Nepal Engineering Council · Electronics & Communication Engineering · Chapter 9
Wireless and Telecommunication System
Tap an option to check it. Wrong picks show the right answer and the hint.
187 questions in 6 syllabus topics.
9.1 Telecommunication and its evolution
31 questions · AExE0901
1. Who was granted the 1876 patent for the telephone?
Show hintHide hint
Think of the inventor whose name became a telephone company.
Show answerHide answer
Answer: A. Alexander Graham Bell
Alexander Graham Bell received the US telephone patent in 1876. Morse is linked with the telegraph, Marconi with wireless telegraphy and Strowger with the automatic exchange.
2. Guglielmo Marconi is best known in telecommunication history for
Show hintHide hint
His system needed no wires between the stations.
Show answerHide answer
Answer: B. Developing practical long-distance wireless (radio) telegraphy
Marconi developed practical radio telegraphy and sent signals across the Atlantic in 1901.
3. Almon B. Strowger's invention (patented 1891) was
Show hintHide hint
It replaced the operator at the switchboard.
Show answerHide answer
Answer: B. The first automatic electromechanical telephone exchange
Strowger invented the step-by-step automatic exchange, which removed the need for human operators to connect calls.
4. The 1948 paper 'A Mathematical Theory of Communication', which founded information theory, was written by
Show hintHide hint
His name is on the channel-capacity formula C = B log2(1 + S/N).
Show answerHide answer
Answer: A. Claude Shannon
Claude Shannon published it in 1948 at Bell Labs. It defined information entropy and channel capacity.
5. Charles K. Kao received the 2009 Nobel Prize in Physics for his pioneering work on
Show hintHide hint
His work concerns a guided transmission medium made of glass.
Show answerHide answer
Answer: C. Transmission of light in fibres for optical communication
In 1966 Kao showed that fibre attenuation could fall below 20 dB/km if impurities were removed from the glass, which made optical fibre communication possible.
6. First-generation (1G) cellular systems such as AMPS used
Show hintHide hint
1G came before digital speech coding was used in mobile phones.
Show answerHide answer
Answer: D. Analog FM voice with FDMA
1G systems carried analog frequency-modulated voice, and each user had a separate frequency channel (FDMA).
7. Digital voice transmission and the Short Message Service (SMS) first became common with
Show hintHide hint
It was the first digital generation.
Show answerHide answer
Answer: B. 2G systems such as GSM
2G (GSM, IS-95) was the first generation with digital speech. GSM introduced SMS.
8. GPRS, which added packet-switched data to GSM networks, is usually classed as
Show hintHide hint
It is an enhancement of a 2G system, not a new generation.
Show answerHide answer
Answer: A. 2.5G
GPRS sits between 2G and 3G because it adds packet data to the existing GSM network. It is called 2.5G.
9. The air interface used by UMTS (3G) is
Show hintHide hint
The 3G air interface separates users by codes.
Show answerHide answer
Answer: B. WCDMA
UMTS uses Wideband CDMA with 5 MHz carriers and direct-sequence spreading.
10. LTE uses SC-FDMA rather than OFDMA on the uplink mainly because SC-FDMA
Show hintHide hint
Think about the battery-powered transmitter in the handset.
Show answerHide answer
Answer: C. Has a lower peak-to-average power ratio, so the handset power amplifier is more efficient
SC-FDMA has a lower PAPR than OFDMA, so the mobile's power amplifier can run more efficiently and the battery lasts longer.
11. The ITU umbrella name for the requirements of 5G systems is
Show hintHide hint
The name contains the year the target was set for.
Show answerHide answer
Answer: B. IMT-2020
IMT-2000 is 3G, IMT-Advanced is 4G and IMT-2020 is 5G.
12. Which of the following is NOT one of the three main 5G usage scenarios defined by ITU?
Show hintHide hint
One of the options is an older 3G-era technology.
Show answerHide answer
Answer: B. HSPA (High Speed Packet Access)
The three IMT-2020 scenarios are eMBB, URLLC and mMTC. HSPA is a 3.5G enhancement of UMTS.
13. A key architectural feature of 4G (LTE/EPC) networks compared with 2G/3G is
Show hintHide hint
Think of how voice is carried in VoLTE.
Show answerHide answer
Answer: A. An all-IP packet-switched core with no circuit-switched domain
The LTE Evolved Packet Core is entirely IP based. Voice is carried as VoIP (VoLTE) or handled by fallback to an older network.
14. Which frequency range is widely researched for 6G systems to obtain very wide bandwidths?
Show hintHide hint
Bandwidth available grows with carrier frequency.
Show answerHide answer
Answer: A. Sub-terahertz and terahertz bands (about 100 GHz to 10 THz)
Very wide contiguous bandwidths are only available at very high carrier frequencies, so 6G research looks at sub-THz and THz bands.
15. The conductors in a twisted-pair cable are twisted mainly to
Show hintHide hint
Think of noise induced equally in both wires.
Show answerHide answer
Answer: D. Reduce crosstalk and electromagnetic interference
Twisting makes the noise picked up by the two wires nearly equal, so it cancels at the differential receiver. This reduces crosstalk and EMI.
16. Which of the following is NOT an advantage of optical fibre over copper cable?
Show hintHide hint
Glass is an insulator.
Show answerHide answer
Answer: D. It can carry electrical power to remote equipment
Glass fibre is a dielectric, so it cannot carry electrical power. The other three are real advantages of fibre.
17. The typical core diameter of a single-mode optical fibre is about
Show hintHide hint
The core is small enough to allow only one mode.
Show answerHide answer
Answer: D. 8 to 10 µm
Single-mode fibre has a core of about 8 to 10 µm. Multimode cores are 50 or 62.5 µm. 125 µm is the cladding diameter of both.
18. Silica optical fibre has its lowest attenuation (about 0.2 dB/km) near the wavelength
Show hintHide hint
It is the longest of the standard telecom windows.
Show answerHide answer
Answer: C. 1550 nm
The minimum loss of silica fibre is in the 1550 nm window. 1310 nm gives about zero dispersion but slightly higher loss, and 850 nm is the high-loss first window.
19. Long-distance HF (3 to 30 MHz) radio communication beyond the horizon mainly depends on
Show hintHide hint
The signal goes up before it comes back down.
Show answerHide answer
Answer: D. Sky-wave reflection from the ionosphere
HF waves are refracted back to Earth by the ionosphere. This sky-wave mode allows communication over thousands of kilometres.
20. Terrestrial microwave relay links require
Show hintHide hint
These towers are placed on hilltops for a reason.
Show answerHide answer
Answer: A. A clear line-of-sight path between antennas
At microwave frequencies the waves travel in nearly straight lines and pass through the ionosphere. Repeaters are therefore placed on towers or hilltops with a clear line of sight.
21. What is the free-space wavelength of a 900 MHz GSM carrier?
Show hintHide hint
Use λ = c/f.
Show answerHide answer
Answer: C. 0.333 m
λ = c/f = 3×10⁸ / 900×10⁶ = 0.333 m.
22. A step-index fibre has core index 1.48 and cladding index 1.46. Its numerical aperture is about
Show hintHide hint
Take the square root of the difference of the squares.
Show answerHide answer
Answer: C. 0.243
NA = √(n1² − n2²) = √(2.1904 − 2.1316) = √0.0588 ≈ 0.243.
23. For a fibre with core index 1.50 and cladding index 1.45, the critical angle at the core-cladding boundary is about
Show hintHide hint
Snell's law with a refraction angle of 90°.
Show answerHide answer
Answer: D. 75.2°
θc = sin⁻¹(n2/n1) = sin⁻¹(1.45/1.50) = sin⁻¹(0.9667) ≈ 75.2°.
24. An optical power of 1 mW is launched into a 40 km fibre with attenuation 0.25 dB/km. Ignoring other losses, the received power is
Show hintHide hint
Find the total loss in dB first, then convert it to a ratio.
Show answerHide answer
Answer: C. 0.1 mW
Total loss = 0.25 × 40 = 10 dB, which is a factor of 10. So Pout = 1 mW / 10 = 0.1 mW.
