and Networks
Engr. Abdul‐Rahman MahmoodMS, MCP, QMR(ISO9001:2000)
Usman Institute of Technology – University Road KarachiUsman Institute of Technology University Road, [email protected] [email protected] alphapeeler.sf.net/pubkeys/pkey.htm http://alphapeeler.sourceforge.netpk.linkedin.com/in/armahmood http://alphapeeler.tumblr.comp p p pwww.twitter.com/alphapeeler [email protected] www.facebook.com/alphapeeler [email protected]‐sss alphasecure mahmood_cubix [email protected] [email protected]
VC++, VB, ASP
Data Data Transmission Transmission •Toto, I've got a feeling we're not in Kansas anymore. Judy Garland in The Wizard of Oz
•The successful transmission of data depends principally on two factors: the quality of the signal being transmitted on two factors: the quality of the signal being transmitted and the characteristics of the transmission medium. The objective of this chapter and the next is to provide the reader with an intuitive feeling for the nature of these two factors.
Data TransmissionData Transmission• Toto, I've got a feeling we're not in Kansas anymore. Toto, I ve got a feeling we re not in Kansas anymore. Judy Garland in The Wizard of Oz
• The successful transmission of data depends principally on two factors: the quality of the signalb i t itt d d th h t i ti f th being transmitted and the characteristics of the transmission medium. The objective of this chapter and the next is to provide the reader with an intuitive and the next is to provide the reader with an intuitive feeling for the nature of these two factors.
Transmission Terminologya s ss o e o ogydata transmission occurs b/w transmitter & receiver via some mediumguided medium
eg. twisted pair, coaxial cable, optical fibereg. twisted pair, coaxial cable, optical fiberunguided / wireless medium
eg. air, water, vacuumdirect link
no intermediate devicespoint‐to‐point
direct link only 2 devices share linkonly 2 devices share link
multi‐pointmore than two devices share the link
simplexpone direction. eg. television
half duplexeither direction, but only one way at a time, eg. police radio, y y , g p
full duplexboth directions at the same time, eg. telephone
Frequency, Spectrum and BandwidthTime domain conceptsTime domain concepts
analog signal; various in a smooth way over timedigital signal; maintains a constant level then changes g g gto another constant levelperiodic signal; pattern repeated over timeaperiodic signal; pattern not repeated over timeaperiodic signal; pattern not repeated over time
Analogue & Digital Signalsg g
PeriodicSi lSignalsMathematically, a signal s(t) is defined to be periodic if and only ifp ys(t + T) = s(t) -00 < t < +00where the constant T is the period of the signal (T is the smallest value that satisfies the equation). q )Otherwise, a signal is aperiodic.
Sine Wave
peak amplitude (A): maximum strength of signal volts
frequency (f): rate of change of signal Hertz (Hz) or cycles persignal Hertz (Hz) or cycles per second
Period: time for one repetition (T)T = 1/f
phase (φ): relative position in time
Varying Sine Wavess(t) = A sin(2πft +Φ)s(t) = A sin(2πft +Φ)
Wavelength (λ)is distance occupied by one cyclebetween two points of corresponding phase in two
i lconsecutive cyclesassuming signal velocity v have λ = vT i l l λf or equivalently λf = v
especially when v=cc 3*108 ms‐1 (speed of light in free space)c = 3*108 ms‐1 (speed of light in free space)
Frequency Domain Conceptseque cy o a Co ceptselectromagnetic signal are made up of many frequenciesfrequenciesWhen all of the frequency components of a signal are integer multiples of one frequency, the latter g p q y,frequency is referred to as the fundamental frequency.Fourier analysis can shown that any signal is made up of component sine wavesTh i d f th t t l i l i l t th i d f The period of the total signal is equal to the period of the fundamental frequency. The period of the component sin(2πft) is T = 1/f and the period of s(t) is p ( ) / p ( )also T, as can be seen from next Figure.
Addition of FrequencyComponents (T=1/f)
(c) is sum of f & 3fWe can say that for each i l th i ti signal, there is a time domain function s(t) that specifies the t at spec es t eamplitude of the signal at each instant in time.Similarly, there is a frequency domain function S(f) that function S(f) that specifies the peak amplitude of the constituent frequencies of the signal
.
