Digital Communication multiple choce objective questions



T305: Digital Communication


5.     Golay code and Reed-Solomon codes are examples of _____________ that are commonly used in communication system.

  1. convolutional error-correcting codes
  2. tree error-correcting codes
  3. parity error-correcting codes
  4. block error-correcting codes
6.     Viterbi decoding is one of the most commonly used technique in modern systems that is used to decode the data encoded by ___________________.

  1. block coding
  2. Hamming coding
  3. convolutional coding
  4. CRC coding

8.     In a linear system, if an input x1(t) produces an output y1(t), and an input x2(t) produces an output y2(t), then an input x1(t) + x2(t)  produces an output y1(t) + y2(t). This property of the linear system obeys ____________________.

  1. frequency preservation property
  2. orthogonal property
  3. principle of superposition
  4. amplification property

9.     Applications like Digitally coded voice signals, high quality audio signals and constant bit rate video signals are used in ATM network where the preservation of a constant bit-rate is the most significant. These applications are supported by the ATM Adaptation layer of _____________.

  1. Type 1
  2. Type 2
  3. Type ¾
  4. Type 5

10.  On-off keying is the modulation scheme used for the majority of optical-fiber communication systems. This scheme is an example of ____________________.

  1. binary frequency shift keying
  2. binary phase shift keying
  3. binary continuous-phase frequency shift keying
  4. binary amplitude shift keying

11.  Light is an electromagnetic wave similar to a radio signal with a frequency __________________.

  1. very much slower than frequency of a radio signal
  2. very much higher than frequency of a radio signal
  3. identical to the frequency of a radio signal
  4. very similar to the frequency of a radio signal
12.  When a specific physical transmission channel is provided for the exclusive and continuous use of each path through a network, this is called _____________.

  1. packet switching
  2. circuit switching
  3. network switching
  4. virtual switching

13.  Packets can be transmitted through a packet-switched network as independent packets. This kind of transmission is known as _________________.

  1. datagrams
  2. virtual circuits
  3. asynchronous
  4. synchronous

14.  The most common addressing scheme used in internet is IP version 4. This IP version is composed of __________.

  1. 24 bits
  2. 36 bits
  3. 32 bits
  4. 64 bits

15.  Differential coding produces an output in which the information is contained in differences between successive bits such that the output changes state if the input bit is a 1 otherwise the output remains the same. Assuming the initial condition for the output is 0, the differential coding for the input sequence 1100101 is ___________.

  1. 1011101
  2. 1010101
  3. 1000110
  4. 1111101

16.  The routing tables within the router are used for routing the packets. The need for routing tables can be avoided by using simple strategies. An example of such a simple strategy is ___________.

  1. dynamic alternative routing
  2. least-cost routing
  3. flooding routing
  4. static routing

17.  One way of implementing a time switch is to use two memory devices: one for storing the incoming data and the other stores the order in which the octets are sent to the outgoing lines. The memory device that is used for storing incoming data is called ______________.

  1. connection store
  2. speech store
  3. buffer store
  4. space store

18.  In optical fiber systems, there are two sorts of light sources: light emitting diodes (LEDs) and laser diodes.  The optical line-width of a laser diode is _______________________.

  1. narrower than that of a LED
  2. wider than that of a LED
  3. equal to that of a LED
  4. similar to that of a LED
19.  In a communication system, when two finite-power waveforms x (t) and y (t) having the properties:      <x y> =0 and < (x + y)2 >= <x>2 + <y>2 , then these waveforms are said to be __________.

  1. identical
  2. overlap
  3. similar
  4. orthogonal

20.  An important impairment to digital signals in a communication system is the irregularities in timing caused by imperfections in clock extraction and waveform regeneration. This effect is known as __________________.

  1. jitter
  2. aliasing
  3. fading
  4. attenuation

21.  When a gap exists in optical fiber cable, a reflection of the signal passing in the fiber will occur. This effect is know as __________________.

  1. aliasing reflection
  2. fresnel reflection
  3. wireless reflection
  4. glass reflection

22.  Optical amplifiers are devices which are used to amplify the signals passing in optical fiber cables. One of the technologies used for manufacturing optical amplifiers is ________________.

  1. Jaccop’s amplifiers
  2. copper amplifiers
  3. iron amplifiers
  4. Raman amplifiers

23.  In ATM network, each message is divided into a number of equal sized packets called cells. Each cell carries a number of octets equal to __________.

a.     32
b.    48
c.     53
d.    64

24.  In ATM network, the main task of the ATM layer is to   __________________.

  1. organize the multiplexing and switching of the cells
  2. split data into a number of octet chunks to fit the size of a cell and pass these to lower layer
  3. minimize the possibility of cells with corrupted headers being passed on to upper layer
  4. all of the above

25.  Because ATM is a packet-switched network, several cells destined for the same output could arrive from different inputs within the same cell time slot. The simultaneous arrival of more than one cell needing to be directed to the same output is known as ______________.