25. A coaxial cable has outer-to-inner conductor diameter ratio D/d = 3.5 and a dielectric with εr = 2.25. Using Z0 = (60/√εr) ln(D/d), its characteristic impedance is about
Show hintHide hint
Do not forget to divide by √εr.
Show answerHide answer
Answer: A. 50 Ω
Z0 = (60/1.5) × ln 3.5 = 40 × 1.253 ≈ 50 Ω.
26. The signal propagation velocity in a coaxial cable with a solid polyethylene dielectric (εr = 2.25) is
Show hintHide hint
Divide c by the square root of the relative permittivity.
Show answerHide answer
Answer: A. 2 × 10⁸ m/s
v = c/√εr = 3×10⁸ / 1.5 = 2×10⁸ m/s, a velocity factor of about 0.67.
27. Using d ≈ 4.12(√h1 + √h2) km with antenna heights in metres, the maximum radio line-of-sight distance between antennas 100 m and 16 m high is about
Show hintHide hint
Take the square root of each height before adding.
Show answerHide answer
Answer: D. 57.7 km
d = 4.12(√100 + √16) = 4.12 × (10 + 4) ≈ 57.7 km.
28. A channel with 1 MHz bandwidth and a 30 dB signal-to-noise ratio has a Shannon capacity of about
Show hintHide hint
Convert the dB value to a ratio before using the formula.
Show answerHide answer
Answer: B. 10 Mbit/s
S/N = 10³ = 1000, so C = B log2(1 + S/N) = 10⁶ × log2(1001) ≈ 9.97 Mbit/s ≈ 10 Mbit/s.
29. A geostationary satellite is about 35,786 km above the equator. The one-way earth-satellite-earth propagation delay for a sub-satellite station is about
Show hintHide hint
The signal makes an up-link and a down-link trip.
Show answerHide answer
Answer: B. 0.24 s
Delay = 2 × 35,786 km / (3×10⁵ km/s) ≈ 0.239 s, which is about 0.24 s.
30. The propagation delay through 100 km of optical fibre with core index 1.5 is
Show hintHide hint
Light in glass travels slower than c by the factor n.
Show answerHide answer
Answer: C. 0.5 ms
v = c/n = 2×10⁸ m/s, so t = 100×10³ / 2×10⁸ = 0.5 ms.
31. A microwave link has 30 dBm transmit power, 15 dBi antennas at each end, a 120 dB path loss and 2 dB total feeder loss. The received power is
Show hintHide hint
Add the gains and subtract the losses, all in dB.
Show answerHide answer
Answer: C. −62 dBm
Pr = 30 + 15 + 15 − 120 − 2 = −62 dBm.
9.2 Cellular network
32 questions · AExE0902
32. In the Friis free-space model, how does received power vary with the transmitter-receiver distance d?
Show hintHide hint
Think of power spreading over the surface of a sphere.
Show answerHide answer
Answer: A. It is proportional to 1/d²
Pr = Pt Gt Gr λ² / ((4π)² d²). Power falls as 1/d², which is 20 dB per decade of distance.
33. The free-space path loss at 900 MHz over a distance of 1 km, using FSPL(dB) = 32.44 + 20 log d(km) + 20 log f(MHz), is about
Show hintHide hint
log 1 = 0, so only the frequency term is left to add.
Show answerHide answer
Answer: A. 91.5 dB
FSPL = 32.44 + 20 log 1 + 20 log 900 = 32.44 + 0 + 59.08 ≈ 91.5 dB.
34. A 10 W transmitter with unity-gain antennas operates at 900 MHz. In free space, the power received at 1 km is about
Show hintHide hint
Convert 10 W to dBm before subtracting the path loss.
Show answerHide answer
Answer: D. −51.5 dBm
10 W = 40 dBm. The free-space loss at 1 km and 900 MHz is about 91.5 dB, so Pr = 40 − 91.5 = −51.5 dBm (about 7×10⁻⁹ W).
35. In a log-distance path loss model with exponent n = 4, doubling the distance increases the path loss by about
Show hintHide hint
Use 10·n·log10(d2/d1).
Show answerHide answer
Answer: D. 12 dB
ΔPL = 10 n log(2) = 10 × 4 × 0.301 ≈ 12 dB.
36. A received power of −50 dBm is measured at 1 km in an area with path loss exponent n = 3. The predicted received power at 2 km is about
Show hintHide hint
The extra loss is 10·n·log(2).
Show answerHide answer
Answer: C. −59 dBm
Pr(2 km) = −50 − 10 × 3 × log 2 = −50 − 9.03 ≈ −59 dBm.
37. A mobile in the shadow region behind a hill still receives a signal mainly because of
Show hintHide hint
This is described by Huygens' principle at an obstructing edge.
Show answerHide answer
Answer: C. Diffraction
Diffraction lets waves bend around sharp edges such as hilltops and building edges. This puts some energy into the geometric shadow region.
38. In mobile radio propagation, scattering mainly occurs when the wave meets
Show hintHide hint
Compare the size of the object with λ.
Show answerHide answer
Answer: D. Rough surfaces and objects small compared with the wavelength, such as foliage and lamp posts
Scattering happens at rough surfaces and small objects, and it spreads energy in many directions. Large smooth surfaces cause reflection, and sharp edges cause diffraction.
39. In the two-ray ground reflection model at large distances, the received power
Show hintHide hint
Look at the large-distance approximation of the two-ray formula.
Show answerHide answer
Answer: A. Falls as 1/d⁴ and does not depend on frequency
For large d, Pr ≈ Pt Gt Gr ht² hr² / d⁴. This is a fourth-power distance law with no λ term.
40. For a 3 GHz link of total length 2 km, the radius of the first Fresnel zone at the midpoint is about
Show hintHide hint
Use r1 = √(λ d1 d2 / d) with d1 = d2 = 1 km.
Show answerHide answer
Answer: B. 7.07 m
λ = 0.1 m and r1 = √(λ d1 d2 / (d1 + d2)) = √(0.1 × 1000 × 1000 / 2000) = √50 ≈ 7.07 m.
41. Small-scale (fast) fading in a mobile channel is caused mainly by
Show hintHide hint
The distance scale is about half a wavelength.
Show answerHide answer
Answer: D. Constructive and destructive interference of multipath components over distances of a few wavelengths
Small-scale fading is the rapid change of signal level over distances of the order of a wavelength. It is caused by multipath waves adding with different phases.
42. When a strong line-of-sight component is present together with many scattered components, the received envelope follows a
Show hintHide hint
Its parameter K is the ratio of the dominant power to the scattered power.
Show answerHide answer
Answer: D. Rician distribution
A dominant LOS (specular) component turns Rayleigh statistics into Rician statistics, described by the K-factor. Rayleigh applies when there is no dominant component.
43. For a Rayleigh-fading envelope, the probability that the received signal falls more than 10 dB below its rms value is about
Show hintHide hint
Use the Rayleigh CDF with the threshold power 0.1 times the mean power.
Show answerHide answer
Answer: A. 9.5%
P(r < R) = 1 − exp(−R²/Pr,rms). With R²/Prms = 0.1, P = 1 − e^(−0.1) ≈ 0.095.
44. A mobile moving at 72 km/h directly toward a 900 MHz base station sees a maximum Doppler shift of
Show hintHide hint
Convert km/h to m/s first.
Show answerHide answer
Answer: C. 60 Hz
v = 20 m/s and λ = 1/3 m, so fd = v/λ = 20 / 0.333 = 60 Hz.
45. A channel shows frequency-selective fading when
Show hintHide hint
Compare the bandwidth of the signal with the bandwidth over which the channel is flat.
Show answerHide answer
Answer: B. The signal bandwidth is greater than the coherence bandwidth of the channel
When the signal bandwidth exceeds the coherence bandwidth (equivalently, the symbol period is shorter than the delay spread), different frequency components fade differently. This causes ISI.
46. A channel is said to be fast fading when
Show hintHide hint
Fast and slow fading are classified by the Doppler spread, which is a time-variation effect.
Show answerHide answer
Answer: A. The coherence time is smaller than the symbol period
Fast fading means the channel impulse response changes within one symbol, so Tc < Ts (Doppler spread larger than the signal bandwidth).
