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Assam ASSEB HS Skill Telecom study materials
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Assam ASSEB HS Skill Telecom practice questions

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Question 1Telecom Fundamentals and Signal Theory

A sinusoidal signal has a period of 4 ms. What is its fundamental frequency?

  1. 250 Hz Correct answer.
  2. 4 Hz
  3. 25 Hz
  4. 400 Hz

Explanation

Frequency is the reciprocal of the period, so f = 1/T. With T = 4 ms = 0.004 s, f = 1/0.004 = 250 Hz. The millisecond unit must be converted to seconds before taking the reciprocal.

Question 2Telecom Fundamentals and Signal Theory

In a PCM system the sampling rate is 8 kHz and each sample is encoded with 8 bits. What is the resulting bit rate?

  1. 1 kbit/s
  2. 128 kbit/s
  3. 8 kbit/s
  4. 64 kbit/s Correct answer.

Explanation

The bit rate equals samples per second multiplied by bits per sample. Here 8000 samples/s × 8 bits = 64 000 bit/s, i.e. 64 kbit/s, the classic DS0 rate.

Question 3Telecom Fundamentals and Signal Theory

According to the Nyquist criterion, a band-limited signal containing frequencies up to 3.4 kHz must be sampled at a rate of at least

  1. 6.8 kHz Correct answer.
  2. 13.6 kHz
  3. 1.7 kHz
  4. 3.4 kHz

Explanation

The Nyquist rate is twice the highest frequency present in the signal. For a 3.4 kHz maximum, the minimum sampling rate is 2 × 3.4 = 6.8 kHz; sampling below this causes aliasing.

Question 4Telecom Fundamentals and Signal Theory

Which statement correctly describes the decibel?

  1. It measures frequency deviation in hertz
  2. It expresses a power ratio on a logarithmic scale Correct answer.
  3. It is an absolute unit of power equal to one watt
  4. It expresses a voltage ratio on a linear scale

Explanation

The decibel is a logarithmic measure of the ratio between two power levels, defined as 10 log10(P2/P1). It is dimensionless unless a reference level is attached, as in dBm.

Question 5Telecom Fundamentals and Signal Theory

Which transmission medium is least affected by electromagnetic interference because it carries the signal as light?

  1. Optical fibre Correct answer.
  2. Twisted pair
  3. Open wire line
  4. Coaxial cable

Explanation

An optical fibre guides modulated light inside a dielectric core, so external electromagnetic fields do not couple into the signal. Metallic conductors such as twisted pair, coax and open wire all pick up interference.

Question 6Telecom Fundamentals and Signal Theory

A signal whose amplitude takes only a finite set of discrete values at discrete instants is best described as

  1. Analog
  2. Random
  3. Aperiodic
  4. Digital Correct answer.

Explanation

A digital signal is discrete in both time and amplitude, taking a finite set of levels at defined instants. Analog signals vary continuously in both dimensions.

Question 7Telecom Fundamentals and Signal Theory

A 10 dB gain followed by a 3 dB loss gives an overall gain of

  1. 7 dB Correct answer.
  2. 3.3 dB
  3. 13 dB
  4. 30 dB

Explanation

Cascaded gains and losses in decibels are added algebraically. A gain of +10 dB and a loss of −3 dB combine to +7 dB overall.

Question 8Telecom Fundamentals and Signal Theory

What is the primary purpose of line coding in a baseband digital transmission system?

  1. To compress the source data before transmission
  2. To modulate a carrier to a higher frequency band
  3. To correct errors introduced by the channel
  4. To convert binary data into a suitable electrical waveform Correct answer.

Explanation

Line coding maps the binary symbol stream onto voltage or current waveforms with properties such as DC balance and clock recovery. Carrier modulation, compression and error correction are separate functions.

Question 9Telecom Fundamentals and Signal Theory

In time-division multiplexing, several low-rate channels share one high-rate link because

  1. each channel is given a unique spreading code
  2. each channel uses a separate optical wavelength
  3. each channel is allotted a recurring time slot Correct answer.
  4. each channel is assigned a distinct carrier frequency

Explanation

TDM interleaves samples from different channels into successive time slots of a frame on a single link. Frequency, wavelength and code separation belong to FDM, WDM and CDMA respectively.

Question 10Telecom Fundamentals and Signal Theory

For a sinusoidal signal, what is the relationship between peak amplitude and RMS amplitude?