  1. queuing
  2. buffering
  3. lacking
  4. contention


27.  Interleaving is a technique used to allow error-correcting codes to protect against bursts. If we are interleaving K code words from a block error correcting code with code words of length L bits (using a matrix with dimensions K rows and L columns), this will introduce ______________.

  1. a total delay = (K x L)
  2. a total delay= 2 (K x L)
  3. a total delay= 4 (K x L)
  4. no delay

28.  One type of optical fiber is the ___________________ that is cheap, easy to use and suitable for short-distance communication.

  1. glass fiber
  2. iron fiber
  3. copper fiber
  4. plastic fiber

29.  In optical fiber, the data carrying capacity of a single fiber can be increased by a special type of multiplexing known as __________________.

  1. code division multiplexing
  2. frequency division multiplexing
  3. amplitude division multiplexing
  4. phase division multiplexing

30.  A useful spectral model of many types of noise encountered in communication systems is White noise. An important property of a White noise is that it has ____________________.

  1. a decreasing power density for all frequencies
  2. an increasing power density for all frequencies
  3. a constant power density for all frequencies
  4. an increasing power density for high frequencies only



Part II: Short essay questions (18 marks)

This part consists of 8 questions carrying a WEIGHT OF 3 marks each. You must answer only 6 (any) of the following questions (Suggested Time about 45 minutes).


31. 



With reference to the header of an ATM cell shown below, explain the purpose of the cell loss priority (CLP).

The cell loss priority (CLP) field is a single field that indicates whether a cell may be discarded or not in th event of network congestion.

For example, when traffic is policed to check for compliance with a traffic contract, non-compliant cells might be flagged rather than being discarded immediately.
32.  Explain the principle of operation of CRC error checking. Include in your answer an example using denary notation for the message M = 135792 and the generator G = 99.

At the encoder the message (M) is divided by a shorter known divisor (G)
135792/99 = 1371.6363

A remainder can be calculated by taking away from the message the largest number that can be divided exactly by the divisor.

135792 – 135729 = 63
The remainder is sent with the message to the decoder, where the remainder is recalculated.

If there are no errors, the remainder will be the same

If there are errors, the remainder is likely to be different.

For example, suppose the first digit is changed from 1 to 2, the remainder will be:
235792/99 = 2381.737
235792 – 2381 x 99 = 73
since 73 is different from 63 the receiver knows there must have been errors.

33.  What do the three parameters (M, M, and 1) indicate for a queue described by the expression M/M/1?
In M/M/1, the first M means that the arrivals are a Markov process, i.e., a poisson arrival distribution

The second M means that the service is also a Markov process: a Poisson/exponential service distribution

The 1 is the number of servers


34.  Explain advantages and disadvantages of using a flooding routing strategy.

Advantages:
    No routing tables are required at each node. A node only needs to know which links it has, not where the links are going.
    Very robust in cases of network failure.
    New links and nodes can be added to a network without any administrative overheads.

Disadvantages:
    For each packet sent, there can be many copies made and transmitted. This increases the traffic load on the links and nodes.
    Possible instability if a loop is formed which does not include the source or destination unless some additional control is imposed, such as a time-to-live counter.

35.  Explain the differences in the transfer of packets in a packet-switched network in which virtual circuits are used compared with one in which datagrams are used.

Virtual circuits must be set up before any data packets are transferred, whereas datagrams can be transferred without any prior exchange of protocol messages.

Virtual circuits allow greater control over the quality of service in the form of flow control, error control and sequence control.

Once a virtual circuit has been set up, switching (forwarding) is simpler, and this can lead to shorter delay compared with datagrams. Routing (forwarding) tables are smaller and simpler for virtual circuits (once a circuit has been set up) than for datagrams.

The same route is used between the source and destination for all data packets unless a fault occurs in a virtual circuit, but may vary for each datagram to balance the traffic loading of the network.

Datagrams contain the full addresses, but data packets carried by virtual circuits need to contain addresses of virtual circuits.

36.  Describe the role of any stores used in the incoming time switch, and how they are kept synchronized.

The incoming time switch has a speech store and a connection store. Incoming octets are stored in the speech store. The connection store is used to determine the order in which octets are read out of the speech store. Both stores are kept in synchronization using a counter.


37.  For each of the following modulation schemes briefly describe how it works, sketch a possible waveform for the sequence 1101.

(i)         On-off keying: A sinusoidal carrier is switched on (for a data 1) and off (for a data 0). Waveform similar to the following with different data sequence.



(ii)        Frequency shift keying: uses segments of sinusoids of different frequency. Waveform similar to the following with different data sequence.



(iii)      



Binary phase shift keying: uses two different phases of a segment of a sinusoidal waveform are used. Waveform similar to the following with different data sequence.


38.  What is meant by wavelength division multiplexing (WDM) in the context of optical fibre communication? Why is it used?