47. A channel has rms delay spread 1 µs. Using Bc ≈ 1/(5στ), its coherence bandwidth is about
Show hintHide hint
Put στ into the given approximation.
Show answerHide answer
Answer: B. 200 kHz
Bc ≈ 1/(5 × 1×10⁻⁶) = 200 kHz.
48. For hexagonal cells the cluster size is N = i² + ij + j² with non-negative integers i and j. Which of the following cannot be a cluster size?
Show hintHide hint
Try small integer pairs (i, j).
Show answerHide answer
Answer: C. 10
The allowed values are 1, 3, 4, 7, 9, 12, 13, ... (for example 3 = 1,1; 7 = 2,1; 12 = 2,2). No integer pair gives 10.
49. In a hexagonal system with cell radius 2 km and cluster size N = 7, the co-channel reuse distance is about
Show hintHide hint
Use D = R√(3N).
Show answerHide answer
Answer: C. 9.17 km
D = R√(3N) = 2 × √21 ≈ 2 × 4.583 ≈ 9.17 km.
50. For a cluster size N = 7, path loss exponent 4 and six first-tier co-channel cells, the approximate signal-to-co-channel interference ratio is
Show hintHide hint
Use S/I = (D/R)ⁿ / i0.
Show answerHide answer
Answer: B. 18.7 dB
S/I = (√(3N))⁴ / 6 = (√21)⁴ / 6 = 441/6 = 73.5, which is about 18.7 dB.
51. A system has 33 MHz of spectrum. It uses two 25 kHz simplex channels for each full-duplex voice channel and a 4-cell reuse pattern. How many duplex channels are available per cell?
Show hintHide hint
Find the total number of duplex channels first, then share them among the cells of a cluster.
Show answerHide answer
Answer: D. 165
Each duplex channel uses 50 kHz, so there are 33 MHz / 50 kHz = 660 channels in total. Per cell: 660/4 = 165.
52. For a fixed cell size, reducing the cluster size N in a cellular system
Show hintHide hint
Think about what happens to D/R.
Show answerHide answer
Answer: C. Increases capacity but also increases co-channel interference
A smaller N means more channels per cell, so capacity rises. But the reuse distance D = R√(3N) shrinks, so co-channel interference increases.
53. In dynamic channel assignment (DCA),
Show hintHide hint
Think of a shared pool versus fixed allotments.
Show answerHide answer
Answer: D. Channels are not permanently allotted to cells; the MSC assigns them from a common pool when a call is requested
In DCA the MSC assigns channels on demand from a central pool, while keeping co-channel reuse constraints. This improves flexibility at the cost of more computation.
54. Reserving a few guard channels in each cell exclusively for handoff requests is done to
Show hintHide hint
Which is worse for a user: a blocked call or a dropped call?
Show answerHide answer
Answer: D. Give priority to handoffs so that ongoing calls are less likely to be dropped
Users find a dropped call more annoying than a blocked new call. Guard channels give handoffs priority, at the cost of slightly higher blocking for new calls.
55. Soft handoff, in which a mobile communicates with two or more base stations at the same time, is a characteristic of
Show hintHide hint
This needs all cells to share the same carrier frequency.
Show answerHide answer
Answer: B. CDMA systems
Neighbouring CDMA cells use the same frequency, so the mobile can combine signals from several base stations before leaving the old one. FDMA and TDMA systems use hard handoff.
56. In mobile-assisted handoff (MAHO), used in GSM,
Show hintHide hint
Who makes the measurements is in the name.
Show answerHide answer
Answer: A. The mobile measures the signal strength of neighbouring base stations and reports it to the network
In MAHO the mobile reports measurements of the surrounding base stations, and the network makes the handoff decision. This speeds up handoff compared with network-only measurement.
57. Setting the handoff margin (handoff threshold minus minimum usable signal level) too small mainly leads to
Show hintHide hint
Think of a mobile near a cell boundary with a fluctuating signal.
Show answerHide answer
Answer: B. Too many unnecessary handoffs (ping-pong effect)
A small margin starts handoff even on brief signal dips, so the mobile keeps switching between base stations. A large margin can make handoff too late, so calls drop.
58. An umbrella cell (large macrocell overlaying microcells) is mainly used to
Show hintHide hint
Consider a car crossing many small cells in a few seconds.
Show answerHide answer
Answer: B. Serve fast-moving users and so reduce the number of handoffs
Fast mobiles are kept on the large umbrella cell so that they do not hand off rapidly between small microcells. Slow users are served by the microcells.
59. If a cell is split into new cells of half the original radius, the number of cells (and so roughly the capacity) in the same area increases by a factor of
Show hintHide hint
Area scales with the square of the radius.
Show answerHide answer
Answer: A. 4
Cell area ∝ R², so halving R gives four times as many cells in the same area. Capacity rises about fourfold.
60. After cell splitting to half radius, with path loss exponent n = 4, the transmit power of the new cells must be reduced by about how much to keep the same received power at the cell edge?
Show hintHide hint
Equate the edge powers: Pt1 R⁻⁴ = Pt2 (R/2)⁻⁴.
Show answerHide answer
Answer: B. 12 dB
Pr ∝ Pt R⁻⁴, so Pt2 = Pt1 / 2⁴ = Pt1/16. This is 10 log 16 ≈ 12 dB.
61. Using 120° sectoring in a 7-cell reuse system reduces the number of first-tier co-channel interferers from 6 to
Show hintHide hint
Only interferers inside the sector's beam count.
Show answerHide answer
Answer: C. 2
With three 120° directional sectors, only 2 of the 6 first-tier co-channel cells radiate toward a given sector.
62. A disadvantage of cell sectoring is
Show hintHide hint
A small group of channels is less efficient than one large group.
Show answerHide answer
Answer: A. More handoffs and lower trunking efficiency because each cell's channels are split among sectors
Sectoring improves S/I, but each sector has only part of the channels. This lowers trunking efficiency and adds inter-sector handoffs.
63. A low-power, customer-installed base station that serves a home or small office over the user's broadband connection is called a
Show hintHide hint
It is the smallest named cell type.
Show answerHide answer
Answer: C. Femtocell
Femtocells are very small, low-power cells installed by the user and connected to the operator's core over broadband. They improve indoor coverage and capacity.
9.3 Signal and system
32 questions · AExE0903
64. The main purpose of an equalizer in a mobile radio receiver is to
Show hintHide hint
Think about symbols overlapping in time.
Show answerHide answer
Answer: A. Compensate for intersymbol interference caused by multipath time dispersion
Multipath delay spread smears each symbol into its neighbours, which causes ISI. An equalizer applies an approximate inverse of the channel response to remove it.
65. Equalization becomes essential in a digital mobile link when the rms delay spread is
Show hintHide hint
Compare the echo duration with the symbol duration.
Show answerHide answer
Answer: A. Comparable to or larger than the symbol period
When the delay spread is a significant fraction of the symbol period (or larger), echoes overlap the following symbols, so ISI must be equalized.
66. The main drawback of a zero-forcing linear equalizer is that it
Show hintHide hint
Inverting a very small number gives a very large one.
Show answerHide answer
Answer: C. Enhances noise at frequencies where the channel response has deep nulls
Zero forcing inverts the channel, so its gain is very large where the channel response is small. Noise at those frequencies is strongly amplified.
67. A decision feedback equalizer (DFE) differs from a linear transversal equalizer in that it
Show hintHide hint
The 'feedback' comes from the decision device.
Show answerHide answer
Answer: D. Uses previously detected symbols to cancel the ISI they cause on the current symbol
A DFE has a feedforward filter plus a feedback filter driven by past decisions. Subtracting their ISI contribution makes the DFE non-linear and gives less noise enhancement.
68. Maximum likelihood sequence estimation (MLSE) equalization is usually implemented with the
Show hintHide hint
The same algorithm is used to decode convolutional codes.
Show answerHide answer
Answer: B. Viterbi algorithm
MLSE searches for the most likely transmitted sequence through a trellis of channel states. The Viterbi algorithm does this search efficiently.
69. Compared with the RLS algorithm, the LMS adaptive equalizer algorithm
Show hintHide hint
There is a trade-off between complexity and convergence speed.
