[et/en/ej/ex/ed/ei] digital communication rvidyalankar.org/upload/3_dc_soln71114155950890.pdf ·...

1114/TY/Pre_Pap/Elec/2014/CP/Class/DC 1

T.Y. Diploma : Sem. V [ET/EN/EJ/EX/ED/EI]

Digital Communication Time: 3 Hrs.] Prelim Question Paper [Marks : 100

Q.1(a) Attempt any THREE of the following : [12]Q.1(a) (i) Give the advantages and disadvantages of digital communication. [4](A) Advantages of digital communication i) Noise immunity is more than analog communication. ii) Digital communication supports error detection and correction techniques. iii) It is easy to regenerate the digital signal than analog. iv) Digital signals are easy to store and manipulate. v) Digital communication is computable with advance data processing

technique like digital technique process image processing. vi) Cost of digital communication system is low.

Disadvantages i) The data rate of digital communication is very high. ii) Loss of information. Q.1(a) (ii) List and explain various properties of Hamming Codes. [4](A) Properties of Hamming Code i) Hamming codes are linear. ii) Number of bits in codeword is equal to n = 2q 1 where q = number of extra bit eg. if Q = 3 then n = ? 23 1 n = 7 iii) Number of message bit K should be always equal to 2q 1 q iv) For the hamming code the minimum distance dmin = 3. v) Hamming codes can generate systematic as well as non – systematic code

words. vi) Hamming codes are used to detect burst error. Q.1(a) (iii) What is companding? Draw the companding curves for PCM

system. [4]

(A) Companding By keeping step size constant we can perform non uniform quantization. This

can be achieved by amplifying low level signals and alternating high level signal is called as compression at the receiver side, the signal is alternated at low level and amplified at high level is called as expansion.

[12][1mmunication. [4]ation. [4]

n. and correction techniques. rection techniques.

han analog. an analog. pulate.

e with advance data processh advance data pcess image processing. age processing.

stem is low. low.

ommunication is very high. ication is very high

various properties of Hamming s properties of Hamming Code Code

odes are linear. linear. of bits in codeword is equal to n in codeword is equ

e q = number of extra bit er of extra bit if Q = 3 then n = ? if Q = 3 then n = ?

2 233 1 n = 7 n = 7 iii) Number of message bit K siii) Number of message bit Kiv) For the hamming code thiv) For the hamming code th

v) Hamming codes can gv) Hamming codeswords. words

vi) Hamming codes vi) Hamming

Q.1(a)Q.1(a)

rks

(iii) What is (iii) Wsystem

CompandBy kee

r

Vidyalankar : T.Y. Diploma DC


The compression of signal at transmitter and expansion at receiver is combinely called as companding.

Q.1(a) (iv) Draw the waveforms of ASK, FSK, PSK for binary input

101101110. [4]

(A) Given bit stream = 101101110

1 0 0 0 1 1 1 1 1

VC

VASK

VASK

VFSK

VPSK

with carr

1CV

with carr

1CV & 2CV

with carr

1CV

1 0 1 1 0 1 1 1 0

fH fLfH fHfL

Alternated

Vin

Expansion at receiver

Compressionat transmitter

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ar PSK for binary inputSK for binary input [4]

n1 1

VAS

VFSK

VPS

th arr arr

1C1VC

with carith

11CC11VVCC & &

2CVVC

1 0 11 0 1

n


Q.1(b) Attempt any ONE of the following : [6]Q.1(b) (i) Discuss ASCII codes with suitable examples. [6](A) ASCII Code : (American standard code for information interchange) Number of bits used to represent every symbol is 7. Number of different

bit patterns = 128. 30H 0 0 1 1 0 0 0 0 31H 1 0 1 1 0 0 0 1 32H 2 0 1 1 0 0 1 0 : : 39H 9 0 1 1 1 0 0 1 41H A 1 0 0 0 0 0 1 42H B 1 0 0 0 0 1 0 : : : : 5A Z 1 0 1 1 0 1 0 61H a 1 1 0 0 0 0 1 62H b 1 1 0 0 0 1 0 : : 7AH z 1 1 1 1 0 1 0 Example BABA 1 0 0 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 1 0 0 0 0 0 1 C 1 0 0 0 0 1 1 Q.1(b) (ii) Draw the block diagram for the generation of PAM. State the

function of each block. [6]

(A) Block diagram for the generation of PAM Pulse amplitude modulation is an analog to digital technique were the

amplitude variations at input analog signal are converted into corresponding output digital signal. Hence here we make use of sampling, the block diagram is as shown below :

1 0 0 0 0 1 1

+A

A

Bipolar NRZ

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ff

1 1 0 1

1 1 1 1 0 1 0 0 1 0

ABA 0 1 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0 1 1 0 0

1 0 0 0 0 1 1 1 0 0 0 0 1 1

Q.1(b)Q.1(b) (ii) Draw the b(ii) Dfunction f

) Block diag Pulse a

amp

dydy1 0 00y

+A

AA

Bipolar Bipolar NRZNRZ



Q.2 Attempt any FOUR of the following : [16]Q.2(a) List various properties required for line codes. [4](A) Properties of line codes

i) DC content ii) Signal power i) For a good line code DC content should be as low as possible. Split phase

Manchester consist of very low DC component where as NRZ L and unipolar NRZ consists of large DC contents.

If there is channel is less noisy then we use split phase Manchester. ii) For a good line code signal power should be as high as possible so that

noise required to corrupt the large. NRZ L and unipolar NRZ has high signal power than split phase Manchester and RZ.

