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通信考试代写-A28560
时间：2021-04-10
SCHOOL OF ENGINEERING
Examination Feedback
Session 2018/19
Module title : Satellite, Mobile and Optical Communications
Assessment Code: A28560
Module Code:
Level: M
Duration 3: hours
General Comments:
Students generally found this hard. Some were well prepared for some but not
all of the three areas covered in the module.
Comments on individual questions
Question No : 1
Outline answer:
Part (a) (i). Candidates were expected to give a brief explanation of spatial diversity
based on what they learnt in the final lecture in the mobile comms section, and to
explain its importance in mitigating dep fades in a multipath environment.
Part (a)(ii). Candidates were expected to describe in their own words, and if
preferred using diagrams, selection combining, equal gain combining and maximal
ratio combining, and to explain their advantages (ie their relative performance) and
their disadvantages(mainly in terms of relative cost and complexity.) This was
covered in the final lecture in the mobile comms section.
Part (b)(i). The hint suggested looking for an economic aspect and a system
performance aspect. For the economic aspect you could mention the high cost of
providing highly stable oscillators (eg atomic clocks), which can be done
economically in a base station serving multiple users, but not in individual
handsets. For the performance aspect, you could mention the need for the
oscillators in the MS and the BS to remain in tune to stay in tune to avoid errors in
demodulation.
Part (b) (ii). See the lecture notes outlining the structure and purpose of the GSM
bursts. The key point arises from the fact that the modulation is GMSK, a form of
FSK.
Part (b)(iii). In this question candidates were expected to analyse the information
given and use it to work out how far the phase and frequency would drift before the
next FCCH burst arrived to synchronise the oscillator. If the phase drift was not
<<π/4 then there could be a problem. One way to approach this is to work out the
drift rate that would cause a problem. Assuming that the frequency drifts linearly
and monotonically during this time at a rate df/dt, than the phase error building up
between corrections is
.
To make the excess phase shift during one bit of an MSK waveform , we
need
Modules A and B therefore appear to be fit for the purpose. Since we are looking for
a low cost solution, other things such as reliability being assumed equal, we would
choose module B.
General comments:
Sections 1(a)(i) to 1(b)(ii) were designed to test knowledge and understanding
of some features of mobile comms in general and GSM in particular. Section
(b)(iii) required candidates to think carefully about the problem using their
knowledge of GSM, and analyse given information to arrive at a suitable answer
to a design question taking costs into account.
Marks were lost for:
In part (a), not reading the question properly or not understanding it. Some
candidates discussed multiple access schemes or GSM frame structures
instead of diversity.
Some missed out aspects of the question. For example, in part (a)(ii), listing the
diversity combining methods but nor describing them (only a very brief
description was expected, for the marks allocated.) For another example,
candidates did not give advantages and/or disadvantages of each type.
In part (b)(iii) some candidates made up overcomplicated and incorrect
methods, using a figure of merit based on cost per drift or similar. This does not
help, because the most expensive one is clearly going to come out best in this
case. Like many engineering design calculations, this problem required you to
work out a binary decision regarding each competing module, and then you
choose the cheapest one that does the job.
Most students could have improved their mark had they:
• Read the question more carefully
• Answered the question that was asked, not the one they had
hoped would be asked.
Question No : 2
Model answer:
Part (a). Candidates were expected to describe FDMA, TDMA, CDMA and
SDMA, and to briefly list at least one advantage and disadvantage of each.
Part (b)(i). This has been covered extensively in lecture examples.
Ru is the long side of the triangle formed by shorter sides of length
and with internal angle 120o between them. Candidates were
expected to include a diagram, and then use the cosine rule (twice) to show that :
Part (b)(ii)
Part (c)(i)
From previous parts, the total area of a cluster is given by
Note that the number of cells per cluster cancels out when you put in an expression
for R in terms of Ru. Within the cluster the number of users supported is
8 x 124 = 992 (strictly 124 frequencies, not 125, because the outer 100 kHz on
each side is normally a guard band).