Show answerHide answer
Answer: A. Is simpler to compute but converges more slowly
LMS needs about 2N+1 operations per iteration but converges slowly. RLS converges much faster at a higher computational cost.
70. Adaptive equalizers in TDMA systems such as GSM adjust their coefficients mainly using
Show hintHide hint
The receiver already knows part of the burst.
Show answerHide answer
Answer: D. A known training sequence transmitted in each burst
GSM puts a known 26-bit training sequence in the middle of each normal burst. The receiver uses it to estimate the channel and set the equalizer.
71. GSM has a symbol period of about 3.69 µs. If the multipath delay spread is 15 µs, the ISI extends over about how many symbols?
Show hintHide hint
Divide the delay spread by the symbol duration.
Show answerHide answer
Answer: D. 4
Delay spread / symbol period = 15 / 3.69 ≈ 4 symbols, which the equalizer must handle.
72. The basic principle of diversity reception is to
Show hintHide hint
Think of the probability that all branches fail together.
Show answerHide answer
Answer: B. Receive several independently fading copies of the signal, which are unlikely to fade deeply at the same time
If the branches fade independently, the chance that all of them are in a deep fade together is small. Combining or selecting them therefore reduces fade depth and duration.
73. For space diversity at a mobile handset, the antennas should be at least about λ/2 apart. At 900 MHz this spacing is about
Show hintHide hint
Find λ, then halve it.
Show answerHide answer
Answer: A. 16.7 cm
λ = 3×10⁸ / 9×10⁸ = 0.333 m, so λ/2 ≈ 0.167 m = 16.7 cm.
74. In a two-branch selection-diversity receiver with independent Rayleigh fading, each branch falls below a threshold 10 dB under its mean SNR with probability 0.095. The probability that both branches are below the threshold is about
Show hintHide hint
Both independent events must happen together.
Show answerHide answer
Answer: B. 0.9%
Pout = (1 − e^(−0.1))² = 0.095² ≈ 0.0091, which is about 0.9%.
75. For M-branch selection diversity in Rayleigh fading, the mean output SNR is Γ(1 + 1/2 + ... + 1/M). The improvement factor for M = 3 is about
Show hintHide hint
Add the harmonic series up to 1/3.
Show answerHide answer
Answer: B. 1.83
Σ 1/k for k = 1 to 3 is 1 + 0.5 + 0.333 = 1.83. This is less than the factor 3 that MRC gives.
76. A maximal ratio combiner has three branches with instantaneous SNRs of 10, 5 and 3 (linear). The combined output SNR is
Show hintHide hint
MRC adds the branch SNRs.
Show answerHide answer
Answer: C. 18
For MRC the output SNR is the sum of the branch SNRs: 10 + 5 + 3 = 18 (about 12.6 dB).
77. In maximal ratio combining, each branch signal is
Show hintHide hint
Stronger branches should count for more.
Show answerHide answer
Answer: B. Co-phased and weighted in proportion to its signal amplitude divided by its noise power before summing
MRC co-phases the branches and weights each by its signal-to-noise amplitude. This gives the maximum possible output SNR among linear combiners.
78. Time diversity in digital mobile systems is commonly obtained by
Show hintHide hint
The copies are separated in time, not in space or frequency.
Show answerHide answer
Answer: C. Channel coding with interleaving, so that coded bits are separated by more than the coherence time
Interleaving spreads adjacent coded bits over a time longer than the coherence time. A deep fade then causes scattered errors that the code can correct.
79. The RAKE receiver used in CDMA systems
Show hintHide hint
Each 'finger' collects one echo.
Show answerHide answer
Answer: A. Resolves multipath components separated by more than one chip and combines them as a form of diversity
Each RAKE finger correlates with the PN code at a different delay. Paths separated by more than one chip period are separated and then combined, often with MRC.
80. In direct-sequence spread spectrum (DSSS), the data signal is spread by
Show hintHide hint
The chip rate is much higher than the bit rate.
Show answerHide answer
Answer: A. Multiplying it by a high-rate pseudo-noise (PN) chip sequence
In DSSS each data bit is multiplied by many PN chips, so the bandwidth grows by the chip-rate to bit-rate ratio.
81. An IS-95 CDMA system has chip rate 1.2288 Mchip/s and data rate 9.6 kbit/s. The processing gain is about
Show hintHide hint
Take the ratio of chip rate to bit rate, then convert to dB.
Show answerHide answer
Answer: B. 21.1 dB
Gp = 1.2288×10⁶ / 9.6×10³ = 128, and 10 log 128 ≈ 21.1 dB.
82. A spread spectrum system has processing gain 30 dB, needs Eb/N0 = 10 dB, and has 2 dB of implementation losses. Its jamming margin is
Show hintHide hint
Subtract both the required Eb/N0 and the losses from the processing gain.
Show answerHide answer
Answer: C. 18 dB
Jamming margin = Gp − (Eb/N0)req − losses = 30 − 10 − 2 = 18 dB.
83. A maximal-length PN sequence is generated by a 10-stage linear feedback shift register. Its period is
Show hintHide hint
One state of the register is never used.
Show answerHide answer
Answer: C. 1023 chips
An m-sequence from an n-stage LFSR has period 2ⁿ − 1 = 2¹⁰ − 1 = 1023 chips. The all-zero state is excluded.
84. Which property holds for a maximal-length (m-) sequence over one period?
Show hintHide hint
The period is an odd number.
Show answerHide answer
Answer: C. The number of 1s exceeds the number of 0s by exactly one
This is the balance property: an m-sequence of period 2ⁿ − 1 contains 2ⁿ⁻¹ ones and 2ⁿ⁻¹ − 1 zeros.
85. In fast frequency hopping spread spectrum,
Show hintHide hint
Compare the hop rate with the symbol rate.
Show answerHide answer
Answer: A. The carrier hops several times during each data symbol
In fast FH the hop rate is greater than the symbol rate, so each symbol is spread over several frequencies. In slow FH several symbols are sent per hop.
86. Classic Bluetooth uses which spread spectrum technique in the 2.4 GHz band?
Show hintHide hint
Its channels are 1 MHz wide and the radio changes channel very often.
Show answerHide answer
Answer: B. Frequency hopping over 79 channels of 1 MHz
Bluetooth BR/EDR hops pseudo-randomly over 79 channels of 1 MHz at a nominal 1600 hops per second.
87. The near-far problem in a direct-sequence CDMA uplink is overcome mainly by
Show hintHide hint
All users should arrive at the base station equally strong.
Show answerHide answer
Answer: A. Fast transmit power control of the mobiles
A nearby mobile can overpower distant ones at the base station receiver. Closed-loop power control makes all mobiles arrive at about the same power.
88. A GSM operator has 25 MHz in each direction, uses 200 kHz carriers with one carrier left as a guard band, and has 8 time slots per carrier. The number of traffic channels is
Show hintHide hint
Remove the guard carrier before multiplying by the slots.
Show answerHide answer
Answer: D. 992
25 MHz / 200 kHz = 125 carriers, minus 1 guard = 124 carriers. 124 × 8 = 992 channels.
89. A GSM TDMA frame lasts 4.615 ms and has 8 time slots. Each time slot lasts about
Show hintHide hint
Divide the frame by the number of slots.
Show answerHide answer
Answer: C. 0.577 ms
Slot duration = 4.615 / 8 ≈ 0.577 ms (156.25 bit periods).
90. A single-cell CDMA system has W = 1.25 MHz, R = 9.6 kbit/s and required Eb/N0 = 7 dB. Ignoring other-cell interference and voice activity, N ≈ 1 + (W/R)/(Eb/N0) gives about
Show hintHide hint
Convert 7 dB to a linear ratio first.
Show answerHide answer
Answer: D. 27 users
W/R = 130.2 and Eb/N0 = 10^0.7 = 5.01, so N ≈ 1 + 130.2/5.01 ≈ 27.
91. Which statement about FDMA is correct?
Show hintHide hint
The resource being divided is frequency.
Show answerHide answer
Answer: D. Each user is given its own narrowband frequency channel and transmits continuously
In FDMA each user has a separate frequency channel for the whole call and transmits continuously. Duplexers are needed for FDD operation.
92. A feature of TDMA that is NOT found in FDMA is that
Show hintHide hint
Think about what a handset does between its own bursts.
