T305:
Digital Communication
5.
Golay
code and Reed-Solomon codes are examples of _____________ that are commonly
used in communication system.
- convolutional error-correcting codes
- tree error-correcting codes
- parity error-correcting codes
- 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 ___________________.
- block coding
- Hamming coding
- convolutional coding
- 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
____________________.
- frequency preservation property
- orthogonal property
- principle of superposition
- 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 _____________.
- Type 1
- Type 2
- Type ¾
- 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 ____________________.
- binary frequency shift keying
- binary phase shift keying
- binary continuous-phase frequency shift keying
- binary amplitude shift keying
11.
Light
is an electromagnetic wave similar to a radio signal with a frequency
__________________.
- very much slower than frequency of a radio signal
- very much higher than frequency of a radio signal
- identical to the frequency of a radio signal
- 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 _____________.
- packet switching
- circuit switching
- network switching
- virtual switching
13.
Packets
can be transmitted through a packet-switched network as independent packets.
This kind of transmission is known as _________________.
- datagrams
- virtual circuits
- asynchronous
- synchronous
14.
The
most common addressing scheme used in internet is IP version 4. This IP version
is composed of __________.
- 24 bits
- 36 bits
- 32 bits
- 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
___________.
- 1011101
- 1010101
- 1000110
- 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 ___________.
- dynamic alternative routing
- least-cost routing
- flooding routing
- 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 ______________.
- connection store
- speech store
- buffer store
- 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 _______________________.
- narrower than that of a LED
- wider than that of a LED
- equal to that of a LED
- 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 __________.
- identical
- overlap
- similar
- 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 __________________.
- jitter
- aliasing
- fading
- 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 __________________.
- aliasing reflection
- fresnel reflection
- wireless reflection
- 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 ________________.
- Jaccop’s amplifiers
- copper amplifiers
- iron amplifiers
- 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 __________________.
- organize the multiplexing and switching of the cells
- split data into a number of octet chunks to fit the size of a cell and pass these to lower layer
- minimize the possibility of cells with corrupted headers being passed on to upper layer
- 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 ______________.
- queuing
- buffering
- lacking
- 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 ______________.
- a total delay = (K x L)
- a total delay= 2 (K x L)
- a total delay= 4 (K x L)
- no delay
28.
One
type of optical fiber is the ___________________ that is cheap, easy to use and
suitable for short-distance communication.
- glass fiber
- iron fiber
- copper fiber
- 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 __________________.
- code division multiplexing
- frequency division multiplexing
- amplitude division multiplexing
- 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
____________________.
- a decreasing power density for all frequencies
- an increasing power density for all frequencies
- a constant power density for all frequencies
- 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).
 |
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.
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.
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
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.
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.
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.
(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.
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).
 |
(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)
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
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
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.
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
-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.
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).
(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
(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.
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 _________________.
- home location register
- visitor location register
- equipment identity register
- 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 ______________.
- internal link blocking
- external link blocking
- input link blocking
- 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 ____________.
- the arrival process
- the service process
- the number of servers
- 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+x-2+ x-4
- 1+x-3+ x-5
- 1+x-2+ x-5
- 1+x-3+ x-4
5.
Golay
code and Reed-Solomon codes are examples of _____________ that are commonly
used in communication system.
- convolutional error-correcting codes
- tree error-correcting codes
- parity error-correcting codes
- 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 ___________________.
- block coding
- Hamming coding
- convolutional coding
- CRC coding
7.
 |
Given the following binary modulation scheme shown below:
The above modulation
scheme is an example of _____________________.
- phase shift keying
- frequency shift keying
- amplitude shift keying
- 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
____________________.
- frequency preservation property
- orthogonal property
- principle of superposition
- 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 _____________.
- Type 1
- Type 2
- Type ¾
- 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 ____________________.
- binary frequency shift keying
- binary phase shift keying
- binary continuous-phase frequency shift keying
- binary amplitude shift keying
11.
Light
is an electromagnetic wave similar to a radio signal with a frequency
__________________.
- very much slower than frequency of a radio signal
- very much higher than frequency of a radio signal
- identical to the frequency of a radio signal
- 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 _____________.
- packet switching
- circuit switching
- network switching
- virtual switching
13.
Packets
can be transmitted through a packet-switched network as independent packets.
This kind of transmission is known as _________________.
- datagrams
- virtual circuits
- asynchronous
- synchronous
14.
The
most common addressing scheme used in internet is IP version 4. This IP version
is composed of __________.
- 24 bits
- 36 bits
- 32 bits
- 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
___________.
- 1011101
- 1010101
- 1000110
- 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 ___________.
- dynamic alternative routing
- least-cost routing
- flooding routing
- 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 ______________.
- connection store
- speech store
- buffer store
- 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 _______________________.
- narrower than that of a LED
- wider than that of a LED
- equal to that of a LED
- 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 __________.
- identical
- overlap
- similar
- 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 __________________.
- jitter
- aliasing
- fading
- 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 __________________.
- aliasing reflection
- fresnel reflection
- wireless reflection
- 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 ________________.
- Jaccop’s amplifiers
- copper amplifiers
- iron amplifiers
- 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 __________________.
- organize the multiplexing and switching of the cells
- split data into a number of octet chunks to fit the size of a cell and pass these to lower layer
- minimize the possibility of cells with corrupted headers being passed on to upper layer
- 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 ______________.
- queuing
- buffering
- lacking
- 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
___________________.
- crosstalk
- aliasing
- base-band signal modulation
- 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 ______________.
- a total delay = (K x L)
- a total delay= 2 (K x L)
- a total delay= 4 (K x L)
- no delay
28.
One
type of optical fiber is the ___________________ that is cheap, easy to use and
suitable for short-distance communication.
- glass fiber
- iron fiber
- copper fiber
- 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 __________________.
- code division multiplexing
- frequency division multiplexing
- amplitude division multiplexing
- 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
____________________.
- a decreasing power density for all frequencies
- an increasing power density for all frequencies
- a constant power density for all frequencies
- 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).
|
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.
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.
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
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.
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.
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.
(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.
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).
|
(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)
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
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
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.
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
-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.
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).
(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
(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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