Nepal Engineering Council · Electronics, Communication & Information Engineering · Chapter 9
Telecommunication and Soft Switching
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187 questions in 6 syllabus topics.
9.1 Telecommunication and wireless communication
31 questions · AEiE0901
1. Who was granted the 1876 patent for the telephone, the starting point of switched voice telephony?
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Think of the inventor of the telephone, not the telegraph or radio.
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Answer: A. Alexander Graham Bell
Alexander Graham Bell received the US telephone patent in 1876. Morse is associated with the telegraph, Marconi with radio and Strowger with the automatic exchange.
2. The first transatlantic wireless (radio) signal, sent in 1901, is credited to:
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He is called the father of practical radio.
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Answer: B. Guglielmo Marconi
Marconi sent the first transatlantic radio signal (the letter S in Morse code) in 1901. Maxwell predicted electromagnetic waves and Hertz demonstrated them in the lab.
3. First-generation (1G) cellular systems such as AMPS and NMT were characterised by:
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1G was the only analog generation.
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Answer: B. Analog FM voice with FDMA
1G systems carried analog FM voice, with each user given a separate frequency channel (FDMA). Digital voice arrived with 2G.
4. GPRS, often called a 2.5G technology, added which main capability to GSM networks?
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The 'G' in GPRS stands for General Packet ...
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Answer: A. Packet-switched data service
GPRS added a packet-switched core (SGSN/GGSN) on top of circuit-switched GSM, so data could be sent without holding a dedicated circuit.
5. 4G LTE uses which multiple-access schemes on the downlink and uplink, respectively?
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The uplink scheme is chosen to save handset battery power.
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Answer: C. OFDMA and SC-FDMA
LTE uses OFDMA on the downlink and single-carrier FDMA on the uplink, because SC-FDMA has a lower peak-to-average power ratio, which suits battery-powered handsets.
6. Which of the following is NOT one of the three ITU usage scenarios defined for 5G (IMT-2020)?
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Three of the options are 5G buzzwords. One belongs to 1G.
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Answer: D. Analog circuit-switched voice (AMPS)
IMT-2020 defines eMBB, URLLC and mMTC as the three 5G usage scenarios. Analog circuit-switched voice belongs to 1G.
7. Which of the following is an unguided (wireless) transmission medium?
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Which one has no physical conductor or waveguide?
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Answer: B. Terrestrial microwave link
Unguided media carry signals through free space, as radio, microwave and infrared links do. Cable and fibre are guided media.
8. In twisted-pair cable, the conductors are twisted together mainly to:
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Think about how noise induced in the two wires behaves.
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Answer: A. Reduce crosstalk and electromagnetic interference
Twisting makes the noise coupled into both wires nearly equal, so it cancels in the differential signal. This reduces crosstalk and EMI.
9. An optical fibre has a core refractive index of 1.50 and a cladding index of 1.45. Its numerical aperture is approximately:
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Use the difference of the squares of the two indices.
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Answer: B. 0.38
NA = √(n1² − n2²) = √(2.25 − 2.1025) = √0.1475 ≈ 0.38.
10. Compared with multimode fibre, single-mode fibre has a very small core (about 8–10 µm) mainly to:
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Only one ray path propagates.
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Answer: C. Eliminate modal (intermodal) dispersion
With only one propagating mode, single-mode fibre has no intermodal dispersion, so it supports much higher bandwidth over longer distances.
11. What is the free-space path loss of a 900 MHz signal over 10 km? Use FSPL(dB) = 32.44 + 20 log d(km) + 20 log f(MHz).
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Remember that 20 log 10 = 20.
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Answer: B. 111.5 dB
FSPL = 32.44 + 20 log 10 + 20 log 900 = 32.44 + 20 + 59.08 ≈ 111.5 dB.
12. A transmitter radiates 10 W at 900 MHz through unity-gain antennas. Using the Friis free-space equation, what is the received power at 1 km?
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First find λ, then apply Pr = Pt·Gt·Gr·λ²/(4πd)².
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Answer: B. About −51.5 dBm
λ = 1/3 m. Pr = Pt·λ²/(4πd)² = 10 × 0.111/(12566)² ≈ 7.0 × 10⁻⁹ W ≈ −51.5 dBm.
13. In the free-space propagation model, doubling the distance between transmitter and receiver increases the path loss by about:
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Received power varies as the inverse square of distance.
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Answer: A. 6 dB
Free-space received power varies as 1/d², so doubling d reduces power by a factor of 4, which is 10 log 4 ≈ 6 dB.
14. The Friis free-space equation is valid only in the far field. For an antenna of largest dimension 1 m operating at 900 MHz, the Fraunhofer distance (2D²/λ) is:
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Find the wavelength first.
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Answer: A. 6 m
λ = 3×10⁸/9×10⁸ = 0.333 m, so df = 2D²/λ = 2 × 1/0.333 = 6 m.
15. Scattering in mobile radio propagation occurs mainly when the wave meets:
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Compare the object size with λ.
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Answer: D. Objects small compared with the wavelength, or rough surfaces
Scattering comes from objects whose size is comparable to or smaller than λ, such as foliage, lamp posts and rough ground, and spreads energy in many directions.
16. Which propagation mechanism lets a radio signal reach a receiver in the geometric shadow behind a hill, and is explained by Huygens' principle?
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Think of the knife-edge model.
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Answer: D. Diffraction
Diffraction bends waves around the edges of obstacles, as Huygens' principle of secondary wavelets explains. The knife-edge model is used to estimate it.
17. Reflection, as a large-scale propagation mechanism, occurs when a radio wave strikes an object whose dimensions are:
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Contrast this with scattering.
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Answer: D. Very large compared with the wavelength
Reflection occurs at surfaces such as the earth, building walls and the ground, which are much larger than λ. Small objects cause scattering instead.
18. In the two-ray ground reflection model, at large distances the received power varies with distance d as:
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The path-loss exponent is larger than in free space.
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Answer: C. 1/d⁴
At large d, the direct and ground-reflected rays nearly cancel, giving Pr ≈ Pt·Gt·Gr·ht²·hr²/d⁴. Path loss therefore rises by 40 dB per decade of distance.
19. A hexagonal cellular layout uses shift parameters i = 2 and j = 1. What is the cluster size N?
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Use N = i² + ij + j².
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Answer: D. 7
N = i² + ij + j² = 4 + 2 + 1 = 7.
20. For a cellular system with cluster size N = 7, path-loss exponent n = 4 and six first-tier co-channel interferers, the approximate signal-to-interference ratio is:
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Use S/I = (√(3N))ⁿ / i0, with i0 = 6.
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Answer: A. 18.7 dB
S/I = (√(3N))ⁿ/i0 = (√21)⁴/6 = 441/6 = 73.5, which is about 18.7 dB.
21. A system has 33 MHz of total spectrum and uses two 25 kHz simplex channels per duplex channel. With a 7-cell reuse pattern, how many duplex channels does each cell get (approximately)?
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First find the total number of duplex channels.
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Answer: C. 94
Duplex channel bandwidth = 50 kHz, so total channels = 33 MHz/50 kHz = 660. Per cell = 660/7 ≈ 94.
22. In which channel assignment strategy does the MSC allocate a channel from a common pool on each call request, giving lower blocking at the cost of more computation?
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The channels are not tied permanently to any cell.
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Answer: C. Dynamic channel assignment
In dynamic channel assignment, channels are not permanently allocated to cells. The MSC assigns them on demand based on current usage and interference, which improves trunking efficiency.
