data and computer transmission terminology …resource.mitfiles.com/cse/iii year/v...

9/1/2014

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Data and Computer Communications

Chapter 3 – Data Transmission

Eighth Editionby William Stallings

Lecture slides by Lawrie Brown

Transmission TerminologyData transmission occurs between a transmitter & receiver via some mediumTransmission media:Guided medium

The electromagnetic waves guided along physical patheg. twisted pair, coaxial cable, optical fiberg p p

Unguided / wireless mediumProvide a means for transmitting electromagnetic waves but do not guide themeg. air, water, vacuum

In both cases, communication is in the form of electromagnetic waves

9/1/2014 Vivekanand Bhat Dept CSE MIT Manipal 2

Transmission Terminology

direct linkno intermediate devices

point-to-pointdirect link only 2 devices share link

multi-pointmore than two devices share the link



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Transmission Terminology

simplexone direction

• eg. television

half duplexhalf duplexeither direction, but only one way at a time

• eg. police radio

full duplexboth directions at the same time

• eg. telephone


Frequency, Spectrum and BandwidthThe signal is a function of time, but it can also be expressed as a function of frequency; Frequency-domain view of a signal is far more important to understand data transmission than a time-domain view.time domain concepts

analog signal• various in a smooth way over time

digital signal• maintains a constant level then changes to another

constant levelperiodic signal

• pattern repeated over timeaperiodic signal

• pattern not repeated over time9/1/2014 Vivekanand Bhat Dept CSE MIT Manipal 6

Analogue & Digital Signals


PeriodicSignals


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Sine WaveA general sine wave can be represented by three parameters: amplitude (A), frequency (f), and phase (φ).peak amplitude (A)

maximum value or strength of signal over timemeasured in volts

frequency (f)is the rate ( in cycles per second, or Hertz (Hz) at which the signal repeatsg pperiod (T)= the amount of time it takes for one repetition T = 1/f

phase (φ)is the measure of the relative position in time within a single period of a signal (the initial angle of a sinusoidal function at its origin)

Varying Sine Wavess(t) = A sin(2πft +Φ)9/1/2014 Vivekanand Bhat Dept CSE MIT Manipal 9




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Wavelength (λ)

It is the distance occupied by a single cycleIn other words the distance between two points of corresponding phase of two p p g pconsecutive cyclesAssuming signal velocity v have λ = vTor equivalently λf = vEspecially when v=c

c = 3*108 ms-1 (speed of light in free space)9/1/2014 Vivekanand Bhat Dept CSE MIT Manipal 15

Wavelength (λ)


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Frequency Domain Concepts

Signal are made up of many frequenciesFor example, the signal

The components this signal are just sine waves of frequencies f1 and 3f1Individual components shown below





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Frequency Domain ConceptsThere are several interesting points that can be madeabout this figure:

The second frequency is an integer multiple of the firstfrequency, the latter frequency is referred to as theq y, q yfundamental frequency.

The period of the total signal is equal to the period of thefundamental frequency. The period of the component sin(2πf1t) is T = 1/f1 and the period of s(t) is also T.


Frequency Domain ConceptsTime domain function s(t) specifies the amplitude ofsignal at each instant of timeThe frequency domain function S(f) specifies thepeak amplitude of constituent frequencies of thesignalThe figure next shows frequency domain function forThe figure next shows frequency domain function forthe last figureS(f) is discrete



Frequency Domain ConceptsThe next figure shows the frequency domain function fora single square pulse

s(t) = 1 -X/2 ≤ t ≤ X/20 otherwise

S(f) is found symmetric around f=0 and so has negativefrequencies


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Frequency Domain Representations


Spectrum & Bandwidthspectrum

range of frequencies contained in signalEx: In fig C above spectrum extends from f to 3f

absolute bandwidthwidth of spectrumEx: In fig C above the bandwidth is 2fEx: In fig C above the bandwidth is 2f

effective bandwidthoften just bandwidthnarrow band of frequencies containing most energy

DC (direct current) Componentcomponent of zero frequency


Data Rate and BandwidthData rate: The number of bits sent in 1s. Often measured in bits per second (bps)

Any transmission system has a limited band of frequencies. This limits the data rate that can be carried

Single square pulse have infinite components and henceSingle square pulse have infinite components and hence infinite bandwidth. But most energy in first few components.

