Examination Paper
Examination Session:
Exam Code: ENGI4507-WE01/ ENGI44G10-WE01
Title: (MEng/MSc)
Radio and Digital Communications 4 Time Allowed: 2 hours
Additional Material provided:
Yes, table included as part of exam paper
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Answer ALL questions.
Please answer each question on a new page.
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This is an online open-book exam, submission is time-limited to 4 hours within a 24 hrs window.
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Candidates are expected to answer all questions in their own words. Should close paraphrasing, direct quotation be necessary candidates must acknowledge the source of any material used in their answers (this includes text, images, diagrams, charts, tables and/or graphs). Full referencing is not expected, but a brief indication of the source of the material (e.g. name of author) must be given.
程序代写 CS代考 加QQ: 749389476
Page 2 of 4 ENGI4507/ENGI44G10
(a) An amplitude shift keying signal (ASK) is represented by
fm(t)=cosω1t 0 ≤t
Page 3 of 4 ENGI4507/ENGI44G10
Suppose that a satellite in geosynchronous orbit (36,000 km above the earth’s surface) radiates 100 W of power at a carrier frequency of 4 GHz and has a transmit antenna with 17 dBi gain. The earth station employs a parabolic antenna with a 3 m diameter and efficiency factor η=0.5.
(a) Given that for a parabolic antenna, the gain G is related to its diameter D and efficiency by
𝐺𝐺=𝜂𝜂(𝜋𝜋𝜋𝜋/𝜆𝜆)2
where 𝜆𝜆 is the wavelength
Determine the following:
(i) The effective radiated power in dBm
(ii) The earth station antenna gain in dBi
(iii) The free space path loss in dB for unit gain transmit and receive antennas (iv) The received power in dBm
[10%] [10%] [20%] [10%]
(b) Suppose the receiver has noise temperature To which is equal to 300o K. Given that the noise power density No = kBTo W/Hz where kB is Boltzman’s constant (1.38×10-23 W.s/K)
(i) Determine the noise power density in dBW/Hz [8%]
(ii) Determine the ratio of the received power to the noise power in dB [7%]
(iii) If the required energy per bit, 𝐸𝐸𝑏𝑏, to noise ratio is 10 dB, determine the data rate, R, that can be transmitted over the link. Use can be made of the following relationship
𝐸𝐸𝑏𝑏/𝑁𝑁𝑜𝑜=𝑃𝑃𝑅𝑅 /(𝑁𝑁𝑜𝑜𝑅𝑅)
(c) Discuss the concept of space division multiple access.
Page 4 of 4
Error function and the complementary error function
ENGI4507/ENGI44G10
erf c(𝑥𝑥) = 2 �∞𝑒𝑒−𝑢𝑢2 𝑑𝑑𝑑𝑑 √𝜋𝜋 𝑥𝑥
0.0659921 0.0477149 0.0338949 0.0236516 0.0162095 0.0109095 0.0072096 0.0046777 0.0029795 0.0018628 0.0011432 0.0006885 0.0004070 0.0002360 0.0001343 0.0000750 0.0000411 0.0000221 0.0000116 0.0000060 0.0000031 0.0000015 0.0000007
erf(𝑥𝑥) = 2 �𝑥𝑥𝑒𝑒−𝑢𝑢2 𝑑𝑑𝑑𝑑 √𝜋𝜋 0
Table of Error function and complementary error function
0.00 0.0000000 0.05 0.0563720 0.10 0.1124629 0.15 0.1679960 0.20 0.2227026 0.25 0.2763264 0.30 0.3286268 0.35 0.3793821 0.40 0.4283924 0.45 0.4754817 0.50 0.5204999 0.55 0.5633234 0.60 0.6038561 0.65 0.6420293 0.70 0.6778012 0.75 0.7111556 0.80 0.7421010 0.85 0.7706681 0.90 0.7969082 0.95 0.8208908 1.00 0.8427008 1.10 0.8802051 1.20 0.9103140
1.0000000 0.9436280 0.8875371 0.8320040 0.7772974 0.7236736 0.6713732 0.6206179 0.5716076 0.5245183 0.4795001 0.4366766 0.3961439 0.3579707 0.3221988 0.2888444 0.2578990 0.2293319 0.2030918 0.1791092 0.1572992 0.1197949 0.0896860
1.30 0.9340079 1.40 0.9522851 1.50 0.9661051 1.60 0.9763484 1.70 0.9837905 1.80 0.9890905 1.90 0.9927904 2.00 0.9953223 2.10 0.9970205 2.20 0.9981372 2.30 0.9988568 2.40 0.9993115 2.50 0.9995930 2.60 0.9997640 2.70 0.9998657 2.80 0.9999250 2.90 0.9999589 3.00 0.9999779 3.10 0.9999884 3.20 0.9999940 3.30 0.9999969 3.40 0.9999985 3.50 0.9999993
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