shannon capacity basics

8/18/2019 shannon capacity basics

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Computer Communication &Networks

Lecture # 04Physical Layer: Signals & Digital Transmission

Nadeem Majeed [email protected]

mailto:[email protected]:[email protected]


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Physical Layer Topics to

Cover Signals Digital Transmission

Analog Transmission Multiplexing

Transmission Media


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Analog & Digital Data

Data can be analog or digital. The termanalog data re ers to in ormation that iscontinuous! digital data re ers to in ormationthat has discrete states. "nalog data take oncontinuous #alues. Digital data take ondiscrete #alues.


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To be transmitted, data must betransformed to electromagnetic signals.

Note


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Signals can be analog or digital.Analog signals can have an infinitenumber of values in a range; digital

signals can have only a limitednumber of values.

Note


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Analog Vs Digital


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Analog Signals


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Sine Wave


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Cont’

Three parameters to describe a sine $a#e

%. Peak amplitude. 're(uency and time period

). Phase


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The bandwidth of a composite signal isthe difference between the

highest and the lowest frequenciescontained in that signal.

Note


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Digital Signals


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Digital Signals

*n addition to being represented by an analogsignal+ in ormation can also be representedby a digital signal. 'or example+ a % can beencoded as a positi#e #oltage and a , as-ero #oltage. " digital signal can ha#e morethan t$o le#els. *n this case+ $e can send

more than % bit or each le#el.


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Digital Signal


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Bit Rate & Bit nterval!contd"#


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Bit nterval and Bit Rate

Example ExampleA digital signal has a bit rate of 2000 bps. What is theduration of each bit (bit interval)

SolutionSolution

The bit interval is the inverse of the bit rate.

Bit interval = 1/ 2000 s = 0.000 00 s= 0.000 00 ! 10 " µ s = 00 µ s


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The bit rate and the bandwidth areThe bit rate and the bandwidth are

proportional to each other. proportional to each other.

Note


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Analog Vs Digital


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Low Pass & Band Pass


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Data Rate Li%its


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Data Rate Li%its

A ver# i$portant consideration in dataco$$unications is ho% fast %e can send data& in bits

per second& over a channel. 'ata rate depends on

three factors1. The band%idth available2. The level of the signals %e use

. The *ualit# of the channel (the level of noise)


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Example 8 Example 8

+onsider the sa$e noiseless channel& trans$itting a signal%ith four signal levels (for each level& %e send t%o bits).The $a!i$u$ bit rate can be calculated as

Bit ate = 2 ! 000 ! logBit ate = 2 ! 000 ! log 22 = 12&000 bps = 12&000 bps


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Increasing the levels of a signal mayreduce the reliability of the system.

Note


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Example Example

+onsider an e!tre$el# nois# channel in %hich the valueof the signal to noise ratio is al$ost -ero. n other %ords&the noise is so strong that the signal is faint. or thischannel the capacit# is calculated as

+ = B log+ = B log 22 (1 3 45 ) = B log(1 3 45 ) = B log 22 (1 3 0)(1 3 0)

= B log= B log 22 (1) = B(1) = B ×× 0 = 00 = 0


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Example Example

We can calculate the theoretical highest bit rate of aregular telephone line. A telephone line nor$all# has a

band%idth of 000. The signal to noise ratio is usuall#1"2. or this channel the capacit# is calculated as

+ = B log+ = B log 22 (1 3 45 ) = 000 log(1 3 45 ) = 000 log 22 (1 3 1"2)(1 3 1"2)= 000 log= 000 log 22 ( 1" )( 1" )

+ = 000+ = 000 ×× 11."2 = &6"0 bps11."2 = &6"0 bps


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Example ExampleWe have a channel %ith a 1 7,- band%idth. The 45for this channel is " 8 %hat is the appropriate bit rate andsignal level9

SolutionSolution

C = B logC = B log 22 (1 + SNR) = 10(1 + SNR) = 10 66 loglog 22 (1 + 63) = 10(1 + 63) = 10 6

6 loglog 22 (64) = 6 Mbps(64) = 6 Mbps

Then we use the Nyquist formula to nd thenumber of signal levels.

66 Mbps = 2Mbps = 2 1 MHz1 MHz loglog 22 L L L = 8 L = 8

irst& %e use the 4hannon for$ula to find our upperirst& %e use the 4hannon for$ula to find our upperli$it.li$it.


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The Shannon capacity gives us theupper limit; the Nyquist formula tells us

how many signal levels we need.

Note


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Trans%ission %pair%ents


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Trans%ission %pair%ents

Signals tra#el through transmission media+$hich are not per ect. The imper ectioncauses signal impairment. This means that

the signal at the beginning o the medium isnot the same as the signal at the end o themedium. 7hat is sent is not $hat is recei#ed.Three causes o impairment are attenuationattenuation ,,distortiondistortion ,, and noisenoise ..


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Deci)el8sed to signal gained or lost strength

d0 2 %, log P 9P% 4Po$er6

d !oltage " #$ log %!&'!#(d )urrent " #$ log %I&'I#(

The d0 is not an absolute (uantity+ it is al$ays a *ATI+ ot$o (uantities. The unit can be used to express po$ergain 4P P%6+ or po$er loss 4P ;P%6


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*+a%ple

Suppose a signal tra#els throughtransmission medium and its po$er isreduced to one hal =. /alculate attenuation

loss>


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Signal Distortion

attenuation

distortion

noise


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Per,or%ance

0and$idthThroughputLatency 4Delay60and$idth


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Thro$ghp$t


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Latency

Latency 2 propagation time5

transmission time 5

(ueuing time 5

processing time


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Propagation Ti%e


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The bandwidth delay product definesthe number of bits that can fill the lin-.

Note


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Analog To DigitalConversion


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Sa%pling

Pulse Code Modulation Pulse Code ModulationSampling Rate: Nyquist TheoremSampling Rate: Nyquist Theorem


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PC-

$anti/ation & *ncoding


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.$anti/ation & *ncodingSa%ples


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According to the Nyquist theorem, thesampling rate must be

at least # times the highest frequencycontained in the signal.

Note


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shannon capacity basics

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