Show answerHide answer
Answer: D. Transmission is discontinuous, so the handset can take measurements in idle slots and needs strict time synchronization
TDMA users transmit in bursts. This needs synchronization and guard times, and the idle slots allow mobile-assisted handoff measurements.
93. LTE uses OFDMA with 15 kHz subcarrier spacing. The useful OFDM symbol duration (excluding the cyclic prefix) is about
Show hintHide hint
Orthogonality needs T = 1/Δf.
Show answerHide answer
Answer: C. 66.7 µs
The useful symbol time is the reciprocal of the subcarrier spacing: 1/15 kHz ≈ 66.7 µs.
94. Space division multiple access (SDMA) serves several users on the same frequency and time by using
Show hintHide hint
The users are separated by where they are.
Show answerHide answer
Answer: B. Smart (adaptive or directional) antennas that point separate beams at different users
SDMA uses the spatial separation of users. Adaptive antenna beams let the same channel be reused for users in different directions.
95. On the IS-95 CDMA forward link, the traffic channels within a cell are separated by
Show hintHide hint
These codes are mutually orthogonal when time-aligned.
Show answerHide answer
Answer: D. 64 orthogonal Walsh codes
The IS-95 forward link spreads each channel with one of 64 Walsh functions. These are orthogonal when synchronized, so users in the same cell are separated.
9.4 Switching systems
31 questions · AExE0904
96. A Strowger step-by-step exchange is an example of
Show hintHide hint
The switches move in step with the dial pulses.
Show answerHide answer
Answer: A. Direct (progressive) control, where each dialled digit directly drives a switch stage
In step-by-step switching the dial pulses of each digit step a selector directly. No common control equipment stores the number.
97. A crossbar exchange differs from a Strowger exchange mainly because it uses
Show hintHide hint
Control is shared by many calls instead of being built into each switch.
Show answerHide answer
Answer: D. Common control equipment (registers and markers) to set up connections
Crossbar systems store the dialled digits in registers. Markers then find and set a path through the crossbar matrix, which is common control.
98. In a stored program control (SPC) exchange,
Show hintHide hint
The control logic is software, not hardwired.
Show answerHide answer
Answer: C. Switching operations are controlled by a program running on a processor, so new services can be added by changing software
SPC uses processors that run call-processing software. This makes features such as call forwarding and abbreviated dialling easy to add.
99. In space-division switching,
Show hintHide hint
The word 'space' refers to physical paths.
Show answerHide answer
Answer: C. Each connection gets its own physical path through the switch for the whole call
Space-division switches such as crossbars dedicate a separate metallic or electronic path to each connection.
100. The basic difference between analog switching and digital switching is that a digital switch
Show hintHide hint
Think about what is being switched: a waveform or coded samples.
Show answerHide answer
Answer: B. Switches PCM samples in time slots instead of continuous analog signals
Digital exchanges switch 8-bit PCM samples in TDM time slots, using time and space stages. Analog switches connect continuous voice-frequency paths.
101. A time slot interchanger (TSI) switches calls by
Show hintHide hint
It moves data from one time slot to another.
Show answerHide answer
Answer: A. Writing incoming samples into a speech memory and reading them out in a different time-slot order
A TSI stores one frame of samples. It reads them out in an order set by the control memory, so channel i is moved to time slot j.
102. A TSI handles 1024 channels in a 125 µs frame. Each slot needs one write and one read. The required memory access time is about
Show hintHide hint
Count both the read and the write for every channel.
Show answerHide answer
Answer: C. 61 ns
There are 2 × 1024 = 2048 memory accesses per frame, so the access time is 125 µs / 2048 ≈ 61 ns.
103. A TSI uses a memory with a 50 ns cycle time and needs one read and one write per channel per 125 µs frame. The maximum number of channels it can switch is
Show hintHide hint
Each channel uses two memory cycles per frame.
Show answerHide answer
Answer: C. 1250
n = 125 µs / (2 × 50 ns) = 1250 channels.
104. The speech memory of a TSI for a 32-channel system with 8-bit samples must store at least
Show hintHide hint
One frame's worth of samples.
Show answerHide answer
Answer: A. 256 bits
One full frame must be stored: 32 samples × 8 bits = 256 bits.
105. A single-stage square crossbar switch connecting 100 inlets to 100 outlets needs how many crosspoints?
Show hintHide hint
There is one crosspoint at every inlet-outlet intersection.
Show answerHide answer
Answer: C. 10,000
An N × N crossbar needs N² crosspoints: 100² = 10,000.
106. A three-stage Clos network with n inlets per first-stage switch is strictly non-blocking when the number of middle-stage switches k satisfies
Show hintHide hint
Consider the worst case on both the inlet side and the outlet side.
Show answerHide answer
Answer: D. k ≥ 2n − 1
Clos showed that k = 2n − 1 middle switches guarantee a free path in the worst case. Up to n − 1 paths can be busy from each side.
107. A non-blocking three-stage Clos switch has N = 128 lines, n = 8 and k = 15. Using Nx = 2Nk + k(N/n)², the number of crosspoints is
Show hintHide hint
Calculate the outer stages and the middle stage separately, then add.
Show answerHide answer
Answer: B. 7680
2 × 128 × 15 = 3840 and 15 × (128/8)² = 15 × 256 = 3840. The total is 7680, against 128² = 16,384 for a single stage.
108. A T-S-T digital switch consists of
Show hintHide hint
Read the letters in order.
Show answerHide answer
Answer: C. A time stage, then a space stage, then a time stage
T-S-T has time slot interchangers on the input and output sides, with a time-multiplexed space switch in between. It is widely used in large digital exchanges.
109. In a switching network, each path between two points needs two links in series. There are three such paths in parallel, and each link is busy with probability 0.3. Using a Lee graph, the blocking probability is about
Show hintHide hint
A path is free only if both of its links are free.
Show answerHide answer
Answer: A. 0.133
One path is blocked with probability 1 − (0.7)² = 0.51. All three are blocked with probability 0.51³ ≈ 0.133.
110. The bit rate of a 32-time-slot E1 PCM system with 8-bit samples taken at 8 kHz is
Show hintHide hint
Multiply slots, bits per slot and frames per second.
Show answerHide answer
Answer: A. 2.048 Mbit/s
32 × 8 bits × 8000 frames/s = 2.048 Mbit/s.
111. A T1 frame carries 24 channels of 8 bits plus one framing bit. The number of bits per frame is
Show hintHide hint
Do not forget the framing bit.
Show answerHide answer
Answer: C. 193
24 × 8 + 1 = 193 bits. At 8000 frames/s this gives 1.544 Mbit/s.
112. The defining idea of a softswitch is that it
Show hintHide hint
Control and bearer are split apart.
Show answerHide answer
Answer: D. Separates call control (software on a server) from the media path carried by media gateways
A softswitch (media gateway controller) runs call-control software. It directs media gateways that carry the voice, for example as RTP over IP.
113. The protocol that a softswitch (media gateway controller) uses to control media gateways is typically
Show hintHide hint
Its name contains 'media gateway'.
Show answerHide answer
Answer: D. MGCP or H.248/Megaco
MGCP and its successor H.248 (Megaco) are master-slave protocols. The MGC uses them to tell gateways to create, modify and delete media connections.
114. In SIP-based VoIP, a call (session) is started by sending a
Show hintHide hint
The caller is asking the other party to join.
Show answerHide answer
Answer: A. INVITE request
INVITE starts a session. REGISTER binds a user to a location, BYE ends a session and ACK confirms the final response to an INVITE.
115. The main function of a media (trunking) gateway in a next-generation network is to
Show hintHide hint
It sits between two different kinds of network.
Show answerHide answer
Answer: B. Convert between circuit-switched TDM voice and packetized voice on an IP network
Media gateways connect the old PSTN TDM trunks to the IP core. They packetize voice (for example into RTP) and do the reverse in the other direction.
116. In DTMF signalling, pressing the key '5' sends the tone pair
Show hintHide hint
Find key 5's row and column on the 4 × 3 keypad.