23. Soft handoff, in which a mobile communicates with two or more base stations at the same time during handover, is characteristic of:
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All cells share the same carrier frequency in this system.
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Answer: D. CDMA systems
In CDMA, all cells use the same frequency, so a mobile can combine signals from several base stations (make-before-break). FDMA and TDMA systems use hard handoff.
24. Reserving a fraction of the channels in each cell only for handoff requests is called the:
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The reserved channels guard ongoing calls.
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Answer: C. Guard channel concept
Guard channels give priority to handoffs. Dropping an ongoing call is worse than blocking a new one, at the cost of slightly more blocking for new calls.
25. A mobile moves at 72 km/h and receives a 900 MHz carrier. What is the maximum Doppler shift?
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Convert km/h to m/s first.
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Answer: A. 60 Hz
v = 20 m/s and λ = 1/3 m, so fd = v/λ = 60 Hz.
26. A channel has an RMS delay spread of 1 µs. Using Bc ≈ 1/(5στ), its coherence bandwidth is approximately:
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Plug the delay spread into Bc ≈ 1/(5στ).
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Answer: A. 200 kHz
Bc ≈ 1/(5 × 1 µs) = 200 kHz.
27. A signal undergoes flat fading when:
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Compare the signal bandwidth with the channel coherence bandwidth.
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Answer: D. Its bandwidth is much smaller than the channel coherence bandwidth
Flat fading requires Bs ≪ Bc (equivalently Ts ≫ στ), so all spectral components of the signal fade together. The other conditions describe frequency-selective or fast fading.
28. A power delay profile has three paths with relative (linear) powers 1, 0.5 and 0.5 at excess delays of 0, 1 and 2 µs. What is the RMS delay spread?
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Find the power-weighted mean and mean-square delays, then take √(second moment − mean²).
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Answer: B. About 0.83 µs
Mean delay = (0 + 0.5 + 1)/2 = 0.75 µs. Mean square delay = (0 + 0.5 + 2)/2 = 1.25 µs². στ = √(1.25 − 0.5625) ≈ 0.83 µs.
29. The received envelope of a multipath channel follows a Rayleigh distribution when:
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With a strong direct path, the distribution becomes Rician.
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Answer: B. There is no dominant line-of-sight component
Rayleigh fading arises from many scattered components with no dominant path. A dominant LOS component gives Rician fading instead.
30. In a Rayleigh fading channel, P(r ≤ R) = 1 − exp(−ρ²), where ρ = R/R(rms). What is the probability that the envelope is more than 10 dB below its RMS value?
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10 dB below means ρ² = 0.1.
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Answer: C. About 9.5%
ρ = −10 dB gives ρ² = 0.1, so P = 1 − e^(−0.1) ≈ 0.095, or about 9.5%.
31. For a Rayleigh fading signal with a maximum Doppler frequency of 60 Hz, what is the level crossing rate at the RMS level (ρ = 1)? Use N = √(2π)·fm·ρ·e^(−ρ²).
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Put ρ = 1 into the formula and remember e^(−1) ≈ 0.368.
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Answer: C. About 55 crossings per second
N = 2.507 × 60 × 1 × e^(−1) = 2.507 × 60 × 0.368 ≈ 55 crossings per second.
9.2 Equalization and diversity techniques
30 questions · AEiE0902
32. What is the main purpose of an equalizer in a digital radio receiver?
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Think about what time dispersion does to adjacent symbols.
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Answer: A. To compensate for intersymbol interference caused by multipath
Time dispersion in a multipath channel spreads each symbol into its neighbours, which causes ISI. The equalizer compensates for the channel's frequency response to reduce that ISI.
33. A zero-forcing equalizer forces ISI to zero by inverting the channel response. What is its main drawback?
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Look at what 1/H(f) does near a spectral null.
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Answer: A. Severe noise enhancement at frequencies where the channel has deep nulls
A ZF equalizer has gain 1/H(f), so wherever H(f) is small it applies very large gain and greatly amplifies the noise.
34. Which equalizer criterion minimizes the combined effect of residual ISI and noise, rather than forcing ISI to exactly zero?
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Its name describes what it minimizes.
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Answer: D. Minimum mean square error (MMSE)
An MMSE equalizer minimizes E[|e|²], the error between the equalizer output and the desired symbol. It balances ISI removal against noise enhancement.
35. A decision feedback equalizer (DFE) differs from a linear transversal equalizer because it:
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The word 'feedback' refers to earlier decisions.
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Answer: B. Uses past detected symbols in a feedback filter to cancel ISI
A DFE is nonlinear. Its feedback filter takes earlier symbol decisions and subtracts the post-cursor ISI they cause from the current symbol.
36. Which equalizer uses the Viterbi algorithm to find the most likely transmitted sequence, and so is optimal in minimizing sequence error probability?
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Look for the sequence estimator.
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Answer: C. Maximum likelihood sequence estimation (MLSE)
An MLSE equalizer tests possible data sequences against a channel model and uses the Viterbi algorithm to pick the most likely one. GSM receivers commonly use it.
37. The most common structure for a linear equalizer is a:
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Think of delayed copies of the input, each multiplied by a weight.
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Answer: C. Tapped delay line (transversal FIR) filter with adjustable tap weights
A linear transversal equalizer is an FIR filter: delayed samples are weighted by adjustable tap coefficients and summed.
38. A channel has an RMS delay spread of 1 µs. If ISI is negligible only when the symbol period is at least 10 times the delay spread, what is the highest symbol rate that needs no equalizer?
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First find the minimum symbol period.
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Answer: B. 100 ksymbols/s
Ts ≥ 10 × 1 µs = 10 µs, so the symbol rate must be at most 1/10 µs = 100 ksymbols/s.
39. GSM transmits at 270.833 kbit/s. Roughly how many bit periods must its equalizer span to handle a 15 µs delay spread?
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Divide the delay spread by the bit period.
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Answer: D. About 4 bit periods
Tb = 1/270.833 kHz ≈ 3.69 µs, and 15/3.69 ≈ 4 bit periods.
40. An adaptive equalizer normally works in two modes. These are:
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One mode uses a known sequence; the other follows the channel.
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Answer: B. Training mode and tracking mode
In training mode, a known sequence lets the taps converge. In tracking mode, decision-directed adaptation follows slow channel changes during data.
41. Why does a TDMA burst usually contain a known training sequence?
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The receiver already knows these bits.
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Answer: D. So the adaptive equalizer can estimate the channel and set its tap weights
The receiver knows the training bits in advance. Comparing them with what arrives lets the equalizer estimate the channel and adjust its taps.
42. What is the main advantage of the LMS algorithm over the RLS algorithm for adaptive equalization?
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LMS is known for being simple rather than fast.
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Answer: D. Much lower computational complexity
LMS needs only about 2N+1 operations per iteration, so it is simple to implement. Its drawback is slow convergence compared with RLS.
43. Compared with LMS, the recursive least squares (RLS) algorithm:
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There is a trade-off between speed and complexity.
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Answer: A. Converges faster but needs more computation
RLS uses the inverse correlation matrix and converges much faster than LMS. The price is higher complexity per iteration.
44. An LMS equalizer tap has weight w = 0.5. With step size µ = 0.1, error e = 0.2 and input sample x = 1.0, what is the updated weight using w(new) = w + µ·e·x?
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Multiply µ, e and x, then add the product to w.