Q.2(b) Discuss Multiplexing and its types. [4](A) Multiplexing and its type

Multiplexing is the technique which allows the simultaneous transmission of multiple signal data link. Multiplexing device is a physical line or a medium or a into logical segment called as channels.

Example cable TV.

MU X

DE M U X

Common communication LinkMultiplexing

analog digital

FDM WDMTDM

Synchronous TDM Asynchronous TDM

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hould be as low as possible. Split e as low as possibw DC component where as NRZ omponent where

DC contents. ntents. noisy then we use split phase Ma then we use split p

gnal power should be as high asgnal power should be as rrupt the large. NRZ pt the large. N L and un L

split phase Manchester and RZ. e Manchester an

exing and its types. nd its types and its type s type

ng is the technique which allows technique whic signal data link. signal data link.

plexing device is a physical line exing device is a pled as channels. channels.

Example Example d cable TV. ble TVy


Q.2(c) What is need for delta modulation? Give its advantages &disadvantages and applications.

[4]

(A) Delta Modulation (DM) Need : In PCM ‘N’ number of bits are transmitted per quantized sample which

asks for large channel Bandwidth and signaling rate. This disadvantage can be overcome by using DM. DM transmits only one bit per sample instead of ‘N’ therefore it

extensively reduces signaling rate and channel Bandwidth.

Advantages i) One bit codeword for output. ii) Simplicity of design for transmitter and receiver. iii) Low signaling rate iv) Low channel Bandwidth

Disadvantage i) Slope overload present ii) Granular Noise

Application i) Satellite transmission System ii) Digital Communication

Q.2(d) Write bandwidth requirement for DPSK, QAM, QPSK, BPSK. [4](A) Bandwidth requirement of

(i) DPSK Symbol duration = Ts = 2Tb

1Fs

= 2fb

fs = fb2

Bandwidth = fb fb2 2

= fb

Bandwidth = fb

(ii) QAM Bandwidth = fs ( fs) = 2fs

= 2Ts

= 2N.Tb

Bandwidth = 2FbN

fb/2 +fb/2

2fs fs +fs +2fs

2fs

0Vidy

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mple whi whi

f ‘N’ therefore it therefore it width.

eceiver. ver.

n System ystemtion

h requirement for DPSK, QAM,rement for DPSKquirement of ent of

bol duration = Ts = 2Tb = Ts = 2T

y1FsFs

= y 2fb

fs fs

Bandwidth = Bandwidth fbfb2fb22

Bandwidth = Band

(ii) QAM (ii) Bandwidth B



(iii) QPSK Bandwidth = fb/2 ( fb/2) bandwidth = fb

(iv) BPSK

Bandwidth = (f0 + fb) (f0 fb) = f0 + fb f0 + fb

Babdwidth = 2fb

Q.2(e) Compare QPSK and QASK (4 points). [4](A) Comparison between QPSK and QASK

Parameters QPSK QASK i) Type of Modulation Quadrature Phase

Modulation Quadrature Phase and Amplitude Modulation

ii) Location of signal points

On the circumference of circle

Equally spaced and placed symmetrical about origin.

iii) Distance between the signal points

d = b2 E for N = 2 d = b2 0.4E for N = 4 or N = 16

iv) Noise immunity Better than QASK More than QPSK v) Probability of error Less than QASK More than QPSK vi) Type of

Demodulation Synchronous Synchronous

vii) System complexity Less complex than QASK

More complex than QPSK

fb/2 fb/2

( f0 fb) f0 ( f0 + fb) (f0 fb) f0 (f0 + fb)

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arSK (4 points). (4 points).

QPSK and QASK d QAS

ters QPSK QPSKalaa Modulation Quadrature Phase n Quadrature PhModulation Modulation alyaation of signal nal

oints oints On the circumfOn the ccircle lyadya

iii) Distance between istance between the signal points he signal points

d = d 2dyiidyiv) Noise immunity iv) Noise immunit

Vidy

ViVidyv) Probability of erro Probability

Vid

ViVidvi) Type of vi) Type o

DemodulatioVid

ViVid

vii) System covii) VVVVV

ananknka

nka b) (f0 (f0 fb) f0 fb)


Q.3 Attempt any FOUR of the following : [16]Q.3(a) Explain the unipolar, polar and bipolar line codes. [4](A) Unipolar Format This is also known as on-off signaling. In this, symbol 1 is represented by

switching off the pulse. When the pulse occupies the full duration of symbol the unipolar format is said to be of the non-return to zero (NRZ) type.

When it occupies one half of the symbol duration, it is said to be of the

return to zero (RZ) type. When it occupies one half of the symbol duration, it is said to be of the return to zero (RZ) type.

The unipolar format offers simplicity of implementation. Polar Format In this, a positive pulse is transmitted for symbol 1 and negative pulse for

symbol 0. It can be of the NRZ or RZ type. A polar wave form has no dc component, provided that the 0s and 1s in the input data occur in equal proportion.

Bipolar Format This is also known as pseudoternary signaling. In this positive and negative

pulses are used alternately for the transmission of 1s, and no pulses for transmission of 0s. It can be of RZ or NRZ type.

Binary

+ 1

t 0

0 0 1 1 0 0 1

t

+ 1

1

0 0 1 1 0 0 1 0 1 1

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ntn of symbmb

Z) type. e.

said to be of the to be of the half of the symbol of the symbol

type.

ers simplicity of implementation. simplicity of implement

ve pulse is transmitted for syme is transmitted f can be of the NRZ or RZ type. e of the NRZ or R

, provided that the 0s and 1s ed that the 0s ion.

Bipolar FoThis is pulse

anaaananannknkkank 0 1 1 1

Vid

Vid

Vidydyy

Vidyid+ 1 +

1 1

0 0 1 1 0 0 0 1 1

dy



* Important feature of bipolar format is that, in the absence of a dc component, even though the input binary data may contain long strings of 0s and 1s.