Hence the density is
Part (c)(ii) The number of cells per cluster is a design decision and a compromise. A
large number means that the powers transmitted by the handset and the base
station can be kept low, as each cell is small. On the other hand, a larger number of
base stations is then required, increasing the system infrastructure cost, and the
control overhead is increased because handovers will occur more frequently as
users cross cell boundaries.
When a call is initiated, it is necessary for the base station to determine the time
delay between the MS and the BS. The timing correction limit of 100 μs sets the
maximum cell size at R = 30 km.
From 2(b)(i), if we initially take R = 30 km, then
The next biggest whole number of cells achievable is 4, with i=2 and j=0.
Rearranging to find the actual value of R gives:
General comments:
This question tested fundamental knowledge and understanding of some
principles of cellular comms, and required that understanding to be applied to a
simple design problem.
Marks were lost for:
In part (a), not reading the question properly or not understanding it. Some
candidates discussed multiple access schemes or GSM infrastructures instead
of multiple access. Also, simply listing TDMA, FDMA, CDMA and SDMA,
without responding to the part of the question that said “Briefly describe how
they work” and “explain their relative advantages and disadvantages.
Despite feedback given on this point against last year’s paper, some candidates
included diagrams that were far too small. Being an engineer is often about
communicating ideas clearly. Exam answer books are free. Use as much
space as is necessary to get your message across clearly for old examiners
with poor eyesight!
Not completing part (c)(i), apparently because of missing that the number of
cells per cluster could be cancelled.
Not attempting part (c)(ii).
Most students could have improved their mark had they:
Read the question more carefully and attempted all parts.
Making sure in a question asking for verbal descriptions or explanations that all
aspects are covered: in 2(a)(i): List….., Describe how they work…..Explain
advantages and disadvantages.
Question No : 3
Model answer:
(a)(i)
2 ( ) / 7.4633 km/ss ev r h Tπ= + =
(ii) The key to success here is to sketch the satellite- observer system as shown below
where S is the satellite point, O is the observation point
The component of velocity toward the observer is the radial velocity toward the receiver
cosr sv v θ=
Therefore
6.649 km/ser s
e
rv v
r +h
= =
(iii)
cos( ) 35.5 kHzrd
vvf θ
λ λ
= = =
(iv)
The result of (iii) demonstrates very large Doppler shift with regards to typically narrow
bandwidth of such satellites as Iridium, which has about 8 MHz operating bandwidth
and channels, which occupies a bandwidth of 31.5 kHz, need to be spaced by 41.7 kHz
to allow space for Doppler shifts.
Fast tracking receivers are one solution.
(This answer was not required to be illustrated with numbers, but basic
explanation is expected)
Part (b)
(i)
( )(dB) dB 10log( );G G T
T
= −
22 9
8
3.14 30 11 100.75 8947587,
3 10
G(dB)=70 (dB);
G/T(dB/K)=70-10log (80)=70-20 =50 (dB/K)
A
DG πη
λ
⋅ ⋅ ⋅ = = = ⋅
ii)
( )210 log 10log(4 ) 20log( );4
( ) ( ) ( )
t t
t t
PGFD EIRP dB d
d
EIRP dB P dB G dB
π
π
= = − −
= +
For 15 W of the output power Pt(dB)=12 dBW; Therefore EIRP=12+35=47 dBW.
re
re d
h
O
S
vs
θ
θ
( )2
6 2
10 log 10log(4 ) 20log( )
4
47 11 20 log(38.5 10 ) 36 151 115 dBW/m ;
t tPGFD EIRP dB R
R
π
π
= = − − =
− − ⋅ ⋅ ≈ − = −
General comments:
Most students did fairly well on this paper unless they had fundamental
confusions over how to deal with dB and linear units.
Part (a)Students who drew the diagram correctly mostly got this right.