WDM increases the capacity of a single optical fibre by putting several channels on a single fibre, distinguishing them by operating each at a different wavelength (optical frequency).

It is used to get more data down a single fibre without having to operate at very high signaling rates (which is the alternative way of increasing the capacity of the fibre).



Part III: Numerical problems (36 marks)

This part consists of 8 questions carrying a WEIGHT of 6 marks each. You must answer only 6 (any) of the following questions (Suggested Time about 60 minutes).

39.  The figure below shows the encoder of a convolutional code. After each bit enters the encoder two code digits are transmitted to the channel.



 









Convolutional encoder


(a)   what is the rate and input constraint length for this code? Explain how you arrive at these values.
(b)   Draw a state diagram and trellis diagram for the code.
(c)   Assume that the encoder starts off with 0 in the storage element. What would the data 110100 be encoded to? (The first bit to be encoded is on the left)

(a) The rate is the ratio between the number of input bits in a frame (1) and the number of code digits transmitted per frame (2), so the rate is ½. The input constraint length is the number of bits involved in the encoding of any one frame, in this case 2.

(b)



(c)


40.  Part of a router in packet-switching system consists of a single-server queue with an infinite buffer. Packet arrivals can be assumed to be a Poisson process with an arrival rate l of 2000 packets per second. The service of 0.4ms is the same for all packets. Determine the mean waiting time in the buffer.

A deterministic system must be assumed with ts = 0.4ms and l= 2000 jobs per second
Thus r = l * ts = 2000 * 0.4* 10-3 = 0.8 and using equation 4.7 of the Reference Book


41.  A single-server queue has a Poisson arrival rate of 12 jobs per second and a negative exponential service rate of 20 jobs per second. The buffer can hold 8 jobs.
b)    Write out the Kendall notation for this queue.
c)     Calculate the probability of blocking.

a)     M/M/1/9
b)    ts = 1/20
A = l ts = 12/20 = 0.6



So A R+1 = 0.6 10 << 1
            = (1-0.6) 0.6 9 = 0.004

42.  Table below shows the parameters of an optical fibre communications link. Draw up a power budget to determine the maximum link length that can be used with this system.

Parameter
Value
Maximum launch power
–10 dBm
Receive sensitivity
–24 dBm
Allowance for connector and splice losses
1.5 dB
Maximum fibre attenuation
1.5 dB/km
Link power penalties
4 dB

The maximum loss allowed is 14 dB, calculated from
-10        dBm
-(-24)     dBm
14         dB
From this we subtract the allowance for connector and splice losses (1.5 dB) and the link power penalities (4dB), leaving 8.5 dB.

The fibre attenuation is 1.5 dB/km, so the allowed distance is 5.7 km.

43.  An M/M/1 queue is to be designed such that the probability that there should be at most five jobs waiting in the system (buffer + server) is 0.95. Determine the maximum acceptable level of server utilization for this queue.

Equation 4.13 from the Reference Book is:
P(number in system £ n) = 1 - rn+1
Substituting values
0.95 = 1 - r6
Rearranging
r6 = 1 - 0.95 = 0.05
taking logs
6 log r = log (0.05) = -1.3
log r = -0.217
r = 10-0.217 = 0.607


44.  A metallic cable transmits electrical signals in a way that approximates closely the ‘Öf model’ of attenuation. A sinusoidal 10 kHz signal with a transmitted amplitude of 6 V at the near end of the cable is found to be attenuated to 1.2 V over a distance of 10 km.
(i)             What is the attenuation of the cable in dB/km?
(ii)    If the frequency of the signal were increased to 160 kHz, with the near-end amplitude unchanged, what would be the received amplitude after traveling 10 km?

i)
attenuation over 10 km is given by 20 log (6/1.2) = 14 dB
that is 14/10 = 1.4 dB / km

ii)
the frequency has increased by a factor of 160/10 = 16, so the attenuation in dB/km increases by a factor of Ö16 = 4 to 1.4 * 4 = 5.6 dB/km, so the attenuation over 10 km is 10 * 5.6 = 56 dB. The voltage ratio is therefore 10 56/20 = 631, and the output voltage 6/631 = 9.5 x 10-3 V = 9.5 mV

45.  A signal at a level of –15 dBm has band-limited noise accompanying it in the range 40 to 60 kHz. The power density of the noise is 12 pW/ Hz. Estimate the Signal-to-Noise ratio as a ratio.

-15 dBm converts to a power level of 10 –1.5 mW = 0.0316 mW
The noise power is 12 *10-9 * 20 * 10 3 mW = 0.0024 mW

And thus the SNR is 131 or 21.2 dB

46.  A component has an exponential reliability function and its mean time to failure (MTTF) is 1000 hours.
(i)     Write down an expression for the reliability function, R(t), and calculate the probability that the component will have failed by 500 hours.
(ii)    A system consists of two of the components described above, connected in series from a reliability point of view. What is the probability that the system will have failed by 500 hours?