* The pulse alternation property of bipolar format provides a capacity for in service performance monitoring such that, any isolated error which causes the deletion or creation of a pulse, will violet this property.

* Bipolar format eliminates the ambiguity that may arise because of polarity inversion during the course of transmission.

For these reasons, the bipolar format is used in the T1 carrier systems for digital telephony.

Q.3(b) Discuss Shannon’s theorem in brief. [4](A) Shannon’s Theorem

Also known as source coding theorem. Statement Given a discrete memoryless source of entropy H, the

average code word length for any source coding is bounded as L H. Explanation : For any source encoder, code efficiency is given as

= H 100%L

H entropy, H = M

ii 1

p log2 i

1p

L average code word length, L = M

ii 1

p i

i is length of the code in bits

As such if L H, minimum value of L i.e. Lmin. = H. Under such circumstances, efficiency will be maximum. To increase efficiency variable length coding is done.

eg. Huffman code.

t 0

+ 1

1

0 0 1 1 0 0 1 0 1 1

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cit rror whi whi

operty. rise because of ause o

he T1 carrier systems carrier systems

theorem in brief. em in brief. rem

wn as source coding theorem. ource coding theoreent Given a discrete memor Given a discrete

rage code word length for any sourage code word length for xplanation : For any source encodplanation : For any s

= = dy

H 100%%LL

H H entropy, Hentropy, H

L L averag

As such Acircuva

nknnknknkk 0 1 1 0 1 1

lanka


Q.3(c) Describe WDM in detail. [4](A) Wavelength Division Multiplexing (WDM) (i) Channels having different frequency ranges can be multiplexed on a long

fiber. (ii) The only difference with electrical FDM is that optical system is

completely passive. (iii) Reason WDM is popular is that the energy on a signal factor is a few

gigahertz width because it is impossible to convert. (iv) Hence wavelength division multiplexing is explained. Q.3(d) Compare between FHSS and DSSS (4 points). [4](A) Comparison between DSSS and FHSS

Parameters DSSS FHSS i) Definition PN sequence of large

bandwidth is multiplied with narrow band data signal.

Data bits are transmitted in different frequency slots which are changed by PN sequence.

ii) Chip rate It is fixed, Rc = c

1T

Rc = max (Rh, Rs)

iii) Modulation technique

BPSK M-ary FSK

iv) Processing gain PG = b

c

TT

= N PG = 2t

Fiber 1 Spectrum

Powe

r Fiber 2 Spectrum

Powe

r

Fiber 3 Spectrum

Powe

r

Fiber 2

Shared fiber Fiber 3

Fiber 4

Fiber 1

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ar frequency ranges can be multiplency ranges can b

e with electrical FDM is thatwith electrical FDM ie.

s popular is that the energy onular is that the enedth because it is impossible to cocause it is impossib

velength division multiplexing is eth division multiplex

e between FHSS and DSSS (4 e between FHSS and DSSarison between DSSS and FHSrison between DSS

Parameters Parameters dydyi) Definition PN sequDefinition PN sebandwndwwitdy

ViVidy

ii) Chip rate ii) ViViViiii) Modulatiii) techVVV

iv)

VVV

kar

kar

nkakannka

nkaFiber 3 er 3

Fiber 4 F



v) Error probability

Pe = b

c

E1 erfc2 JT

Pe = rbc

1 e R / 22

vi) Acquisition time Long Short vii) Effect of

distance This system is distance relative.

Effect of distance is less.

Q.3(e) Explain multiplexing hierarchy (AT&T) for FDM system. [4](A) Multiplexing Hierarchy in FDM

Consider example at telephony in which each voice channel is having range of 300Hz to 3.4 KHz.

Here we need to multiplex such ‘n’ no of voice channel by modulating it with different subcarriers.

Multiplexing hierarchy goes as follows. Level 1 : Basic Group : [12 voice channels multiplied together] Level 2 : Super Group : [Upto 5 B.G mux together i.e.: upto 12 5 = 60 channels] Level 3 : Master Group : [Upto 10 S.Cr mux together i.e. upto 600 V.C.] Level 4 : Jumbo Group : [Upto 6 M.G. maximum together i.e. upto 3600 V.C.]

Q.4(a) Attempt any THREE of the following : [12]Q.4(a) (i) List and explain different types of errors in data communication. [4](A) Types of error : i) Single bit error or one bit error ii) Burst error Single bit error : if only 1 bit in transmitted bit sequence is changed

because of noise then it is called as single bit error.

S1 1

S1 2

S1 12

F

D

M 1 2

5

F

D

M 1 2

5

F

D

M 12

5

F

D

M

12 rc

4 KHz

B Group

SG

MG

2 4 = 48 KHz

48 5 = 240 K

240 10 = 2400K

2400 6 = 14.400K

JG 3600 channel

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r[4][4]

channel is having range of s having range of

oice channel by modulating it ice channel by modulatin

.

nnels multiplied together] ultiplied together] G mux together i.e.: upto 12her i.e.: upto 12 5 = 60

to 10 S.Cr mux together i.e. upto S.Cr mux togetheto 6 M.G. maximum together i.e. uto 6 M.G. maximum togeth

Q.4(a) (a) Attempt any Attempt an THREE oQ.4(a)) (i) List and explain(i) List and (A) Types of error (A) Typ i) Single biti) S ii) Burst e

