cs307 model 3 - studoob.instudoob.in/app/q/cse/s5/cs307/cs307_model_3.pdf · b=fh-fl 20= 60 –fl...

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Reg No.:_______________ Name:__________________________

APJ ABDUL KALAM TECHNOLOGICAL UNIVERSITY

THIRD SEMESTER B.TECH DEGREE EXAMINATION, DECEMBER 2017

Course Code: CS307

Course Name: DATA COMMUNICATION

Max. Marks: 100 Duration: 3 Hours

PART A

Answer all questions, each carries 3 marks. Marks

1 Differentiate half duplex and full duplex transmission with example. (3)

2 A signal has the bandwidth of 20Hz. Highest frequency is 60Hz. What is the

lowest frequency? Draw the spectrum if the signal contains all integral

frequencies of same amplitude.

(3)

3 Differentiate STP and UTP cables (3)

4 Describe Line-of-Sight Propagation (3)

PART B

Answer any two full questions, each carries 9 marks.

5 a) Differentiate attenuation distortion and Delay distortion (4)

b) Define white noise. Given an amplifier with an effective noise temperature of

10,000K and a 10MHz bandwidth, what thermal noise level, in dBW, may expect

at its output.

(5)

6 a) Describe the factors depends on channel capacity. (4)

b) Define Shannon capacity formula.

Assume that the TV picture is to be transmitted over a channel with 4.5MHz

bandwidth and a 35dB signal-to-noise ratio. Find the capacity of the channel(bps)

(5)

7 a) List out the benefits of optical fiber compared to twisted pair and coaxial cable. (4)

b) Describe about Satellite Microwave. (5)

PART C

Answer all questions, each carries 3 marks.

8 For the bit stream 101011100 .Sketch the wave form using

a)NRZ-L b)NRZ-I

(3)

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9 State the advantages of SSB over DSBTC (3)

10 Describe Wavelength division multiplexing. (3)

11 Illustrate the working of a cable modem (3)

PART D

Answer any two full questions, each carries 9 marks.

12 a) For the bit stream 01001110 , Sketch the waveform using

a) NRZ-I b) Bipolar-AMI c) Pseudoternary

Assume that the signal level for the preceding bit for NRZ-I was high, the most

recent preceding 1 bit (AMI) has a negative voltage and the most recent

preceding 0 bit (pesudoternary) has a negative voltage.

(5)

b) Analyse the reasons for preferring DM over PCM for encoding analog signals

that represent digital data.

(4)

13 a) A signal is quantised using 10 bit PCM. Find the signal-to-quantization noise

ratio.

(4)

b) Given the bit pattern 01100 , encode this data using ASK,BFSK, and BPSK

where frequency is 2Hz.

(5)

14 a) Outline the concept of CDMA. (5)

b) Describe about SONET/SDH (4)

PART E

Answer any four full questions, each carries 10 marks.

15 a) Differentiate Synchronous and Asynchronous Transmission (6)

b) Describe error detection process (4)

16 a) For P= 110101, M= 1010001101, find CRC (10)

17 a) Define hamming distance with example. (3)

b) Demonstrate block code principle with help of an example (7)

18 a) List the three benefits of spread spectrum (3)

b) Summarise the working of FHSS (7)

19 a) Illustrate the process of DHSS (6)

b) Describe the principles of packet switching network (4)

20 a) Compare and contrast circuit switching and packet switching (6)

b) Distinguish the significance of packet size in a packet switching network (4)


****


Total Pages: Reg No.:_______________ Name:__________________________

APJ ABDUL KALAM TECHNOLOGICAL UNIVERSITY THIRD SEMESTER B.TECH DEGREE EXAMINATION, DECEMBER 2017

Course Code: CS307 Course Name: DATA COMMUNICATION

Max. Marks: 100 Duration: 3 Hours PART A

Answer all questions, each carries 3 marks. Marks

1 Differentiate half duplex and full duplex transmission with example.

Definition(1)

Diagram(1)

Example(1)

In half-duplex mode, each station can both transmit and receive, but not at the same time. : When one device is sending, the other can only receive, and vice versa .