Marks were lost for:
Not visualising the question clearly and drawing diagram.to help visualise the
problem.
Confusion with dB and linear units.
Most students could have improved their mark had they:
Drawn a clear diagram for part (a) and been careful with basic trigonometry.
Practised more beforehand in the use of dB units and linear units,
understanding how to handle them, combine them and convert from one to the
other correctly.
Question No : 4
Model answer:
a)
Rayleigh scattering. See the lecture notes for a detailed graph with fibre loss vs.
wavelength considering intrinsic and extrinsic loss mechanisms.
b)
The loss (per unit length) due to absorption and scattering can be extracted
directly from the straight section BC. With this info, one can extract the loss due
to radiation for the bend sections.
The total loss between AB will be the absorption and scattering loss per unit
length multiplied by the total length of the bend + the loss due to radiation.
c)
The student were expected to know that loss is due to reflection, and this can be
quantified via the Fresnel reflection coefficient. Notice that the broken link
creates two interfaces: fibre- air & air-fibre.
d) Eight different advantages are listed in the introductory section of Opt.
Comm. Systems.
General comments:
The question was designed primarily to test the understanding of students
regarding the different loss mechanisms in an optical communication system.
This question was attempted by less than half of the students. The average
mark was better than Q5 (attempted by all, but one student), but worse than Q6.
Marks were lost for:
misunderstanding the question (e.g. loss vs. dispersion), omitting the fact that a
broken fibre leads to two interfaces with associated Fresnel reflection losses
and not being able to articulate clearly the advantages of an optical fibre
(technical and economical).
Most students could have improved their mark had they:
Read the question more carefully
Question No : 5
Model answer:
a)
Knowing the minimum number of photons, one can calculate the minimum
power required at the receiver end as done in the problem sheet (with
solutions).
From power budget analysis (Power_source (dB) – attenuation in the fibre (dB)
= Power_receiver), one can estimate then straightforwardly the maximum length
of the fibre.
b)
Knowing that the first window corresponds to ~800 nm, then it is evident from
the graph the only material operating at such wavelength.
c) See lecture notes.
d) See lecture notes.
General comments:
The question was designed to test the understanding of students regarding: (i)
power budget analysis, (ii) optical comm. system terminology (first window ~
800 nm), and (iii) operation mechanisms of detector and modulators. It also
tests the ability of the students to interpret information represented in (seen)
standard plots.
All, but one student attempted this question. The average marks was the lowest
among Q4-Q6.
Marks were lost for:
omitting the key information provided (first window = 800 nm operation
wavelength) to tackle part (a) and understand the graph in part (b), missing the
circuit aspect in part (c), answering something completely different for part (d)
Most students could have improved their mark had they:
Read the question more carefully and understand all info provided.
Reflected on the meaning of the different graphs seen in class.
Question No : 6
Model answer:
a)
See lecture notes. A detailed block diagram is used at the beginning of each
block (optical fibre, transmitter, receiver…) to highlight the aspect of the optical
communication system to be discussed.
b)
The solution is 1 ps/km, which can be calculated easily from the given dispersion
parameter and the laser spectral linewidth.
c)
Inputting the info given and the calculated chromatic dispersion from part (b) in
the equation provided, leads to 280 km
d)
The key here is to explicitly state the relationship between the electrical and
optical bandwidth: f3dB,elect = (sqrt(2)/2)* f3dB,opt
General comments:
The question was designed to test the student’s understanding of an optical
communication system from a system point of view. It also probed the student’s
understanding regarding fibre dispersion.
This question was attempted by ~75% of students, obtaining the highest average
mark with significant difference compared to Q4 and Q5.
Marks were lost for:
overcomplicating the calculation of the chromatic dispersion and omitting the
relationship between electrical and optical bandwidth.
Most students could have improved their mark had they:
Read the question more carefully and understand all info provided.