Limited bandwidth increases distortion have a direct relationship between data rate & bandwidth

To explain these relationships, consider the square wave9/1/2014 Vivekanand Bhat Dept CSE MIT Manipal 27

Data Rate and Bandwidth


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Suppose that we let a positive pulse representbinary 1 and a negative pulse represent binary0.Then, the waveform represents the binaryThen, the waveform represents the binarystream 1010. . . .The duration of each pulse is 1/2f1; hence, thedata rate is 2f 1 bits per second (bps).What are the frequency components of thissignal? To answer this question, consider againFigure



Data Rate and BandwidthBy adding together sine waves at frequencies f1 and 3f1,we get a waveform that resembles the square wave.


Data Rate and BandwidthLet us continue this process by adding a sine wave offrequency 5f1


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Data Rate and Bandwidththen adding a sine wave of frequency 7f1


Data Rate and BandwidthAs we add additional odd multiples of f, suitably scaled,the resulting waveform approaches that of a squarewave more and more closely.It can be shown that the frequency components of thesquare wave can be expressed as follows:

This waveform has an infinite number of frequencycomponents and, hence, an infinite bandwidth.However, the amplitude of the kth frequencycomponent,kf1, is only 1/k, so most of the energy in thiswaveform is in the first few frequency components.


Data Rate and BandwidthSuppose that we are using a digital transmission system that iscapable of transmitting signals with a bandwidth of 4 MHz.Let us attempt to transmit a sequence of alternating 1s and 0s asthe square wave


Data Rate and BandwidthWhat data rate can be achieved?We look at the three cases:Case 1:Let us approximate our square wave withthe waveformAl h h hi f i "di d"Although this waveform is a "distorted" square wave,it is sufficiently close to the square wave that areceiver should be able to discriminate between abinary 0and a binary 1.


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Now, if we let fl = l06 cycles/second = 1 MHz, then thebandwidth of the signal is (5 X l06) - l06 = 4 MHz.N h f f 1 MH i d f h f d l fNote that for fl = 1 MHz, period of the fundamental frequency

is T = 1/ l06 = l0-6 = 1 µs.Thus, if we treat this waveform as a bit string of 1s and Os,one bit occurs every 0.5 µs, for a data rate of 2 X l06 = 2 Mbps.Thus, for a bandwidth of 4 MHz, a data rate of 2 Mbps isachieved.


Data Rate and BandwidthCASE 2: Now suppose that we have a bandwidth of 8 MHz


Data Rate and BandwidthIf fl = 2 MHz, using the same line of reasoning as before, thebandwidth of the signal is (5 x 2 x l06 ) - (2 x l06 ) = 8 MHz.But in this case T =1/ fl = 0.5 µs. As a result, one bit occursevery 0.25 µs for a data rate of 4 Mbps.Th h hi b i l b d bli h b d id hThus, other things being equal, by doubling the bandwidth, wedouble the potential data rate.


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Data Rate and BandwidthCase3--


Data Rate and BandwidthNow, let fl = 2 MHz. Using the same line ofreasoning as before, the bandwidth of the signal inFigure is (3 X 2 X l06 ) -(2 X l06 ) = 4 MHz.

But, in this case, T = 1/ fl = 0.5 µs. As a result, onebit occurs every 0.25 µs, for a data rate of 4 Mbps.

Thus, a given bandwidth can support various datarates depending on the requirements of the receiver.


SUMMARIZECASE1– Bandwidth=4MHz; data rate= 2MbpsCASE2– Bandwidth=8MHz; data rate= 4MbpsCASE3– Bandwidth=4MHz; data rate= 4Mbps

Limiting the bandwidth creates the greater distortions,there by interpreting the received signal more difficult


Effect of bandwidth on digital signalA digital bit stream with a data rate of 2000 bits persecond. With a bandwidth of 1700 to 2500 Hz, therepresentation is quite good.