Eg: Walkie-talkies and CB (citizens band) radiosare both half-duplex systems. In full-duplex mode(also called duplex), both stations can transmit and receive simultaneously. Eg: One common example of full-duplex communication is the telephone network.When two people are communicating by a telephone line, both can talk and listen at thesame time.

(3)

2 A signal has the bandwidth of 20Hz. Highest frequency is 60Hz. What is the lowest frequency?

Draw the spectrum if the signal contains all integral frequencies of same amplitude.

Solution(2)

Diagram(1)

Letfh be the highest frequency,fl the lowest frequency, and B the bandwidth. Then

(3)


B=fh-fl

20= 60 –fl

Fl= 60-20

=40Hz

3 Differentiate STP and UTP cables

1 mark for each point

unshielded Twisted Pair (UTP)

ordinary telephone wire

cheapest

easiest to install

suffers from external EM interference

shielded Twisted Pair (STP)

metal braid or sheathing that reduces interference

more expensive

harder to handle (thick, heavy)

(3)

4 Describe Line-of-Sight Propagation.

Description(2)

Figure (1)

Above 30 MHz, neither ground wave nor sky wave propagation modes operate, and communication must

be by line of sight. For satellite communication, a signal above 30 MHz is not reflected by the ionosphere

and therefore a signal can be transmitted between an earth station and a satellite overhead that is not

(3)


beyond the horizon. For ground-based communication, the transmitting and receiving antennas must be

within an effective line of sight of each other. The term effective is used because microwaves are bent or

refracted by the atmosphere. The amount and even the direction of the bend depends on conditions, but

generally microwaves are bent with the curvature of the earth and will therefore propagate farther than the

optical line of sight

PART B Answer any two full questions, each carries 9 marks.

5 a) Differentiate attenuation distortion and Delay distortion

Figure (1)

Description(3)

Attenuation

where signal strength falls off with distance

depends on medium

received signal strength must be:

strong enough to be detected

sufficiently higher than noise to receive without error

so increase strength using amplifiers/repeaters

is also an increasing function of frequency

so equalize attenuation across band of frequencies used

eg. using loading coils or amplifiers

Delay Distortion

only occurs in guided media

propagation velocity varies with frequency

hence various frequency components arrive at different times

particularly critical for digital data

since parts of one bit spill over into others

causing intersymbol interference

(4)


b) Define white noise. Given an amplifier with an effective noise temperature of 10,000K and a

10MHz bandwidth, what thermal noise level, in dBW, may expect at its output. (k= 1.38*10-23

J/K) Definition (2)

Solution N = 10 log k + 10 log T + 10 log B

(5)


= –228.6 dBW + 10 log 104 + 10 log 107

= –228.6 + 40 + 70

= –118.6 dBW

6 a) Describe the factors depends on channel capacity.

4 factor each Carrie 1 mark.

Data rate: The rate, in bits per second (bps), at which data can be communicated

Bandwidth: The bandwidth of the transmitted signal as constrained by the transmitter and the

nature of the transmission medium, expressed in cycles per second, or Hertz

Noise: The average level of noise over the communications path

Error rate: The rate at which errors occur, where an error is the reception of a 1 when a 0 was

transmitted or the reception of a 0 when a 1 was transmitted

(4)

b) Define Shannon capacity formula.

Assume that the TV picture is to be transmitted over a channel with 4.5MHz bandwidth and a

35dB signal-to-noise ratio. Find the capacity of the channel(bps)

Definition(2)

Solution(3) For a given level of noise, we would expect that greater signal strength would improve the ability to

receive data correctly in the presence of noise. The key parameter involved in this reasoning is the signal-

to-noise ratio (SNR, or S/N), which is the ratio of the power in a signal to the power contained in the noise

that is present at a particular point in the transmission.