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Answer: B. 0.52
w(new) = 0.5 + 0.1 × 0.2 × 1.0 = 0.5 + 0.02 = 0.52.
45. Taking LMS complexity as 2N + 1 operations per iteration, how many operations does an 11-tap LMS equalizer need per iteration?
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Substitute N = 11.
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Answer: C. 23
2N + 1 = 2 × 11 + 1 = 23 operations per iteration.
46. In the LMS algorithm, increasing the step size µ generally:
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Bigger steps reach the target sooner but overshoot more.
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Answer: A. Speeds up convergence but increases misadjustment and risks instability
A larger µ makes bigger corrections, so the taps converge faster. They also fluctuate more around the optimum, and too large a µ makes the algorithm diverge.
47. An equalizer that adapts without any training sequence, for example using the constant modulus algorithm, is called a:
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It works 'without seeing' known data.
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Answer: C. Blind equalizer
Blind algorithms such as CMA use known properties of the signal, such as constant envelope, instead of a training sequence.
48. Diversity improves performance on fading channels because:
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Think about the probability of several independent events happening together.
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Answer: C. Independently fading branches are unlikely to be in a deep fade at the same time
With several uncorrelated copies of the signal, the chance that all of them fade deeply at once is much smaller, so the receiver can use the best one or combine them.
49. Selecting the strongest of several base stations, as opposed to several antennas at one site, is called macroscopic diversity. It mainly combats:
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Shadowing has a large spatial scale.
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Answer: B. Large-scale shadowing by terrain and buildings
Macroscopic diversity uses widely separated sites so that the signals are not shadowed by the same obstacle. Microscopic diversity combats small-scale fading.
50. With selection diversity on M Rayleigh branches, the mean SNR improvement factor is Σ(1/k) for k = 1 to M. For M = 4, the improvement is about:
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Add the harmonic series up to 4, then convert to dB.
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Answer: D. 3.2 dB
1 + 1/2 + 1/3 + 1/4 = 2.083, and 10 log 2.083 ≈ 3.2 dB.
51. A maximal ratio combiner receives two branches with instantaneous SNRs of 10 and 5 (linear). What is the output SNR?
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For MRC, the branch SNRs add.
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Answer: A. About 11.8 dB
MRC output SNR is the sum of the branch SNRs: 10 + 5 = 15, and 10 log 15 ≈ 11.8 dB.
52. Compared with selection combining, maximal ratio combining:
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It is the optimum linear combiner.
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Answer: A. Co-phases and weights each branch by its signal amplitude and gives the highest output SNR
MRC co-phases all branches and weights each in proportion to its signal-to-noise level before summing. This is the optimum linear combiner.
53. Each branch of a selection diversity receiver has P(SNR < threshold) = 0.095. With two independent branches, the probability that the output SNR is below threshold is about:
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Both branches must fail together.
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Answer: D. 0.009
Selection fails only if both branches are below threshold: 0.095² ≈ 0.009.
54. A RAKE receiver in a CDMA system improves performance by:
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Each finger handles one delayed path.
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Answer: B. Resolving multipath components with separate fingers and combining them
Each RAKE finger correlates with one delayed multipath copy. The outputs are then combined, often by MRC, so multipath becomes a source of diversity.
55. For space diversity at a mobile handset in a rich-scattering environment, antennas spaced about λ/2 give nearly uncorrelated fading. At 900 MHz, this spacing is about:
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Find λ, then halve it.
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Answer: A. 16.7 cm
λ = 3×10⁸/9×10⁸ = 0.333 m, so λ/2 ≈ 16.7 cm.
56. Why do base-station space-diversity antennas need much wider spacing (often tens of wavelengths) than antennas on a mobile?
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Where are most of the scatterers located?
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Answer: D. Scatterers are near the mobile, so signals arrive at the base station over a narrow angle
Local scatterers surround the mobile, not the elevated base station. The small angular spread at the base station gives slow spatial decorrelation, so the antennas must be far apart.
57. Polarization diversity at a base station typically uses:
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Diversity is obtained without any physical spacing.
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Answer: C. Two co-located antennas with orthogonal (e.g., ±45°) polarizations
Scattering depolarizes the transmitted wave, so the two orthogonally polarized branches fade nearly independently. The antennas can share one mount, with no spatial separation needed.
58. For frequency diversity, the carrier frequencies must be separated by more than the:
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Fading at two frequencies is uncorrelated beyond a certain bandwidth.
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Answer: B. Coherence bandwidth of the channel
Signals separated by more than the coherence bandwidth fade nearly independently, which gives diversity gain.
59. What is the main practical cost of frequency diversity compared with space or polarization diversity?
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Think about which resource is duplicated.
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Answer: C. It needs extra spectrum, and a separate receiver for each carrier
Frequency diversity sends the same information on two or more widely separated carriers. It therefore uses additional bandwidth and more receiver hardware.
60. Time diversity is obtained by sending repeated (or interleaved) versions of a message separated in time by more than the channel's:
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This parameter is set by the Doppler spread.
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Answer: A. Coherence time
Samples separated by more than Tc see nearly independent fading. That is why interleaving combined with channel coding gives time diversity.
61. A mobile channel has a maximum Doppler shift of 100 Hz. Using Tc ≈ 0.423/fm, what is the minimum spacing between repeated transmissions for time diversity?
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Compute the coherence time.
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Answer: B. About 4.2 ms
Tc ≈ 0.423/100 = 4.23 ms, so repetitions should be at least about 4.2 ms apart.
9.3 Switching systems and Traffic engineering
31 questions · AEiE0903
62. Which was the first automatic electromechanical telephone exchange, in which the dialled pulses directly stepped selectors?
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It is named after its inventor, an undertaker.
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Answer: A. Strowger step-by-step system
Almon Strowger's step-by-step (direct progressive control) switch was the first automatic exchange. Each dialled digit stepped the selector switches directly.
63. The main feature that distinguished crossbar exchanges from Strowger exchanges was the use of:
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The control equipment is shared and not tied to each selector.
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Answer: B. Common control (registers and markers) separate from the switching network
In crossbar systems, common control equipment receives the digits, finds a path and then sets up the crosspoints. In Strowger, each selector step followed the dial pulses directly.
64. In a stored program control (SPC) exchange, switching functions are controlled by:
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The name says where the program lives.
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Answer: C. A digital computer executing software stored in memory
SPC uses a processor and stored programs to control the switching network. New features can then be added by changing software instead of rewiring.
65. How many crosspoints does a single-stage, non-blocking square space switch need to connect 100 inlets to 100 outlets?
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Every inlet needs a crosspoint to every outlet.
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Answer: C. 10,000
A single-stage N × N matrix needs N² = 100² = 10,000 crosspoints.
66. A strictly non-blocking three-stage Clos network has N = 128 lines, n = 8 inputs per first-stage switch and k = 2n − 1 middle-stage switches. Using 2Nk + k(N/n)², how many crosspoints does it need?
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First compute k = 2n − 1 and N/n.
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Answer: B. 7,680
k = 15 and N/n = 16. Crosspoints = 2 × 128 × 15 + 15 × 16² = 3840 + 3840 = 7,680, compared with 16,384 for a single stage.
67. An E1 frame of 125 µs has 32 time slots. In a time-slot interchange (TSI) switch, how much time is available to handle each slot?
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Divide the frame duration by the number of slots.
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Answer: D. About 3.9 µs
125 µs / 32 slots ≈ 3.9 µs per time slot.
68. The basic idea of soft switching is to:
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The control plane and the bearer plane are split.