(i) The component has an exponential reliability function and its mean time to failure (MTTF) is 1000 hours, so R(t) = Exp(-t/1000) (where t is in hours).
The probability of surviving to 500 hours is therefore  R(500) = Exp(-500/1000) = 0.61
The probability of failing by 500 hours is 1-0.61 = 0.39.

(ii) For components with exponential reliability functions the failure rate is 1/MTTR, and when such components are connected in series their failure rates add.  So the reliability function for the system is given by: R(t) = Exp(-2t/1000)
The probability of surviving to 500 hours is therefore  R(500) = Exp(-2 x 500/1000) = 0.37
The probability of failing by 500 hours is 1-0.37 = 0.63.
Alternatively, using the result from part (ii), the probability of them both surviving is 0.612 and therefore the probability of at least one failing is 1 - 0.612 = 0.63.


Part IV:  Longer problems (16 marks)

This part consists of 2 questions carrying a WEIGHT of 8 marks each. You must answer ALL of the following questions (Suggested Time about 30 minutes).


47.  Figure below shows the links and link costs of a network.

(i)     Use Dijkstra’s algorithm to find the least-cost paths in the network, with node A as the source node. Your answer should clearly show the least-cost list, the candidate list and the cost.

(ii)    Construct a routing table based on the least-cost paths for node A.















(i) The following table can be constructed using the Dijkstra algorithm:

Path
Initial
1st
2nd
3rd
4th
5th
6th
Least –cost list
A
A
A
A
A
A
A



B
B
B
B
B




C
C
C
C





E
D
D






E
E







F
Candidate list

B
C
D
D
F



D
D
E
F




E
E
F



DAA
¥
0
0
0
0
0
0
DAB
¥
2
2
2
2
2
2
DAC
¥
¥
3
3
3
3
3
DAD
¥
8
8
5
5
5
5
DAE
¥
4
4
4
4
4
4
DAF
¥
¥
¥
7
6
6
6

(ii)
Routing table for A
Destination       Next node
A                      -
B                      B
C                      B
D                      B
E                      E
F                      E


48.  A base-band digital transmission system using metallic cables is to be upgraded so that the bit-rate is increased by a factor of 4. The noise at the receiver is assumed to be white and Gaussian, and the same for the original and the upgraded links. The average power transmitted over the upgraded links must remain the same.

(i)     Suppose that the original link operates by transmitting 5 V rectangular pulses which are attenuated to 50 m V at the receiver. The maximum length of the link for the given error specification is 4 km. What is the attenuation in decibels per kilometre?

(ii)    Estimate the attenuation in decibels per kilometre at the higher bit-rate.

(iii)   The energy in a rectangular pulse height of V volts and duration T seconds is V2 T. To maintain the same error rate, what must be the minimum height of the pulses received over the higher-speed link?

(iv)  What is the maximum length of the higher-speed link if the received pulses are to have a sufficient magnitude?

(a) The attenuation in decibels over 4 km is:

   20 Log (5 / 5x10-2 ) = 20 log 100 = 40.

 In other words, it is 10 dB km-1. 

(b) Increasing the bit-rate by a factor of 4 will increase all frequency components by a factor of 4. Since attenuation in decibels per kilometre in metallic cables is roughly proportional to the square root of frequency, the attenuation of the higher-speed link can be estimated at twice that of the original: 20 dB km-1.

(c) To maintain the same error rate with the noise characteristics unchanged, the pulse energy must remain the same at the higher bit rate. Yet at the higher rate the received pulse width, T, will have been reduced by a factor of 4 in comparison with the original. To maintain the pulse energy, V2 T, constant, the value of V2 must increase by a factor of 4. The received voltage, V, must therefore increase by a factor √4 = 2, to 100 mV or 0.1 V.

(d) An attenuation from 5 V to 0.1 V is 20 log (5/0.1) = 34 dB. At an attenuation of 20 dB km-1, the link therefore has a maximum length of 34/20 = 1.7 km. Increasing the bit-rate by a factor of 4 has thus decreased the maximum link length to 1.7/4 = 0.425 of its original value.


Spring 2006/2007


Answers-FoemA-Final-Part II

Part I: Multiple Choice questions (30 marks)

This part consists of 30 questions carrying a WEIGHT OF 1 mark each.


1.     In Mobile telecommunication systems, a number of databases are used to store information about mobile users. One of these databases contains data records about mobile stations that can be used to check for stolen mobile stations. This type of database is known as _________________.

  1. home location register
  2. visitor location register
  3. equipment identity register
  4. signaling level register

2.     The banyan network is an example of a blocking network. There are two types of blocking that occur in the banyan network: one is called the output port blocking and the other is called ______________.

  1. internal link blocking
  2. external link blocking
  3. input link blocking
  4. buffer link blocking

3.     A convenient way for specifying a queuing system is to use Kendall notation which takes the form: A/B/N/R. In Kendall notation, Symbol B is used to represent ____________.