SNRdB = 10 log10

The signal-to-noise ratio is important in the transmission of digital data because it sets the upper bound on

the achievable data rate. Shannon’s result is that the maximum channel capacity, in bits per second, obeys

the equation, 1 We use the formula: C = B log2 (1 + SNR)

B = 4.5 × 106 MHz = bandwidth, and

SNRdB = 35 = 10 log10 (SNR),hence

SNR = 10 35/10 = 10 3.5,

and therefore C = 4.5 × 106 log2 (1 + 10 3.5)

= 4.5 × 106 × log2 (3163)

C = (4.5 × 106 × 11.63)

= 52.335 × 106 bps

(5)


7 a) List out the benefits of optical fiber compared to twisted pair and coaxial cable.

4 points each carries 1 mark • Greater capacity: The potential bandwidth, and hence data rate, of optical fiber is immense; data rates of

hundreds of Gbps over tens of kilometers have been demonstrated. Compare this to the practical

maximum of hundreds of Mbps over about 1 km for coaxial cable and just a few Mbps over 1 km or up to

100 Mbps to 10 Gbps over a few tens of meters for twisted pair.

• Smaller size and lighter weight: Optical fibers are considerably thinner than coaxial cable or bundled

twisted-pair cable—at least an order of magnitude thinner for comparable information transmission

capacity. For cramped conduits in buildings and underground along public rights-of-way, the advantage of

small size is considerable. The corresponding reduction in weight reduces structural support requirements.

• Lower attenuation: Attenuation is significantly lower for optical fiber than for coaxial cable or twisted

pair and is constant over a wide range.

• Electromagnetic isolation: Optical fiber systems are not affected by external electromagnetic fields.

Thus the system is not vulnerable to interference, impulse noise, or crosstalk. By the same token, fibers do

not radiate energy, so there is little interference with other equipment and there is a high degree of security

from eavesdropping. In addition, fiber is inherently difficult to tap.

• Greater repeater spacing: Fewer repeaters mean lower cost and fewer sources of error. The performance

of optical fiber systems from this point of view has been steadily improving. Repeater spacing in the tens

of kilometers for optical fiber is common, and repeater spacings of hundreds of kilometers have been

demonstrated. Coaxial and twisted-pair systems generally have repeaters every few kilometers.

(4)

b) Describe about Satellite Microwave.

Description (4)

Figure(1)

satellite is relay station

receives on one frequency, amplifies or repeats signal and transmits on another frequency

eg. uplink 5.925-6.425 GHz & downlink 3.7-4.2 GHz

typically requires geo-stationary orbit

height of 35,784km

spaced at least 3-4° apart

typical uses

television

long distance telephone

private business networks

(5)


global positioning

PART C Answer all questions, each carries 3 marks.

8 For the bit stream 01001100011 .Sketch the wave form using

a)NRZ-L b)NRZ-I

a)1.5 mark

b) 1.5 mark

(3)


9 State the advantages of SSB over DSBTC

SSB (1 mark)

Advantage(2 mrak) A popular variant of AM, known as single sideband (SSB), takes advantage of this fact by sending only

one of the sidebands, eliminating the other sideband and the carrier.

The principal advantages of this approach are as follows:

• Only half the bandwidth is required, that is, BT=B where B is the bandwidth of the original signal. For

DSBTC, BT=2B

• Less power is required because no power is used to transmit the carrier or the other sideband.