Singl

yyyal

yal

1 2 2

55 ayaya

F

D

Mal 1

B Group p

SG G

ya2 4 = 48 KHz 48 KH

4848 5 = 240 K


Received bit seq. 1 0 0 1 0 1 1 code of C Burst error : If 2 or more than 2 bits are changed because of noise then it

is called as Burst error. Q.4(a) (ii) Explain Baudot code with suitable example. [4](A) Baudot code In baudot coding every symbol is represented by using 5 binary bits. 2 special codes are used : i) figure shift code. 1 1 1 1 1 ii) letter shift code 1 1 0 1 1 By using 5 bit code we can generate only 32 different binary bit sequence

but the number of characters in English text are more than 32 SO 2 special codes are used. Now codes attached like this :

A 1 1 0 0 0 B ? 1 0 0 1 1 : 1 1 1 1 1 : 1 1 1 1 1 Z ” 1 0 0 0 1 Example A B? Q.4(a) (iii) Define sampling theorem. List types of sampling techniques.

Draw the naturally sampled signal. [4]

(A) Sampling Theorem "A bandlimited signal can be reconstructed exactly if it is sampled at a rate

atleast twice the maximum frequency component in it."

1 1 0 1 1 1 1 0 0 0 1 1 1 1 1 1 1 0 0 0

1 1 0 1 1 1 0 0 1 1 1 1 1 1 1 1 0 0 1 1

Letter shift special code

Figure shift code

transmitted symbol

1 0 0 0 0 1 1

+A

A

C

cause of noise then it of noise then it

e. [4]

nted by using 5 binary bits. by using 5 binary bits

n generate only 32 different binn generate only 32 differacters in English text are more trs in English text are m

odes attached like this : ached like this :

1 1 1 1

0 0 0 1

xample A A B?

Q.4(a)a) (iii) Define (iii) DDraw

SampliA b

idyidyidyidy1 1 01 1 1 0 1 1

1 1 0

VididVid

Vid

VidLetter shift ter sh

special code co

r



There are two types of sampling technique : (1) Natural sampling (2) Flat top sampling

Q.4(a) (iv) Explain cyclic redundenry check with suitable diagram. Explain

CRC for data 100100 to be transmitted with division 1101. [4]

(A) Cyclic redundancy cheek technique is more powerful than parity check & check sum error detection. It is based on binary division. A sequence of redundant called CRC or CRC remainder is appended at the

end of data unit such as byte. The resulting data unit after adding CRC remainder becomes exactly

divisible by another predetermined binary no. At the receiver, this data unit is divided by the same binary no. There is no error if this division does not yield any remainder. But a non-zero remainder indicates presence of error in received data unit Such an erroneous bit is then rejected.

Procedure : Divide the data unit by predetermined divisor Obtain the remainder which is ‘CRC’ CRC should have exactly bit less than divisor Append the CRC to the end

of the data unit and then divide it with predetermined divisor.

Natural Sampled output

Data 00 ….. 0

Divisor

CRC

Data CRC

n bit

n bitcode word

suitable diagram. Explain diagram. Expltted with division 1101. with division 1101.

[4

s more powerful than parity ch powerful than p

called CRC or CRC remainder is ap CRC or CRC remain

s byte. s by unit after adding CRC remaindet after adding CRC re

er predetermined binary no. ermined binary nr, this data unit is divided by the data unit is divided

error if this division does not yie f this division does zero remainder indicates presence emainder indicates p

n erroneous bit is then rejected.ous bit is then r

dure : ure :Divide the data unit by predetede the data unit by Obtain the remainder which btain the remainder whCRC should have exactly bRC should have exactly Append the CRC to thAppend the CRof the data unit of the data divide it with previde itdivisor. r


Data : 100100 Divisor : 1101 n = no. of divisor bits 1

= 4 1 = 3 divided = 100100000

Divisor = 1101

1111011101 100100000

1101100011011010110111101101

1001101

001

CRC = 001 Code word = Data (+) CRC

= 100100001 Q.4(b) Attempt any ONE of the following : [6]Q.4(b) (i) Discuss Frequency Hopping (FH) spread spectrum. [6](A) Frequency Hopping (FH) spread spectrum is a FM or FSK technique while DS

spread spectrum, described in sec.2, is an AM (or PSK) technique. The signal to be frequency hopped is usually a BFSK signal although M-ary FSK, MSK or TFM can be employed. Considering that BFSK is used, we have that the original signal, before spread spectrum is applied, is

s(t) = s 02P cos ( t d(t) t )

where d(t) is the data to be transmitted. The FH modulation is then applied by varying the carrier frequency so that the resulting FH spread spectrum is

v(t) = s 02P cos ( t d(t) t )

Here the FH signal has a carrier frequency fi = i/2 which changes at the hopping rate fH, i.e., the carrier frequency fi changes each TH seconds. The frequency chosen each TH is selected in a pseudo-random manner from a specified set of frequencies. Typically 32-500 different frequencies are used to form this set.