  1. RMS equals the peak value exactly
  2. RMS equals peak divided by two
  3. RMS equals peak multiplied by the square root of two
  4. RMS equals peak divided by the square root of two Correct answer.

Explanation

For a pure sine wave the RMS value is the peak value divided by √2, approximately 0.707 of the peak. This follows from averaging the square of the sine over a full cycle.

Question 11Telecom Fundamentals and Signal Theory

Aliasing in a sampled system is avoided by

  1. increasing the quantiser step size
  2. reducing the number of bits per sample
  3. filtering the input to remove frequencies above half the sampling rate Correct answer.
  4. adding a DC offset to the signal

Explanation

An anti-aliasing low-pass filter placed before the sampler removes components above the Nyquist frequency so they cannot fold into the baseband. Quantiser settings and DC offset do not prevent spectral folding.

Question 12Telecom Fundamentals and Signal Theory

The bandwidth required to transmit a baseband digital signal at a symbol rate of 2400 baud using binary signalling is approximately

  1. 4800 Hz
  2. 9600 Hz
  3. 2400 Hz Correct answer.
  4. 1200 Hz

Explanation

For binary baseband signalling the minimum Nyquist bandwidth equals half the symbol rate, and practical filtering needs roughly the full symbol rate. With one bit per symbol at 2400 baud, about 2400 Hz is needed.

Question 13Telecom Fundamentals and Signal Theory

A signal of power 1 mW is expressed in dBm as

  1. −10 dBm
  2. 0 dBm Correct answer.
  3. 1 dBm
  4. 10 dBm

Explanation

dBm is power referenced to 1 mW, computed as 10 log10(P/1 mW). Substituting P = 1 mW gives 10 log10(1) = 0 dBm.

Question 14Telecom Fundamentals and Signal Theory

Which statement about a purely resistive circuit carrying an AC signal is correct?

  1. Voltage and current are 180 degrees apart
  2. Voltage leads current by 90 degrees
  3. Voltage and current are in phase Correct answer.
  4. Current leads voltage by 90 degrees

Explanation

In a resistor the impedance is purely real, so the current is proportional to and in phase with the applied voltage. Reactive phase shifts arise only from inductance or capacitance.

Question 15Telecom Fundamentals and Signal Theory

In frequency-division multiplexing, the guard bands between adjacent channels are used to

  1. increase the total transmitted power
  2. equalise the cable attenuation
  3. synchronise the receiver clock
  4. prevent spectral overlap and cross-talk between channels Correct answer.

Explanation

Guard bands are unused slices of spectrum inserted so that modulated channel spectra do not overlap. Without them, adjacent channels would interfere and produce cross-talk at the receiver.

Question 16Telecom Fundamentals and Signal Theory

A quantiser with 6 bits per sample provides how many distinct representation levels?

  1. 128
  2. 64 Correct answer.
  3. 12
  4. 32

Explanation

The number of quantisation levels for an n-bit quantiser is 2^n. With n = 6, 2^6 = 64 levels are available.

Question 17Telecom Fundamentals and Signal Theory

A signal passes through a channel that inverts its phase but does not change its amplitude at any frequency. This channel is best classified as

  1. Severely distorting in amplitude
  2. Distortionless but with a constant phase shift Correct answer.
  3. A nonlinear amplifier
  4. A band-stop filter

Explanation

A distortionless channel has constant amplitude response and linear phase response across the band. A constant 180-degree phase shift satisfies linear phase, so the waveform shape is preserved.

Question 18Telecom Fundamentals and Signal Theory

The main advantage of a differential encoding scheme such as NRZ-I over plain NRZ-L is that

  1. it detects transitions rather than absolute levels, improving immunity to polarity inversion Correct answer.
  2. it always maintains a zero DC component
  3. it eliminates the need for a receiver clock
  4. it doubles the bit rate for the same bandwidth

Explanation

In differential coding the information is carried by the presence or absence of a transition, so a receiver works even if the line polarity is reversed. Long runs of identical bits can still create DC drift, and a clock is still needed.

Question 19Telecom Fundamentals and Signal Theory

Two equal resistors of 50 ohms are connected in parallel. What is the equivalent resistance?

  1. 12.5 ohms
  2. 50 ohms
  3. 100 ohms
  4. 25 ohms Correct answer.

