54032603 ec 1301 communication theory

8/3/2019 54032603 EC 1301 Communication Theory

1/42

NOORUL ISLAM COLLEGE OF ENGG, Kumaracoil DEPARTMENT OF ECE

2 MARKS & QUESTION- ANSWERS

EC 1301 Communication Theory Class : S5 ECE Prepared by :E.P.Kannan ECE


2/42

DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING SUBJECT CODE: EC 1301

SUBJECT: COMMUNICATION THEORY (FOR FIFTH SEMESTER ECE) TWO MARKS QUESTIONS

1. Define modulation? Modulation is a process by which some characteristics of high frequency carrier signal is varied in accordance with the instantaneous value of the modulating signal.

2.What are the types of analog modulation? Amplitude modulation. Angle Modulation 1. Frequency modulation 2. Phase modulation. 3.Define depth of modulation. Itis defined as the ratio between message amplitude to that of carrier amplitude.m=Em/Ec 4. What are the degrees of modulation? Under modulation. m1 5.What is the need for modulation? Needs for modulation: Ease of transmission Multiplexing Reduced noise Narrow bandwidth Frequency assignment Reduce the equipments limitations.


3/42

6.What are the types of AM modulators? There are two types of AM modulators. They are Linear modulators Non-linear modulators Linear modulators are classified as follows Transistor modulator There are three types of transistor modulator. Collector modulator Emitter modulator Base modulator Switching modulators Non-linear modulators are classified as follows Square law modulator Product modulator Balanced modulator 7.Give the classification of modulation. There are two types of modulation. They are Analog modulation Digital modulation Analog modulation is

classified as follows Continuous wave modulation Pulse modulation Continuous wave modulation is classified as follows Amplitude modulation Double side band suppressed carrier Single side band suppressed carrier Vestigial side band suppressed carrier Angle modulation Frequency modulation Phase modulation Pulse modulation is classified as follows


4/42

Pulse amplitude modulation Pulse position modulation Pulse duration modulation Pulse code modulation Digital modulation is classified as follows Amplitude shiftkeying Phase shift keying Frequency shift keying

8.What is single tone and multi tone modulation? If modulation is performed fora message signal with more than one frequency component then the modulation is called multi tone modulation. If modulation is performed for a message signal wit

h one frequency component then the modulation is called single tone modulation.

9 .The antenna current of an AM transmitter is 8A when only carrier is sent. Itincreases to 8.93A when the carrier is modulated by a single sine wave. Find thepercentage modulation. Solution: Given: Ic =8A It=8.93A

m=0.8

Formula:

It=Ic (1+m2/2) 8.93=8(1+m2/2) m=0.701 It=8 (1+0.82/2) It=9.1A


5/42

10.Compare AM with DSB-SC and SSB-SC.

AM signal Bandwidth=2fm

DSB-SC Bandwidth=2fm

SSB-SC Bandwidth=fm

Contains USB, LSB, carrier

Contains USB,LSB

Contains LSB or USB

More power is required for transmission

Power required is less than that of AM.

Power required is less than AM &DSB-SC

11 What are the advantages of VSB-AM? 1.It has bandwidth greater than SSB but less than DSB system. 2.Power transmission greater than DSB but less than SSB system. 3.No low frequency component lost. Hence it avoids phase distortion.

12 Compare linear and non-linear modulators. Linear modulators 1.Heavy filteringis not required. 2.These modulators are used in high level modulation. 3.The carrier voltage is very much greater than modulating signal voltage. 2.These modulators are used in low level Modulation. 3.The modulating signal voltage is verymuch greater than the carrier signal voltage. Non-linear modulators 1.Heavy filtering is required

13.How will you generating DSBSC-AM ? There are two ways of generating DSBSC-AM

such as 1.balanced modulator 2.ring modulators


6/42

14. What are advantages of ring modulator? 1.Its output is stable. 2. It requires no external power source to activate the diodes. 3.Virtually no maintenance. 4. Long life.

15. Define demodulation. Demodulation or detection is the process by which modulating voltage is recovered from the modulated signal. It is the reverse processof modulation.

16. What are the types of AM detectors? 1. Nonlinear detectors 2. Linear detectors

17.What are the types of linear detectors? 1.Synchronous or coherent detector. 2.Envelope or non coherent detector.

18.draw the block diagram of coherent detector.

modulated signal

Product modulator

LPFOUTPUT

Carrier Signal


7/42

19.Define multiplexing.

Multiplexing is defined as the process of transmitting several message signals simultaneously over a single channel.

20.Define sensitivity. It is defined as a measure of its ability to receive weaksignals.

21.Define selectivity. Selectivity of a receiver is defined as its ability to select the desired signals among the various signals.

22. Define stability. It is the ability of the receiver to deliver a constant amount of output for a given a given period of time.

23.Define super heterodyne principle. It can be defined as the process of operation of modulated waves to obtain similarly modulated waves of different frequency. This process uses a locally generated carrier wave, which determines the change of frequency.

24.A transmitter supplies 8 Kw to the antenna when modulated. Determine the total power radiated when modulated to 30%. m=0.3;Pc=8 kw Pt=Pc(1+m2/2) =8.36 kw

25.What are the drawbacks of emitter modulator? 1.The amplifier is operated in class A mode, thus the efficiency is low. 2.The output power is very small. Thusit is not suitable for generating high level modulation.


8/42

26. Define frequency modulation. Frequency modulation is defined as the processby which the frequency of the carrier wave is varied in accordance with the instantaneous amplitude of the modulating or message signal.

27. Define modulation index of frequency modulation. It is defined as the ratioof maximum frequency deviation to the modulating frequency. = f/fm

28. What

o you meant

y multitone mo

ulation? Mo

ulation

one for the message signal with more than one frequency component is calle

multitone mo

ulation.