In general, if the data rate of the digital signal is W bps,then a very good representation can be achieved with ay g pbandwidth of 2W Hz;

However, unless noise is very severe, the bit pattern canbe recovered with less bandwidth than this.


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Observation


Relationship between data rate and bandwidth

The higher the data rate of a signal, the greater is its effective bandwidth.

Looked at other way, the greater the bandwidth of atransmission system, the higher is the data rate that can bet itt d th t ttransmitted over that system.

The higher the center frequency, (bandwidth of a signal asbeing centered about some frequency) the higher the potentialbandwidth & therefore the higher the potential data rate.


Analog and Digital Data Transmission

Analog and digital correspond to continuous and discrete, respectively. These 2 terms are used in at least 3 contexts ( data, signalling & transmission ) in digital communications

Data E titi th t i i f tiEntities that convey meaning, or information

SignalsElectric or electromagnetic representations of data,

SignallingPhysical propagation of signals along a suitable medium

Transmissioncommunication of data by propagation and processing of signals9/1/2014 Vivekanand Bhat Dept CSE MIT Manipal 47

Analog and Digital DataAnalog data take on continues values in some interval.

For example, audio (speech) and video are continuously varying patterns of intensity. Most data collected by sensors, such as temperature and pressure are continuous valued

Frequency components of typical speech may be found between approximately 100 Hz and 7 kHz, and has a dynamic range of about 25 dB (a shout is approx 300 times louder than whisper).

Digital data take on discrete vales; examples text and integers9/1/2014 Vivekanand Bhat Dept CSE MIT Manipal 48

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Acoustic Spectrum (Analog) for human speech


Analog and Digital DataAnother common example of analog data is video

To produce a picture on the screen, an electron beamscans across the surface of the screen from left to rightand top to bottom.To achieve adequate resolution, the beam produces atotal of 483 horizontal lines at a rate of 30 completescans of the screen per second.Tests have shown that this rate will produce a

sensation of flicker rather than smooth motion.The flicker is eliminated by a process of interlacing (theodd numbered scan lines and the even numbered scanlines are scanned separately, with odd and even fieldsalternating on successive scans )9/1/2014 Vivekanand Bhat Dept CSE MIT Manipal 50


Analog and Digital SignalsLike the data signals can either analog or digitalAn analog signal is a continuously varying electromagnetic wave that may be propagated over a variety of media, depending on spectrum

Examples of wire media, such as twisted pair, coaxial bl fib ti bl & id d di hcable, fiber optic cable & unguided media , such as

atmosphere or space propagation

A digital signal is a sequence of voltage pulses that maybe transmitted over a wire medium;For example, a constant positive voltage level mayrepresent binary 1, and a constant negative voltage levelmay represent binary 0.9/1/2014 Vivekanand Bhat Dept CSE MIT Manipal 52

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Audio Signals

Freq range 20Hz-20kHz (speech 100Hz-7kHz)Easily converted into electromagnetic signalsVarying volume converted to varying voltageCan limit frequency range for voice channel toCan limit frequency range for voice channel to 300-3400Hz


Video SignalsTo produce a video signal, a TV camera, which performs similar functions to the TV receiver, is used.

USA - 483 lines per frame, at frames per sechave 525 lines but 42 lost during vertical retrace

525 lines x 30 scans = 15750 lines per sec63 5 s per line63.5μs per line11μs for retrace, so 52.5 μs per video line

Max frequency if line alternates black and whiteHorizontal resolution is about 450 lines giving 225 cycles of wave in 52.5 μsMax frequency of 4.2MHz


Digital Data

As generated by computers etc.Has two dc componentsbandwidth depends on data rate


Analog Signals


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Digital Signals


Advantages & Disadvantages of Digital Signals

cheaperless susceptible to noisebut greater attenuationdigital now preferred choice


Analog and Digital Transmission

• Analog transmission is a means of transmitting analog signalswithout regard to their content;

• The signals may represent analog data (e.g., voice) or digitaldata (e.g., binary data that pass through a modem).

• In either case, the analog signal will become weaker (attenuate)after a certain distance. To achieve longer distance, amplifiersare usedare usedDigital transmission, in contrast, is concerned with the contentof the signal.A digital signal can be transmitted only a limited distance beforeattenuation endangers the integrity of the data.To achieve greater distances, repeaters are used.A repeater receives the digital signal, recovers the pattern of 1sand 0s, and retransmits a new signal, thereby overcoming theattenuation.