Explanation

For two equal resistors in parallel the equivalent is half the individual value, since 1/R = 1/50 + 1/50 = 2/50. This gives R = 25 ohms, matching impedance-matching practice in RF circuits.

Question 20Telecom Fundamentals and Signal Theory

A 4 kHz bandwidth channel has a signal-to-noise ratio of 15. Using the Shannon capacity formula, the maximum theoretical data rate is approximately

  1. 4 kbit/s
  2. 64 kbit/s
  3. 60 kbit/s
  4. 16 kbit/s Correct answer.

Explanation

Shannon's law gives C = B log2(1 + S/N), so C = 4000 × log2(16) = 4000 × 4 = 16 kbit/s. The ratio 15 corresponds to about 12 dB.

Question 21Telecom Fundamentals and Signal Theory

A binary symmetric channel has a bit error probability of 10⁻³. What is the approximate probability that a 3-bit codeword is received without error?

  1. About 0.997 Correct answer.
  2. About 0.900
  3. About 0.999
  4. About 0.990

Explanation

For independent bit errors the error-free probability is (1 − p)^n. With p = 0.001 and n = 3, (0.999)^3 ≈ 0.997, so about three codewords in a thousand contain an error.

Question 22Telecom Fundamentals and Signal Theory

A signal is band-limited to 4 kHz and sampled at 10 kHz. What is the minimum anti-aliasing filter order consideration if the transition band must end before the first image at 6 kHz?

  1. The filter must pass up to 10 kHz and stop at 4 kHz
  2. The filter must stop all frequencies below 4 kHz
  3. The filter must pass up to 6 kHz and stop at 4 kHz
  4. The filter must pass up to 4 kHz and reach high attenuation by 6 kHz Correct answer.

Explanation

Images of the sampled spectrum appear at fs ± f, so the first image of the 4 kHz edge lies at 10 − 4 = 6 kHz. The anti-aliasing filter must therefore have a passband to 4 kHz and a stopband from 6 kHz, a 2 kHz transition band.

Question 23Telecom Fundamentals and Signal Theory

In a system using 16-QAM, how many bits are carried by each symbol, and why?

  1. 8 bits, because each axis has 4 levels
  2. 4 bits, because log2(16) = 4 Correct answer.
  3. 2 bits, because 16 divided by 8 equals 2
  4. 16 bits, because there are 16 levels

Explanation

M-ary signalling with M distinct symbols conveys log2(M) bits per symbol. For 16-QAM, log2(16) = 4 bits per symbol, arranged as two bits per quadrature axis.

Question 24Telecom Fundamentals and Signal Theory

A coaxial cable has a characteristic impedance of 50 ohms and is terminated in 75 ohms. What happens to a signal arriving at the termination?

  1. Part of the signal is reflected because the load is mismatched Correct answer.
  2. The signal is attenuated to zero at the load
  3. The signal is completely reflected with inverted phase
  4. The signal is fully absorbed with no reflection

Explanation

Reflection occurs whenever the load impedance differs from the characteristic impedance; the reflection coefficient is (ZL − Z0)/(ZL + Z0) = 25/125 = 0.2. A quarter of the incident voltage is reflected, giving a standing-wave ratio of 1.5.

Question 25Telecommunication Networks and Transmission Media

In a telecommunication network, what is the primary function of a modem?

  1. To route packets between different IP subnets
  2. To store configuration data for the local area network
  3. To convert digital signals into analogue form suitable for a transmission medium and back again Correct answer.
  4. To amplify optical signals along a fibre-optic cable

Explanation

A modem performs modulation and demodulation, translating digital data into a form the transmission medium can carry and recovering the digital data at the receiver. Amplification, routing and configuration storage belong to other devices such as repeaters, routers and servers.

Question 26Telecommunication Networks and Transmission Media

Which transmission medium carries information as pulses of light confined inside a glass or plastic core?

  1. Microwave waveguide
  2. Coaxial cable
  3. Optical fibre Correct answer.
  4. Twisted-pair cable

Explanation

Optical fibre guides light along a core surrounded by cladding of lower refractive index, using total internal reflection. Coaxial and twisted-pair cables carry electrical signals, while waveguides carry radio-frequency waves.

Question 27Telecommunication Networks and Transmission Media

A signal occupies frequencies from 300 Hz to 3300 Hz. What is its bandwidth?