29. Define phase mo

ulation. Phase mo

ulation is

efine

as the process of changing the phase of the carrier signal in accor

ance with the instantaneous amplitu

e of the message signal.

30. How FM wave can

e converte

to PM wave? Differentiator FM Signal PM

Message signal

Carrier

31. How PM wave can

e converte

to FM wave? Message Signal Integrator Phase Mo

ulator FM Signal


9/42

Carrier

32. What are the types of Frequency Mo

ulation? Base

on the mo

ulation in

ex FMcan

e

ivi

e

into types. They are Narrow

an

FM an

Wi

e

an

FM. If the mo

ulation in

ex is greater than one then it is wi

e

an

FM an

if the mo

ulationin

ex is less than one then it is Narrow

an

FM

33. What is the

asic

ifference

etween an AM signal an

a narrow

an

FM signal? In the case of sinusoi

al mo

ulation, the

asic

ifference

etween an AM signal an

a narrow

an

FM signal is that the alge

raic sign of the lower si

e frequency in the narrow

an

FM is reverse

.

34. What are the two metho

s of pro

ucing an FM wave? Basically there are two metho

s of pro

ucing an FM wave. They are, i) Direct metho

In this metho

the transmitter originates a wave whose frequency varies as function of the mo

ulatingsource. It is use

for the generation of NBFM ii) In

irect metho

In this metho

the transmitter originates a wave whose phase is a function of the mo

ulation.Normally it is use

for the generation of WBFM where WBFM is generate

from NBFM

35 . Compare WBFM an

NBFM. WBFM Mo

ulation in

ex is greater than 1 Frequency

eviation 75 KHz Ban

wi

th 15 times NBFM Noise is more suppresse

NBFM Mo

ulationin

ex less than 1 Frequency

eviation 5 KHz Ban

wi

th 2fm Less suppressing of noise


10/42

36. List the properties of the Bessel function. The properties of the Bessel function is given

y, i) ii) Jn ()=(-1)nJ-n() for all n,

oth positive an

negative.For small values of the mo

ulation in

ex ,we have J0 ()=1 J1 ()=/2 Jn()=0,n>2. iiiJ2n ()=1 n = - 37. Give the average power of an FM signal. The amplitude of the frequency modulated signal is constant .The power of the FM signal is same as thatof the carrier power. P=1/2 Ec2 . 38. Define phase deviation. The maximum phasedeviation of the total angle from the carrier angle is called phase deviation.

39. Define frequency Deviation. The maximum departure of the instantaneous frequency from the carrier frequency is called frequency deviation.

40. State the Carsons rule. An approximate rule for the transmission

andwidth ofan FM Signal generated

y a single tone-modulating signal of frequency fm is defined as B =2 f(1+1/ )

41.Define the deviation ratio D for non-sinusoidal modulation.


11/42

The deviation ratio D is defined as the ratio of the frequency deviation f, whichcorresponds to the maximum possi

le amplitude of the modulation signal m(t),tothe highest modulation frequency . D = f / fm

42.What is the use of crystal controlled oscillator? The crystal-controlled oscillator always produces a constant carrier frequency there

y enhancing frequencysta

ility.

43. What are the disadvantages of FM system? 1. A much wider channel is required

y FM. 2. FM transmitting and receiving equipments tend to

e more complex andhence it is expensive

44. How will you generate message from frequency-modulated signals? First the frequency-modulated signals are converted into corresponding amplitude-modulated signal using frequency dependent circuits. Then the original signal is recoveredfrom this AM signal.

45. What are the types of FM detectors? Slope detector and phase discriminator.

46. What are the types of phase discriminator? Foster seely discriminator and ratio detector.

47. What are the disadvantages of

alanced slope detector? 1. Amplitude limitingcannot

e provided 2. Linearity is not sufficient 3. It is difficult to align

ecause of three different frequency to which various tuned circuits to

e tuned.4. The tuned circuit is not purely

and limited.


12/42

48. Define pro

a

ility. The pro

a

ility of occurrence of an event A is defined as,

P(A)= num

er of possi

le favora

le outcomes Total num

er of equal likely outcomes

49.What are mutually exclusive events? Two possi

le outcomes of an experiment ar

e defined as

eing mutually exclusive if the occurrence of one outcome precludesthe occurrence of the other.

50. Define pro

a

ility density function. Pro

a

ility density function is definedas fx(x) is defined in terms of cumulative distri

ution function Fx (x) as fx(x) = d Fx (x) dx 51. Define noise. Noise is defined as any unwanted form of energy, which tends to interfere with proper reception and reproduction of wanted signal.

52. Give the classification of noise. Noise is

roadly classified into two types. They are External noise and internal noise.

53. What are the types of External noise? External noise can

e classified into1. Atmospheric noise 2. Extraterrestrial noises 3. Man made noises or industrialnoises

54. What are types of internal noise?


13/42

Internal noise can

e classified into 1. Thermal noise 2. Shot noise 3. Transittime noise 4. Miscellaneous internal noise

55. What are the types of extraterrestrial noise and write their origin? The twotype of extraterrestrial noise are solar noise and cosmic noise Solar noise isthe electrical noise emanating from the sun. Cosmic noise is the noise receivedfrom the center part of our galaxy, other distant galaxies and other virtual poi

nt sources.

56. Define transit time of a transistor. Transit time is defined as the time taken

y the electron to travel from emitter to the collector.

57. Define flicker noise. Flicker noise is the one appearing in transistors operating at low audio frequencies. Flicker noise is proportional to the emitter current and junction temperature and inversely proportional to the frequency.

58. State the reasons for higher noise in mixers. 1. Conversion transconductanceof mixers is much lower than the transconductance of amplifiers. 2. If image frequency rejection is inadequate, the noise associated with the image frequency a

lso gets accepted.

59. Define signal to noise ratio. Signal to noise ratio is the ratio of signal power to the noise power at the same point in a system.


