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 CS6311 University College Cork 1 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 CS6311 University College Cork 2 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? CS6311 University College Cork 3 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 CS6311 University College Cork 4 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 CS6311 University College Cork 5 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) CS6311 University College Cork 6