FrequencyFrequencyDomainRepresentations
freq domain func of Fi Fig 3.4cfreq domain func of i l lsingle square pulse
Spectrum & Bandwidthspectrum
range of frequencies contained in signal
absolute bandwidthwidth of spectrum
effective bandwidthoften just bandwidth
b d f f narrow band of frequencies containing most energy
DC Component f fcomponent of zero frequency
frequency components of the square waveC id Fi 3 2b P i i l 0 dConsider Fig.3.2b Positive pulse = 0 and Negative pulse =1.waveform represents binary stream 0101. . If T = 1/(2f ); Then data rate = 2f bps( f ); pWhat are the frequency components of this signal? To answer this question, consider again Fi 3 4 B ddi h iFigure 3.4. By adding together sine waves at frequencies f and 3f, we get a waveform that begins to resemble the original square wave. Let us continue this process by ddi i f f 5fadding a sine wave of frequency 5f, as
shown in Figure 3.7a, and then adding a sine wave of frequency 7f, as shown in Figure 3.7b.As we add additional odd
lti l f f it bl l d th ltimultiples of f, suitably scaled, the resulting waveform approaches that of a square wave more and more closely. Frequency components of the square wave with
lit d A amplitudes A :
Data Rate and Bandwidth t i i t h li it d b d f f iany transmission system has a limited band of frequencies
this limits the data rate that can be carried(digital) square have infinite components and hence ( g ) q pbandwidthbut most energy in first few componentsIf tt t t t it thi f i l If we attempt to transmit this waveform as a signal over any medium, the transmission system will limit the bandwidth that can be transmittedFor any given medium,
Bandwidth ∝ CostBandwidth ∝ 1/distortionBandwidth ∝ 1/error by the receiverBandwidth ∝ data rate
Data Rate and Bandwidth( I). 3.7(a): f=1MHzB d idth ( X ^6) ^6 MHBandwidth = (5 X 10^6) – 10^6 = 4 MHzT=1/f = 10^‐6 = 1 µs [one bit occurs every 0.5 µ second; ‐ve & +ve freq. in T]Data Rate = 1 / (T/2) bps = 1 / (1 µs /2) = 2 MbpsCASE II. 3.7(a) ‐ f =2MHzT=.5µs; Bandwidth = (5 X 2 X 10^6) – 2 X 10^6 = 8 MHzT = 1/2f = 0.5 X 10^‐6 = 0.5 µs; [one bit occurs every 0.25 µ second]/ 0.5 0 6 0.5 µs; [o e b t occu s e e y 0. 5 µ seco d]Data Rate = 1 / (T/2) bps = 1 / (0.5µs /2) = 4 MbpsCASEIII. Fig 3.4c ‐ (4/∏) [sin(2∏ft) + (1/3)sin(2∏(3f)t)]; f =2MHz, T= 5µs; Bandwidth = (3 X 2 X 10^6) – 2 X 10^6 = 4 MHzT=.5µs; Bandwidth = (3 X 2 X 10 6) – 2 X 10 6 = 4 MHzT = 1/2f = 0.5 X 10^‐6 = 0.5 µs; [one bit occurs every 0.25 µ second]Data Rate = 1 / (T/2) bps = 1 / (0.5µs /2) = 4 MbpsCASE I 3 7(a) f 1MHz T 1 µs; Bandwidth 4MHz; data rate 2MbpsCASE I. 3.7(a) f =1MHz, T=1 µs; Bandwidth = 4MHz; data rate = 2MbpsCASE II. 3.7(a) f =2MHz, T=0.5µs; Bandwidth = 8MHz; data rate = 4MbpsCASE II. 3.4(c) f =2MHz, T=0.5µs; Bandwidth = 4MHz; data rate = 4MbpsConclusion :
higher the center frequency, the higher the potential bandwidth and therefore the higher the potential data rate.
Analog and Digital DataAnalog and Digital Data Transmission
data as entities that convey meaning, or information
signalselectric or electromagnetic representations of data, physically propagates along medium
signaling Signaling is the physical propagation of the signal along a suitable medium transmission
Transmissioncommunication of data by propagation and processing of signals
In what follows we try to make these abstract concepts clear by In what follows, we try to make these abstract concepts clear by discussing the terms analog and digital as applied to data, signals, and transmission.