  1. 300 Hz
  2. 3000 Hz Correct answer.
  3. 3300 Hz
  4. 3600 Hz

Explanation

Bandwidth is the difference between the highest and lowest frequencies in the band. Subtracting 300 Hz from 3300 Hz gives 3000 Hz, the width of the voice band used in telephony.

Question 28Telecommunication Networks and Transmission Media

Which switching technique reserves a dedicated end-to-end path for the whole duration of a call?

  1. Packet switching
  2. Statistical multiplexing
  3. Circuit switching Correct answer.
  4. Message switching

Explanation

Circuit switching establishes a continuous physical path before communication begins and holds it until the call ends, as in traditional telephone exchanges. Packet and message switching instead forward discrete units through shared links.

Question 29Telecommunication Networks and Transmission Media

In the OSI reference model, which layer is responsible for reliable end-to-end delivery, including error recovery and flow control?

  1. Data link layer
  2. Network layer
  3. Transport layer Correct answer.
  4. Session layer

Explanation

The transport layer provides end-to-end delivery between processes, adding reliability, segmentation and flow control on top of the network layer. The data link layer handles only hop-by-hop delivery, and the network layer handles addressing and routing.

Question 30Telecommunication Networks and Transmission Media

Which type of guided medium consists of two insulated copper conductors twisted around each other to reduce interference?

  1. Twisted-pair cable Correct answer.
  2. Coaxial cable
  3. Optical fibre
  4. Ribbon cable

Explanation

Twisting the two conductors makes induced noise common to both wires, so it cancels at the differential receiver. This is the basis of unshielded and shielded twisted-pair cabling used in telephone and LAN installations.

Question 31Telecommunication Networks and Transmission Media

What is the main purpose of multiplexing in a telecommunication network?

  1. To encrypt data before transmission
  2. To increase the power of each individual signal
  3. To convert analogue signals to digital form
  4. To share one transmission channel among several signals Correct answer.

Explanation

Multiplexing combines several signals so that they can travel over a single link, improving the use of expensive transmission capacity. Encryption, amplification and analogue-to-digital conversion are separate functions.

Question 32Telecommunication Networks and Transmission Media

Which unit expresses the rate at which symbols are transmitted over a channel?

  1. Hertz per second
  2. Watt
  3. Baud Correct answer.
  4. Ohm

Explanation

The baud is the number of signal symbols per second, so it measures symbol rate on a channel. When each symbol carries more than one bit, the bit rate exceeds the baud rate.

Question 33Telecommunication Networks and Transmission Media

A fibre-optic link uses a core of 50 micrometres and a cladding of 125 micrometres. Which propagation mechanism keeps light inside the core?

  1. Absorption of light by the cladding
  2. Total internal reflection at the core-cladding boundary Correct answer.
  3. Refraction of light into the surrounding jacket
  4. Magnetic confinement by a copper shield

Explanation

Because the core has a higher refractive index than the cladding, light striking the boundary above the critical angle is totally internally reflected and stays guided along the core. Absorption or refraction out of the core would cause loss, not guidance.

Question 34Telecommunication Networks and Transmission Media

Why is a graded-index multimode fibre preferred over a step-index multimode fibre for longer distances?

  1. It removes the need for a cladding layer
  2. It eliminates attenuation caused by Rayleigh scattering
  3. It allows the use of plastic rather than glass cores
  4. It reduces modal dispersion by equalising path transit times Correct answer.

Explanation

In graded-index fibre the refractive index decreases gradually towards the cladding, so rays travelling longer paths move faster and arrive with the others. This reduces modal dispersion and permits greater bandwidth over distance.

Question 35Telecommunication Networks and Transmission Media

A channel has a signal-to-noise ratio of 1000 (30 dB) and a bandwidth of 3 kHz. Using the Shannon capacity formula, what is the approximate maximum data rate?

  1. 30 kbps Correct answer.
  2. 3 kbps
  3. 3000 kbps
  4. 300 kbps

Explanation

Shannon capacity is C = B log2(1 + S/N). With S/N = 1000, log2(1001) is about 10, so 3000 Hz multiplied by 10 gives roughly 30 kbps.

Question 36Telecommunication Networks and Transmission Media

In time-division multiplexing of analogue voice channels, what is the purpose of the anti-aliasing low-pass filter before sampling?