14/42

60. Define noise figure. S/ N at the input Nose figure F = S/ N at the output S/N = Signal power / Noise Power

61. Explain thermal noise. Thermal noise is the name given to the electrical noise arising from the random motion of electrons in a conductor.

62. Give the expression for noise voltage in a resistor. The mean square value of

thermal noise voltage is given

y Vn2 = 4 K T B R K Boltz man constant R resistance T o

solute temperature B - Bandwidth

63. Explain White Noise. Many types of noise sources are Gaussian and have flatspectral density over a wide frequency range. Such spectrum has all frequency components in equal portion, and is therefore called white noise. The power spectral density of white noise is independent of the operating frequency.

64. What is narrow

and noise? The receiver of a communication system usually includes some provision for preprocessing the received signal. The preprocessing may take the form of a narrow

and filter whose

andwidth is large enough to pass modulated component of the received


15/42

signal essentially undistorted

ut not so large as to admit excessive noise through the receiver. The noise process appearing at the output of such filter is called narrow

and noise.

65. Give the representation of narrow

and noise in terms of envelope and phase components. Narrow

and noise in terms of envelope and phase components as n(t) =r(t) cos (2fct + (t) ) r(t) = (nI 2(t) + nQ 2(t)) (t) = tan -1(nQ(t) / nI (t )) Th

e function r (t) and (t) are called envelo

e and

hase of n(t).

66. Give the ex ression for equivalent noise tem erature in terms of hy othetical tem

erature. The ex

ression for equivalent noise tem

erature in terms of hy

othetical tem erature is Te = (

- 1) T0 Where,

is the noise figure and T0 absolute tem erature.

67. Give the

riss formula in terms of noise tem

erature. The

riss formula in terms of noise tem erature is Te = T1 + T2 / G1 + T3 / G1 G2 + G1 , G2 , Gain of am

lifiers 68. What is called image frequency? Image frequency is defined as the signal frequency lus twice the intermediate frequency. This has the effect of twostations being received simultaneously and hence it is undesirable. fsi = fs +

2 fi


16/42

fsi - image frequency It can be eliminated by roviding adequate image signal selectivity between antenna and mixer in ut.

69. What is intermediate frequency? Intermediate frequency (I

) is defined as the difference between the signal frequency and the oscillator frequency. I

= fs fo I

= fo fs when fs > fo when fo > fs (or)

70. Define Partition noise. In an electron tube having one or more

ositive grids, this noise is caused by irratic artition of the cathode current among the ositive electrodes. In a transistor, the artition noise is created from the random fluctuation in the division of current between the collector and base.

71. Give the ex ression for noise voltage when several sources are cascaded. Enr= Sqrt (4 KTB (R1 + R2 + ..) ) Where R1 , R2 --- are the resistances of the noise resistors. K Boltz man constant T absolute tem

erature B Bandwidth

72. Define random variable Random variable is defined as a rule or ma ing fromthe original sam le s ace to a numerical sam le s ace subjected to certain constraints. Random variable is also defined as a function where domain is the set of

outcomes s and

hos

rang

is R, is th

r

al lin

.73. D fin Random proc ss.


17/42

A Random proc ss X (s,t) is a function that maps ach l m nt of a sampl s spac

into a tim function call d sampl function. Random proc ss is a coll ction oftim functions.

74. Giv th La s of probability. Additiv la of probability Cas i Wh n v ntsar mutually xclusiv , P (AB) = P(AB) = P(A) + P(B) Cas ii Wh n v nts ar notmutually xclusiv P(AB) = P(A) + P(B) - P (AB) Multiplication la of probability:

Cas

i Wh

n

v

nts ar

ind

p

nd

nt P(AB) = P(A) P(B) Cas

ii Wh

n

v

nts ar

d

p nd nt P(AB) = P(A) P(B/A) = P(B) P(A/B) 75. What is fr qu ncy translation? Suppos that a signal is band limit d to th fr qu ncy rang xt nding from a fr qu ncy f1 to a fr qu ncy f2. Th proc ss of fr qu ncy translation is on in hich th original signal is r plac d ith a n signal hos sp ctral rang xt nds from f1 to f2 and hich n signal b ars, in r cov rabl form th sam information as as born by th original signal.

76. What ar t o situations id ntifi d in fr qu ncy translations? Th t o situations id ntifi d in fr qu ncy translation ar i Up conv rsion In this cas th translat d carri r fr qu ncy is gr at r than th incoming carri r fr qu ncy ii Do

n conv rsion


18/42

In this cas th translat d carri r fr qu ncy is small r than th incoming carri r fr qu ncy. Thus, a narro band

M signal r quir s ss ntially th sam transmission band idth as th AM signal. 77. D fin Tracking. Tracking is th proc ss of corr ctly tuning a numb r of tunabl circuits in a r c iv r. 78. What is TR

r c iv r? Tun d Radio

r qu ncy is also call d straight r c iv r.H r th r c iv

r op rat s in straight for ard mann r ithout fr qu ncy conv rsion. 79.What ar th advantag s of sup rh t rodyn r c iv r ov r TR

? Th advantag s of sup rh t

rodyn

r

c

iv

r ov

r TR

ar

high s

l

ctivity ,improv

d s

nsitivity throughout th carri r fr qu ncy band..It liminat s imag fr qu ncy. 80.What is th figur of m rit of DSBSC syst m ? Th figur of m rit of DSBSC signal is unity

81.Compar th nois p rformanc of an AM and

M syst m? Th figur of m rit ofAM syst m is 1/3 h n th modulation is 100 p rc nt and 2 that of

M is(3/2)mf .Th us of

M off rs improv d nois p rformanc ov r AM h n (3/2)mf2 > 1/3.mf modulation ind x in

M.

82.What is Captur ff ct? Wh n th int rf r nc signal and

M input ar of qual str ngth, th r c iv r fluctuat s back and froth b t n th m .This ph nom nonis kno n as th captur ff ct.