The primary advantage of FH is that it enables the transmitter to change its

carrier frequency and thereby avoid an otherwise in-band interfering signal. For example, consider that the FH signal spends an equal time at each of 1000

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the following : ng : uency Hopping (FH) spread spec Hopping (FH) spr

ping (FH)H) spread spectrum is a F spread spectrumrum, described in sec.2, is an AM escribed in sec.2, is

quency hopped is usually a BFSK spped is usually aan be employed. Considering than be employed. Conside

nal signal, before spread spectru signal, before spre s(t) = ) = dydys 0s2P cos ( t d(t) t2P cos ( t d(ts 0s t d(t) tt d(t0

where d(t) is the data to be twhere d(t) is the data to be by varying the carrier freqby varying the carrier freq

v(t) = v(t) =idids 02P cos ( t2P coss 0t0

Here the FH signal Here the FHhopping rate fhop H, ifrequency chofrequspecified sespecito form t

he



frequencies f1, f2, …… , f1000. Also assume that BFSK is employed. Then if the bit rate of the data is fb, the bandwidth used by the signal, at any carrier frequency fi, is B = 4fb. Now assume that there is an interfering signal having a bandwidth B = 4fb and a fixed center frequency fj. If frequency hopping were not employed and the interference were located at the transmitted signal's carrier frequency, i.e., j = i, then if the interfering signal power were sufficiently large the probability of error would be Pe = 0.5 since under these circumstances we could do no better than to guess. (In a military system, the interferer determines the signal's carrier frequency and purposely transmits at that frequency to block or jam the communications.) By employing FH and using say 1000 frequencies, the probability of the same interferer causing an error is reduced to Pe = 1/2000 = 5 10 4 (note that thermal noise is ignored since it is considered to be a second-order effect when "jamming" is present).

Q.4(b) (ii) Explain the properties of Hamming Code.

Define : (1) Hamming distance. (2) Minimum distance. (3) Weight of code words. (4) Error detection capacity. (5) Error correction capacity.

[6]

(A) Properties of Hamming Code i) Hamming codes are linear. ii) Number of bits in codeword is equal to n = 2q 1 where q = number of extra bit eg. if Q = 3 then n = ? 23 1 n = 7 iii) Number of message bit K should be always equal to 2q 1 q iv) For the hamming code the minimum distance dmin = 3. v) Hamming codes can generate systematic as well as non – systematic code

words. vi) Hamming codes are used to detect burst error. i) Hamming Distance (d) : It is defined as the number of bits in which 2

valid codewords differ from each other. e.g. X1 = 1 0 1 1 0 0 1 X2 = 1 1 0 1 0 1 1 d = 3. ii) Minimum distance (dmin) : It is smallest hamming distance between any 2

valid codewords.

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ingted signagna

power were r wence under these er the

military system, the y system, the purposely transmits at ely transmits at

By employing FH and using loying FH and using nterferer causing an error is nterferer causing an error

ermal noise is ignored since it is mal noise is ignored since "jamming" is present). g" is present).

ng Code. e. ce.

ance. code words. words.

etection capacity. etection capacity. r correction capacity. rrection capacity.

ng Code are linear. near.

bits in codeword is equal to n = 2 odeword is equal t q

number of extra bit er of extra bit Q = 3 then n = ? then n = ?

2 23 1 1 n = 7 n = 7

) Number of message bit K shoulber of message bit iv) For the hamming code the mr the hamming code thv) Hamming codes can generv) Hamming codes can gene

words. words. vi) Hamming codes are vi) Hamming code

i) Hamming Disti) Hamminvalid codew

e.g. X e


iii) Weight of codewords (W) : It is number of man zero elements of codeword provided. Codeword is not all zero codeword.

dmin = Min [W(x)]

where X 0 0 0 0 ……0.

Error detection Capability : It is calculated by this, dmin S + 1 Where dmin = Minimum distance. S = Number of bits in error that can be detected. e.g. dmin = 3 3 S + 1 3 1 S S 2 upto 2 bits in the error can be detected.

Error correction capability : it is calculated by this expression. dmin 2t + 1 where t = number of bits in error that can be corrected.

eg. dmin = 3 3 2t + 1 3 1 2t 2 2t t 1 this coding technique can correct upto 1 bit. Q.5 Attempt any FOUR of the following : [16]Q.5(a) Draw the waveforms for 1010011 for :

(i) Split phase monchester. (ii) Differential monchester line code formats

[4]

(A) For 1010011

1 0 1 0 0 1 1

Split phase manchester t

differential manchester t

+ve

ve 0

+ve

ve 0

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1

ed.

ated by this expression. y this expressi

hat can be corrected. be corrected.

ique can correct upto 1 bit. an correct upto 1 b

y FOUR of the following : of the following : e waveforms for 1010011 for :ms for 101001

plit phase monchester. it phase monchest Differential monchester line coerential monchest

or 1010011 10011

Vidy

VViViViVViVii

1 0 1 0 1 0 00 0

VV+ve +ve

ve ve 0 0

+ve



Q.5(b) What do you understand by channel capacity? How can it beincreased?

[4]

(A) Channel Capacity Data Rate R = n.fs Where, n = Number of bits/sample. fs = Number of sample/sec.

Channel capacity is defined as maximum possible data rate is with which information can be convey on channel with minimum error.

Q.5(c) List the different error detecting methods. Describe checksum

method with suitable example. [4]

(A) Different error detection schemes Repetition codes Parity bits Checksums Cyclic redundancy checks (CRCs) Cryptographic hash functions

Checksums Most of error detection techniques uses a process known as checksum to generate an error-detection character. The character results from summing all the bytes of a messageM together, discarding and carry over from the addition. Again, the process is repeated at the receiver and the two checksums are compared. A match between receiver checksum and transmitted checksum indicates good data. A mismatch indicates an error has occurred. This method, like CRC, is capable of detecting single or multiple errors in the message. The major advantage of checksum is that it is simple to implement in either hardware or software. The drawback to checksum is that, unless you use a fairly large checksum (16- or 32-bit instead of 8-bit), there are several data-bit patterns that could produce the same checksum result, thereby decreasing its effectiveness. It is possible that if enough errors occur in a message that a checksum could be produced that would be the same as a good message. This is why both checksum and CRC error-detection methods do not catch 100% of the errors that could occur, they both come pretty close.

Example : What is the checksum value for the extended ASCII message “Help!”?