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Answer: D. Separate call-control software from the media (bearer) switching hardware
A softswitch runs call control as software on general-purpose servers. Media gateways carry the voice traffic, and open protocols connect the two.
69. In hierarchical alternate routing, traffic that finds all circuits of a high-usage trunk group busy is:
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The last-choice route is engineered to carry the overflow.
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Answer: A. Overflowed to the next choice route, ending with the final route
High-usage routes are tried first. Overflow traffic goes to alternate routes and finally to a final route, which is dimensioned for a low grade of service.
70. In DTMF (touch-tone) signalling, each dialled digit is represented by:
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'Dual tone' is in the name.
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Answer: A. One tone from a low-frequency group plus one from a high-frequency group
DTMF sends two tones at once: one from the low group (697–941 Hz) and one from the high group (1209–1633 Hz). Each pair identifies one key.
71. In telephone signalling, which type of signal carries supervisory information such as seizure, answer and clear-down?
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Supervision concerns the state of the line or trunk.
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Answer: C. Line signalling
Line signals carry the supervisory states of a circuit (seize, answer, clear). Register signalling carries the address digits between exchanges.
72. One erlang of traffic corresponds to:
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It is a measure of average occupancy.
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Answer: A. One circuit continuously occupied for one hour
Traffic intensity in erlangs is the average number of simultaneously busy circuits. One erlang means one circuit kept busy continuously.
73. During the busy hour, a trunk group carries 120 calls with an average holding time of 3 minutes. What is the traffic intensity?
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Express the holding time in hours.
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Answer: C. 6 E
A = (calls per hour × holding time in hours) = 120 × 3/60 = 6 erlangs.
74. A traffic intensity of 5 erlangs, expressed in CCS (hundred call-seconds per hour), is:
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An hour has 3600 seconds.
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Answer: B. 180 CCS
1 E = 3600 call-seconds per hour = 36 CCS, so 5 E = 180 CCS.
75. An exchange serves 200 subscribers, each generating 0.05 E in the busy hour. The total offered traffic is:
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Multiply the per-user traffic by the number of users.
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Answer: B. 10 E
Total traffic = 200 × 0.05 = 10 erlangs.
76. Using the Erlang B formula, what is the blocking probability when 2 E of traffic is offered to 4 trunks?
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Sum Aᵏ/k! for k = 0 to N in the denominator.
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Answer: C. About 0.095
B = (A⁴/4!) / Σ(Aᵏ/k!) for k = 0 to 4 = (16/24) / (1 + 2 + 2 + 1.333 + 0.667) = 0.667/7 ≈ 0.095.
77. A trunk group is offered 10 E and has a grade of service of 0.02. What traffic is actually carried?
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Carried traffic = offered traffic − lost traffic.
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Answer: C. 9.8 E
Lost traffic = A × B = 10 × 0.02 = 0.2 E, so carried traffic = 10 − 0.2 = 9.8 E.
78. In a busy hour, 1500 calls are offered to a route and 30 of them are blocked. The grade of service is:
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Divide lost calls by offered calls.
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Answer: B. 0.02
GoS = blocked calls / offered calls = 30/1500 = 0.02 (2%).
79. The 'busy hour' of a telephone exchange is the:
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Equipment is dimensioned for this period.
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Answer: A. Continuous 60-minute period of the day with the highest traffic
Exchanges are dimensioned for the busy hour, the uninterrupted one-hour period with the maximum traffic intensity.
80. The Erlang B formula is based on which assumption about calls that find all trunks busy?
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Erlang C is the queuing version.
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Answer: A. They are cleared (lost) from the system
Erlang B models a lost-calls-cleared system with Poisson arrivals. Erlang C models a lost-calls-delayed (queuing) system.
81. Trunking efficiency in cellular systems means that, for the same grade of service, a large pool of channels:
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Compare the Erlang B capacity of 5 channels with that of 50.
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Answer: A. Carries more traffic per channel than a small pool
Under Erlang B, the traffic carried per channel at a fixed GoS rises as the number of trunked channels grows. Splitting channels into small groups (for example by sectoring) reduces efficiency.
82. To route an incoming call to a GSM mobile, the gateway MSC first interrogates the:
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This register always knows which VLR the subscriber is in.
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Answer: B. Home location register (HLR)
The GMSC asks the subscriber's HLR. The HLR obtains a roaming number (MSRN) from the visited MSC/VLR, and the call is then routed there.
83. In GSM call routing, the Mobile Station Roaming Number (MSRN) is:
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It exists only while the call is being routed.
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Answer: A. A temporary number allocated by the visited MSC/VLR to route the call
The visited VLR allocates the MSRN for the duration of call setup. The GMSC then uses it to route the call to the visited MSC.
84. In common channel signalling (CCS), signalling information is:
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'Common' means one channel serves many circuits.
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Answer: B. Carried on a separate data link shared by many speech circuits
CCS separates signalling from speech. Messages for many trunks travel on a dedicated signalling network, for example SS7 links.
85. In an SS7 network, the node that acts as a packet switch, routing signalling messages between other nodes, is the:
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Think of the node that 'transfers' signals.
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Answer: C. Signal transfer point (STP)
STPs route SS7 messages. SSPs are the exchanges that originate or terminate signalling, and SCPs hold service databases.
86. Compared with in-band channel-associated signalling, an important advantage of common channel signalling is:
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Users cannot reach the separate signalling path.
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Answer: D. Faster call setup and immunity from fraud by tones injected by users
Users cannot reach the signalling path, so tone fraud is not possible. Call setup is also faster, and signalling can continue during a call.
87. An ISDN Basic Rate Interface (2B + D) has a total user bit rate of:
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Add the two B channels and the D channel.
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Answer: D. 144 kbps
2 × 64 kbps B channels + 16 kbps D channel = 144 kbps. With framing overhead, the line rate is 192 kbps.
88. In the European ISDN Primary Rate Interface (30B + D), what is the rate of the D channel?
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The BRI and PRI D channels have different rates.
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Answer: D. 64 kbps
In PRI, the D channel is a full 64 kbps time slot. Only the BRI D channel is 16 kbps.
89. For personal communication networks (PCN) carrying bursty data, packet switching is generally preferred over circuit switching because it:
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Think about an idle dedicated channel during silent periods.
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Answer: C. Shares channel capacity statistically and uses no resources between bursts
Bursty sources leave a dedicated circuit idle most of the time. Packet switching multiplexes many users statistically, which gives much better channel utilisation.
90. For a pure ALOHA network, throughput is S = G·e^(−2G). What is the throughput at an offered load of G = 0.5?
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With G = 0.5, the exponent is −1.
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Answer: D. About 0.184
S = 0.5 × e^(−1) = 0.5 × 0.368 ≈ 0.184. This is also the maximum for pure ALOHA.
91. What is the maximum throughput of slotted ALOHA, and at what offered load does it occur?
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Differentiate G·e^(−G).
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Answer: D. 0.368 at G = 1
For slotted ALOHA, S = G·e^(−G), which is maximum at G = 1, where S = e^(−1) ≈ 0.368.
92. Packet reservation multiple access (PRMA), proposed for voice in PCN, works by letting a talkspurt:
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It combines contention with reservation.
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Answer: B. Reserve the same slot in subsequent frames after one successful contention
In PRMA, terminals contend for a slot as in slotted ALOHA. After a success, a voice terminal keeps that slot in later frames until its talkspurt ends.