83.What is thr shold ff ct? As th input nois po r is incr as d th carri r to nois ratio is d cr as d th r c iv r br aks and as th carri r to nois ratiois r duc d furth r crackling sound is h ard and th output SNR cannot b pr dict d by th quation. This ph nom non is kno n as thr shold ff ct.

84.Ho is thr shold r duction achi v d in

M syst m? Thr shold r duction is achi v d in

M syst m by using an

M d modulator ith n gativ f dback or by usinga phas lock d loop d modulator.


19/42

85. What is Pr - mphasis? Th pr modulation filt ring in th transistor, to rais th po r sp ctral d nsity of th bas band signal in its upp r-fr qu ncy rang is call d pr mphasis (or pr distortion) Pr mphasis is particularly ff ctiv in

M syst ms hich ar us d for transmission of audio signals.

86. D fin d - mphasis. Th filt ring at th r c iv r to undo th signal pr - mphasis and to suppr ss nois is call d d - mphasis. 87.D fin Sampling th or m. A

band limit

d signal of finit

n

rgy,

hich has no fr

qu

ncy compon

nts high

rthan fm H rtz may b compl t ly r cov r d from a kno l dg of its sampl s tak nat th rat of 2fm sampl s p r s cond.

88.What do you inf r from th r c iv r output of a coh r nt d t ctor? Th output quation y(t)=1/2Cacm(t) +1/2nI(t) indicat s that th m ssag signal and in-phas nois compon nt of th filt r d nois app ar additiv ly at th r c iv r output. Th quadratur compon nt of th narro band nois is compl t ly r j ct d by th coh r nt d t ctor.

89.Wh n is th figur of m rit of SSBSC syst m 1?

or th sam av rag transmitt d signal po r and th sam av rag nois po r in th m ssag band idth ,an SS

B r

c

iv

r

ill hav

xactly th

sam

output signal to nois

ratio as a DSB-SC r c iv r h n both r c iv rs us coh r nt d t ction for th r cov ry of th m ssag signal. 90.Compar th nois p rformanc of AM r c iv r ith that of DSB-SC r c iv r. Th figur of m rit of DSB-SC or SSB-SC r c iv r using coh r nt d t ction is al ays unity,th figur of m rit of AM r c iv r using nv lop d t ction is al ays l ss than unity.Th r for nois p rformanc of AM r c iv r is al ays inf rior to that of DSBSC du to th astag of po r for transmitting th carri r. 91.What is th figur of m rit of a AM syst m ith 100 p rc nt modulation? Th

figur of m rit of a AM syst m ith 100 p rc nt modulation is 1/3.This m ans that oth r factors b ing qual an AM syst m must transmit thr tim s as much av rag po r as a suppr ss d syst m in ord r to achi v th sam quality of nois p rformanc .


20/42

92.What ar th charact ristics of a r c iv r? Th charact ristics of a r c iv rar s nsitivity, s l ctivity, fid lity, signal to nois ratio.

96.Why is quival nt nois t mp ratur us d for nois m asur m nt?

or lo nois

d vic s th nois figur is clos to unity, hich mak s th comparison difficult and h nc it is pr f rabl to us quival nt nois t mp ratur .

97.What is th

function of amplitud

limit

r in

M syst

m? Th

function of amplitud limit r in

M syst m is us d to r mov th amplitud variations by clippingth modulat d av at th filt r output almost to th z ro axis .Th r sultant

av is round d off by anoth r BP

that is an int gral part of th limit r th r

by suppr ssing th harmonics of th carri r fr qu ncy.

98.What ar compon nts in a fr qu ncy discriminator?

r qu ncy discriminator hasgot t o compon nts .Slop d t ctor or diff r ntiator ith a pur ly imaginary fr

qu ncy r spons that vari s lin arly ith fr qu ncy. It produc s output h r th amplitud and fr qu ncy vary ith th m ssag signal. An nv lop d t ctor that r cov r th amplitud variations and produc s m ssag signal.

99.What is a post d

t

ction filt

r? Th

post d

t

ction filt

r nam

d asbas

-band lo pass filt r has a band idth that is just larg nough to accommodat th high

st fr qu ncy compon nt of th m ssg signal.

100. D fin lossl ss chann l. Th chann l d scrib d by a chann l matrix ith only on nonz ro l m nt in ach column is call d a lossl ss chann l. In th lossl

ss chann l no sourc s information is lost in transmission.

101. D fin D t rministic chann l


21/42

A chann l d scrib d by a chann l matrix ith only on nonz ro l m nt in ach ro is call d a d t rministic chann l and this l m nt must b unity.

102. D fin nois l ss chann l. A chann l is call d nois l ss if it is both lossl ss and d t rministic. Th chann l matrix has only on l m nt in ach ro and in ach column and this l m nt is unity. Th input and output alphab ts ar of th sam siz .

103. Prov that

I (xi xj) = I(xi) + I(xj) if xi and xj ar ind p nd nt. If xi and xj ar ind p nd nt.

P (xi xj) = P(xi) P(xj) I (xi xj) = log1/P(xi xj) = log 1/ P(xi) P(xj) = I(xi) +I(xj) 104. Explain Shannon-

ano coding. An ffici nt cod can b obtain d by th follo ing simpl proc dur , kno n as Shannon-

ano algorthim. 1. List th sourc symbols in ord r of d cr asing probability. 2. Partition th s t into t o s ts that ar as clos to quiprobabl as possibl , and sign 0 to th upp r s t and1 to th lo r s t. 3. Continu this proc ss, ach tim partitioning th s ts

ith as n

arly

qual probabiliti

s as possibl

until furth

r partitioning is notpossibl .

