Signals and Data Transfer

8/6/2019 Signals and Data Transfer

http://slidepdf.com/reader/full/signals-and-data-transfer 1/73

UNIT - I

Basic concepts



SDM,CA/SoC 2

Topics

Introduction

Internet Model

Signals

Transmission Media



Signals



SDM,CA/SoC 4

Signals and Data Transfer

One of the maj or func tions of the physical la yer is tomove da ta in the for m of signals ac ross atr ansm ission med ium.The da ta need to be d igitiz ed for stor age andinter pr etation by computer The d igit al da ta has to be converted into signals inor der to tr ansm it it over the med ium (o ptical ,elec tromagnetic signals).

The signals a r e class ified asAnalog SignalDigit al Signal



SDM,CA/SoC 5

A nalog Signal

Analog signalis a c ontinuou s wa vefor m

has sm ooth tr ans itio ns.It holds infinite valueswithin a bou nda ry.

Examples: na tur allyoccurri ng music and voice



SDM,CA/SoC 6

Digital Signal

Digit al signalis a d iscr ete or non-

continuou s wa vefor mis the wa ve for m w ithins tan taneous tr ans itio ns.

It can hold only a se t of values w ithi n a bou nda ryExamples: a udio ,vi deo or anythi ng in digit al for m



SDM,CA/SoC 7

P eriodic A nd A periodicSignals

Another classification of signals is pe riodic s ignalsape riodic s ignals .



SDM,CA/SoC8

P eriodic Signal

The wa ve comple tes a pa tter n within ameas ur a ble time fr ame and the pa tter n isr epea tedThe pa tter n r epea ted is called as Cycle.The time d ur ation is called as Pe riod.

The pe riod is constant for a Sig nal.



SDM,CA/SoC9

AP eriodic Signal

W hen a wa ve does n ot exhibit any pa tter n ,over a pe riod of time then it is called asAPeriodic signal



A nalog Signals



SDM,CA/SoC11

Types of A nalog Signals

Simple Sig nalComposite Sig nal



Simple Signals



SDM,CA/SoC13

Sine WaveThe Simple Signal consists of only one signalComponent.Example : Sine wa ve



SDM,CA/SoC14

Sine Wave

Simple pe riodic anal og signalS(t) = A Sin(2 Tht+ N)

W her eS- ins tan taneous ampl itude (Ampl itut e at time t)A- Peak Ampl itude

h- Fr equenc yN- Phaset-T ime



SDM,CA/SoC15

A mplitude& P eak A mplitude

A mplitude:It is the vertical d istance from a p oint on the curv e to the

horizo ntal ax is.P eak A mplitude

Peak ampl itude is the a bsolute longest vertical d istance and it occur s at the peak / crust of the curv e.



SDM,CA/SoC17

P eriod and F requencyFr equenc y periods a r e inver sely pro portio nalFr equenc y = num ber of periods / Cycle pe r sec ond.

f = 1 / T.W hen the s ignal tr ans its ins tan tane ous (jump) , then

the fr equenc y is infinite.W hen ther e is no change in s ignal s tate then the

fr equenc y = 0.



SDM,CA/SoC18

Units of P eriods and F requenciesUnits of P eriods and F requencies

Unit Equivalent Unit Equivalent

Seconds (s) 1 s hertz (Hz) 1 Hz

Milliseconds (ms) 10 ±3 s kilohertz (KHz) 10 3 Hz

Microseconds ( s) 10 ±6 s megahertz (MHz) 10 6 Hz

Nanoseconds (ns) 10 ±9 s gigahertz (GHz) 10 9 Hz

P icoseconds (ps) 10 ±12 s terahertz (THz) 10 12 Hz



SDM,CA/SoC19

Example 1Example 1

Exp r ess a pe riod of 100 ms in m icrosec onds , andexp r ess the corr esp ond ing fr equenc y in k ilohertz .

Solution Solution100 ms Now we use the inver se r ela tionshi p to find thefr equenc y

f = 1 /100 KH z = 10 -2 KH z



SDM,CA/SoC20

P eriod and F requency

Change in a short span of time means highfr equenc yChange over a long span of time means l owfr equenc y.If a signal d oes n ot change at all , its fr equenc y

is zeroIf a signal c hanges instantane ously, its

fr equenc y is infinite



SDM,CA/SoC21

P hase

P hase describes the position of the waveformrelative to time zero .



SDM,CA/SoC22

P hase

A change in phase can be an y num ber of angles between 0 and 360 de gr ees.Phase c hanges often occur on comm on an gles ,such as 45 , 90, 135 , etc.Phase is meas ur ed in de gr ees or r adiansA phase s hift of 360° c orr esponds to a shift of

a comple te pe riodPhase s hift of 180° c orr esponds to a shift of one-half of a pe riod



SDM,CA/SoC 23

Ex ample 2 Ex ample 2A sine wa ve is off set one-s ixth of a cycle w ithr espec t to t ime zero. W hat is its phase indegr ees and r adians?