9.4 Data communication switching techniques
31 questions · AEiE0904
93. In the usual taxonomy of switched networks, packet-switched networks are further divided into:
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One subtype is connectionless and the other connection-oriented.
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Answer: A. Datagram networks and virtual-circuit networks
Switched networks are classed as circuit-switched, packet-switched and message-switched. Packet-switched networks are then either datagram or virtual-circuit networks.
94. Which switching method stores each complete message at intermediate nodes before forwarding it, as in the old telegraph networks?
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The whole message is the unit that is stored.
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Answer: D. Message switching
Message switching is store-and-forward at the level of the whole message. There is no call setup and no packetisation.
95. In the layered view of switching, circuit switching takes place at which layer?
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What is actually reserved is a physical channel.
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Answer: C. Physical layer
A circuit-switched network reserves physical resources (time slots or frequency bands) on each link. The switching is therefore done at the physical layer.
96. Which statement about a circuit-switched network is correct?
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Think of a traditional telephone call.
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Answer: C. Resources are reserved at setup and held for the whole connection, even when idle
A dedicated path (a channel on each link) is reserved during setup and remains allocated until teardown, whether or not data is flowing.
97. Communication over a circuit-switched network takes place in which three phases?
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Think of a phone call: dialling, talking, hanging up.
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Answer: B. Connection setup, data transfer and connection teardown
A circuit is first set up end to end, then data is transferred, and finally the circuit is torn down to release the resources.
98. A circuit takes 0.2 s to set up. A 1 MB (8 × 10⁶ bit) file is then sent at 2 Mbps, with 5 ms propagation delay. Ignoring teardown, the total time is about:
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Add the setup time, the transmission time and the propagation delay.
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Answer: A. 4.2 s
Total = setup + transmission + propagation = 0.2 + 8×10⁶/2×10⁶ + 0.005 = 0.2 + 4 + 0.005 ≈ 4.2 s.
99. During the data-transfer (transmission) phase of a circuit-switched connection:
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The path was fixed in the setup phase.
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Answer: C. Data flows continuously along the reserved path without per-unit routing at the switches
Once the circuit exists, the switches have already been configured. Data passes through them with no addressing or store-and-forward processing.
100. In the transmission phase, how long does it take to put a 1 MB (8 × 10⁶ bit) file onto a 10 Mbps circuit?
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Remember that 1 byte = 8 bits.
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Answer: C. 0.8 s
Transmission time = bits/rate = 8×10⁶/10×10⁶ = 0.8 s.
101. In a datagram network:
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There is no connection, so each packet is on its own.
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Answer: A. Each packet is routed independently and packets may arrive out of order
Datagram networks are connectionless. Each packet is forwarded on its own destination address, so packets can take different routes and arrive out of order.
102. Which phases does datagram packet switching skip?
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Datagram networks are connectionless.
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Answer: C. Connection setup and teardown
A datagram network has no setup or teardown phase. Packets are simply sent, and each switch forwards them using its routing table.
103. A 1000-bit packet crosses 3 store-and-forward links of 1 Mbps each. Ignoring propagation and processing delays, what is its end-to-end delay?
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Each node must receive the whole packet before forwarding it.
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Answer: D. 3 ms
Each link needs 1000/10⁶ = 1 ms, and the packet must be fully received before it is forwarded. Total = 3 × 1 ms = 3 ms.
104. A 3000-bit message is split into three 1000-bit packets, which cross 3 store-and-forward links of 1 Mbps. Ignoring propagation and processing, when is the last packet delivered?
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Packets travel through the links like a pipeline.
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Answer: B. 5 ms
With pipelining, delay = (packets + links − 1) × packet time = (3 + 3 − 1) × 1 ms = 5 ms.
105. In a datagram network, a switch decides where to send a packet by looking up which field in its routing table?
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Where is the packet trying to go?
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Answer: C. The destination address
Each datagram carries the full destination address, which the switch matches in its routing table to choose the output port.
106. As a datagram travels from source to destination, its destination address:
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Compare this with a VCI.
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Answer: A. Remains the same at every hop
In datagram networks, the destination address is global and stays unchanged end to end. Only virtual-circuit identifiers are local and rewritten at each hop.
107. Which is true of a virtual-circuit network?
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It combines features of circuit and datagram switching.
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Answer: A. It has setup, data-transfer and teardown phases, and all packets follow the same path
A VC network combines features of both methods. A path is set up first, and the switches keep state, so all packets follow it and arrive in order.
108. The virtual circuit identifier (VCI) in a virtual-circuit network has:
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It is a local label, not an address.
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Answer: C. Link (local) scope and may change at every switch
A VCI is a small number meaningful only on one link. Each switch maps the incoming VCI to an outgoing VCI, so the identifier can change hop by hop.
109. A VC switch table has the entry: incoming (port 1, VCI 14) → outgoing (port 3, VCI 22). A frame arrives on port 1 with VCI 14. It will leave:
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Both the port and the label come from the outgoing side of the entry.
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Answer: D. On port 3 carrying VCI 22
The switch matches the incoming port and VCI, then forwards the frame on the outgoing port with the VCI rewritten to the outgoing value.
110. Which of the following are examples of virtual-circuit packet networks?
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Look for connection-oriented packet technologies.
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Answer: D. X.25, Frame Relay and ATM
X.25, Frame Relay and ATM set up virtual circuits identified by local labels (LCN, DLCI, VPI/VCI). IP is a datagram network.
111. ISDN services are grouped into which three categories?
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One group just carries information, one is a complete application, and one adds features.
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Answer: A. Bearer services, teleservices and supplementary services
ITU-T groups ISDN services into bearer services (information transfer), teleservices (complete end-to-end services such as telephony and fax) and supplementary services that add features to them.
112. ISDN bearer services:
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They only 'bear' the information.
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Answer: B. Transfer user information between users without the network changing its content
Bearer services cover only the lower layers (1–3). They transfer information transparently, while teleservices include higher-layer functions.
113. Calling line identification presentation (CLIP) and call waiting are examples of ISDN:
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They add something to a basic call.
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Answer: D. Supplementary services
Supplementary services modify or add to a basic bearer service or teleservice. Examples are CLIP, call waiting, call forwarding and conference calling.
114. An IS-95 CDMA system spreads 9.6 kbps data with a 1.2288 Mcps PN code. What is the processing gain?
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Find the chip-to-bit rate ratio, then convert to dB.
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Answer: D. About 21 dB
Gp = Rc/Rb = 1.2288×10⁶/9.6×10³ = 128, and 10 log 128 ≈ 21 dB.
115. In direct-sequence spread spectrum (DSSS), the signal bandwidth is spread by:
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'Direct sequence' refers to the code sequence.
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Answer: B. Multiplying the data by a high-rate pseudo-noise chip sequence
In DSSS, each data bit is multiplied by a PN code whose chip rate is much higher than the bit rate. This widens the spectrum by the processing gain.
116. In fast frequency hopping spread spectrum:
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Compare the hop rate with the symbol rate.
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Answer: A. The carrier changes frequency several times during each symbol
Fast FH means the hop rate is higher than the symbol rate. In slow FH, several symbols are sent on each hop.
117. A spread-spectrum system has a processing gain of 30 dB, needs Eb/N0 = 10 dB and has 2 dB of implementation losses. Its jamming margin is:
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Subtract what the receiver needs from what spreading gives.
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Answer: B. 18 dB
Jamming margin = Gp − (Eb/N0)required − losses = 30 − 10 − 2 = 18 dB.