105. What ar th typ s of Corr lation? Th typ s of Corr lation ar Cross Corr

lation and Auto Corr lation

106. What is th diff r nc b t n Corr lation and Convolution? 1. In Corr lation physical tim t is dummy variabl and it disapp ars aft r solution of an int gral. But in convolution s a dummy var able. 2. Convolut on s a funct on of delayparameter t but convolut on s a funct on of t. 3. Convolut on s commutat ve but correlat on s noncom mutat ve.


22/42

107. Def ne S gnal. A s gnal s def ned as any phys cal quant ty carry ng nformat on that var es w th t me. The value of s gnal may be real or complex. The types of s gnal are cont nuous s gnal and d screte t me s gnal.

108.Def ne entropy. Entropy s the measure of the average nformat on content per second. It

s g

ven by the express

on H(X)=I P(x

)log2P(x

) b

ts/sample. 109.De

f ne mutual nformat on. Mutual nformat on I(X,Y) of a channel s def ned by I(

X,Y)=H(X)-H(X/Y) b ts/symbol H(X)- entropy of the source H(X/Y)- cond t onal entropy of Y.

110.State the propert es of mutual nformat on. 1. I(X,Y)=I(Y,X) 2. I(X,Y)>=0 3.I(X,Y)=H(Y)-H(Y/X) 4. I(X,Y)=H(X)+H(Y)-H(X,Y).

111.Gve the relat

on between the d

fferent entrop

es. H(X,Y)=H(X)+H(Y/X) =H(Y)+

H(X/Y) H(X)- entropy of the source,H(Y/X),H(X/Y)-cond t onal entropy H(Y)-entropy of dest nat on H(X,Y)- Jo nt entropy of the source and dest nat on

112.Def ne nformat on rate. If the t me rate at wh ch source X em ts symbols sr symbols per second. The nformat on rate R of the source s g ven by R=r H(X)

b

ts/second H(X)- entropy of the source 113.What

s data compact

on? For eff

c

ent s gnal transm ss on the redundant nformat on must be removed from the s gnal pr or to transm ss on .Th s nformat on w th no loss of nformat on s ord nar

ly


23/42

performed on a s gnal n d g tal form and s referred to as data compact on or lossless data compress on.

114.State the property of entropy. 1.0< H(X) < log2K , s the rad x of the alphabet X of the source. 115.What s d fferent al entropy? The average amount of nformat

on per sample value of x(t)

s measured by H(X)= - fx(x)log fx(x)dx b

t/sa

mple H(X) d fferent al entropy of X.

116.Whats the channel capac

ty of a d

screte s

gnal? The channel capac

ty of a

d screte s gnal C= max I(X,Y) P(x ) I(X,Y)-mutual nformat on.

117. What s source cod ng and entropy cod ng? A convers on of the output of a DMS

nto a sequence of b

nary symbols

s called source cod

ng. he des

gn of a var

able length code such that ts average cod word length approaches the entropy of the DMS s often referred to as entropy cod ng.

118.State Shannon Hartley theorem. The capac ty C of a add t ve Gauss an no se channel s C=B log2 (1+S/N) B= channel bandw dth ,S/N=s gnal to no se rat o.

119.What s the entropy of a b nary memory-less source? The entropy of a b narymemory-less source H(X)=-p0 log2p0-(1-p0)log2(1-p0) p0-probab l ty of symbol 0,p1=(1- p0 ) =probab l ty of transm tt ng symbol 1

120.How s the eff c ency of the cod ng techn que measured? Eff c ency of the code =H(X) /L


24/42

L=p(x )l average code word length .l =length of the code word.

121.What happens when the number of cod ng alphabet ncreases? When the number of cod ng alphabet ncreases the eff c ency of the cod ng techn que decreases.

122.Whats channel d

agram and channel matr

x? The trans

t

on probab

l

ty d

agr

am of the channel s called the channel d agram and ts matr x representat on s

called the channel matr x.

123.Whats

nformat

on theory? Informat

on theory deals w

th the mathemat

cal m

odel ng and analys s of a commun cat on system rather than w th phys cal sourcesand phys cal channels

124.Whats the channel capac

ty of a BSC and BEC? For BSC the channel capac

ty

C=1+plog2 p +(1-p)log2(1-p). For BEC the channel capac ty C=(1-p)

PART B 1. Der ve the express on for AM & ts Power and Eff c ency calculat on: AM Def n t on Let m(t) = Em Cos mt c(t) = Ec Cos ct EAM = Ec + Em Cos mt = Ec [ 1+ (Em/Ec) Cos mt ] m= Em/Ec s(t)AM = EAM Cos ct = Ec (1 + m Cosmt) Cosct = Ec Cosct + mE

c Cosmt Cosct


25/42

= Ec Cosct + mEc [Cos(c + m)t + Cos(c m)t] 2 = Carri r + USB + LSB Po r r lation iAM: Total Po r, Pt = PC + PUSB + PLSB PC - Carri r po r

PUSB - Upp r Sid Band po r PLSB - Lo r Sid Band po r PC = E2 rms R = Ec2 2RPUSB = PLSB = m2Ec2 8R Pt = PC ( 1 + m2/2 )

Curr nt r lation in AM: Pt = I t2 R P C = I c2 R I t = I c Sqrt (1 + m2/2) Effic

i

ncy: % = Power in side band Total Power = m2 2+m2 m=1 = 33.33 % x 100 x 100


26/42

2. Describe t

e generation of AM wave by Non-Linear modulators : Square law modulator: AM wave Summer Non-Linear Element BPF

Summer

to add carrier & modulating signal.