  1. the arrival process
  2. the service process
  3. the number of servers
  4. the maximum number of jobs in the queue

4.    



Consider the following self-synchronizing scrambler:

      The tap polynomial that represents the above self-synchronizing scrambler is ____________

  1. 1+x-2+ x-4
  2. 1+x-3+ x-5
  3. 1+x-2+ x-5
  4. 1+x-3+ x-4

5.     Golay code and Reed-Solomon codes are examples of _____________ that are commonly used in communication system.

  1. convolutional error-correcting codes
  2. tree error-correcting codes
  3. parity error-correcting codes
  4. block error-correcting codes
6.     Viterbi decoding is one of the most commonly used technique in modern systems that is used to decode the data encoded by ___________________.

  1. block coding
  2. Hamming coding
  3. convolutional coding
  4. CRC coding

7.    



Given the following binary modulation scheme shown below:

The above modulation scheme is an example of _____________________.

  1. phase shift keying
  2. frequency shift keying
  3. amplitude shift keying
  4. continuous-phase frequency shift keying

8.     In a linear system, if an input x1(t) produces an output y1(t), and an input x2(t) produces an output y2(t), then an input x1(t) + x2(t)  produces an output y1(t) + y2(t). This property of the linear system obeys ____________________.

  1. frequency preservation property
  2. orthogonal property
  3. principle of superposition
  4. amplification property

9.     Applications like Digitally coded voice signals, high quality audio signals and constant bit rate video signals are used in ATM network where the preservation of a constant bit-rate is the most significant. These applications are supported by the ATM Adaptation layer of _____________.

  1. Type 1
  2. Type 2
  3. Type ¾
  4. Type 5

10.  On-off keying is the modulation scheme used for the majority of optical-fiber communication systems. This scheme is an example of ____________________.

  1. binary frequency shift keying
  2. binary phase shift keying
  3. binary continuous-phase frequency shift keying
  4. binary amplitude shift keying

11.  Light is an electromagnetic wave similar to a radio signal with a frequency __________________.

  1. very much slower than frequency of a radio signal
  2. very much higher than frequency of a radio signal
  3. identical to the frequency of a radio signal
  4. very similar to the frequency of a radio signal
12.  When a specific physical transmission channel is provided for the exclusive and continuous use of each path through a network, this is called _____________.

  1. packet switching
  2. circuit switching
  3. network switching
  4. virtual switching

13.  Packets can be transmitted through a packet-switched network as independent packets. This kind of transmission is known as _________________.

  1. datagrams
  2. virtual circuits
  3. asynchronous
  4. synchronous

14.  The most common addressing scheme used in internet is IP version 4. This IP version is composed of __________.

  1. 24 bits
  2. 36 bits
  3. 32 bits
  4. 64 bits

15.  Differential coding produces an output in which the information is contained in differences between successive bits such that the output changes state if the input bit is a 1 otherwise the output remains the same. Assuming the initial condition for the output is 0, the differential coding for the input sequence 1100101 is ___________.

  1. 1011101
  2. 1010101
  3. 1000110
  4. 1111101

16.  The routing tables within the router are used for routing the packets. The need for routing tables can be avoided by using simple strategies. An example of such a simple strategy is ___________.

  1. dynamic alternative routing
  2. least-cost routing
  3. flooding routing
  4. static routing

17.  One way of implementing a time switch is to use two memory devices: one for storing the incoming data and the other stores the order in which the octets are sent to the outgoing lines. The memory device that is used for storing incoming data is called ______________.

  1. connection store
  2. speech store
  3. buffer store
  4. space store

18.  In optical fiber systems, there are two sorts of light sources: light emitting diodes (LEDs) and laser diodes.  The optical line-width of a laser diode is _______________________.

  1. narrower than that of a LED
  2. wider than that of a LED
  3. equal to that of a LED
  4. similar to that of a LED
19.  In a communication system, when two finite-power waveforms x (t) and y (t) having the properties:      <x y> =0 and < (x + y)2 >= <x>2 + <y>2 , then these waveforms are said to be __________.

  1. identical
  2. overlap
  3. similar
  4. orthogonal

20.  An important impairment to digital signals in a communication system is the irregularities in timing caused by imperfections in clock extraction and waveform regeneration. This effect is known as __________________.

  1. jitter
  2. aliasing
  3. fading
  4. attenuation

21.  When a gap exists in optical fiber cable, a reflection of the signal passing in the fiber will occur. This effect is know as __________________.

  1. aliasing reflection
  2. fresnel reflection
  3. wireless reflection
  4. glass reflection

22.  Optical amplifiers are devices which are used to amplify the signals passing in optical fiber cables. One of the technologies used for manufacturing optical amplifiers is ________________.