118. IEEE 802.11 DSSS spreads each bit with an 11-chip Barker code. The processing gain is about:
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Take 10 log of the number of chips per bit.
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Answer: B. 10.4 dB
Gp = 11 chips per bit, and 10 log 11 ≈ 10.4 dB.
119. An FDMA system has 12.5 MHz of spectrum with 30 kHz channels and a 10 kHz guard band at each edge. How many channels can it support?
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Subtract both guard bands before dividing.
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Answer: A. 416
N = (12.5 MHz − 2 × 10 kHz)/30 kHz = 12.48 MHz/30 kHz = 416.
120. A TDMA/FDMA system has 25 MHz of spectrum, 200 kHz carriers, a 100 kHz guard at each edge and 8 time slots per carrier. How many simultaneous users can it support?
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Find the number of carriers first, then multiply by the slots.
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Answer: C. 992
Carriers = (25 MHz − 2 × 100 kHz)/200 kHz = 124. Users = 124 × 8 = 992.
121. A single-cell CDMA system has processing gain 128 and needs Eb/N0 = 7 dB. With N ≈ 1 + Gp/(Eb/N0), about how many users can it support?
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Convert 7 dB to a linear ratio first.
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Answer: D. 26
Eb/N0 = 7 dB = 5.01, so N ≈ 1 + 128/5.01 ≈ 26.5. That gives about 26 users.
122. In a CDMA uplink, the near-far problem is mainly overcome by:
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Make every signal arrive at the base station at about the same level.
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Answer: B. Accurate power control of the mobile transmitters
A nearby mobile's strong signal can swamp distant users that share the band. Fast closed-loop power control makes all signals arrive at the base station at similar power.
123. Space division multiple access (SDMA) separates users mainly by using:
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The users are separated by direction.
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Answer: B. Directional or smart antennas that steer beams to different users
SDMA reuses the same channel for users in different spatial directions. Beamforming or smart antennas provide the spatial separation.
9.5 IP switching
32 questions · AEiE0905
124. Ipsilon IP switching, introduced in the mid-1990s, combines:
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It aimed to get ATM speed with IP intelligence.
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Answer: C. ATM switching hardware with IP routing software in a switch controller
An Ipsilon IP switch keeps the ATM switch fabric but drops the ATM signalling and routing software. An attached IP switch controller runs standard IP routing and decides which flows get cut-through ATM paths.
125. In an Ipsilon IP switch, which protocol does the IP switch controller use to manage the attached ATM switch (set up and tear down connections in its fabric)?
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One protocol manages the switch hardware; the other talks to neighbours.
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Answer: A. General Switch Management Protocol (GSMP)
GSMP is the master–slave protocol the controller uses to configure the ATM switch's VC connections and ports. IFMP runs between neighbouring IP switch nodes.
126. In Ipsilon IP switching, IFMP (Ipsilon Flow Management Protocol) is used to:
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It manages flows between neighbours.
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Answer: C. Tell an adjacent node to send a particular flow on a specified new ATM VC
IFMP runs between neighbouring IP switch nodes. A node sends a redirect message that binds a flow to a dedicated VC label, so later packets of that flow can be switched in ATM hardware.
127. Before a flow is selected for cut-through in an Ipsilon IP switch, its packets are:
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Ordinary router behaviour is the starting point.
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Answer: A. Sent on a default VC, reassembled and forwarded hop by hop by the controller's IP software
By default, cells go on a default channel to the switch controller. The controller reassembles the packets and routes them in software, like a conventional router, until it decides to switch the flow.
128. Ipsilon IP switching is described as 'data-driven' (traffic-driven) because:
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What triggers the setting up of the cut-through path?
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Answer: B. Label/VC bindings are created when actual traffic of a flow is observed
In data-driven schemes, a flow's arrival triggers the setting up of a switched path. Control-driven schemes, such as tag switching and MPLS, bind labels from routing information before any traffic arrives.
129. An important advantage of Ipsilon IP switching over classical IP over ATM is that it:
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Think about which ATM machinery it gets rid of.
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Answer: D. Avoids ATM signalling and ATM address resolution, using only IP routing
IP switching runs standard IP routing and uses IFMP for labels. This removes the need for ATM UNI signalling, ATM addressing, ARP servers and LANE overlays.
130. In IP switch flow classification, which type of traffic is the best candidate for a switched (cut-through) path?
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The set-up overhead must be repaid by many packets.
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Answer: B. Long-lived flows with many packets, such as FTP transfers or media streams
Setting up a cut-through path costs some overhead, which is recovered only when many packets follow. Long-lived flows are therefore switched, while short transactions stay hop by hop.
131. Short-lived traffic such as a single DNS query is normally handled by an IP switch by:
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Is it worth setting up a path for one packet?
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Answer: B. Forwarding it hop by hop through the controller without setting up a flow
A single DNS query is one or two packets, so setting up a cut-through path is not worth it. The controller simply routes it in software.
132. A port-pair (application) flow in IP switching is identified by:
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Think of the five fields that identify one application session.
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Answer: B. Source and destination IP addresses, protocol and source and destination port numbers
Fine-grained flows are defined by the address pair, the transport protocol and the TCP/UDP port numbers. Coarser host-pair flows use only the source and destination addresses.
133. A typical flow-classification policy in an IP switch decides to switch a flow when:
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It decides whether the flow is long-lived.
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Answer: D. A configured number of packets of that flow is seen within a time interval
The classifier counts packets per flow. A flow that exceeds a threshold, for example N packets within T seconds, is judged long-lived and redirected to a dedicated VC.
134. The basic service model of the IP network layer is:
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Reliability is left to TCP.
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Answer: C. Best-effort, connectionless datagram delivery with no delivery guarantee
IP makes no guarantees of delivery, order, delay or duplicate suppression. Reliability, where needed, comes from higher layers such as TCP.
135. The IPv4 header checksum covers:
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IP leaves payload checking to upper layers.
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Answer: B. Only the IP header
The IPv4 checksum protects only the header. Payload integrity is left to transport-layer checksums.
136. In the Differentiated Services (DiffServ) model, packets are marked using:
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It marks traffic classes in the IP header itself.
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Answer: D. A 6-bit DSCP in the former Type of Service byte
DiffServ reuses the ToS byte as the DS field, with a 6-bit DSCP that selects a per-hop behaviour such as Expedited Forwarding for voice. No per-flow state is kept in core routers.
137. Which statement about the Integrated Services (IntServ) model is correct?
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Per-flow signalling is the key feature, and also the weakness.
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Answer: D. It reserves resources per flow using RSVP, which limits scalability in large cores
IntServ gives per-flow guaranteed or controlled-load service by signalling with RSVP. Every router must keep per-flow state, which scales poorly in large networks.
138. In the TCP/IP protocol suite, IP belongs to which layer?
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It is the layer that routes between networks.
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Answer: B. Internet (network) layer
IP is the internet-layer protocol that routes datagrams between networks. TCP and UDP sit above it in the transport layer.
139. A received IPv4 datagram has the Protocol field set to 17. Its payload belongs to:
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TCP is 6; this is the other common transport protocol.
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Answer: D. UDP
Common protocol numbers are 1 for ICMP, 6 for TCP, 17 for UDP and 89 for OSPF.
140. When a web request is sent over Ethernet, the correct encapsulation order, from innermost to outermost, is:
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Follow the layers from the top down.
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Answer: C. HTTP data → TCP segment → IP datagram → Ethernet frame
Each layer adds its own header around the data from the layer above: application data inside TCP, inside IP, inside the link-layer frame.