Exploited the units of the different parameters to do the maths.
学霸联盟
Examination Feedback
Session 2018/19
Module title : Satellite, Mobile and Optical Communications
Assessment Code: A28560
Module Code:
Level: M
Duration 3: hours
General Comments:
Students generally found this hard. Some were well prepared for some but not
all of the three areas covered in the module.
Comments on individual questions
Question No : 1
Outline answer:
Part (a) (i). Candidates were expected to give a brief explanation of spatial diversity
based on what they learnt in the final lecture in the mobile comms section, and to
explain its importance in mitigating dep fades in a multipath environment.
Part (a)(ii). Candidates were expected to describe in their own words, and if
preferred using diagrams, selection combining, equal gain combining and maximal
ratio combining, and to explain their advantages (ie their relative performance) and
their disadvantages(mainly in terms of relative cost and complexity.) This was
covered in the final lecture in the mobile comms section.
Part (b)(i). The hint suggested looking for an economic aspect and a system
performance aspect. For the economic aspect you could mention the high cost of
providing highly stable oscillators (eg atomic clocks), which can be done
economically in a base station serving multiple users, but not in individual
handsets. For the performance aspect, you could mention the need for the
oscillators in the MS and the BS to remain in tune to stay in tune to avoid errors in
demodulation.
Part (b) (ii). See the lecture notes outlining the structure and purpose of the GSM
bursts. The key point arises from the fact that the modulation is GMSK, a form of
FSK.
Part (b)(iii). In this question candidates were expected to analyse the information
given and use it to work out how far the phase and frequency would drift before the
next FCCH burst arrived to synchronise the oscillator. If the phase drift was not
<<π/4 then there could be a problem. One way to approach this is to work out the
drift rate that would cause a problem. Assuming that the frequency drifts linearly
and monotonically during this time at a rate df/dt, than the phase error building up
between corrections is
.
To make the excess phase shift during one bit of an MSK waveform , we
need
Modules A and B therefore appear to be fit for the purpose. Since we are looking for
a low cost solution, other things such as reliability being assumed equal, we would
choose module B.
General comments:
Sections 1(a)(i) to 1(b)(ii) were designed to test knowledge and understanding
of some features of mobile comms in general and GSM in particular. Section
(b)(iii) required candidates to think carefully about the problem using their
knowledge of GSM, and analyse given information to arrive at a suitable answer
to a design question taking costs into account.
Marks were lost for:
In part (a), not reading the question properly or not understanding it. Some
candidates discussed multiple access schemes or GSM frame structures
instead of diversity.
Some missed out aspects of the question. For example, in part (a)(ii), listing the
diversity combining methods but nor describing them (only a very brief
description was expected, for the marks allocated.) For another example,
candidates did not give advantages and/or disadvantages of each type.
In part (b)(iii) some candidates made up overcomplicated and incorrect
methods, using a figure of merit based on cost per drift or similar. This does not
help, because the most expensive one is clearly going to come out best in this
case. Like many engineering design calculations, this problem required you to
work out a binary decision regarding each competing module, and then you
choose the cheapest one that does the job.
Most students could have improved their mark had they:
• Read the question more carefully
• Answered the question that was asked, not the one they had
hoped would be asked.
Question No : 2
Model answer:
Part (a). Candidates were expected to describe FDMA, TDMA, CDMA and
SDMA, and to briefly list at least one advantage and disadvantage of each.
Part (b)(i). This has been covered extensively in lecture examples.
Ru is the long side of the triangle formed by shorter sides of length
and with internal angle 120o between them. Candidates were
expected to include a diagram, and then use the cosine rule (twice) to show that :
Part (b)(ii)
Part (c)(i)
From previous parts, the total area of a cluster is given by
Note that the number of cells per cluster cancels out when you put in an expression
for R in terms of Ru. Within the cluster the number of users supported is
8 x 124 = 992 (strictly 124 frequencies, not 125, because the outer 100 kHz on
each side is normally a guard band).