Solution : The checksum value is found by adding up the bytes representing the Help! characters:

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ar rate is with which is with which ror.

s. Describe checksum. Describe checksum [4]

CRCs) tions

ion techniques uses a process kniques uses a prdetection character. The charaction character. The c

a messageM together, discardinsageM together, din, the process is repeated ae process is repe

are compared. A match beompared. A matted checksum indicates good dum indicates g

ccurred. This method, like CRcurred. This methltiple errors in the message. The errors in the mess

is simple to implement in eitheple to implement in echecksum is that, unless youchecksum is that, unless yinstead of 8-bit), there arenstead of 8-bit), there aresame checksum result, tme checksum rethat if enough errorsthat if enouthat would be the that would CRC error-deteCRCoccur, they boccur

Example “Help!”


01001000 H 01100101 e 01101100 l 01110000 p 00100001 ! 00010000 Checksum checksum error-detection process that uses the sum of the data stream in bytes. The hardware solution relies once more on exclusive OR gates, which perform binary-bit addition. Each 8-bits of data are exclusive ORed with the accumulated total of all previous 8-bit groups. The final accumulated total is the checksum character.

Q.5(d) Explain FDMA system with schematic diagram. Compare FDMA and

TDMA. [4]

(A) FDMA (Frequency division multiple access) This technique is based an FDM tech In FDMA available bandwidth is shared by all the station Each station is allocated with a particular frequency band to send its data.

Example

In above diagram frequency band f0 to f1 is reserved for station 1, f2, to f3 is reserved for s2 etc.

Guard bands are also provided in between the adjacent frequency slot to avoid crosstalk.

Features Overall channel band width is shared by multiple users therefore no of

users can transmit their information simultaneously. Guard band are provides because :

i) To avoid cross talk ii) Impossible to achieve ideal filtering to separate different users

Power efficiency different users. Synchronization is not required

Station1

Station2

Station3

Station 4

Guard Bands

f0 f1 f2 f3 f4 f5

freq

Available B.W.

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e data stream in ream iive OR gates, which R gates, which

xclusive ORed with the e ORed with the final accumulated total is ccumulated total is

gram. Compare FDMA and ompare FDMA a [4

ccess) M tech

h is shared by all the station ared by all the statth a particular frequency band to articular frequency

In above diagram frequenc above diagram frequenis reserved for sis reserved for s2 etc. Guard bands are alsGuard bands aavoid crosstalk. oid cr

Features FeaOverall chOusers cauGuari)

yal

yalStation ll

yaalard Bandsnds

f1 f2 ff3aladdy

Availa



Advantages : All stations can operate continuously and simultaneously. Power required for transmission depends on the no of channels being

transmitted. SNR is improved because of Fm Synchronization is not required.

Disadvantages : Each channel on earth station can used only a part of total satellite B.W. Inspite having guard bands adjacent channel interference is present. Because of the use of Fm, Bw required & therefore less no of channels

can be accommodated in available Bandwidth.

Parameter FDMA TDMA Technique Sharing of overall B.W. of

satellite transponder Sharing of overall time os satellite transponder

Syndrome Not required Required Code word Not required Not required Power efficiency

Less Full power n is poss

Guard time & Guard band

Guard band required Guard time required

Q.5(e) Describe working of DSSS system in detail. [4](A) DSSS system (Direction Sequence (DS) Spread Spectrum)

A Direct Sequence (DS) spread spectrum signal is one in which the amplitude of an already modulated signal is amplitude modulated by a very high rate NRZ binary stream of digits. Thus, if the original signal is s(t), where s (t) = s2P d(t) cos 0t … (1)

(a binary PSK signal), the DS spread spectrum signal is

v (t) = g(t) s(t) = s2P g(t) d(t) cos 0t … (2)

where g(t) is a pseudo-random noise (PN) binary sequence having the values 1. The characteristics of g(t) are extremely interesting and are discussed in some detail in Ranging using DS spread spectrum. Here we merely assume that g(t) is a binary sequence as is the data d(t). The sequence g(t) is generated in a deterministic manner and is repetitive. However, the sequence length before repetition is usually extremely long and to all intents and purposes, and without serious error, we can assume that the sequence is

of total satellite B.W. l satellite B.W. erference is present. nce is present.

erefore less no of channels erefore less no of channeh.

TDMA TDMA kn B.W. of onder

Sharing of overall timeSharing of ovesatellite transpondeite trnkan

Required Reqananred Not required Not rananFull power Fulananuard band required Guard required laalarking of DSSS system in detail of DSSS system in

tem (Direction Sequence (DS) Section Sequencect Sequence (DS) spread spectruct Sequence (DS) spread s

n already modulated signal is am lready modulated sRZ binary stream of digits. Thusnary stream of digits

where s (t) = where s (t) = dydys2P d(t) co(t) s

(a binary PSK signal), the Da binary PSK signa

v (t) = g(t)

where g(t) is a psewhere g(t)

The charactThe some detthat gen


truly random, i.e., there is no correlation at all between the value of a particular bit and the value of any other bits. Furthermore, the bit rate fc of g(t) is usually much greater than the bit rate fb of d(t). As a matter of fact the rate of g(t) is usually so much greater than fb, we say that g(t) "chops the bits of data into chips", and we call the rate of g(t) the chip rate fc, retaining the words, bit rate, to represent fb. To see that multiplying the BPSK sequence s(t) by g(t) spreads the spectrum we refer to figure 1 which shows a data sequence d(t), a pseudo-random (often called a pseudo-noise or PN) sequence g(t) and the product sequence g(t) d(t). Note that (as is standard practice) the edges of g(t) and d(t) are aligned, that is, each transition in d(t) coincides with a transition on g(t). The product sequence is seen to be similar to g(t) indeed if g(t) were truly random, the product sequence would be another random sequence g'(t) having the same chip rate fc as g(t). Since the bandwidth of the BPSK signal s(t) is nominally 2fb the bandwidth of the BPSK spread spectrum signal v(t) is 2fc and the spectrum has been spread by the ratio fc/fb. Since the power transmitted by s(t) and v(t) is the same, i.e., Ps, the power spectral density Gs(t) is reduced by the factor fb/fc.