141. Which protocol works alongside IP at the internet layer to report errors such as 'destination unreachable' and 'time exceeded'?
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Ping uses it.
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Answer: A. ICMP
ICMP messages are carried inside IP datagrams and report delivery problems back to the source. Ping and traceroute rely on them.
142. A TCP segment is carried in a 1500-byte IP datagram. If the IP and TCP headers are both 20 bytes with no options, the maximum TCP payload (MSS) is:
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Subtract both headers.
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Answer: C. 1460 bytes
MSS = 1500 − 20 (IP) − 20 (TCP) = 1460 bytes.
143. The first byte of an IPv4 header is 0x46. What is the header length?
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IHL counts 4-byte words.
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Answer: B. 24 bytes
The high nibble 4 is the version and the low nibble 6 is the IHL in 32-bit words, so the header is 6 × 4 = 24 bytes (it includes 4 bytes of options).
144. An IPv4 datagram has Total Length = 1500 and IHL = 5. How many bytes of data does it carry?
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Convert IHL to bytes first.
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Answer: C. 1480 bytes
Header = 5 × 4 = 20 bytes, so data = 1500 − 20 = 1480 bytes.
145. The maximum possible size of an IPv4 datagram (header plus data) is limited by the 16-bit Total Length field to:
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Compute the largest 16-bit unsigned number.
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Answer: D. 65,535 bytes
A 16-bit field can count up to 2¹⁶ − 1 = 65,535 bytes.
146. A 4000-byte IPv4 datagram (20-byte header, 3980 bytes of data) must cross a link with an MTU of 1500 bytes. Into how many fragments is it split?
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Each fragment also needs its own 20-byte header.
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Answer: A. 3
Each fragment carries at most 1480 data bytes (a multiple of 8). 3980 = 1480 + 1480 + 1020, so 3 fragments are needed.
147. A datagram's payload is fragmented so that each full fragment carries 1480 data bytes. What Fragment Offset value does the second fragment carry?
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The offset counts 8-byte blocks.
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Answer: C. 185
The offset is in units of 8 bytes. The second fragment starts at data byte 1480, so offset = 1480/8 = 185.
148. A datagram with 3980 bytes of data and a 20-byte header is fragmented for a 1500-byte MTU (1480 data bytes per full fragment). What is the Total Length of the last fragment?
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Don't forget the header.
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Answer: D. 1040 bytes
The remaining data is 3980 − 2960 = 1020 bytes. Adding the 20-byte header gives a total length of 1040 bytes.
149. A fragment has a Fragment Offset field value of 100. Its data begins at which byte of the original datagram's payload?
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Multiply by the unit size.
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Answer: A. 800
The offset is in 8-byte units: 100 × 8 = 800.
150. A router receives a 2000-byte IPv4 datagram with the DF (Don't Fragment) bit set, and the outgoing link MTU is 1500 bytes. The router will:
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DF forbids the obvious fix.
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Answer: B. Drop the datagram and send an ICMP 'fragmentation needed' message
With DF set, fragmentation is not allowed. The router discards the packet and returns ICMP Destination Unreachable (fragmentation needed), which path-MTU discovery relies on.
151. A packet leaves a host with TTL = 64 and passes through 10 routers. What TTL does it have when it reaches the destination host?
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Each router subtracts one.
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Answer: A. 54
Each router decrements TTL by 1: 64 − 10 = 54.
152. The source address field of an IPv4 header contains the hexadecimal value C0A80001. In dotted-decimal notation, this is:
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Convert each pair of hex digits to decimal.
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Answer: A. 192.168.0.1
C0 = 192, A8 = 168, 00 = 0, 01 = 1, giving 192.168.0.1.
153. Why must an IPv4 router recompute the header checksum of each packet it forwards?
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Which field changes at every router?
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Answer: D. Because the TTL field (and possibly other fields) changes at every hop
Decrementing the TTL changes the header, so the checksum must be updated at every hop. IPv6 dropped the header checksum partly for this reason.
154. Compared with the IPv4 header, the IPv6 base header:
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IPv6 simplified router processing.
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Answer: C. Has a fixed length of 40 bytes and no header checksum
IPv6 uses a fixed 40-byte base header with 128-bit addresses. It removes the checksum and leaves fragmentation to the source, using an extension header.
155. When IP packets are switched over ATM using AAL5 (8-byte trailer, 48-byte cell payloads), how many ATM cells does a 1500-byte IP packet need? Ignore LLC/SNAP encapsulation.
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Add the trailer, then round up to whole cells.
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Answer: A. 32
1500 + 8 = 1508 bytes, padded to a multiple of 48. 1508/48 = 31.4, which rounds up to 32 cells.
9.6 Soft switching
32 questions · AEiE0906
156. In a softswitch architecture, the element that holds the call-control intelligence and directs the media gateways is the:
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Which element 'controls' the gateways?
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Answer: D. Media gateway controller (call agent)
The media gateway controller, also called the call agent, is the softswitch core. It processes signalling, makes routing decisions and tells the media gateways what connections to make.
157. The main function of a media gateway (MG) in a softswitch network is to:
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It handles the voice itself, not the call logic.
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Answer: D. Convert media between circuit-switched TDM streams and IP packets (RTP)
The MG terminates PSTN trunks or lines. It packetises the voice into RTP streams (and the reverse) under the control of the media gateway controller.
158. Which softswitch element relays SS7 signalling between the PSTN and the IP-based call controller?
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Look for the element named after signalling.
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Answer: C. Signalling gateway (SG)
The signalling gateway terminates SS7 links from the PSTN. It carries the messages over IP (for example with SIGTRAN) to the media gateway controller.
159. Which protocol is used between a media gateway controller and the media gateways it controls, in a master–slave relationship?
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Its name has 'gateway control' in it.
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Answer: C. H.248/Megaco (or MGCP)
MGCP and its successor H.248/Megaco let the controller create, modify and delete connections and terminations in the media gateways.
160. SIGTRAN transports SS7 signalling over IP networks using which transport protocol?
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It is neither TCP nor UDP.
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Answer: D. SCTP (Stream Control Transmission Protocol)
SIGTRAN adaptation layers such as M3UA and SUA run over SCTP. SCTP gives reliable, message-oriented, multi-streamed transport suited to signalling.
161. In a softswitch network, which element typically plays announcements, collects DTMF digits and provides conference bridging?
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It processes media for services.
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Answer: D. Media server
Media servers provide media-processing resources such as announcements, IVR and conferencing, under the control of the softswitch or application server.
162. ENUM (E.164 Number Mapping) allows a telephone number to be mapped to Internet services (URIs) by using:
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It reuses the Internet's existing naming system.
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Answer: B. The Domain Name System (DNS)
ENUM stores records for telephone numbers in DNS under the e164.arpa domain. A query returns URIs (such as SIP addresses) for that number.
163. Under ENUM, which domain name corresponds to the E.164 number +977 1 4412345?
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The digits are reversed, one per label.
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Answer: D. 5.4.3.2.1.4.4.1.7.7.9.e164.arpa
Remove non-digits, reverse the digits, separate each with a dot and append e164.arpa: 97714412345 becomes 5.4.3.2.1.4.4.1.7.7.9.e164.arpa.
164. ENUM stores the URIs for a telephone number in which DNS resource record type?
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It is a 'naming authority pointer'.
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Answer: D. NAPTR
ENUM uses Naming Authority Pointer (NAPTR) records, which can return a rewritten URI (for example sip: or mailto:) for each service.