Show answerHide answer
Answer: D. 770 Hz and 1336 Hz
Key 5 is in row 2 (770 Hz) and column 2 (1336 Hz). 697+1209 is '1', 852+1477 is '9' and 941+1336 is '0'.
117. Compared with rotary pulse dialling, DTMF dialling
Show hintHide hint
Think of entering digits during a call to a phone-banking menu.
Show answerHide answer
Answer: D. Is faster and can be sent end-to-end through a connected voice path
DTMF tone pairs are in the voice band, so they are sent quickly and pass through an established connection, for example to an IVR system. Pulse dialling interrupts the DC loop.
118. In loop-disconnect (rotary) dialling at the standard rate, the digit 0 is sent as
Show hintHide hint
A digit cannot be sent as no pulses.
Show answerHide answer
Answer: B. 10 pulses
Each digit n is sent as n loop breaks, and 0 is sent as 10 breaks, at about 10 pulses per second.
119. In the CCITT out-of-band signalling system for analog FDM channels, the signalling tone is at
Show hintHide hint
It lies above 3400 Hz but below 4000 Hz.
Show answerHide answer
Answer: A. 3825 Hz
Out-of-band signalling uses 3825 Hz, which is inside the 4 kHz channel but above the 300 to 3400 Hz speech band. 2600 Hz is an in-band single-frequency tone.
120. In a 30-channel E1 PCM system using channel associated signalling (CAS), the signalling bits are carried in
Show hintHide hint
This slot lies between the two groups of 15 speech channels.
Show answerHide answer
Answer: B. Time slot 16
TS0 carries frame alignment, and TS16 carries CAS signalling (multiframed) for the 30 speech channels. Robbed-bit signalling in the LSB is used in T1.
121. Which of the following is a supervisory (line) signal rather than an address signal?
Show hintHide hint
Supervisory signals describe the condition of the line.
Show answerHide answer
Answer: B. Answer signal (called party goes off-hook)
Supervisory signals show the state of a circuit: seizure, answer, clear-forward or clear-back. Address signals carry the dialled number.
122. In hierarchical alternate routing in the PSTN, traffic that cannot find a free circuit on a high-usage trunk group
Show hintHide hint
High-usage groups are deliberately sized to overflow.
Show answerHide answer
Answer: A. Overflows to an alternate route, finally reaching the final (low-blocking) trunk group
High-usage groups are designed to carry most of the traffic. Overflow traffic tries alternate routes up the hierarchy, and the final route is designed for low blocking.
123. Dynamic non-hierarchical routing (DNHR) differs from fixed hierarchical routing mainly in that
Show hintHide hint
The word 'dynamic' refers to the time dimension.
Show answerHide answer
Answer: C. Route choices change with time of day (or network load) to use idle capacity elsewhere
DNHR uses routing patterns that change with the time of day, or with real-time load. It exploits differences in busy hours across the network instead of a fixed hierarchy.
124. The OSPF link-state routing protocol computes shortest paths using
Show hintHide hint
OSPF stands for 'Open Shortest Path First'.
Show answerHide answer
Answer: D. Dijkstra's algorithm
Each OSPF router builds a full topology map from link-state advertisements. It then runs Dijkstra's shortest-path-first algorithm.
125. How many trunk groups are needed to connect 10 exchanges in a fully meshed network?
Show hintHide hint
Count each pair of exchanges once.
Show answerHide answer
Answer: B. 45
A full mesh needs n(n − 1)/2 = 10 × 9 / 2 = 45 trunk groups. This rapid growth is why tandem exchanges are used.
126. A line concentrator connects 200 subscriber lines to 20 trunks into the exchange. Its concentration ratio is
Show hintHide hint
Divide the number of lines by the number of trunks.
Show answerHide answer
Answer: B. 10:1
Concentration ratio = inlets / outlets = 200 / 20 = 10:1.
9.5 Traffic engineering
31 questions · AExE0905
127. One erlang of telephone traffic is equivalent to
Show hintHide hint
It is a dimensionless average occupancy.
Show answerHide answer
Answer: A. One circuit kept continuously busy for one hour
Traffic intensity in erlangs is the average number of simultaneously busy circuits. One erlang means one circuit occupied all the time.
128. A trunk group receives 120 calls in the busy hour with a mean holding time of 3 minutes. The offered traffic is
Show hintHide hint
Express the holding time in hours.
Show answerHide answer
Answer: D. 6 erlang
A = λh = 120 calls/h × (3/60) h = 6 E.
129. A traffic load of 180 CCS (hundred call-seconds) per hour is equal to
Show hintHide hint
How many hundred-call-seconds are there in one hour?
Show answerHide answer
Answer: A. 5 erlang
1 E = 3600 call-seconds per hour = 36 CCS, so 180/36 = 5 E.
130. The busy hour of a telephone network is
Show hintHide hint
Networks are dimensioned for peak load.
Show answerHide answer
Answer: B. The continuous 60-minute period during which the traffic is highest
Networks are sized for the busy hour, the continuous one-hour period of maximum traffic, because a network that meets its GoS then meets it at all other times.
131. In a busy hour, 20 out of 1000 call attempts are lost because no trunk is free. The grade of service is
Show hintHide hint
GoS is the fraction of offered calls that are lost.
Show answerHide answer
Answer: C. 0.02
GoS = calls lost / calls offered = 20/1000 = 0.02 (2% blocking).
132. Using the Erlang B formula, the blocking probability for 2 E offered to 4 trunks is about
Show hintHide hint
Find the ratio of the last term to the sum of all terms.
Show answerHide answer
Answer: C. 0.095
B(N, A) = (Aᴺ/N!) / Σ(Aᵏ/k!) = (16/24) / (1 + 2 + 2 + 1.333 + 0.667) = 0.667/7 ≈ 0.095.
133. Using the Erlang B formula, the blocking probability for 5 E offered to 10 trunks is about
Show hintHide hint
With twice as many trunks as erlangs, blocking is low but not negligible.
Show answerHide answer
Answer: A. 0.018
With the Erlang B recursion B(k) = A·B(k−1) / (k + A·B(k−1)) starting at B(0) = 1, B(10, 5) ≈ 0.0184.
134. 5 E is offered to a trunk group with blocking probability 0.018. The traffic actually carried is about
Show hintHide hint
Carried traffic is offered traffic minus lost traffic.
Show answerHide answer
Answer: B. 4.91 E
Carried traffic = A(1 − B) = 5 × (1 − 0.018) ≈ 4.91 E. The lost traffic is about 0.09 E.
135. The Erlang B formula assumes that
Show hintHide hint
What happens to a call that finds all trunks busy?
Show answerHide answer
Answer: B. Blocked calls are cleared (lost) and call arrivals are Poisson from an infinite source
Erlang B is a lost-calls-cleared model with Poisson arrivals from an effectively infinite population. Erlang C is the model for blocked calls that are delayed.
136. For a delay system with 2 E offered to 3 servers, the Erlang C probability that a call must wait is about
Show hintHide hint
Erlang C can be computed from Erlang B.
Show answerHide answer
Answer: D. 0.444
Erlang B(3, 2) = 0.2105. Then C = N·B / (N − A(1 − B)) = 3 × 0.2105 / (3 − 2 × 0.7895) ≈ 0.444.
137. In the same system (A = 2 E, N = 3, P(wait) = 0.444) with mean holding time 180 s, the average waiting time over all calls is
Show hintHide hint
Use W = C·h / (N − A).
Show answerHide answer
Answer: D. 80 s
W = P(wait) × h / (N − A) = 0.444 × 180 / (3 − 2) ≈ 80 s.
138. Trunking efficiency means that, for the same grade of service,
Show hintHide hint
Compare 1 group of 20 circuits with 2 groups of 10.
Show answerHide answer
Answer: C. A large trunk group can carry more traffic per circuit than several small groups with the same total number of circuits
In Erlang B tables, traffic per trunk at a fixed GoS rises with group size. Pooling circuits into one large group is more efficient.
139. A cell can carry 10 E at the required GoS. Each subscriber generates 0.025 E in the busy hour. The number of subscribers the cell supports is
Show hintHide hint
Divide the cell's capacity by each user's load.
Show answerHide answer
Answer: C. 400
Users = total traffic / traffic per user = 10 / 0.025 = 400.