Non-Linear element active element (Diode) BPF - extracting desired modulating products

To operate diode under Non-Linear region, magnitude of carrier component is

ig

er during positive cycle of t

e modulating voltage and lesser during negative

alf cycle of t

e modulating voltage. T

e resulting current is, Io = a1 V1 + a2 V12 + V1 = Em Cosmt + Ec Cosct N gl cting s cond and high r ord r t rms, Io = a1 Em Cosmt + a1 Ec Cosct + 2a2 EmEc Cosmt Cosct Aft r passing through BP

, Io = a1 Ec Cosct+a2 EmEc Cosct Cosmt

Balanc d Modulator: Diagram R f r Book Input to th transistor T1, Vb 1 = m(t) +c(t)


27/42

Input to th transistor T2, Vb 2 = m(t) c(t) m(t) = Em Cosmt c(t) = Ec Cosct ic1 =a1 Vb 1 + a2 Vb 12 + . ic2 = a1 Vb 2 + a2 Vb 22 +. Th output is giv n by, Vo = K[ic1 ic2 ] Vo = K 2 [ a1 Ec Cosct + 2a2 Em Ec Cosct Cosmt ] m = ( 2a2 Em/a1 ) = modulation ind x

3.Writ short not s on D Modulation of AM av : D Modulation d finition Typ s:1. Env lop d t ctor 2. Squar la d t ctor

Env lop d t ctor: R quir m nts to us nv lop d t ctor: 1. AM av has to b narro band 2. P rc ntag of modulation should b l ss than 100%

Diagram - R f r book


28/42

During positiv half cycl of th input signal, diod is for ard bias d and capacitor C charg s to p ak valu , h n th input signal fall b lo th p ak valu ,th diod is r v rs bias d and th capacitor C discharg s. Th discharging proc ss continu s until th n xt positiv half cycl .

Wh n th input signal is gr at r than th voltag across th capacitor, th diod conducts again and th proc ss is r p at d. Th charging tim constant ( rf +

RS )


29/42

or larg r carri r, [ m(t)/A ]2 is n gl ct d Blocking capacitor ill suppr ss th dc t rm A2/2 Th n th output is A m(t) 4. Compar th Synchronous D t ctor andth Costas PLL D t ctor. Synchronous or Coh r nt D t ctor. Th block diagram ofsynchronous d t ctor Op ration All typ s of lin ar modulation can b d t ct d by using Synchronous d t ctor.Th incoming signal is first multipli d ith locally g n rat d carri r signal and th n nt r d via a LP

.Th LP

band idth is usually sam as th m ssag BW or som tim s larg r. It is assum d that th local osci

llator is

xactly synchroniz

d

ith th

carri

r in both phas

and v

locity,h

nc

th nam . Writ th xpr ssion for m ssag signal from th modulat d signals using coh r nt d t ctor. Thus th synchronous d t ctor is capabl of d modulatingDSB-SC and SSB-SC AM Costas PLL D t ctor (

or DSB-SC-AM) Th Block diagram of costas PLL d t ctor Op ration It consists of t o synchronous d t ctor. 1)In phas coh r nt d t ctor or I chann l:- This d t ctor is suppli d ith DSB-SC-AM and alocally g n rat d carri r hich is in phas ith th transmitt d carri r. 2)Quadratur phas coh r nt d t ctor or Q chann l:- This d t ctor is f d ith DSB-SC-AM and a locally g n rat d carri r hich is in phas ith th transmitt d carri r. Assum th local oscillator carri r signal is prop rly synchroniz d ith transmitting carri r.So IChann l output contains th d sir d d modulat d signal h r

as Q-chann l output is z ro. If th r is a phas shift of ; between local oscill

ator carrier and transmitting carrier then I-channel output will remain in the same value but qchannel output contains some signal . Thus combining the I and Qchannel


30/42

outputs in phase discrimination a dc signal is obtained that automatically corrects the phase errors in VCO. Comparison In Synchronous detector the detection ise

ective only when locally generated carrier is properly synchronized with the

modulated signal Any shi t in phase or requency o the locally generated carrier results in phase or delay distortion. To avoid this distortions a pilot carrier is inserted. Costas PLL detector consists o

two coherent detectors. These tw

o detectors are coupled together to orm a negative eedback system designed in

such a way to maintain the local oscillator synchronise with the carrier.

5. Compare Phase shit method and modi

ied Phase shi

t method Phase Shi

t Method

:(SSB-SC-AM) Block diagram or phase shi t method Analysis The undesired side band is removed by generating two side band components out o phase.I the undesired sideband is LSB then the two LSB are generated such that they are 180 out o phase with each other.So that the USBs get added and the LSBs cancel each other. Show the analysis using Expressions Thus one o the sideband is cancelled whereasthe other is rein orced. Modi ied Phase Shi t Method or Weavers method. Block diagram Operation Here a phase shi t is applied to AF carrier requency only and a ter the resulting vge has been applied to BM1 and BM2. Show the analysis using Expressions Comparison Phase shi t Method: It can switch rom one Side Band to ot

her Low audio modulating

requency is used


31/42

Generates SSB at any requency It does not require a SB ilter or any wide bandaudio phase shi t n/w Low requency signals r used. Side bands may be easily switched & System is complex

Weavers Method:

6.Explain the Narrow Band Frequency Modulation Frequency Modulation. Frequency modulation is de

ined as the process o

changing the

requency o

the carrier wav

e in accordance with respect to the message signal keeping the amplitude and phase as a constant one. Narrow Band Frequency Modulation: De inition I the modulation index is less than one, then the FM signal is narrow and requency modulated signal. BLOCK DIAGRAM coswmt Integrato r Product Modulator

Summe r

S(t)NBFM

-90 degree phase shi

t network

Carrier(coswct)

S(t)FM=Ec cos[wct+(K Em/wm)sinwmt]; Here m(t)=Em coswmt. Modulation index()=Kf Em/wm. S(t)FM=Ec cos[wct+sinwmt] Derivation Since B


32/42

7. Explain the Wide Band Frequency Modulation

Definition: If the modulation index is greater than one, then the FM signal is wide

and frequency modulated signal.