165. In an H.323 VoIP network, which component translates aliases and telephone numbers into transport (IP) addresses and controls admission?
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It 'keeps the gate' of the zone.
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Answer: B. Gatekeeper
The H.323 gatekeeper provides address translation, admission control and bandwidth management using RAS signalling.
166. The FCAPS model used for softswitch and telecom network management stands for:
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One of the letters stands for billing-related management.
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Answer: A. Fault, Configuration, Accounting, Performance and Security
The ITU-T TMN model groups management functions into five areas: Fault, Configuration, Accounting, Performance and Security.
167. For billing, a softswitch produces records of each call's parties, start time and duration. These are called:
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These records describe each call in detail.
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Answer: A. Call detail records (CDRs)
CDRs are produced by the call agent for accounting management. They are later rated and billed.
168. A softswitch route has 1000 trunk seizures in an hour, and 600 of them are answered. What is its answer-seizure ratio (ASR)?
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Divide answered calls by seizures.
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Answer: B. 60%
ASR = answered calls / seizures = 600/1000 = 60%. It is a key performance indicator for route quality.
169. In a VoIP network, voice samples are carried in RTP packets over UDP rather than TCP because:
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Think about what retransmission does to delay.
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Answer: A. Retransmitting late packets is useless for real-time speech, and TCP adds delay
For conversational voice, a late packet is as bad as a lost one. TCP's retransmission and congestion control add delay, so RTP uses lightweight UDP and relies on timestamps and sequence numbers.
170. A G.711 (64 kbps) call uses 20 ms packets with 40 bytes of IP/UDP/RTP headers per packet. Ignoring layer-2 overhead, what is the IP bandwidth per direction?
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Find the payload per packet, add the headers and multiply by the packet rate.
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Answer: A. 80 kbps
Payload = 64 000 × 0.02/8 = 160 bytes per packet. Each packet is then 200 bytes, sent 50 times a second: 200 × 8 × 50 = 80 kbps.
171. A G.729 (8 kbps) call uses 20 ms packets with 40 bytes of IP/UDP/RTP headers per packet. Ignoring layer-2 overhead, what is the IP bandwidth per direction?
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The header is bigger than the voice payload.
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Answer: C. 24 kbps
Payload = 8000 × 0.02/8 = 20 bytes. Each packet is then 60 bytes, sent 50 times a second: 60 × 8 × 50 = 24 kbps.
172. For a G.729 call with 20-byte voice payloads and 40-byte IP/UDP/RTP headers, what fraction of each packet is header overhead?
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Divide the header size by the total packet size.
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Answer: A. About 67%
Overhead = 40/(40 + 20) = 0.667, or about 67%.
173. With G.711 (64 kbps) and a 30 ms packetisation interval, how many bytes of voice payload does each RTP packet carry?
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Multiply the bit rate by the time, then divide by 8.
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Answer: C. 240 bytes
64 000 bit/s × 0.030 s = 1920 bits = 240 bytes.
174. An enterprise softswitch has a 2 Mbps WAN link, and each G.729 call needs 24 kbps of IP bandwidth. About how many simultaneous calls can the link carry?
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Divide the link capacity by the per-call bandwidth.
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Answer: A. 83
2000 kbps / 24 kbps ≈ 83.3, so about 83 calls.
175. In SIP-based VoIP, which request does a caller send to start a session?
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It 'invites' the other party.
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Answer: B. INVITE
INVITE starts a session and carries the SDP offer. REGISTER binds a user to a location, BYE ends a session and ACK confirms the final response.
176. A VoIP path has 25 ms codec delay, 70 ms network delay and a 40 ms jitter buffer. What is the one-way delay, and does it meet the ITU-T G.114 recommended limit of 150 ms?
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Add up all the delay components.
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Answer: B. 135 ms, within the limit
Total = 25 + 70 + 40 = 135 ms, which is below 150 ms.
177. VoLTE provides voice on 4G LTE networks by using:
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LTE has no circuit-switched core.
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Answer: A. The IP Multimedia Subsystem (IMS) with SIP signalling
LTE has an all-IP core with no circuit-switched domain. VoLTE carries voice as IP packets, controlled by SIP through the IMS core.
178. Before VoLTE was deployed, an LTE handset making or receiving a voice call typically used:
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The phone has to 'fall back' for voice.
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Answer: C. Circuit-switched fallback (CSFB) to a 2G/3G network
With CSFB, the phone leaves LTE and moves to the 2G/3G circuit-switched network for the call, then returns to LTE afterwards.
179. Voice over Wi-Fi (VoWiFi) on an untrusted Wi-Fi access connects the handset securely to the operator's IMS core through the:
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A security gateway is needed for untrusted access.
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Answer: B. ePDG (evolved Packet Data Gateway) using IPsec tunnels
For untrusted non-3GPP access, the handset builds an IPsec tunnel to the ePDG. The ePDG connects to the packet core, so IMS voice works over Wi-Fi.
180. DSL technology provides broadband access by:
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No new cable is needed.
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Answer: C. Using frequencies above the voice band on existing copper telephone pairs
DSL reuses the telephone twisted pair. It sends data at frequencies above the voice band, so analog telephone service can continue on the same line.
181. At the telephone exchange, DSL lines from many subscribers terminate on a:
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Its name says it multiplexes DSL access.
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Answer: D. DSLAM (DSL access multiplexer)
The DSLAM contains the exchange-end DSL modems. It aggregates the subscribers' traffic onto the operator's data network.
182. ADSL uses DMT modulation with a subcarrier spacing of 4.3125 kHz up to 1.104 MHz. How many subcarriers does this give?
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Divide the total bandwidth by the subcarrier spacing.
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Answer: C. 256
1104 kHz / 4.3125 kHz = 256 subcarriers.
183. A DMT modem uses 200 subcarriers with an average of 8 bits per subcarrier at 4000 DMT symbols/s. What is the raw data rate?
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Multiply subcarriers × bits per subcarrier × symbols per second.
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Answer: A. 6.4 Mbps
Rate = 200 × 8 × 4000 = 6 400 000 bit/s = 6.4 Mbps.
184. Why is ADSL called 'asymmetric'?
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Compare download and upload speeds.
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Answer: B. Its downstream data rate is much higher than its upstream data rate
ADSL gives most of the spectrum to downstream traffic, which suits typical Internet use where users download much more than they upload.
185. Which xDSL variant gives symmetric T1/E1 rates and was designed to replace repeatered T1/E1 lines, originally using two or more copper pairs?
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It was developed for leased T1/E1 circuits.
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Answer: B. HDSL
HDSL (High bit-rate DSL) provides symmetric 1.544 or 2.048 Mbps service over multiple pairs without mid-span repeaters.
186. Which member of the xDSL family gives the highest data rates, but only over short copper loops, typically from a nearby cabinet fed by fibre?
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Its first letter stands for 'very high'.
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Answer: C. VDSL/VDSL2
VDSL uses bandwidth up to tens of MHz, which supports very high rates. That bandwidth is attenuated quickly, so it is used on short loops, often with fibre to the cabinet.
187. ADSL2+ extends the downstream band to 2.208 MHz while keeping the 4.3125 kHz subcarrier spacing. How many subcarriers does it use?
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Divide the new bandwidth by the spacing.
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Answer: D. 512
2208/4.3125 = 512 subcarriers, double ADSL's 256. This allows downstream rates up to about 24 Mbps.