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Preferred Mode of TransmissionWhich is the preferred method of transmission?The answer being supplied by the telecommunicationsindustry and its customers is digitalThe reasons :Digital technology: Advent LSI and VLSIData integrity: With use of repeaters the effects of noiseData integrity: With use of repeaters, the effects of noiseis not cumulativeCapacity utilization: Economical to build very highbandwidth channelsSecurity and privacy: Encryption can be readily appliedto digital dataIntegration Treating both analog and digital data digitally


Transmission ImpairmentsSignal received may differ from signal transmitted due to various transmission impairments :

analog signals - degradation of signal qualitydigital signals - bit errors

Most significant impairments areattenuation and attenuation distortiondelay distortionnoise


AttenuationAttenuation means a loss of signal strengthIt depends on transmission mediumFor guided media signal strength falls of exponentially For unguided media it is complex function of distance & atmosphereAttenuation introduces three considerations for transmissionReceived signal strength must be1. strong enough to be detected2. sufficiently higher than noise to receive without error

To overcome above problems increase signal strength using amplifiers/repeaters3. Attenuation is also an increasing function of frequency

To overcome this problem equalize attenuation across band of frequencies

eg. using loading coils or amplifiers9/1/2014 Vivekanand Bhat Dept CSE MIT Manipal 64

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Figure 3.26 Attenuation



Measurement of Attenuation: DecibelsAn important parameter in any transmission system is the signal strengthIt is customary to express gains, losses, and relative levels in decibels, because

1. Signal strength often falls off exponentially, so loss is easily expressed in terms of the decibel, which is a logarithmic unit.

2. Decibel numbers can be added (or subtracted) when we are measuring several points (cascading) instead of just two.

The decibel is a measure of ratio between two signal levels. The decibel gain is given by:

GdB = 10 log10(Pout/Pin)Pin- input signal power level

Pout - output signal power levelA gain of 3 dB is equivalent to doubling the power.A loss of 3 dB (–3 dB) is equivalent to losing one-half the power.


Suppose a signal travels through a transmission medium andits power is reduced to one-half. This means that P2 is(1/2)P1. In this case, the attenuation (loss of power) can becalculated as

Example 3.26

A loss of 3 dB (–3 dB) is equivalent to losing one-half thepower.


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A signal travels through an amplifier, and its power isincreased 10 times. This means that P2 = 10P1 . In this case,the amplification (gain of power) can be calculated as

Example 3.27


One reason that engineers use the decibel to measure thechanges in the strength of a signal is that decibel numberscan be added (or subtracted) when we are measuring severalpoints (cascading) instead of just two. In Figure 3.27 a signaltravels from point 1 to point 4. In this case, the decibel valuecan be calculated as

Example 3.28

can be calculated as


Figure 3.27 Decibels for Example 3.28


Decibel and Signal Strength Table below shows the relationship between the decibels value and power of 10

POWER RATIO dB POWER RATIO dB

101 10 10-1 -10

102 20 10-2 -20

103 30 10-3 -30

104 40 10-4 -40

105 50 10-5 -50

106 60 10-6 -609/1/2014 Vivekanand Bhat Dept CSE MIT Manipal 72

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Decibel and Signal Strength We define a decibel loss as:

LdB= -10 log10 (Pout/Pin)=10 log10 (Pin/Pout)

Example: If a signal with a power level of 10 mW isinserted onto a transmission line and the measuredpower some distance away is 5 mW, the loss can beexpressed asLdB =10 log(10/5) = l0(0.3) = 3 dB

Note that the decibel is a measure of relative, not absolute difference.