Fourier Series Representation(aperiodic). S(t)=Ec cos(wct+sinwmt) =Re{Ec.e^jwct.e^jsinwmt} Let S^(t)=(Ec.e^jsinwmt) S^(t)=Cn.e^j2fmnt Cn= ^(t).e^-j2fmnt. dt olving

Cn=Ec Jn() ^(t)=Ec J().e^j2fmnt (t)WBFM=Ec Jn().cos(wct+nwmt) (Complex Fourier ers representation of periodic function)

8.Explain the method of Generation of FM signal . (1) Direct FM generation (2) Indirect FM generation

(1) Direct FM generation In this method the transmitter originates a wave whosefrequency varies as function of the modulating source. It is used for the generation of NBFM (a)Varactor diode implementation of angle modulation. (

)Reactancetu

e implementation of angle modulation. Diagram - explanation


33/42

Analysis (2) Indirect FM generation In this method the transmitter originates awave whose phase is a function of the modulation. Normally it is used for the generation of WBFM where WBFM is generated from NBFM Diagram - explanation Analysis

9. Write short notes on Foster seeley Discriminator. It is used for FM DetectionCircuit Diagram Operation Advantages 1. It is much easier to design 2. Only two

tuned circuits are necessary and they are tuned to same frequency 3. Linearityis

etter Disadvantages: It requires Amplitude limiting circuit 10. What do youmeant

y Noise. Give the different types of noise explain. Noise Definition Noise Classification 1. Internal Noise 2. External Noise Internal noise - Explanation Internal Noise types a. Thermal nose Explanation

. hot noise - Explanation c. Transist time noise Explanation d. Miscellaneous noise Explanation External Noise Explanation


34/42

1. Natural noise Explanation 2. Man made noise Explanation Natural noise types a. Terrestial noise Explanation

. Extra tyerrestian noise Explanation 11. Explain in detail a

out Noise temperature: Noise performance of any system is indicated

y noise figure. Noise figure is not applica

le when we are dealing with VHF &microwave amprs. Because noise power and signal power o

tained in these devicesare close to unity. Hence comparison is rather difficult.

o Noise temperature

is used. - Diagram - Explanation - Calculation. - Equivalent noise temperature T

e = T(F-1) o

tain an expression for Noise figure when networks are in cascade -Diagram - Calculation F 2 1 F 3 1 F 4 1 + + + ... F = F1 + G1 G1G 2 G1G 2G3

12. Explain the effect of Noise when amplifiers are connected in cascade Fig. shows a number of amplifier stages in cascade each having a resistance at its input and output. We proceed to find the equivalent input noise voltage and its equivalent noise resistance for the complete receiver. Diagram Consider the first two stages of the multi stage amplifier having gains A1 and A2 and input resistances R1 and R2 respectively R3forms the output resistance. The rms noise voltage at the output due to R3 is given by, Vn 3 4kTBR 3 The same noise voltage is present in the output if instead of R3 we have resistance R3 at the input of stage 2,where R3 is given by, Vn 3 Vn 3 = Substituting vn3 and simplify A2 R Hence R3

= 3 A2 2 The noise resistance already present at the input of the second stage is R2. Hence the net noise resistance.


35/42

R3 A2 2 This resistance R2t may be transferred to the input of the first stage.R R 2+ 3 2 R2t R A2 R = 2 + 23 2 R2 = 2 = 2 2 A1 A1 A1 A1 A2 Substitute R2t and simplify. R 2 t = R2 R3 = R 2 +

Hence the not noise resistance at the input of the first stage is given by1 R 0 = R1 + R2 = R1 +

R R2 + 23 2 2 A1 A1 A2

13. Describe the effect of noise in reactive circuits Consider the circuit of Fig. in which a physical resistor R is placed in parallel with a parallel tuned circuit. The tuned circuit is theoretically noiseless. The presence of this tunedcircuit does not affect the noise generated by the resistor R. Diagram Next consider the practical case when this tuned circuit is non

ideal i.e. the inductor L

of the tuned circuit possesses a small resistive element Rs an shown in Fig. This resistive element Rs generates noise. In preceding sections, we have considered physical input resistor as the source of noise. It may, however, be clearly understood that the noise producing resistance need not necessarily be a physicalresistor. Thus the series resistance Rs of the coil forms the noise source gene

rating a noise voltage Vn as shown in Fig. We are required to calculate the noise voltage across the capacitor O. Calculation_

V = 4 kTBRs .

14. Define Noise figure and obtain an expression for Noise figure of an amplifier

The noise figure F is defined aqs the ratio of signalto

noise power supplied to

the input terminals of the system (amplifier or receiver) to the signalto

nois

e power supplied by the system to the output load impedance.


36/42

Thus Noise Figure F =Calculation of Noise Figure Diagram

S/N at the input/ at the output. S/N

i. Determine input signal power Si ii. Determine the input noise power Ni iii. Calculate the input signal to noise power ratio Si/Ni iv. Determine the output si

gnal power So v. Determine the output noise power No vi. Calculate the output signal to noise power ratio So/No vii. From steps (iii) calculate the noise figureF. Calculation.F= RL N o ( Ra + Rt ) 4 k TBA 2 R aT t_

15. Obtain an expression for Noise Figure in Terms of Equivalent Noise Resistance

In order to correlate the noise figure and the equivalent noise resistance we define the term Req as the noise resistance not including Rt.