  1. Jaccop’s amplifiers
  2. copper amplifiers
  3. iron amplifiers
  4. Raman amplifiers

23.  In ATM network, each message is divided into a number of equal sized packets called cells. Each cell carries a number of octets equal to __________.

a.     32
b.    48
c.     53
d.    64

24.  In ATM network, the main task of the ATM layer is to   __________________.

  1. organize the multiplexing and switching of the cells
  2. split data into a number of octet chunks to fit the size of a cell and pass these to lower layer
  3. minimize the possibility of cells with corrupted headers being passed on to upper layer
  4. all of the above

25.  Because ATM is a packet-switched network, several cells destined for the same output could arrive from different inputs within the same cell time slot. The simultaneous arrival of more than one cell needing to be directed to the same output is known as ______________.

  1. queuing
  2. buffering
  3. lacking
  4. contention


26. 



A metallic wire can be made dc blocking by adding a transformer or capacitor to prevent dc signals from passing which affects transmitted signals causing them to drift towards 0 V as shown in the figure below:

This effect is called ___________________.

  1. crosstalk
  2. aliasing
  3. base-band signal modulation
  4. baseline wander

27.  Interleaving is a technique used to allow error-correcting codes to protect against bursts. If we are interleaving K code words from a block error correcting code with code words of length L bits (using a matrix with dimensions K rows and L columns), this will introduce ______________.

  1. a total delay = (K x L)
  2. a total delay= 2 (K x L)
  3. a total delay= 4 (K x L)
  4. no delay

28.  One type of optical fiber is the ___________________ that is cheap, easy to use and suitable for short-distance communication.

  1. glass fiber
  2. iron fiber
  3. copper fiber
  4. plastic fiber

29.  In optical fiber, the data carrying capacity of a single fiber can be increased by a special type of multiplexing known as __________________.

  1. code division multiplexing
  2. frequency division multiplexing
  3. amplitude division multiplexing
  4. phase division multiplexing

30.  A useful spectral model of many types of noise encountered in communication systems is White noise. An important property of a White noise is that it has ____________________.

  1. a decreasing power density for all frequencies
  2. an increasing power density for all frequencies
  3. a constant power density for all frequencies
  4. an increasing power density for high frequencies only



Part II: Short essay questions (18 marks)

This part consists of 8 questions carrying a WEIGHT OF 3 marks each. You must answer only 6 (any) of the following questions (Suggested Time about 45 minutes).


31. 



With reference to the header of an ATM cell shown below, explain the purpose of the cell loss priority (CLP).

The cell loss priority (CLP) field is a single field that indicates whether a cell may be discarded or not in th event of network congestion.

For example, when traffic is policed to check for compliance with a traffic contract, non-compliant cells might be flagged rather than being discarded immediately.
32.  Explain the principle of operation of CRC error checking. Include in your answer an example using denary notation for the message M = 135792 and the generator G = 99.

At the encoder the message (M) is divided by a shorter known divisor (G)
135792/99 = 1371.6363

A remainder can be calculated by taking away from the message the largest number that can be divided exactly by the divisor.

135792 – 135729 = 63
The remainder is sent with the message to the decoder, where the remainder is recalculated.

If there are no errors, the remainder will be the same

If there are errors, the remainder is likely to be different.

For example, suppose the first digit is changed from 1 to 2, the remainder will be:
235792/99 = 2381.737
235792 – 2381 x 99 = 73
since 73 is different from 63 the receiver knows there must have been errors.

33.  What do the three parameters (M, M, and 1) indicate for a queue described by the expression M/M/1?
In M/M/1, the first M means that the arrivals are a Markov process, i.e., a poisson arrival distribution

The second M means that the service is also a Markov process: a Poisson/exponential service distribution

The 1 is the number of servers


34.  Explain advantages and disadvantages of using a flooding routing strategy.

Advantages:
    No routing tables are required at each node. A node only needs to know which links it has, not where the links are going.
    Very robust in cases of network failure.
    New links and nodes can be added to a network without any administrative overheads.

Disadvantages:
    For each packet sent, there can be many copies made and transmitted. This increases the traffic load on the links and nodes.
    Possible instability if a loop is formed which does not include the source or destination unless some additional control is imposed, such as a time-to-live counter.

35.  Explain the differences in the transfer of packets in a packet-switched network in which virtual circuits are used compared with one in which datagrams are used.

Virtual circuits must be set up before any data packets are transferred, whereas datagrams can be transferred without any prior exchange of protocol messages.

Virtual circuits allow greater control over the quality of service in the form of flow control, error control and sequence control.

Once a virtual circuit has been set up, switching (forwarding) is simpler, and this can lead to shorter delay compared with datagrams. Routing (forwarding) tables are smaller and simpler for virtual circuits (once a circuit has been set up) than for datagrams.

The same route is used between the source and destination for all data packets unless a fault occurs in a virtual circuit, but may vary for each datagram to balance the traffic loading of the network.

Datagrams contain the full addresses, but data packets carried by virtual circuits need to contain addresses of virtual circuits.

