CS6311 Sample Exam Questions

CS6311 Sample Exam Questions
(NB: These are provided as an aid to help you study for this course. They are not
intended to be comprehensive or to provide a guide to topics that will be included in
or omitted from examinations).
Signals
1. Draw a diagram to illustrate frequency, phase and amplitude. Explain the
relationships between (i) frequency and period and (ii) frequency and
wavelength.
2. If a signal has a frequency of 1 MHz what is its period? Assuming speed of
light (3x108 meters/second), what is its wavelength?
3. Using Shannon’s Equation, calculate the capacity for a channel between 8
MHz and 10 MHz, with an SNR=30dB.
4. For a channel bandwidth of 2 MHz and a capacity of 8 MB/s, calculate the
number of signaling levels required.
5. Given a channel with an intended capacity of 21 Mb/s and channel bandwidth
of 3 MHz. What SNR is required to achieve this capacity?
Antennas & Propagation
6. What is an Isotropic antenna?
7. Define the term “Antenna Gain”.
8. Determine the Gain for a horn antenna, assuming that its effective area is 0.81
x π m2 and the wavelength is 30cm.
9. Determine the height of an antenna that must be able to reach customers 80
km away.
10. Calculate the effective line of sight (in kilometres) for an antenna with height
10 metres.
11. In a case of two antennas with respective heights of 10m and 100m calculate
the effective line of sight in kilometres.
12. Show that doubling the transmission frequency or doubling the distance
between transmitting antenna and receiving antenna attenuates the power
received by 6dB.
13. Calculate the free space loss on a 3 GHz satellite link over a distance of
35000km and speed of light propagation at 3x108 m/s.
14. For the previous question, calculate the free space loss if the transmitter and
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receiver antenna gains are 35db and 45db, respectively.
15. Explain and compare the terms Signal-to-Noise Ratio (SNR) and Bit Error
Rate (BER).
16. Explain the relationship between Bit Error Rate and the E b/No.
17. Explain using a diagram the three multipath propagation effects of reflection,
diffraction and scattering.
18. Explain how inter-symbol interference happens.
19. What causes fading? Explain the difference between short-term and long-term
fading.
20. Explain the concept of ranges for transmission, detection and interference,
contrasting with the situation when using a wired network.
Multiplexing
21. Using diagrams, define the terms SDM, TDM. FDM and CDM.
22. Define the acronym OFDM. Explain how OFDM can achieve high data rates.
Modulation
23. Explain the relationship between SNR, BER, data rate and bandwidth.
24. Explain the need for both digital and analog modulation in wireless
communication systems.
25. Show, using diagrams, the basic principles of ASK, PSK and FSK.
26. Describe the basic principle of Quadrature Amplitude Modulation (QAM).
27. How can the choice of modulation scheme impact on the achieved BER?
Spread Spectrum
28. Define spread spectrum communication. What is its key benefit and how is it
achieved.
29. What is the relationship between the bandwidth of a signal before and after it
has been encoded using spread spectrum communication?
30. What is the relationship between the bit rate of a signal before and after it has
been encoded using DSSS?
31. A FHSS system employs a total bandwidth of 400 MHz and individual
channel bandwidths of 100Hz. What is the minimum number of bits needed in
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the pseudo-random sequence number used?
Medium Access Control
32. Illustrate using diagrams the problems of hidden/exposed terminals and the
near/far effect.
33. Explain why CSMA/CD is not suitable for wireless networks.
34. Using diagrams explain TDD and FDD.
35. For Simple TDD, calculate the effective data rate assuming a data slot of 100
bits, propagation delay of 5 microseconds, burst transmission time of 5
microseconds, guard time of 1 microsecond.
36. For the figure below, state what type of multiple access scheme is in use.
417 µs 1 2 3 11 12 1 2 3
downlink 11 12 uplink
t 37. Compare TDD and FDD, considering issues such as capacity allocation,
hardware complexity, and spectrum needs.
38. Consider a simple CDMA-based medium access control protocol. There are
two senders, node A using KeyA (010011) and node B using KeyB (110101).
Node A wishes to transmit bit 1 and node B wishes to transmit bit 0. Assume
that we encode binary 1 as 1 and binary 0 as -1 and that the nodes transmit at
equal power levels. Show (a) the spread signal transmitted by each of the two
nodes, (b) the combined signal as received at the destination, and (c) the
calculations used by the receiver to determine the bit sent by Node A and the
bit sent by Node B.