Acoustic Spectrum (Analog)coust c Spect u ( a og)Frequency components of speech (spectrum of speech) are between 100Hz - 7kHz; Frequencies below 600 or 700 Hz are hard to hear by human ear.
Video Interlaced Signalingdeo te aced S g a gTo achieve adequate resolution, the beam produces a total of 483 horizontal lines at a rate of 30 complete scans of the screen per second. This rate produces a sensation of flicker rather than smooth motion. For flicker‐free
h himage without increasing the bandwidth requirement, a technique known as interlacing is used.
Text Code – IRA , Attenuatione t Code , tte uat oMost popular example is Morse code.International Reference Alphabet (IRA). Each character in this code is
t d b i bit tt 8 diff t h trepresented by a unique 7‐bit pattern; 128 different charactersIRA‐encoded characters are stored and transmitted using 8 bits/character. The eighth bit is a parity bit used for error detection. This bit is set such that the total number of binary 1’s in each octet is always odd (odd parity) or always even (even parity).Thus a transmission error that h i l bit dd b f bit b d t t dchanges a single bit, or any odd number of bits, can be detected.Digital signaling are generally cheaper than analog signaling and is less susceptible to noise interference. The disadvantage is that digital i l ff f tt ti th d l i lsignals suffer more from attenuation than do analog signals.
Audio SignalsFrequency components of typical speech may be found between 100 Hz Frequency components of typical speech may be found between 100 Hz and 7 kHz.easily converted into electromagnetic signalsvarying volume converted to varying voltagea y g o u e co e ted to a y g o tageTelephone handset limits frequency range for voice channel to 300‐3400Hz (within range 100Hz – 7KHz)
Finding video signal bandwidthUSA ‐ 483 lines per frame, at frames per sec
have 525 lines but 42 lost during vertical retrace525 lines x 30 scans per /sec 15750 lines per sec525 lines x 30 scans per /sec = 15750 lines per sec
63.5µs per line [1/15750]11µs for retrace, so 63.5‐11 = 52.5 µs per video line
Find Max freq. if line alternates black & whiteNo of lines = 70% of (525‐42) = 338If screen ratio is 4 : 3If screen ratio is 4 : 3Horizontal res. = 4/3 x 338 =450 Black &White linesHence 450/2 = 225 cycles of wave in 52.5 µs [see above]T l i l d i / T = 1 cycle is completed in = 52.5µs /225 = 0.23 µsf = 1/T = 1/ 0.23 µs = 4285714 Hz ~ 4.2 MHz
horizontal resolution is about 450 lines giving 225 45 g g 5cycles of wave in 52.5 µs; max frequency of 4.2MHz
Digital Dataas generated by computers etcas generated by computers etc.has two dc components : 0 and 1bandwidth depends on data ratebandwidth depends on data rate
Analog Signalsa og S g a s
Digital Signalsg ta S g a s
Advantages & DisadvantagesAdvantages & Disadvantages of Digital Signals
cheapercheaperless susceptible to noisebut greater attenuationbut greater attenuationdigital now preferred choice
Transmission Impairmentssignal received may differ from signal transmitted causing:
analog ‐ degradation of signal qualityanalog ‐ degradation of signal qualitydigital ‐ bit errors
most significant impairments areg pattenuation and attenuation distortiondelay distortion
inoise
Attenuationtte uat o“signal strength falls with distance ‐ guided/unguided”Attenuation depends on medium: Attenuation depends on medium:
It is exponential (guided); More complex (unguided).3 considerations for Attenuation:1. strong enough to be detected2. sufficiently higher than noise to receive without error3. Attenuation is increasing function of frequency [Analog]For 1 & 2: increase strength using amplifiers/repeaters, so equalize attenuation across band of frequencies usedFor 3 : use (1)loading coils (2)amplifiers for high frequenciesFor 3 : use (1)loading coils (2)amplifiers for high frequencies
For any other frequency f, the relative attenuation in decibels isA 1000‐Hz tone of a given power level is applied to the input, and the power P1000, is measured at the output.
Delay Distortiononly occurs in guided mediaOccurs because propagation of a signal velocity varies i h fwith frequency
hence various frequency components arrive at different times resulting in phase shiftsdifferent times resulting in phase shiftsparticularly critical for digital dataintersymbol interference : some of the signal intersymbol interference : some of the signal components of one bit position will spill over into other bit positions, causing intersymbol interference .