Solutio nSolutio nW e kn ow that one c omple te cycle is 360degr ees.Ther efor e, 1/6 cycle is

(1/6) 360 = 60 de gr ees = 60 x 2 /360 r ad =1.046 r ad



SDM,CA/SoC 24

Sine wave examples



SDM,CA/SoC 25

Sine wave examples (continued)



SDM,CA/SoC 26

Sine wave examples (continued)



SDM,CA/SoC 28

Time and frequency domains(continued )

time-d omain pl ot shows c hanges in s ignal ampl itude w ith r espec t to time

fr equenc y-domain plot shows the r ela tionshi p betweenampl itude and fr equenc y



SDM,CA/SoC 29

Time and frequency domains

(continued)



SDM,CA/SoC 30

Composite Signal

A single-fr equenc y sine wa ve is not useful indata communications;

we need to change one or mor e of itschar acteristics to make it usefulW henW hen wewe c hangechange oneone or or mor emor e char acteristicschar acteristics

of of aa singlesingle--fr equenc yfr equenc y signal ,signal , itit becomes becomes aacompositecomposite signalsignal mademade of of man yman y fr equenc iesfr equenc ies..



SDM,CA/SoC 31

Composite Signal

The s ignal c onsists of multi ple c omponen ts.Eac h componen t is of diff er ent fr equenc ies /

ampl itude.It can be d ivided into componen ts.Example : V oice Sig nal



SDM,CA/SoC 32

Composite Signal

A ccording to Fourier analysis, any compositesignal can be represented as a combination of simple sine waves with different frequencies,

phases, and amplitudes .



SDM,CA/SoC 33

T hree harmonics



SDM,CA/SoC 34

A dding first three harmonics



SDM,CA/SoC 35

Bandwidth

The bandw idth is a p ro perty of a med iumIt is the diff er ence between the high est and the

lowes t fr equenc ies that the med ium cansatisf actori ly pass.



SDM,CA/SoC 36

Bandwidth

Bandw idth of a med ium is the r ange of fr equenc ies it carries.

Example: A med ium can ca rry from 1000 HZto 5000 HZ.The bandw idth is als o calc ulated as thediff er ence between the high est and l owes t fr equenc ies.Bandw idth = HF ± LF = 5000 ± 1000= 4000



SDM,CA/SoC 37

Bandwidth



SDM,CA/SoC 38

Bandwidth

W hen the fr equenc y spec trum of the s ignal ishigh er than the med ium¶s then the fr equenc y r ate of the s ignal that lies out side the spec trumof the med ium is cho pped off .



SDM,CA/SoC 39

Bit Rate & Bit Interval

Bit rate: Num ber of bits tr ansm itted in one sec ond( b ps).Bit Interval: The time d ur ation for which the s ignalhas to r ema in in a s tate to r epr esen t a bit Val ue.It is the time d ur ation to send a s ingle bit.Example: I f the bit r ate is 2000 b ps , then find the bit

interval.Bit r ate = n um ber of bits/ (B it Interval)Bit Interval= 1/2000 = .0005 sec onds



SDM,CA/SoC 40

Example 3Example 3

If a pe riodic s ignal is dec omposed into fivesine wa ves w ith fr equenc ies of 100 , 300 , 500 , 700 , and 900 H z, what is the bandw idth? Dr awthe spec trum, assuming all c omponen ts have amax imum ampl itude of 10 V.

SolutionSolutionB = hf - lf = 900 - 100 = 800 H zThe spec trum has only five sp ikes , at 100 , 300 , 500 , 700 , and 900



SDM,CA/SoC 41

Example 4

A signal has a bandw idth of 20 H z. Thehigh est fr equenc y is 60 H z. W hat is the lowes tfr equenc y? Dr aw the spec trum if the signalcontains all integr al fr equenc ies of the sameampl itude.SolutionSolutionB = hf B = hf -- lf lf

20 = 6020 = 60 -- f lf lf l = 60f l = 60 -- 20 = 40 H z20 = 40 H z



SDM,CA/SoC 42

Example 5Example 5

A signal has a spec trum w ith fr equenc ies between 1000 and 2000 H z ( bandw idth of 1000Hz). A med ium can pass fr equenc ies from 3000to 4000 H z (a bandw idth of 1000 H z). Can thissignal f aithfu lly pass through thi s med ium?

SolutionSolution

The answe r is de finitely no.The answe r is de finitely no.