Hence the density is
Part (c)(ii) The number of cells per cluster is a design decision and a compromise. A
large number means that the powers transmitted by the handset and the base
station can be kept low, as each cell is small. On the other hand, a larger number of
base stations is then required, increasing the system infrastructure cost, and the
control overhead is increased because handovers will occur more frequently as
users cross cell boundaries.
When a call is initiated, it is necessary for the base station to determine the time
delay between the MS and the BS. The timing correction limit of 100 μs sets the
maximum cell size at R = 30 km.
From 2(b)(i), if we initially take R = 30 km, then
The next biggest whole number of cells achievable is 4, with i=2 and j=0.
Rearranging to find the actual value of R gives:
General comments:
This question tested fundamental knowledge and understanding of some
principles of cellular comms, and required that understanding to be applied to a
simple design problem.
Marks were lost for:
In part (a), not reading the question properly or not understanding it. Some
candidates discussed multiple access schemes or GSM infrastructures instead
of multiple access. Also, simply listing TDMA, FDMA, CDMA and SDMA,
without responding to the part of the question that said “Briefly describe how
they work” and “explain their relative advantages and disadvantages.
Despite feedback given on this point against last year’s paper, some candidates
included diagrams that were far too small. Being an engineer is often about
communicating ideas clearly. Exam answer books are free. Use as much
space as is necessary to get your message across clearly for old examiners
with poor eyesight!
Not completing part (c)(i), apparently because of missing that the number of
cells per cluster could be cancelled.
Not attempting part (c)(ii).
Most students could have improved their mark had they:
Read the question more carefully and attempted all parts.
Making sure in a question asking for verbal descriptions or explanations that all
aspects are covered: in 2(a)(i): List….., Describe how they work…..Explain
advantages and disadvantages.
Question No : 3
Model answer:
(a)(i)
2 ( ) / 7.4633 km/ss ev r h Tπ= + =
(ii) The key to success here is to sketch the satellite- observer system as shown below
where S is the satellite point, O is the observation point
The component of velocity toward the observer is the radial velocity toward the receiver
cosr sv v θ=
Therefore
6.649 km/ser s
e
rv v
r +h
= =
(iii)
cos( ) 35.5 kHzrd
vvf θ
λ λ
= = =
(iv)
The result of (iii) demonstrates very large Doppler shift with regards to typically narrow
bandwidth of such satellites as Iridium, which has about 8 MHz operating bandwidth
and channels, which occupies a bandwidth of 31.5 kHz, need to be spaced by 41.7 kHz
to allow space for Doppler shifts.
Fast tracking receivers are one solution.
(This answer was not required to be illustrated with numbers, but basic
explanation is expected)
Part (b)
(i)
( )(dB) dB 10log( );G G T
T
= −
22 9
8
3.14 30 11 100.75 8947587,
3 10
G(dB)=70 (dB);
G/T(dB/K)=70-10log (80)=70-20 =50 (dB/K)
A
DG πη
λ
⋅ ⋅ ⋅ = = = ⋅
ii)
( )210 log 10log(4 ) 20log( );4
( ) ( ) ( )
t t
t t
PGFD EIRP dB d
d
EIRP dB P dB G dB
π
π
= = − −
= +
For 15 W of the output power Pt(dB)=12 dBW; Therefore EIRP=12+35=47 dBW.
re
re d
h
O
S
vs
θ
θ
( )2
6 2
10 log 10log(4 ) 20log( )
4
47 11 20 log(38.5 10 ) 36 151 115 dBW/m ;
t tPGFD EIRP dB R
R
π
π
= = − − =
− − ⋅ ⋅ ≈ − = −
General comments:
Most students did fairly well on this paper unless they had fundamental
confusions over how to deal with dB and linear units.
Part (a)Students who drew the diagram correctly mostly got this right.