(3)

10 Describe Wavelength division multiplexing.

Description(3marks)

FDM with multiple beams of light at different freq

carried over optical fiber links

commercial systems with 160 channels of 10 Gbps

lab demo of 256 channels 39.8 Gbps

architecture similar to other FDM systems

multiplexer consolidates laser sources (1550nm) for transmission over single fiber

Optical amplifiers amplify all wavelengths

Demux separates channels at the destination

(3)

11 Illustrate the working of a cable modem

dedicate two cable TV channels to data transfer

each channel shared by number of subscribers, using statistical TDM

Downstream

cable scheduler delivers data in small packets

(3)


active subscribers share downstream capacity

also allocates upstream time slots to subscribers

Upstream

user requests timeslots on shared upstream channel

Headend scheduler notifies subscriber of slots to use

PART D Answer any two full questions, each carries 9 marks.

12 a) For the bit stream 01001110 , Sketch the waveform using

a) NRZ-I b) Bipolar-AMI c) Pseudoternary

Assume that the signal level for the preceding bit for NRZ-I was high, the most recent preceding

1 bit (AMI) has a negative voltage and the most recent preceding 0 bit (pesudoternary) has a

negative voltage.

a) 2 marks

b) 2 marks

c) 1 marks

(5)


b) Analyse the reasons for preferring DM over PCM for encoding analog signals that represent

digital data.

Reason(4 mark) Analog signals in the voice band that represent digital data have more high frequency components than

analog voice signals. These higher components cause the signal to change more rapidly over time. Hence,

DM will suffer from a high level of slope overload noise. PCM, on the other hand, does not estimate

changes in signals, but rather the absolute value of the signal, and is less affected than DM.

(4)

13 a) A signal is quantised using 10 bit PCM. Find the signal-to-quantization noise ratio.

Equation(1)

Solution(3) (SNR)db = 6.02 n + 1.76, where n is the number of bits used for quantization.

In this case, n=-10 bit

(SNR)db = 60.2 + 1.76

= 61.96 dB.

(4)

b) Given the bit pattern 01100 , encode this data using ASK,BFSK, and BPSK where frequency is

2Hz.

ASK(1)

BFSK(2)

BPSK(2)

(5)


14 a) Outline the concept of CDMA.

Description(5)

(5)

b) Describe about SONET/SDH

Figure(1)

Description(3)

Synchronous Optical Network (ANSI)

Synchronous Digital Hierarchy (ITU-T)

have hierarchy of signal rates

Synchronous Transport Signal level 1 (STS-1) or Optical Carrier level 1 (OC-1) is

51.84Mbps

carries one DS-3 or multiple (DS1 DS1C DS2) plus ITU-T rates (eg. 2.048Mbps)

multiple STS-1 combine into STS-N signal

ITU-T lowest rate is 155.52Mbps (STM-1) (Synchronous Transport Module)

(4)


PART E

Answer any four full questions, each carries 10 marks.

15 a) Differentiate Synchronous and Asynchronous Transmission

Asynchronous (3)and Synchronous Transmission (3)

timing problems require a mechanism to synchronize the transmitter and receiver

receiver samples stream at bit intervals

if clocks not aligned and drifting will sample at wrong time after sufficient bits are

sent

two solutions to synchronizing clocks

asynchronous transmission

synchronous transmission

(6)


Asynchronous – Behavior

simple

cheap

overhead of 2 or 3 bits per char (~20%)

good for data with large gaps (keyboard

Synchronous Transmission block of data transmitted sent as a frame clocks must be synchronized

can use separate clock line or embed clock signal in data

need to indicate start and end of block use preamble and postamble

more efficient (lower overhead) than async


b) Describe error detection process

Figure (1)

Description(3) Regardless of the design of the transmission system, there will be errors, resulting in the change of one or

more bits in a transmitted frame.