Fig. 1 : (a) The waveform of the data but stream d(t). (b) The chipping waveform g(t).

(c) The waveform of the product g(t) d(t).

To recover the DS spread spectrum signal, the receiver shown in figure 2 first multiplies the incoming signal with the waveform g(t) and then by the carrier

2 cos 0t. The resulting waveform is then integrated for the bit duration and the output of the integrator is sampled, yielding the data d(kTb). We note that at the receiver it is necessary to- regenerate both the sinusoidal carrier of frequency 0 and also to regenerate the PN waveform g(t).

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hae chip ra ra

reads the spectrum the spectrum d(t), a pseudo-random pseudo-random

nd the product sequence product sequencee edges of g(t) and d(t) are edges of g(t) and d(t) a

des with a transition on g(t). es with a transition on to g(t) indeed if g(t) were truly indeed if g(t) we

be another random sequence gother random seque ince the bandwidth of the BPSK e bandwidth of th

of the BPSK spread spectrum s BPSK spread spn spread by the ratio fd by the rat c/ffb. Sinc

s the same, i.e., P same, i.e., s, the power sp, the tor ftor b/fc.

Fig. 1 : (a

To recover tTo remultipliesV

2 c



Fig. 2 : A BPSK communication system incorporating a spread spectrum

technique. Q.6 Attempt any FOUR of the following : [16]Q.6(a) Explain BPSK, draw its waveforms. [4](A) BPSK : Binary phase shift keying

In the BPSK phase of a carrier is modified according to value of the symbol.

b(t) b(t) phase shift transmitted signal 1 +1 0o c2P cos.(2 ft)

0 1 180o c2P cos.(2 ft 180 )

In general transmitted BPSK signal is. S(t) = cb(t) . 2P .cos(2 ft)

+V0o

V

180o

Tb

c2P cos(2 f t)

s(t)

Binary bit

Seq.

b(t)

Working of transmitter of BPSK

Bipolar NRZ

encoderVidy

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ar a spread spectrum ad spectr

[16][4]

ied according to value of the symcording to value o

ransmitted signal ted saa

c2P cos.(2 ft)P cos.(2 ft)cc

lalacc2P cos.(2 ft 180 )2P cos.(2 ft 18ccftft

In general transmitted BPSn general transmitt S(t) = S(t) = b(t) . 2bdyy

aalyal

dydyayaal

dyalV

180180oo aVBinary

bitSeq.


Bipolar NRZ encoder : given information signal will consists of binary bits 1 & 0 this binary bit sequence is converted into bipolar NRZ signal by this block output of bipolar NRZ encoder is b(t).

Where, b(t) = +1 when bit = 1 b(t) = 1 when bit = 0

Modulator : bipolar b(t) is used as modulating signal and c2P cos(2 ft) is used as carrier. The modulator multiplies 2 input signal to generate BPSK signal S(t) = cb(t) . 2P .cos(2 ft) BPSK waveforms :

Q.6(b) Draw the block diagram of M-ary FSK transmitter and receiver. [4](A) M ary Frequency Shift Keying (MFSK)

In MFSK, we use group of N bits, which gives rise to M different symbols. M = 2N

Each symbol duration is TS = N Tb

Each symbol uses different frequency, hence, MFSK uses frequencies f1, f2, … fM. The probability of error is minimized of the frequencies f1, f2, … fM are selected so that M signals are mutually orthogonal. One commonly used arrangement simply provides that the carrier frequency be successive even harmonics of the symbol frequency fS = 1/TS.

1 0 1 1 10

t

t

t

Binary bit

+1

1

b(t)0

0c2P (2 ft)

BPSK signal s(t)

1o 0o180o 1o 0o180o

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ar

yalan

kar

yala

c2P cos(2 ft)ft)cc is al to generate BPSK generate BPSK

b) Draw the block diagram of Mb) Draw the block diagram of ) M) M

Vidary Frequency Shift Keary Frequency Shift Ke

In MFSK, we use group o MFSK, we use g M = 2 N

Each symbol duratEach symbo

Each symbol uEach

The prselec

anlayadyalaalaaaaallallaalalaalalayyayyyayayayayayayaaaaa

t

al t)

11oo 0o18080oo 1o 180o



Thus we have,

1 S

2 S

3 S

f K ff K 2 f

f K 2M 2 f

Now we can write equation of MFSK signal as

MFSK 1 S 1 2 S 2 m S MV t P t 2P cos t P t 2P cos t ... P t 2P cos t

where P1(t), P2(t), … Pm(t) are NRZ unipolar signals out of which only one is +1 for every TS (= NTb) sec. Transmitter In the figure, when N bits of b(t) enter into serial to parallel converter, they appear simultaneously at the output of serial to parallel converter. These bits are given to DAC which generates an analog voltage Vm where m = 1, 2, … M. That is, the output of DAC may have M different levels. This voltage Vm is applied to frequency modulator which generates a sine wave of constaint amplitude but the frequency is dependent on input voltage Vm. Receiver

Vm

m = 1, 2, .. M

b (t) Serial toparallel

converter

Frequency Modulator

VMFSK (t) N bitDAC

12 : N

Select

largest

output

VMFSK(t)

BPF 1

BPF 2

BPF M

Envelopedetector

Envelopedetector

Envelopedetector

.