Decibel and Signal StrengthThe decibel is also used to measure the difference in voltage, taking into account that power is proportional to the square of the voltage:

P=(V2/R)

WhereP = power dissipated across resistance RV= voltage across resistance R

LdB= 10 log(Pin/Pout)=10log((Vin2/R)/(V2

out/R))= 20log(Vin/Vout)


Decibel and Signal Strength• The dBW (decibel-Watt) is used extensively in microwave

applications.• The value of 1 W is selected as a reference and defined to be 0

dBW.• The absolute decibel level of power in dB W is defined as

PowerdBW=10log(PowerW/1W)

For example, a power of 1000 W is 30dBW, and a power of 1mW is -30dBW.Another common unit is the dBm (decibel-milliWatt), which uses I mW as the reference , thus 0dBmW=1mW

PowerdBmW=10log(PowermW/1 mW)The relationship+30 dBmW = 0 dBW0 dBmW = -30dBW9/1/2014 Vivekanand Bhat Dept CSE MIT Manipal 75

DecibelsExample: Consider a series in which the input is at a power level of 4 mW, the first element is a transmission line with a 12-dB loss (-12 dB gain) , the second element is an amplifier with a 35-dB gain, & the third element is transmission line with a 10-dB loss. Calculate output power Pout?The net gain is (-12 + 35 - 10) = 13 dBG = 13 = 10 log (P /4 mW)GdB = 13 = 10 log (Pout/4 mW)

Pout = 4 x 101.3 mW = 79.8 MW


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Decibel and Signal Strength A unit in common use in cable television and broadband LAN applications is the dBmV (decibel-millivolt).This is an absolute unit with 0 dBmV equivalent to 1 mV. Thus,

The voltage levels are assumed to be across a 75-ohm resistance.


Delay DistortionDelay distortion occurs because the velocity of propagation of signal through a guided medium varies with frequency. Hence various frequency components arrive at different times

Delay distortion particularly critical for digital data since parts of one bit spill over into other bit position causing intersymboli t finterference

Equalizing techniques can also be used for delay distortion.

Again using a leased telephone line as an example, Figure shows the effect of equalization on delay as a function of frequency.



NoiseAdditional signals inserted between transmitter and receiver

Noise divided into 4 types:1. Thermal Noise2. Intermodulation noise3. Crosstalk4. Impulse noise

Thermal Noise (also referred as white noise)due to thermal agitation of electronsuniformly distributedIt cannot be eliminated


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Thermal NoiseThe amount of thermal noise to be found in a bandwidthof 1 Hz in any device or conductor is

N0= kT(W/Hz)

whereNo = noise power density in watts per 1Hz of bandwidthNo noise power density in watts per 1Hz of bandwidthk = Boltzmann's constant = 1.38 X10-23 J/KT = temperature, in kelvin ( K is used to represent 1 kelvin)


Thermal NoiseThe noise is assumed to be independent of frequency.Thus, the thermal noise, in watts, present in a bandwidthof B hertz can be expressed as

N= kTBor, in decibel-watts,

N = 10 log k + 10 log T + 10 log BN 10 log k 10 log T 10 log B= -228.6 dBW + 10 log T + 10 log B

Example: Given a receiver with an effective noise temperature of 294 K and a 10 MHz bandwidth, the thermal noise level at the receiver output isN = -228.6 dBW + 10 log 294 + 10 log 107

= -228.6 +24.7 + 70 = -133.9 dbW9/1/2014 Vivekanand Bhat Dept CSE MIT Manipal 82

Intermodulation Noise

When signals at different frequencies share the same transmission medium, the result may be intermodulationnoise. 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 those frequenciesoriginal frequencies, or multiples of those frequencies.

For example, the mixing of signals at frequencies f1 and f2 might produce energy at the frequency f1 + f2.

Intermodulation noise is produced when there is somenonlinearity in the transmitter, receiver, or interveningtransmission system.