For a given frame of bits, additional bits that constitute an error-detecting code are added by the

transmitter. This code is calculated as a function of the other transmitted bits. Typically, for a data block

of k bits, the error-detecting algorithm yields an error-detecting code of bits, where The error-detecting

code, also referred to as the check bits, is appended to the data block to produce a frame of n bits, which is

then transmitted. The receiver separates the incoming frame into the k bits of data and (n-k) bits of the

error-detecting code. The receiver performs the same error-detecting calculation on the data bits and

compares this value with the value of the incoming error-detecting code. A detected error occurs if and

only if there is a mismatch. Thus P3 is the probability that a frame contains errors and that the error-

detecting scheme will detect that fact.P2 is known as the residual error rate and is the probability that an

error will be undetected despite the use of an error-detecting scheme.

(4)

16 a) For P= 110101, M= 1010001101, find CRC (7)


Finding bits ( 2)

Calculation(5)

b) Would you expect that the inclusion of a parity bit with each character would change the

probability of receiving a correct message? The inclusion of a parity bit extends the message length. There are more bits that can be in

(3)


error since the parity bit is now included. The parity bit may be in error when there are no errors in the

corresponding data bits. Therefore, the inclusion of a parity bit with each character would change the

probability of receiving a correct message.

17 a) Define hamming distance with example.

Definition (2)

Example(1) The Hamming distance d (v1,v2) between two n-bit binary sequences v1 and v2 is the number of bits in

which v1 and v2 disagree. For example, if

v1 = 011011 v2 = 110001

then

d(v1 , v2) = 3

(3)

b) Demonstrate block code principle with help of an example

Explanation(3)

Example(5) Suppose we wish to transmit blocks of data of length k bits. Instead of transmitting each block as k bits,

we map each k-bit sequence into a unique n-bit codeword.

The preceding example illustrates the essential properties of a block error correcting code. An (n, k) block

(7)


code encodes k data bits into n-bit codewords. Typically, each valid codeword reproduces the original k

data bits and adds to them (n-k) check bits to form the n-bit codeword. Thus the design of a block code is

equivalent to the design of a function of the form vc=f(vd) where vd is a vector of k data bits and vc is a

vector of n codeword bits.

18 a) List the three benefits of spread spectrum • The signals gains immunity from various kinds of noise and multipath distortion. The earliest

applications of spread spectrum were military, where it was used for its immunity to jamming.

• It can also be used for hiding and encrypting signals. Only a recipient who knows the spreading code can

recover the encoded information.

• Several users can independently use the same higher bandwidth with very little interference. This

property is used in cellular telephony applications, with a technique know as code division multiplexing

(CDM) or code division multiple access (CDMA).

(3)

b) Summarise the working of FHSS

Diagram(2)

Explanation(5)

(7)

19 a) Illustrate the process of DHSS

Diagram(2)

Explanation(4)

(6)

b) Describe the principles of packet switching network.

Figure (1)

Explanation(3)

Data are transmitted in short packets. A typical upper bound on packet length is 1000 octets (bytes). If a

source has a longer message to send, the message is broken up into a series of packets (Figure 10.8). Each

packet contains a portion (or all for a short message) of the user’s data plus some control information. The

control information, at a minimum, includes the information that the network requires to be able to route

the packet through the network and deliver it to the intended destination. At each node en route, the packet

is received, stored briefly, and passed on to the next node

(4)


Consider a packet to be sent from station A to station E.The packet includes control information that

indicates that the intended destination is E. The packet is sent from A to node 4. Node 4 stores the packet,

determines the next leg of the route (say 5), and queues the packet to go out on that link (the 4–5

link).When the link is available, the packet is transmitted to node 5, which forwards the packet to node 6,

and finally to E

20 a) Compare and contrast circuit switching and datagram packet switching (6)


b) What is the significance of packet size in a packet switching network?

Figure(1)

Explanation(3) There is a significant relationship between packet size and transmission time. As a smaller packet size is

used, there is a more efficient "pipelining" effect, as shown in Figure. However, if the packet size

becomes too small, then the transmission is less efficient.

(4)


****


cs307 model 3 - studoob.instudoob.in/app/q/cse/s5/cs307/cs307_model_3.pdf · b=fh-fl 20= 60 –fl...

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