140. In call-centre traffic engineering, a service level such as '80/20' means
Show hintHide hint
It combines a percentage and a time limit.
Show answerHide answer
Answer: D. 80% of calls are answered within 20 seconds
Service level is the percentage of calls answered within a target time. Delay systems like this are dimensioned with Erlang C.
141. For bursty data traffic in a wireless network, packet-switched routing is preferred over circuit-switched routing because
Show hintHide hint
Bursty traffic has long idle periods.
Show answerHide answer
Answer: B. Radio resources are shared and used only when packets are actually sent
Circuit switching holds a channel even when no data is sent, which wastes spectrum on bursty traffic. Packet switching shares the channel among users statistically.
142. In a virtual-circuit packet network,
Show hintHide hint
Think of X.25 or ATM connections.
Show answerHide answer
Answer: B. A route is set up before data transfer and all packets of the connection follow it
Virtual circuits are connection-oriented. A logical path is set up first and packets carry a short VC identifier. Datagrams are routed independently.
143. A 3000-bit message crosses 3 links (2 intermediate nodes) at 1 Mbit/s each. Ignoring propagation, processing and header overhead, how long does packet switching with three 1000-bit packets take?
Show hintHide hint
Packets can be sent on different links at the same time.
Show answerHide answer
Answer: C. 5 ms
One packet takes 1 ms per link. With pipelining the total is (number of packets + hops − 1) × 1 ms = (3 + 3 − 1) × 1 = 5 ms. Message switching would take 9 ms.
144. Compared with circuit switching, a disadvantage of packet switching for a personal communication network (PCN) carrying voice is
Show hintHide hint
Real-time speech needs a steady delay.
Show answerHide answer
Answer: A. Variable delay (jitter) caused by queuing at the nodes
Statistical multiplexing and queuing give random delays, which real-time voice handles badly. Circuit switching gives a constant delay after set-up.
145. Common channel signalling (CCS) means that
Show hintHide hint
The signalling is collected into one common link.
Show answerHide answer
Answer: C. Signalling for many speech circuits is carried as messages on a separate dedicated signalling link
In CCS (for example SS7) a data link separate from the speech trunks carries signalling messages for many circuits. Channel-associated signalling carries it per channel.
146. Which of the following is NOT an advantage of SS7 common channel signalling over channel-associated signalling?
Show hintHide hint
One option describes in-band signalling.
Show answerHide answer
Answer: D. It needs a separate signalling tone in each voice channel
SS7 removes in-band signalling tones altogether. Its advantages include faster set-up, immunity to talk-off and fraud, and support for intelligent-network database services.
147. In an SS7 network, the node that mainly routes signalling messages between other signalling points is the
Show hintHide hint
It works like a router for signalling messages.
Show answerHide answer
Answer: C. Signal Transfer Point (STP)
STPs are packet switches of the SS7 network. SSPs are exchanges that create messages, and SCPs are databases used by intelligent-network services.
148. The SS7 user part that sets up and releases trunk circuits between exchanges is
Show hintHide hint
It is the part for ISDN users.
Show answerHide answer
Answer: C. ISUP
The ISDN User Part handles call set-up and release (IAM, ACM, ANM, REL). TCAP serves transaction and database queries, and SCCP adds connectionless addressing.
149. The combined bit rate of the two B channels and the D channel of an ISDN Basic Rate Interface (2B + D), excluding framing overhead, is
Show hintHide hint
Add both B channels and the D channel.
Show answerHide answer
Answer: A. 144 kbit/s
2 × 64 kbit/s (B) + 16 kbit/s (D) = 144 kbit/s. With framing overhead the S/T interface runs at 192 kbit/s.
150. The European ISDN Primary Rate Interface on a 2.048 Mbit/s link has the structure
Show hintHide hint
It is based on the 32-slot E1 frame.
Show answerHide answer
Answer: D. 30B + D (64 kbit/s D channel)
The E1 PRI carries 30 B channels plus a 64 kbit/s D channel in TS16, with TS0 for framing. The North American T1 PRI is 23B + D.
151. A non-ISDN terminal (TE2), such as an analog telephone, connects to the ISDN S/T interface through a
Show hintHide hint
It adapts an older terminal to the ISDN interface.
Show answerHide answer
Answer: A. Terminal adapter (TA)
TE1 devices connect directly to the S/T reference point. TE2 devices need a terminal adapter at the R reference point.
152. In broadband ISDN based on ATM, each cell is
Show hintHide hint
The cells are small and of fixed size.
Show answerHide answer
Answer: B. 53 bytes long with a 5-byte header and 48-byte payload
ATM uses fixed 53-byte cells (5-byte header and 48-byte payload). This allows fast hardware switching and low delay variation.
153. The maximum throughput of pure ALOHA, reached at offered load G = 0.5, is about
Show hintHide hint
The vulnerable period is two packet times.
Show answerHide answer
Answer: B. 18.4%
S = G e^(−2G), with a maximum of 1/(2e) ≈ 0.184 at G = 0.5.
154. The maximum throughput of slotted ALOHA is about
Show hintHide hint
Slotting doubles the pure ALOHA maximum.
Show answerHide answer
Answer: D. 36.8%
Slotting halves the vulnerable period, so S = G e^(−G), with a maximum of 1/e ≈ 0.368 at G = 1.
155. A pure ALOHA channel has offered load G = 0.25 packets per packet time. Its throughput is about
Show hintHide hint
Use S = G e^(−2G).
Show answerHide answer
Answer: A. 0.152
S = G e^(−2G) = 0.25 × e^(−0.5) = 0.25 × 0.607 ≈ 0.152.
156. IEEE 802.11 wireless LANs use CSMA/CA instead of CSMA/CD mainly because
Show hintHide hint
Can a radio 'listen' while it is transmitting?
Show answerHide answer
Answer: B. A wireless station cannot reliably detect a collision while it is transmitting (for example, because of hidden terminals)
A radio's own transmission drowns out other signals, and hidden terminals may not hear each other. So 802.11 avoids collisions with backoff and RTS/CTS instead of detecting them.
157. In packet reservation multiple access (PRMA), used for voice in packet radio,
Show hintHide hint
It mixes contention with reservation.
Show answerHide answer
Answer: A. A terminal contends for a slot with slotted ALOHA and, if it succeeds, keeps that slot in later frames while its talkspurt lasts
PRMA combines slotted ALOHA contention with TDMA-like reservation. A successful voice terminal keeps its slot until the talkspurt ends, which suits speech with silent gaps.
9.6 Rules and regulations
30 questions · AExE0906
158. The International Telecommunication Union was founded in Paris (as the International Telegraph Union) in the year
Show hintHide hint
It is older than the telephone.
Show answerHide answer
Answer: D. 1865
The ITU was founded on 17 May 1865. It took its present name in 1932 and became a UN specialized agency in 1947.
159. The headquarters of the ITU is located in
Show hintHide hint
Several UN agencies are based in this Swiss city.
Show answerHide answer
Answer: D. Geneva, Switzerland
The ITU has had its headquarters in Geneva since 1948.
160. The ITU is
Show hintHide hint
It belongs to the UN system.
Show answerHide answer
Answer: B. A specialized agency of the United Nations for information and communication technologies
The ITU is the UN specialized agency for ICTs. Its members are states, plus sector members from industry and academia.
161. ITU recommendations such as G.711 (PCM coding of voice) and H.323 are produced by
Show hintHide hint
'T' is for telecommunication standardization.
Show answerHide answer
Answer: C. ITU-T
ITU-T, the Telecommunication Standardization Sector, publishes the G-, H-, V-, X- and other series of recommendations.
162. The main role of the ITU-D sector is to
Show hintHide hint
'D' is for development.
Show answerHide answer
Answer: C. Promote telecommunication/ICT development and help close the digital divide in developing countries
ITU-D (Development Sector) works on capacity building, policy assistance and ICT access in developing countries.
163. The supreme policy-making body of the ITU, which meets every four years, is the
Show hintHide hint
It is attended by plenipotentiaries of all member states.
Show answerHide answer
Answer: C. Plenipotentiary Conference
The Plenipotentiary Conference sets ITU policy, adopts the strategic and financial plans and elects the Secretary-General and other officials.