NoiseCrosstalk

a signal from one line is picked up by anotherImpulse Noise

non continuous consisting of irregular pulses or spikes of short duration & high amplitude

• eg. external electromagnetic interferenceg ga minor annoyance for analog signalsbut a major source of error in digital data

• a noise spike could corrupt many bits


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Channel CapacityThe maximum rate at which data can be transmitted over a given communication channel It is a function of

data rate - in bits per secondbandwidth - in cycles per second or Hertznoise – average level of noise over the communication linknoise average level of noise over the communication linkerror rate - of corrupted bits

limitations due to physical properties of transmission mediumWe want most efficient use of capacity


Nyquist BandwidthConsider noise free channelsIf rate of signal transmission is 2B then a signal with frequencies no greater than B is sufficient to carry to signal rate .

ie. given bandwidth B, highest signal rate is 2BFor binary signals, 2B bps needs bandwidth B HzCan increase rate by using M signal levelsNyquist Formula for capacity of channel: C = 2B log2MWhere M is the number of discrete signal or voltage levels

Example: For M =8 & B = 3100Hz, then C =18600 bpsso increase rate by increasing signals

at cost of receiver complexitylimited by noise & other impairments9/1/2014 Vivekanand Bhat Dept CSE MIT Manipal 87

Shannon Capacity FormulaConsider 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 power /noise power )=10 log10 (SNR)

The max channel capacity is C=B log2(1+SNR) bpstheoretical maximum capacityget lower in practise


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Shannon Capacity FormulaExample: Suppose that the spectrum of a channel is

between 3 MHz and 4 MHz & SNRdB = 24 dB. Compute channel capacity and number of signal levels

Bandwidth is = 4 -3 = 1 MHzSNRdB = 24 dB = 10 log10 (SNR)SNR =251C = 106 x log2 (1 +251) = 106 x 8 = 8 MbpsC =2B log2 M8 x106 = 2 x106 x log2M4 = log2MM =169/1/2014 Vivekanand Bhat Dept CSE MIT Manipal 89

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Expression Eb/NO

A parameter related to SNR that is more convenient fordetermining digital data rates and error rates.

The parameter is the ratio of signal energy per bit to noise-power density per hertz, Eb/NO

Consider a signal, digital or analog, that contains binary digitaldata transmitted at a certain bit rate R.

Recalling that 1 watt = 1 J/s, the energy per bit in a signal is given by

Eb=STb,

Where S is the signal power & Tb is the time required to send one bit. Data rate R is just R= 1/Tb

Expression Eb/NOHence, (Eb / NO) = ((S / R) / NO) = (S / kTR)In decibel notation,

(Eb/NO)dBW = SdBW - 10logR - 10logk - 10logT

(Eb/NO)dBW = SdBW - 10logR + 228.6dBW-10logT

Ratio Eb/NO is important because the bit error rate for digitalRatio Eb/NO is important because the bit error rate for digitaldata is a (decreasing) function of this ratio.

Given a value of Eb/NO needed to achieve a desired error rate,the parameters in the preceding formula may be selected.The bit rate R increases, the transmitted signal power, relativeto noise, must increase to maintain the required Eb/NO

Advantage of Eb/NO over SNR is that latter quantity depends onthe bandwidth

Expression Eb/NOExample: For binary phase shift keying, Eb/NO = 8.4 dB is required for a bit error rate of 10-4 (one bit error out of every 10, 000). If the effective noise temperature is 2900 K (room temperature) and the data rate is 2400 bps. Compute received signal power

(Eb/NO)dBW = SdBW - 10logR + 228.6 dBW-10logT

8.4 = SdBW - 10 log 2400 + 228.6 dBW – 10 log 290

SdBW = -161.8 dBW

Expression Eb/NO

We can relate Eb/NO to SNR as follows. We have

( Eb / NO ) = ( S / NOR )

The Parameter NO is the noise power density in Watts/HertzNoise in a signal with bandwidth B is N = N BNoise in a signal with bandwidth BT is N = NOBT

Substituting we have

( Eb/NO ) = ( S / N ) ( BT / R ) ------ (1)

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Expression Eb/NOAnother formulation of interest relates to Eb/NO spectral frequency. Shannon’s result can be rewritten as

(S/N) =(2C/B -1)Using equation (1) and equating BT with B and R with C, we have

(Eb/NO)=(B/C) (2C/B -1)This is a useful formula that relates the achievable spectralThis is a useful formula that relates the achievable spectral efficiency C / B to Eb / NO

Example: Compute Eb/NO required to achieve a spectral efficiency of 6 bps/Hz.

Eb/NO = (1/6) (26 - 1) = 10.21 dB

data and computer transmission terminology …resource.mitfiles.com/cse/iii year/v...

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