36.  Describe the role of any stores used in the incoming time switch, and how they are kept synchronized.

The incoming time switch has a speech store and a connection store. Incoming octets are stored in the speech store. The connection store is used to determine the order in which octets are read out of the speech store. Both stores are kept in synchronization using a counter.


37.  For each of the following modulation schemes briefly describe how it works, sketch a possible waveform for the sequence 1101.

(i)         On-off keying: A sinusoidal carrier is switched on (for a data 1) and off (for a data 0). Waveform similar to the following with different data sequence.




(ii)        Frequency shift keying: uses segments of sinusoids of different frequency. Waveform similar to the following with different data sequence.




(iii)      



Binary phase shift keying: uses two different phases of a segment of a sinusoidal waveform are used. Waveform similar to the following with different data sequence.


38.  What is meant by wavelength division multiplexing (WDM) in the context of optical fibre communication? Why is it used?

WDM increases the capacity of a single optical fibre by putting several channels on a single fibre, distinguishing them by operating each at a different wavelength (optical frequency).

It is used to get more data down a single fibre without having to operate at very high signaling rates (which is the alternative way of increasing the capacity of the fibre).




Part III: Numerical problems (36 marks)

This part consists of 8 questions carrying a WEIGHT of 6 marks each. You must answer only 6 (any) of the following questions (Suggested Time about 60 minutes).

39.  The figure below shows the encoder of a convolutional code. After each bit enters the encoder two code digits are transmitted to the channel.



 









Convolutional encoder


(a)   what is the rate and input constraint length for this code? Explain how you arrive at these values.
(b)   Draw a state diagram and trellis diagram for the code.
(c)   Assume that the encoder starts off with 0 in the storage element. What would the data 110100 be encoded to? (The first bit to be encoded is on the left)

(a) The rate is the ratio between the number of input bits in a frame (1) and the number of code digits transmitted per frame (2), so the rate is ½. The input constraint length is the number of bits involved in the encoding of any one frame, in this case 2.

(b)



(c)


40.  Part of a router in packet-switching system consists of a single-server queue with an infinite buffer. Packet arrivals can be assumed to be a Poisson process with an arrival rate l of 2000 packets per second. The service of 0.4ms is the same for all packets. Determine the mean waiting time in the buffer.

A deterministic system must be assumed with ts = 0.4ms and l= 2000 jobs per second
Thus r = l * ts = 2000 * 0.4* 10-3 = 0.8 and using equation 4.7 of the Reference Book


41.  A single-server queue has a Poisson arrival rate of 12 jobs per second and a negative exponential service rate of 20 jobs per second. The buffer can hold 8 jobs.
b)    Write out the Kendall notation for this queue.
c)     Calculate the probability of blocking.

a)     M/M/1/9
b)    ts = 1/20
A = l ts = 12/20 = 0.6



So A R+1 = 0.6 10 << 1
            = (1-0.6) 0.6 9 = 0.004

42.  Table below shows the parameters of an optical fibre communications link. Draw up a power budget to determine the maximum link length that can be used with this system.

Parameter
Value
Maximum launch power
–10 dBm
Receive sensitivity
–24 dBm
Allowance for connector and splice losses
1.5 dB
Maximum fibre attenuation
1.5 dB/km
Link power penalties
4 dB

The maximum loss allowed is 14 dB, calculated from
-10        dBm
-(-24)     dBm
14         dB
From this we subtract the allowance for connector and splice losses (1.5 dB) and the link power penalities (4dB), leaving 8.5 dB.

The fibre attenuation is 1.5 dB/km, so the allowed distance is 5.7 km.

43.  An M/M/1 queue is to be designed such that the probability that there should be at most five jobs waiting in the system (buffer + server) is 0.95. Determine the maximum acceptable level of server utilization for this queue.

Equation 4.13 from the Reference Book is:
P(number in system £ n) = 1 - rn+1
Substituting values
0.95 = 1 - r6
Rearranging
r6 = 1 - 0.95 = 0.05
taking logs
6 log r = log (0.05) = -1.3
log r = -0.217
r = 10-0.217 = 0.607


44.  A metallic cable transmits electrical signals in a way that approximates closely the ‘Öf model’ of attenuation. A sinusoidal 10 kHz signal with a transmitted amplitude of 6 V at the near end of the cable is found to be attenuated to 1.2 V over a distance of 10 km.
(i)             What is the attenuation of the cable in dB/km?
(ii)    If the frequency of the signal were increased to 160 kHz, with the near-end amplitude unchanged, what would be the received amplitude after traveling 10 km?

i)
attenuation over 10 km is given by 20 log (6/1.2) = 14 dB
that is 14/10 = 1.4 dB / km

ii)
the frequency has increased by a factor of 160/10 = 16, so the attenuation in dB/km increases by a factor of Ö16 = 4 to 1.4 * 4 = 5.6 dB/km, so the attenuation over 10 km is 10 * 5.6 = 56 dB. The voltage ratio is therefore 10 56/20 = 631, and the output voltage 6/631 = 9.5 x 10-3 V = 9.5 mV

45.  A signal at a level of –15 dBm has band-limited noise accompanying it in the range 40 to 60 kHz. The power density of the noise is 12 pW/ Hz. Estimate the Signal-to-Noise ratio as a ratio.