164. The international treaty called the Radio Regulations is revised by the
Show hintHide hint
It is the radiocommunication conference.
Show answerHide answer
Answer: A. World Radiocommunication Conference (WRC)
WRCs, held every 3 to 4 years under ITU-R, review and revise the Radio Regulations, including the international table of frequency allocations.
165. For frequency allocation, the ITU Radio Regulations divide the world into three regions. Nepal falls in
Show hintHide hint
Asia and Oceania form one region.
Show answerHide answer
Answer: B. Region 3
Region 1 is Europe, Africa, the Middle East and the former USSR. Region 2 is the Americas. Region 3 is most of Asia and Oceania, which includes Nepal.
166. In ITU terminology, authorizing a particular radio station to use a specific frequency under stated conditions is called
Show hintHide hint
Allocation is for services and allotment is for areas. What is the term for a station?
Show answerHide answer
Answer: D. Assignment
Allocation means a band given to a radio service. Allotment means a channel given to a country or area. Assignment means the frequency given to an individual station.
167. In a frequency allocation table, a 'secondary' service allocation means that its stations
Show hintHide hint
Secondary services rank below primary ones.
Show answerHide answer
Answer: B. Must not cause harmful interference to primary services and cannot claim protection from them
Secondary services must accept interference from primary services and must not cause harmful interference to them. In the table they are printed in normal type, not capitals.
168. World Telecommunication and Information Society Day, observed on 17 May, marks
Show hintHide hint
Recall the ITU founding date.
Show answerHide answer
Answer: D. The founding of the ITU
17 May 1865 is the date the International Telegraph Convention was signed and the ITU was founded.
169. Under the ITU-T E.164 international numbering plan, Nepal's country calling code is
Show hintHide hint
Nepal's code is used before every international call to Nepal.
Show answerHide answer
Answer: A. 977
Nepal's code is +977. +975 is Bhutan, +880 is Bangladesh and +91 is India.
170. Under ITU-T Recommendation E.164, the maximum length of an international telephone number (country code included) is
Show hintHide hint
It is more than the familiar 10-digit national numbers.
Show answerHide answer
Answer: B. 15 digits
E.164 limits international public telecommunication numbers to 15 digits, country code included.
171. The ITU call-sign prefix allocated to Nepal (seen in amateur call signs such as 9N1...) is
Show hintHide hint
It begins with a digit.
Show answerHide answer
Answer: A. 9N
The ITU allocated the 9N prefix to Nepal. VU is India, S2 is Bangladesh and A5 is Bhutan.
172. The Nepal Telecommunications Authority (NTA) was established under the
Show hintHide hint
It is the act that governs telecommunications.
Show answerHide answer
Answer: A. Telecommunications Act, 2053 (1997)
The Telecommunications Act, 2053 created the NTA as the autonomous telecom regulator, and the NTA began functioning in 1998.
173. Which of the following is NOT a function of the Nepal Telecommunications Authority?
Show hintHide hint
A regulator does not compete in the market it regulates.
Show answerHide answer
Answer: C. Operating the national fixed-line and mobile network as a service provider
The NTA regulates the sector: licensing, tariffs, interconnection, quality of service and consumer protection. Operators such as Nepal Telecom provide the service.
174. The Rural Telecommunication Development Fund in Nepal is intended mainly to
Show hintHide hint
The purpose is in its name.
Show answerHide answer
Answer: A. Extend telecom services to rural and unserved areas (universal service)
The fund, collected from operators' revenues, finances telecom and internet access in rural and commercially unattractive areas. This supports universal service obligations.
175. In Nepal, approving telecommunication terminal equipment (such as handsets) for use on public networks after technical checks is called
Show hintHide hint
It certifies a model or type of equipment.
Show answerHide answer
Answer: C. Type approval
Type approval certifies that equipment meets technical and safety standards before it is sold or connected to the network. In Nepal this is done by the telecom regulator, NTA.
176. In Nepal's telecom sector, formulating national telecommunication and ICT policy is primarily the role of
Show hintHide hint
Policy comes from the government ministry.
Show answerHide answer
Answer: D. The Ministry of Communication and Information Technology
The ministry formulates government policy, and the NTA regulates and licenses operators within that policy.
177. The ministry of the Government of Nepal responsible for telecommunications, broadcasting and information technology is the
Show hintHide hint
Look for 'communication' in the name.
Show answerHide answer
Answer: B. Ministry of Communication and Information Technology
The Ministry of Communication and Information Technology (MoCIT) covers communication, broadcasting, postal services and IT.
178. In Nepal, the Frequency Policy Determination Committee is mainly responsible for
Show hintHide hint
Look at what its name says it determines.
Show answerHide answer
Answer: B. Making policy on the allocation and use of radio frequencies
The committee decides policy on how the radio spectrum is allocated and used in Nepal, in line with the ITU Radio Regulations.
179. The main purpose of a National Frequency Allocation Plan (table) is to
Show hintHide hint
It is the national version of the ITU frequency table.
Show answerHide answer
Answer: A. Specify which radio services may use each frequency band in the country, consistent with the ITU Radio Regulations
A national frequency allocation table adapts the ITU international table to national needs, showing primary and secondary services in each band.
180. Radio spectrum needs regulation mainly because it is
Show hintHide hint
Two transmitters on the same frequency in the same place cause interference.
Show answerHide answer
Answer: B. A limited natural resource whose uncontrolled use causes harmful interference
Spectrum is finite and shared. Without coordinated allocation and licensing, users interfere with each other. Regulation makes efficient and interference-free use possible.
181. In ITU nomenclature, the VHF band covers
Show hintHide hint
FM broadcasting at about 100 MHz is in this band.
Show answerHide answer
Answer: A. 30 to 300 MHz
HF is 3 to 30 MHz, VHF is 30 to 300 MHz, UHF is 300 MHz to 3 GHz and SHF is 3 to 30 GHz.
182. The wavelength range of the UHF band (300 MHz to 3 GHz) is
Show hintHide hint
Apply λ = c/f at both band edges.
Show answerHide answer
Answer: D. 1 m to 10 cm
λ = c/f gives 3×10⁸/3×10⁸ = 1 m and 3×10⁸/3×10⁹ = 0.1 m, so UHF waves are decimetric waves.
183. ITU band number N covers 0.3 × 10ᴺ Hz to 3 × 10ᴺ Hz. A 2.4 GHz Wi-Fi signal lies in band
Show hintHide hint
Find N for which 0.3×10ᴺ ≤ 2.4×10⁹ < 3×10ᴺ.
Show answerHide answer
Answer: C. 9 (UHF)
Band 9 covers 0.3×10⁹ to 3×10⁹ Hz (300 MHz to 3 GHz, UHF), and 2.4 GHz is inside it.
184. According to the ITU Radio Regulations, 'harmful interference' is interference that
Show hintHide hint
It is more serious than any small noise increase.
Show answerHide answer
Answer: A. Endangers a radionavigation or safety service, or seriously degrades, obstructs or repeatedly interrupts an authorized radio service
This is the RR definition. It goes beyond minor or 'permissible' interference and requires regulatory action.
185. Wi-Fi and Bluetooth devices can operate without individual licences in most countries mainly because they use
Show hintHide hint
The band is also used by microwave ovens.
Show answerHide answer
Answer: B. The 2.4 GHz ISM band under low-power licence-exempt rules
The 2.4 GHz industrial, scientific and medical band is open to licence-exempt low-power devices. These devices must accept interference and must not cause harmful interference.
186. Assigning spectrum licences to the bidders willing to pay the most is called
Show hintHide hint
It is the market-based method.
Show answerHide answer
Answer: C. Spectrum auction
Auctions are a market-based method that assigns spectrum to whoever values it most. A beauty contest assigns spectrum by comparing proposals against set criteria.
187. Nepal's Radio Act, which regulates radio communication equipment through licensing, was enacted in
Show hintHide hint
It is the oldest of these laws.
Show answerHide answer
Answer: D. 2014 BS (1957 AD)
The Radio Act, 2014 regulates radio apparatus through licensing. 2053 is the Telecommunications Act, 2049 the National Broadcasting Act and 2063 the Electronic Transactions Act.