39. What limits the number of simultaneous users in a CDMA system?
40. Why does CDMA require adaptive transmission power control?
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41. Outline the concept of reservation as used in some on-demand medium access
control protocols. Give detailed examples of how a protocol that uses (a)
explicit or (b) implicit reservations, might operate.
42. Under what circumstances might you choose a reservation-based scheme over
a purely random-access scheme?
43. Explain the differences between non-persistent, 1-persistent and p-persistent
CSMA.
44. Describe using state transition diagrams, a simple protocol that uses MACA
(Multiple Access with Collision Avoidance) together with ACKs.
45. Show how MACA solves the hidden terminal problem.
RELEVANT LOG TABLE ENTRIES
log10(0.1) = -1
log10(0.2) = -0.699
log10(0.3) = -0.523
log10(35) = 1.544
log10(35x105) = 6.544
log10(35x106) = 7.544
log10(10) = 1
log10(100) = 2
log10(1000) = 3
log10(10000) = 4
log10(125) = 2.097
log10(126) = 2.100
log10(127) = 2.104
log10(128) = 2.107
log2(125)=6.966
log2(126)=6.977
log2(127)=6.989
log2(128)=7.000
log2(1000)=9.966
log2(1001)=9.967
log2(1002)=9.969
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SELECTED SAMPLE SOLUTIONS
Q2
T = 1/f, so T = 1/1x106 = 1x10-6 seconds
λ = vT = 3x108 x 1x10-6 = 300 metres
Q3
B = 10 – 8 = 2MHz = 2x106 Hz
SNR dB = 30 dB = 10 log10 (SNR ) = 10 log10 (1000)
C = 2x106 x log2(1001) = 2x106 x 9.967 = 19.93 x 106 = 19.93 Mb/s
Q4
8x106 = 2 x 2x106 x log2M = 4x106 x log2M
2 = log2M, so M = 4
Q5
Q8
21x106 = 3x106 x log2(1 + SNR)
7 = log2(1 + SNR), so 1+SNR=128, and SNR=127
SNR dB = 10 log10 (127 ) = 21 .04
G = 4π x 0.81 x π / λ2 = 4π2 x 0.81 / (0.3)2 = 31.53 / 0.09 = 350.33
Q9
802 = 3.572 x 1.33 x h so h = 377.58
Q10
D = 3.57 Kh so D = 3.57
4 x10 = 13 Kms
3
Q11
D = 3.57( Kh 1 + Kh 2 ) = 3.57 ( 4 x10 + 4 x100 ) = 3.57 (3.64 + 11.52)
3
3
= 54.12 Kms
Q12
Pt/Pr = (4πd/λ)2
If we double the frequency, we halve λ, or if we double the distance, we
double d, so the new ratio for either of these events is:
Pt/Pr2 = (8πd/λ)2
Therefore:
10 log (Pr/Pr2) = 10 log (22) = 6 dB
Q13
λ = 3x108 / 3x109 = 0.1 m
LdB = − 20 log(0.1) + 20 log(35 x10 6 ) + 21.98 = 20 + 150.88 + 21.98 = 192.86
Q14
Ldb = 192.86 – 35 – 45 = 112.86
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Q31
Q35
(400 MHz)/(100 Hz) = 4 × 106.
The minimum number of PN bits = the smallest integer value not less than
log2 (4 × 106) = 22
R = 100 / 2 (11x10-6) = 4.5 x 106 bits / second
Q38
Node A: +1 x (-1, +1, -1, -1, +1, +1) = (-1, +1, -1, -1, +1, +1)
Node B: -1 x (+1, +1, -1, +1, -1, +1) = (-1, -1, +1, -1, +1, -1)
Combined = (-2, 0, 0, -2, +2, 0)
To retrieve Node A:
(-2, 0, 0, -2, +2, 0) x (-1, +1, -1, -1, +1, +1) = 2 + 0 + 0 + 2 + 2 + 0 = 6 (>0
therefore bit 1)
To retrieve Node B:
(-2, 0, 0, -2, +2, 0) x (+1, +1, -1, +1, -1, +1) = -2 + 0 + 0 – 2 - 2 + 0 = -6 (<0
therefore bit 0)
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