Marks were lost for:
Not visualising the question clearly and drawing diagram.to help visualise the
problem.
Confusion with dB and linear units.
Most students could have improved their mark had they:
Drawn a clear diagram for part (a) and been careful with basic trigonometry.
Practised more beforehand in the use of dB units and linear units,
understanding how to handle them, combine them and convert from one to the
other correctly.
Question No : 4
Model answer:
a)
Rayleigh scattering. See the lecture notes for a detailed graph with fibre loss vs.
wavelength considering intrinsic and extrinsic loss mechanisms.
b)
The loss (per unit length) due to absorption and scattering can be extracted
directly from the straight section BC. With this info, one can extract the loss due
to radiation for the bend sections.
The total loss between AB will be the absorption and scattering loss per unit
length multiplied by the total length of the bend + the loss due to radiation.
c)
The student were expected to know that loss is due to reflection, and this can be
quantified via the Fresnel reflection coefficient. Notice that the broken link
creates two interfaces: fibre- air & air-fibre.
d) Eight different advantages are listed in the introductory section of Opt.
Comm. Systems.
General comments:
The question was designed primarily to test the understanding of students
regarding the different loss mechanisms in an optical communication system.
This question was attempted by less than half of the students. The average
mark was better than Q5 (attempted by all, but one student), but worse than Q6.
Marks were lost for:
misunderstanding the question (e.g. loss vs. dispersion), omitting the fact that a
broken fibre leads to two interfaces with associated Fresnel reflection losses
and not being able to articulate clearly the advantages of an optical fibre
(technical and economical).
Most students could have improved their mark had they:
Read the question more carefully
Question No : 5
Model answer:
a)
Knowing the minimum number of photons, one can calculate the minimum
power required at the receiver end as done in the problem sheet (with
solutions).
From power budget analysis (Power_source (dB) – attenuation in the fibre (dB)
= Power_receiver), one can estimate then straightforwardly the maximum length
of the fibre.
b)
Knowing that the first window corresponds to ~800 nm, then it is evident from
the graph the only material operating at such wavelength.
c) See lecture notes.
d) See lecture notes.
General comments:
The question was designed to test the understanding of students regarding: (i)
power budget analysis, (ii) optical comm. system terminology (first window ~
800 nm), and (iii) operation mechanisms of detector and modulators. It also
tests the ability of the students to interpret information represented in (seen)
standard plots.
All, but one student attempted this question. The average marks was the lowest
among Q4-Q6.
Marks were lost for:
omitting the key information provided (first window = 800 nm operation
wavelength) to tackle part (a) and understand the graph in part (b), missing the
circuit aspect in part (c), answering something completely different for part (d)
Most students could have improved their mark had they:
Read the question more carefully and understand all info provided.
Reflected on the meaning of the different graphs seen in class.
Question No : 6
Model answer:
a)
See lecture notes. A detailed block diagram is used at the beginning of each
block (optical fibre, transmitter, receiver…) to highlight the aspect of the optical
communication system to be discussed.
b)
The solution is 1 ps/km, which can be calculated easily from the given dispersion
parameter and the laser spectral linewidth.
c)
Inputting the info given and the calculated chromatic dispersion from part (b) in
the equation provided, leads to 280 km
d)
The key here is to explicitly state the relationship between the electrical and
optical bandwidth: f3dB,elect = (sqrt(2)/2)* f3dB,opt
General comments:
The question was designed to test the student’s understanding of an optical
communication system from a system point of view. It also probed the student’s
understanding regarding fibre dispersion.
This question was attempted by ~75% of students, obtaining the highest average
mark with significant difference compared to Q4 and Q5.
Marks were lost for:
overcomplicating the calculation of the chromatic dispersion and omitting the
relationship between electrical and optical bandwidth.
Most students could have improved their mark had they:
Read the question more carefully and understand all info provided.
Exploited the units of the different parameters to do the maths.
学霸联盟