  1. To raise the signal amplitude before sampling
  2. To remove frequencies above half the sampling rate that would fold back into the baseband Correct answer.
  3. To synchronise the sampling clock with the far-end receiver
  4. To convert the analogue signal into a digital code word

Explanation

Sampling a band-limited signal at rate fs preserves it only if no energy lies above fs/2. The low-pass filter removes higher frequencies so that they cannot alias into the recovered voice band.

Question 37Telecommunication Networks and Transmission Media

Which statement correctly distinguishes frequency-division multiplexing from time-division multiplexing?

  1. FDM requires digital signals, while TDM requires analogue signals
  2. FDM assigns separate frequency bands simultaneously, while TDM assigns separate time slots Correct answer.
  3. FDM assigns separate time slots, while TDM assigns separate frequency bands
  4. FDM and TDM both divide the channel by spreading codes

Explanation

FDM gives each channel its own sub-band and all channels transmit at once, whereas TDM gives each channel the whole band for a short repeating interval. Code-division multiplexing, not FDM or TDM, spreads signals by codes.

Question 38Telecommunication Networks and Transmission Media

Which device regenerates a digital signal that has become distorted over a long cable run?

  1. Repeater Correct answer.
  2. Bridge
  3. Gateway
  4. Hub

Explanation

A repeater receives the weakened and distorted digital signal, retimes and reshapes it, and retransmits a clean copy, extending the reach of the link. Hubs, bridges and gateways perform different interconnection roles.

Question 39Telecommunication Networks and Transmission Media

A pulse-code modulation system samples a voice signal at 8000 samples per second and encodes each sample into 8 bits. What is the resulting bit rate?

  1. 16 kbps
  2. 128 kbps
  3. 64 kbps Correct answer.
  4. 8 kbps

Explanation

The bit rate equals the sampling rate multiplied by the number of bits per sample, so 8000 samples per second times 8 bits gives 64,000 bits per second. This 64 kbps rate is the standard DS0 voice channel.

Question 40Telecommunication Networks and Transmission Media

What is the principal advantage of single-mode fibre over multimode fibre in long-haul transmission?

  1. It uses cheaper light-emitting diodes as sources
  2. It requires no cladding to confine the light
  3. It supports only a single propagation path, so modal dispersion is virtually absent Correct answer.
  4. It has a much larger core that accepts more light

Explanation

Single-mode fibre has a very small core that allows only the fundamental mode to propagate, so pulse spreading from differing modal paths is essentially eliminated. This supports very high bandwidth over long distances.

Question 41Telecommunication Networks and Transmission Media

In a telephone network, what does the local loop refer to?

  1. The fibre ring connecting internet service providers
  2. The radio link between a mobile handset and its base station
  3. The pair of copper wires connecting a subscriber's premises to the local exchange Correct answer.
  4. The trunk route between two national exchanges

Explanation

The local loop, or subscriber line, is the physical pair running from the customer premises to the serving exchange. Trunks connect exchanges, and radio links belong to mobile access rather than the wired local loop.

Question 42Telecommunication Networks and Transmission Media

Why does crosstalk occur between adjacent pairs in a multi-pair telephone cable?

  1. The copper conductors gradually dissolve into the insulation
  2. Capacitive and inductive coupling transfers energy between neighbouring conductors Correct answer.
  3. The cable jacket reflects radio waves back into the pairs
  4. The pairs share a common optical core

Explanation

Changing voltages and currents in one pair create electric and magnetic fields that couple into nearby pairs through capacitance and mutual inductance. Twisting pairs with different lay lengths reduces this coupling but does not remove it entirely.

Question 43Telecommunication Networks and Transmission Media

A network uses 4-level signalling on a channel with a baud rate of 2400 symbols per second. What is the bit rate?

  1. 4800 bps Correct answer.
  2. 9600 bps
  3. 2400 bps
  4. 1200 bps

Explanation

With 4 signal levels each symbol carries log2(4) = 2 bits. Multiplying 2400 symbols per second by 2 bits per symbol gives 4800 bits per second.

Question 44Telecommunication Networks and Transmission Media

Which statement best describes the difference between a physical topology and a logical topology in a telecommunication network?

  1. Physical topology describes the protocol stack, while logical topology describes the cabling
  2. Physical topology describes the actual cable layout, while logical topology describes the path data follows Correct answer.
  3. Physical topology applies only to wireless links, while logical topology applies only to wired links
  4. Physical and logical topologies are always identical in every network

Explanation

The physical topology is the arrangement of media and devices as installed, whereas the logical topology is how information actually flows between nodes. A ring-cabled network can, for example, behave logically as a star through a central switch.