.

.

.

.

.

N bit

output

(ADC)

b1

b2

bN

.

.

. Vi

dyala

nkara

S MM2P cos ttS MM2P cossS

t of which only one is +1 ch only one is +1

s of b(t) enter into serial to pa of b(t) enter into serialy at the output of serial to para at the output of serial t

n to DAC which generates an ana AC which generates is, the output of DAC may have output of DAC

applied to frequency modulator w to frequency mod mplitude but the frequency is dee but the freque

ver er

ank

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ankkVm

m = 1, 2, .. M 2, .. M

Frequency Modulator

VMFSKMFSK(t)

BPFB11

BPFBPF2

.

dyy

Envelde

Viy

Vidd

Viddyd


PSD of MFSK signal

VMFSK(t) = 1 S 1 2 2 m S mP t 2P cos t P t cos t ... P t 2P cos t

GMFSK(t) = S1 1 M M

Pf f f f ... f f f f

8

+ 2 2S b1 b 1 b

NP TSa f f NT Sa f f NT ....

8

2 2M b M bSa f f NT Sa f f NT

Q.6(c) Draw the block diagram of ADM transmitter. Describe its working

with waveforms. [4]

(A) Adaptive delta modulation (ADM) In the adaptive delta modulator if there is the rate of change of input signal

is very high we will increase the level. And if the rate of change of input signal is low then we will decrease the level.

GMFSK(f)SP8

S bNP T8

f1 f f fM f

f1 fb / N fM + fb / N

bfB W M 2N

accumulator

x (kts)

x (kts)Quantizer

delay ts

++

+

Step size calculation

e (nts) Vidy

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....

ck diagram of ADM transmittegram of ADM traorms.

e delta modulation (ADM) ulation (ADM)e adaptive delta modulator if the adaptive delta mod

very high we will increase the l igh we will increasignal is low then we will decreas is low then we will dec

anaaaananknkkananknka

anannnala8

fff11 f fff

lalanff11 ffbb / N

B

x

x (ktx s)



ADM Receiver

Advantages : i) It reduces slop overload error by increasing step size. ii) It reduces granular noise by decreasing step size.

Disadvantages i) ADM transmitter receiver are complex and ii) Cost of ADM is high. Q.6(d) State applications of s.s. modulation (any four). [4](A) Applications of Spread Spectrum Techniques (i) For combining the intentional interference. (ii) For reducing the unintentional interference. (iii) For suppressing the interference due to the multipath interference. (iv) In the low probability of intercept application as explained in section. (v) Due to large bandwidth of a spread spectrum signal can be recognized

only by the authorised user. All other receiver consider this as noise. Q.6(e) Explain quantization in detail. [4](A) Quantization Process

It is a process of approximation or rounding off. It converts sampled signal into approximate quantized signal which

consists of finite no. of pre-decided voltage levels.

delay ts Step size calculation

+

+

Receivers sign LPF

time

2

2

2

> <

1 2 3 4 5 6 7 8 9 10 11 12 13 0

ampl

itud

e

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arp overload error by increasing stload error by incr

granular noise by decreasing step noise by decreasi

ges transmitter receiver are compleer receiver are

ost of ADM is high. st of ADM is high.

State applications of s.s. modu applications of s.s. m Applications of Spread Spect Applications of Spread Spe

(i) For combining the inten (i) For combining the inten (ii) For reducing the unin (ii) For reducing t (iii) For suppressing th (iii) For sup (iv) In the low prob (iv) In the lo (v) Due to large (v)

only by tho

6(e) Explain Quant

anaaanandelalala+

++

kkaraaaaararartim

> <

8 9 10 11 12 13 10 11 12


Each sample value at input of quantizer is approximated to nearest predicted voltage level. These standard levels are called quantization levels.

Input signal x(t) is assumed to have peak-peak swing of VL to VH volts.

This entire voltage range has been divided into Q equal intervals each of size 'S'.

S is called step size

S = H LV VQ

i.e. Q = 8 for nw At the center of these ranges the quantization levels q0, q1, q2, , q7

are placed. xq (t) is quantized version of x(t) which is available at the output of q

quantizer. When x(t) is in the range of 0, the corresponding to any value of x(t)

the quantizer output will be equal to 'q0'. Similarly for all the values of x(t) in 1, the quantizer output is q1.

Thus in each range from 0 to 7 signal x(t) is rounded off to nearest quantization level and quantized signal is produced.

In PCM, coded no. is transmitted for each level sampled in modulating signal.

If we do not use quantizer block in PCM then this will need a large no. of bits per word which will increase bit rate and finally bandwidth.

Quantization has effect of reducing this infinite no. of levels to relatively small nos. which can be transmitted without difficulty.

VL q0

q1

q2

q3

q4

q5

q6

q7

s

ss

s

s

s

s

s

Amplitude

Time

Quantization noise

Quantized signal xq(t) Input signal x(t)

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ared to have peak-peak swing of have peak-peak s

ge has been divided into Q equal ge has been divided into Q

ze

L

Q 8 for nw nw

e center of these ranges the qua of these ranges placed. placed.

q (t) is quantized version of x(t (t) is quantized verquantizer. ntizer.

When x(t) is in the range hen x(t) is in the rangthe quantizer output wille quantizer output wilx(t) in x(t) in 11, the quantize, the q

Thus in each rangeThus in eachquantization leventiza

In PCM, code In PCMsignal. s

If we do Ibits p

Qu

nnkkkkkqq0 0

qq1

qq2 2

q3

q4

q5

q6

Time

(t)

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