Req = R eq R t The total equivalent noise resistance of this receiver is then

given by,

Ra Rt Ra + Rt The equivalent noise voltage effective at the input of the receiver is given by,

R = Req +_

Vni = 4 k TBR

Calculate the output noise power and calculate Noise Figure.F = 1 + Req

Ra + Rt R a .R t

extreme condition Rt > Ras the ratio (Ra + Ri) Rt approaches unity and


37/42

Ra 16. Explain in detail how noise figure is measured ?At high frequencies where transit time becomes dominant, such a calculation is not possible. In such a case, it becomes necessary to make measurements which enable us to determine the noise figure. One such method makes use of the diode noise generator.Diagram Calculation

F = 1+

R eq

F=

Ra I a (1.6 x1019 ) = 19.3Ra I p 2 x1.38 x103 x300

16. Super heterodyne receiver Explanation

Definition of receiver Functions of receiver Features of receiver Receiver BlockDiagram Explanation of Block Diagram

17. Prove that the figure of merit of DSB SC system is unity.

Block Diagramrefer Simon Haykins S(t) DSB SC = c Ac co c t m(t) Analysis Input

signal Po r = c2 Ac2 P / 2 Input nois Po r = N0 Calculat (Input signal Po r / Input nois Po r) -------------- (1) Output signal Po r = c2 Ac2 P / 2 Output nois Po r = N0 Calculat (Output signal Po r / Output nois Po r)--------(2) (1) ---- =

igur of m rit


38/42

(2)

18. Nois in AM r c iv rs - AM syst m using nv lop d t ction

Block Diagram - r f r Simon Haykins AM av quation Input signal Po r = Ac2 (1+ ka2 P) / 2 Av rag po r of nois in th m ssag band idth = N0 = Input nois

Po r

ilt r d signal = AM signal + nois signal Analysis Output signal Po r = Ac2 ka2 P/ 2 Output nois Po r = N0 Calculat (Input signal Po r / Input nois Po

r) -------------- (1) Calculat (Output signal Po r / Output nois Po r)--------(2) (1) ---- =

igur of m rit (2)

19. Nois in angl modulation Syst m

Block Diagram R pr s ntation of

M av Analysis Output signal Po r = 2 kf2 P An lysis

2 N0 2 w3 Output noise Power = -------------3A2 Output sign l Power 3kf2 P ------

----------------- = -------Output noise Power w


39/42

OR Figure of merit = (3/2) 2

20. Expl in FM threshold reduction.

Threshold effect As the input noise power is incre sed the c rrier to noise r tio is decre

sed the receiver

reks

nd

s the c

rrier to noise r

tio is reduced

further cr ckling sound is he rd nd the output SNR c nnot

e predicted

y the e

qu tion. This phenomenon is known s threshold effect. Reduction method: Block di gr m Expl n tion .

21. Expl in the procedure of Sh nnon F no Coding Algorithm nd Huffm n Coding lgorithm

Shnnon F

no Coding Algorithm 1. Arr

nge the sym

ol pro

ility in the descending order . 2. P rtition the set into two sets th t re s close to equipro

le s possi

le , nd ssign 0 to the upper set nd 1 to the lower set 3. Continue this process, e ch time p rtitioning the sets with s ne rly equ l pro

ilities s possi

le until further prtitioning is not possi

leHuffm n Coding lgorithm

1. Arrnge the sym

ol pro

ility in the descending order . 2. Com

ine the pro

ilities of the two sym

ols hving the lowest pro

ilitiesnd reorder the resu

lt nt pro

ilities; this step is reduction 1 . This procedure is repe ted untilthere re two ordered pro

ilities rem ining. 3. St rt encoding with the l streduction, which consists of ex

ctly two ordered pro

ilities .Assign 0s the

first digit in the code words forll the source sym

olsssoci

ted with the fir

st pro

ility; ssign 1 to the second pro

ility


40/42

4. Now go

ck nd ssign 0 nd 1 to the second digit for the two pro

ilitiesth t were com

ined in the previous reduction step ,ret ining ll ssignments m de in step 3. 5. Keep regressing this w y until the first column is re ched.

22. .St te nd prove the properties of mutu l inform tion.

1. Mutu l inform tion I (xi yj) is symmetric. i.e. I (xi yj) = I (yj xi) Proof 2

. Mutu l inform tion I (xi yj) is m ximum when p (xi / yj) = 1 Proof 3. I (X,Y) 0 Proof . 4. Mutu l inform tion is rel ted to the joint entropy of the ch nnel input

nd ch

nnel output

y Proof I(X,Y) = H (X) + H (Y) H(X,Y)

23. Expl in the different types of ch nnel.

Loss less Chnnel H (X/ Y) = 0 , I(X,Y) = H (X) Ch

nnel c

p

city = M

x (I(X,Y))

= H (X) Ch nnel di gr m - Expl n tion Deterministic ch nnel H (Y/ X) = 0 I(X,Y)= H (Y) Ch nnel c p city = M x (I(X,Y)) = H (Y) , Ch nnel di gr m - Expl n tionNoise less Ch nnel H (X/ Y) = 0 H (Y/ X) = 0 Ch nnel c p city = M x (I(X,Y)) = H(Y) = H (X)


41/42

Ch nnel di gr m - Expl n tionBin ry Symmetric Ch nnel

Ch nnel c p city = M x (I(X,Y)) = H (Y) = Plog2 P + (1- P) log2 (1-P) Ch nnel di gr m - Expl n tion

24.C lcul te the c p city of g ussi n ch nnel.

St te the Theorem ,C p city = log2 (1 + S/N) bits/ symbol Analysis - r f r not s25..

ind th chann l capacity of binary rasur chann l P (x1 ) =

Dr w the ch nnel di gr m of the

in ry Er sure ch nnel nd get the ch nnel m trix. P (y1/ x1) = 1-P P (y2/ x1) = P P (y2/ x2) = P P (y3/ x2) = 1-P Ch nnel c p city = M

x (I(X,Y)) = H (Y) - H (Y/ X) i. Find H (Y/ X) using formul

H (Y/ X) =

- (Plog2 P + (1- P) log2 (1-P)) ii. Find H (Y) using formul H (Y) = - (1-P) [ log 2 + log2 (1-P) + log 2(1- ) - log 2(1- ) ] - Plog2 P iii . I(X,Y) = (1-P) H (Y) iv. Ch nnel c p city = M x (I(X,Y)) = 1-P


42/42

54032603 ec 1301 communication theory

Documents