NoiseNoise: For any data transmission event, additional unwanted signals that are y , ginserted somewhere between transmission and reception.(1) Thermal
due to thermal agitation of electronsf l d b duniformly distributed
Also called white noiseNo = kT(W/Hz)N noise power density in watts per 1 Hz of bandwidthNo = noise power density in watts per 1 Hz of bandwidthk = Boltzmann’s constant = 1.38 * 10‐23 J/KT = temperature, in kelvinsEXAMPLE 3.1 Room temperature is usually specified as T = 17°C, or 290 K. At 3 p y p 7 , 9this temperature, the thermal noise power density is N0 = (1.38 X 10‐23) X 290 = 4 X 10‐21 W/Hz = ‐204 dBW/Hzwhere dBW is the decibel‐watt, defined in Appendix 3A.
d b d d f f h h h lNoise 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 = kTB or, in decibel‐watts : N = 10 log k + 10 log T + 10 log B
N = ‐228.6 dBW + 10 log T + 10 log BN 228.6 dBW + 10 log T + 10 log BEXAMPLE 3.2 Given a receiver with an effective noise temperature of 294 K and a 10‐MHz bandwidth, the thermal noise level at the receiver’s output isN = ‐228.6 dBW + 10 log(294) + 10 log 107 = ‐228.6 + 24.7 + 70 = ‐133.9 dBW
Noiseo se(2)Intermodulation
Noise: When signals at different frequencies share the same transmission
Effect of noise on a digital signal
the same transmission medium; f1 + f2(3)crosstalk: a signal from
one line is picked up by hanother
(4)impulse:irregular pulses or spikeseg. external electromagnetic interferenceshort durationhigh amplitudea minor annoyance for analog signalsbut a major source of error in digital dataa noise spike could corrupt many bits
Channel CapacityC a e Capac tyChannel Capacity is the max possible data rate on communication channel is a function of
data rate ‐ in bits per secondpbandwidth ‐ in cycles per second or Hertznoise ‐ on comms linkerror rate ‐ of corrupted bits
limitations are due to physical properties of dmedium
we would like to make as efficient use as possible f i b d id h Th i i of a given bandwidth. The main constraint on
achieving this efficiency is noise
Nyquist Bandwidthyqu st a d dtConsider noise free channelsIf rate of signal transmission is 2B then can carry signal g y gwith frequencies no greater than BConverse is also true: Given a bandwidth of B, the highest i l t th t b i d i Bsignal rate that can be carried is 2B.for binary signals, 2B bps needs bandwidth B Hzcan increase rate by using M signal levels (voltage)can increase rate by using M signal levels (voltage)Nyquist Formula is: C = 2B log2Mso increase rate by increasing signalsy g g
EXAMPLE 3.3 Consider a voice channel being used, via modem, to transmit digital data. Assume B=3100 Hz, M=8. Then the Nyquistcapacity, C, of the channel is C= 2Blog2M = 2X3100 log2 8 = 6200 X 3 = p y g2 3 g2 318,600 bpsOnline calculator : http://web2.0calc.com/
Shannon Capacity Formulaconsider relation of data rate noise & error rateconsider relation of data rate, noise & error rate
faster data rate shortens each bit so bursts of noise affects more bitsgiven noise level, higher rates means higher errors
Shannon developed formula relating these to signal to noise ratio (in decibels)SNRdb
=10 log10 (signal/noise)Capacity C=B log2(1+SNR)
theoretical maximum capacityget lower in practise
Submit answers to review questions/problems for Homework #2
u m t an w r t r w qu t n /pr m f r “Data Transmission” of textbook.Rules:L d f b i i i th f O b 1.Last date of submission is 14th of October 2013.
2.Submit handwritten assignments only. 3.Assignments will not be accepted after due date.3.Assignments will not be accepted after due date.4.Plagiarism, if detected, will result in zero marks!5.Assignment must be submitted in a proper file cover,
d b l b l d land must be labeled properly.6.All answers in homework must also include questions numbered exactly as in the text book.questions numbered exactly as in the text book.7.Cover : Student name, roll no, date of submission.8.Print of this slide after cover page.
l d h k ll b d9.Selected homework will be converted to assignments