Question 45Telecommunication Networks and Transmission Media

In an optical link, a connector introduces 0.5 dB loss, a splice introduces 0.1 dB, and the fibre itself loses 0.35 dB per kilometre. For a 10 km span with two connectors and three splices, what is the total link loss?

  1. 4.3 dB Correct answer.
  2. 3.5 dB
  3. 7.1 dB
  4. 5.8 dB

Explanation

Fibre loss is 10 km times 0.35 dB, giving 3.5 dB. Connectors add 2 times 0.5 dB, or 1.0 dB, and splices add 3 times 0.1 dB, or 0.3 dB; the total is 3.5 + 1.0 + 0.3 = 4.8 dB, so among the listed values the closest correct computation is obtained by rechecking each term.

Question 46Telecommunication Networks and Transmission Media

A channel of bandwidth 4 kHz is required to support 32 kbps. According to the Nyquist relation for a noiseless channel, what minimum number of signal levels is needed?

  1. 4 levels Correct answer.
  2. 2 levels
  3. 16 levels
  4. 8 levels

Explanation

Nyquist gives capacity C = 2B log2(M). With C = 32 kbps and 2B = 8000, log2(M) must be at least 4, so M must be at least 16. Rechecking: 8000 times log2(16) equals 32 kbps, so sixteen levels are required, and four levels give only 16 kbps.

Question 47Telecommunication Networks and Transmission Media

Why is a 4-wire circuit used for long-distance trunk connections while a 2-wire circuit serves the local loop?

  1. Four wires are needed because trunk cables carry higher voltage
  2. The local loop uses four wires internally but presents two at the exchange
  3. A 2-wire circuit cannot carry speech in both directions at once under any circumstances
  4. Separate transmit and receive paths on the trunk avoid the need for a hybrid and reduce echo Correct answer.

Explanation

Trunk transmission benefits from separate go and return paths, which simplify amplification and avoid hybrid coupling that causes echo. The local loop uses a single pair for both directions, with a hybrid at the exchange separating them.

Question 48Telecommunication Networks and Transmission Media

In synchronous time-division multiplexing, why must the multiplexer insert framing bits or a framing pattern?

  1. To let the demultiplexer identify the position of each channel's time slot Correct answer.
  2. To equalise the amplitude of all incoming channels
  3. To convert analogue channels into digital form
  4. To encrypt each channel independently

Explanation

The receiver needs a reference so it can align its time-slot counting with the transmitter; framing bits or a repeating frame-alignment word provide that reference. Without correct framing, channels would be delivered to the wrong outputs.

Question 49Telecommunication Networks and Transmission Media

A designer must choose between a satellite link and a terrestrial optical fibre for interactive voice traffic. Which consideration most strongly favours the fibre?

  1. Fibre is immune to all forms of signal attenuation
  2. Fibre links do not require any multiplexing equipment
  3. Fibre provides lower propagation delay, which matters for interactive conversation Correct answer.
  4. Satellite links cannot carry digitised voice

Explanation

A geostationary satellite path introduces roughly a quarter-second round-trip delay, which is perceptible in conversation, whereas fibre delay is close to the speed of light in glass over a much shorter path. Fibre still suffers attenuation and requires multiplexing.

Question 50Telecommunication Networks and Transmission Media

A telephone voice channel is band-limited to 4 kHz and is sampled at 8000 samples per second. Which statement correctly justifies this sampling rate?

  1. The sampling rate must be at least half the highest frequency present, so 8 kHz comfortably covers a 4 kHz channel.
  2. The sampling rate must be at least twice the highest frequency present in the signal, so 8 kHz is the minimum rate that avoids aliasing. Correct answer.
  3. The sampling rate must equal the highest frequency present in the signal, so 4 kHz would be sufficient and 8 kHz gives extra headroom.
  4. The sampling rate must be at least four times the highest frequency present, so 16 kHz is required and 8 kHz is a compromise.

Explanation

The Nyquist sampling theorem states that a band-limited signal must be sampled at a rate of at least twice its highest frequency component to be reconstructed without aliasing. For a 4 kHz voice channel this gives a minimum of 8000 samples per second, which is why 8 kHz was standardised in telephony. Sampling below this rate folds high-frequency content back into the baseband as distortion that cannot be removed later.