SDM,CA/SoC 43

Digital Signals

B it Interval and Bit RateA s a Composite A nalog Signal

T hrough Wide- Bandwidth MediumT hrough Band-Limited MediumVersus A nalog Bandwidth

Higher Bit Rate



SDM,CA/SoC 44

A digital signal



SDM,CA/SoC 45

Ex ample 6 Ex ample 6

A digit al s ignal has a bit r ate of 2000 b ps.W

hat is the dur ation of eac h bit ( bit intervalSolutionSolutionThe bit interval is the inver se of the bit r ate.Bit interval = 1 / 2000 s = 0.000500 s= 0.000500 x 106 ms = 500 ms



SDM,CA/SoC 46

Bit rate and bit interval



SDM,CA/SoC 47

Digital versus analog



SDM,CA/SoC 48

A digital signal is a composite signal with an A digital signal is a composite signal with aninfinite bandwidthinfinite bandwidth

. .

T he bit rate and the bandwidth areT he bit rate and the bandwidth are

proportional to each other . proportional to each other .



SDM,CA/SoC 49

Transmission ImpairmentTransmission Impairment

A ttenuation

D istortion

Noise



SDM,CA/SoC 50

A ttenuation



SDM,CA/SoC 51


Ima gine a s ignal tr avels through a tr ansm issionmed ium and its power is r educed to half . Thismeans that P2 = 1 /2 P1. In this case , the

attenuation (l oss of power ) can be calc ulatedasSolutionSolution

10 l og10 (P2 /P1) = 10 l og10 (0.5P1 /P1) = 1010 l og10 (P2 /P1) = 10 l og10 (0.5P1 /P1) = 10log10 (0.5)log10 (0.5)

= 10(= 10(± ±0.3) =0.3) = ± ±3 dB3 dB



SDM,CA/SoC 52


Ima gine a s ignal tr avels through an ampl ifier and its power is incr eased ten times. T hismeans that P2 = 10 ¥ P1. In this case , theampl ification (gain of power ) can be calc ulatedas10 l og10 (P2 /P1) = 10 l og10 (10P1 /P1)10 l og10 (P2 /P1) = 10 l og10 (10P1 /P1)= 10 l og10 (10) = 10 (1) = 10 dB= 10 l og10 (10) = 10 (1) = 10 dB



SDM,CA/SoC 53

Distortion



SDM,CA/SoC 54

Noise



SDM,CA/SoC 55

T hroughput

P ropagation Speed

P ropagation T ime

Wavelength



SDM,CA/SoC 56

T hroughput



SDM,CA/SoC 57

P ropagation time



SDM,CA/SoC 58

Signal Str engthAll s ignals expe rience l oss (a ttenuation).

Attenuation is den oted as a dec ibel (dB) l oss.Decibel losses (and gains) a r e add itive.



SDM,CA/SoC 59

Wavelength



SDM,CA/SoC 60



SDM,CA/SoC 61



SDM,CA/SoC 62



SDM,CA/SoC 63

Analog to anal og conver sion or modulation

The char acteristics of the ca rrier signal varydepend ing on the char acteristics of an anal og signalor modulating signal.

W hen we need anal og -to-anal og modulation.The da ta / modulating signal is of lower fr equenc ies,hence cann ot be tr ansm itted over long distances.In case of band pass med ium /allotted spec ific r angeis bro ad band , the fr equency need to be shift ed toma tch the med ium¶s fr equenc y.



SDM,CA/SoC 64

The modulation is of 3 ty pesAmpl itude Mo dulation (AM )

Fr equenc y Mo dulation (F M)Phase Mo dulation (P M)



SDM,CA/SoC 65

Ampl itude Mo dulation (AM )

In AM th e ampl itude of the ca rrier signal is varied(keep ing the fr equenc y and p hase unchanged)acc or ding to the ampl itude of the modulating signal.

Usually the out put of AM cir cuit is doub le banded(doub le s ide).The lowe r half is the m irror i mage of the u ppe r half .Thus the ampl itude of the modulating signal ac ts asenvelo pe for th e modulated s ignal.



SDM,CA/SoC 66

Bandwidth requirementBw r =2 * Bw m

W her e

Bw r is the r equir ed bandw idth / bandw idth of the modulated s ignal.Bw m is the bandw idth of the modulating signal

A dvantagesCompa r atively less bandw idth than F M and P M .DisadvantagesFM signals a r e much prone to noise.



SDM,CA/SoC 67

Comparison of A SK and A M

A SK A M

Mo dulating signal is d igit al Mo dulating signal is anal og

Mo dulated s ignal has only 2 level of

ampl itude.

Mo dulated s ignal has finite num ber of

ampl itude le vels.

Ampl itude of carrier signal is variedaccor ding to 0 or 1

Ampl itude of carrier signal variesaccor ding to the ampl itude of modulating signal.

Mo dulation is done for converting

digital into anal og signal

Mo dulation is done to shift the

fr equenc y r ange.

Fr equenc y and p hase r ema ins same F r equenc y and p hase r ema ins same



SDM,CA/SoC 69



SDM,CA/SoC 70



SDM,CA/SoC 71



SDM,CA/SoC 72



SDM CA/SoC 73

Signals and Data Transfer

Documents

Transcript of Signals and Data Transfer