-15 dBm converts to a power level of 10 –1.5 mW = 0.0316 mW
The noise power is 12 *10-9 * 20 * 10 3 mW = 0.0024 mW

And thus the SNR is 131 or 21.2 dB

46.  A component has an exponential reliability function and its mean time to failure (MTTF) is 1000 hours.
(i)     Write down an expression for the reliability function, R(t), and calculate the probability that the component will have failed by 500 hours.
(ii)    A system consists of two of the components described above, connected in series from a reliability point of view. What is the probability that the system will have failed by 500 hours?

(i) The component has an exponential reliability function and its mean time to failure (MTTF) is 1000 hours, so R(t) = Exp(-t/1000) (where t is in hours).
The probability of surviving to 500 hours is therefore  R(500) = Exp(-500/1000) = 0.61
The probability of failing by 500 hours is 1-0.61 = 0.39.

(ii) For components with exponential reliability functions the failure rate is 1/MTTR, and when such components are connected in series their failure rates add.  So the reliability function for the system is given by: R(t) = Exp(-2t/1000)
The probability of surviving to 500 hours is therefore  R(500) = Exp(-2 x 500/1000) = 0.37
The probability of failing by 500 hours is 1-0.37 = 0.63.
Alternatively, using the result from part (ii), the probability of them both surviving is 0.612 and therefore the probability of at least one failing is 1 - 0.612 = 0.63.



Part IV:  Longer problems (16 marks)

This part consists of 2 questions carrying a WEIGHT of 8 marks each. You must answer ALL of the following questions (Suggested Time about 30 minutes).


47.  Figure below shows the links and link costs of a network.

(i)     Use Dijkstra’s algorithm to find the least-cost paths in the network, with node A as the source node. Your answer should clearly show the least-cost list, the candidate list and the cost.

(ii)    Construct a routing table based on the least-cost paths for node A.















(i) The following table can be constructed using the Dijkstra algorithm:

Path
Initial
1st
2nd
3rd
4th
5th
6th
Least –cost list
A
A
A
A
A
A
A



B
B
B
B
B




C
C
C
C





E
D
D






E
E







F
Candidate list

B
C
D
D
F



D
D
E
F




E
E
F



DAA
¥
0
0
0
0
0
0
DAB
¥
2
2
2
2
2
2
DAC
¥
¥
3
3
3
3
3
DAD
¥
8
8
5
5
5
5
DAE
¥
4
4
4
4
4
4
DAF
¥
¥
¥
7
6
6
6

(ii)
Routing table for A
Destination       Next node
A                      -
B                      B
C                      B
D                      B
E                      E
F                      E


48.  A base-band digital transmission system using metallic cables is to be upgraded so that the bit-rate is increased by a factor of 4. The noise at the receiver is assumed to be white and Gaussian, and the same for the original and the upgraded links. The average power transmitted over the upgraded links must remain the same.

(i)     Suppose that the original link operates by transmitting 5 V rectangular pulses which are attenuated to 50 m V at the receiver. The maximum length of the link for the given error specification is 4 km. What is the attenuation in decibels per kilometre?

(ii)    Estimate the attenuation in decibels per kilometre at the higher bit-rate.

(iii)   The energy in a rectangular pulse height of V volts and duration T seconds is V2 T. To maintain the same error rate, what must be the minimum height of the pulses received over the higher-speed link?

(iv)  What is the maximum length of the higher-speed link if the received pulses are to have a sufficient magnitude?

(a) The attenuation in decibels over 4 km is:

   20 Log (5 / 5x10-2 ) = 20 log 100 = 40.

 In other words, it is 10 dB km-1. 

(b) Increasing the bit-rate by a factor of 4 will increase all frequency components by a factor of 4. Since attenuation in decibels per kilometre in metallic cables is roughly proportional to the square root of frequency, the attenuation of the higher-speed link can be estimated at twice that of the original: 20 dB km-1.

(c) To maintain the same error rate with the noise characteristics unchanged, the pulse energy must remain the same at the higher bit rate. Yet at the higher rate the received pulse width, T, will have been reduced by a factor of 4 in comparison with the original. To maintain the pulse energy, V2 T, constant, the value of V2 must increase by a factor of 4. The received voltage, V, must therefore increase by a factor √4 = 2, to 100 mV or 0.1 V.

(d) An attenuation from 5 V to 0.1 V is 20 log (5/0.1) = 34 dB. At an attenuation of 20 dB km-1, the link therefore has a maximum length of 34/20 = 1.7 km. Increasing the bit-rate by a factor of 4 has thus decreased the maximum link length to 1.7/4 = 0.425 of its original value.


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