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WBCHSE HS Vocational Telecom study materials
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WBCHSE HS Vocational Telecom practice questions

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Question 1Basic Electricity and Electronics for Telecom

In a telecom equipment room, a resistor is marked with the colour bands brown, black, red and gold. What is its nominal resistance and tolerance?

  1. 1 kΩ with ±5% tolerance Correct answer.
  2. 10 Ω with ±2% tolerance
  3. 100 Ω with ±5% tolerance
  4. 1 kΩ with ±10% tolerance

Explanation

The four-band code reads brown = 1, black = 0, red = multiplier 100, giving 10 × 100 = 1000 Ω, that is 1 kΩ. A gold fourth band denotes ±5% tolerance, which is the standard value used in telecom circuit boards.

Question 2Basic Electricity and Electronics for Telecom

Which property of a capacitor makes it useful for coupling an AC speech signal between two amplifier stages while blocking the DC bias?

  1. Its reactance falls as frequency rises, so it passes AC but blocks DC. Correct answer.
  2. Its resistance increases with temperature.
  3. It converts alternating current into direct current.
  4. It stores energy in a magnetic field.

Explanation

Capacitive reactance is given by Xc = 1/(2πfC), so it becomes very large at zero frequency (DC) and small at audio frequencies. This lets a coupling capacitor pass the speech signal while isolating the DC operating point of each stage.

Question 3Basic Electricity and Electronics for Telecom

A subscriber line has a loop resistance of 600 Ω and the exchange battery supplies 48 V. Ignoring other losses, what current flows in the loop?

  1. 8 mA
  2. 80 mA Correct answer.
  3. 800 mA
  4. 0.8 A

Explanation

Ohm's law gives I = V/R = 48/600 = 0.08 A, which equals 80 mA. This is the order of loop current a telephone set needs to operate its transmitter and dialling circuit.

Question 4Basic Electricity and Electronics for Telecom

What is the purpose of the dielectric material between the plates of a capacitor?

  1. It increases the plate area available for charge.
  2. It allows the capacitor to store energy as a magnetic field.
  3. It insulates the plates and raises the capacitance compared with air. Correct answer.
  4. It provides a conducting path between the plates.

Explanation

The dielectric is an insulator that prevents direct conduction between the plates while its relative permittivity multiplies the capacitance, since C = ε₀εrA/d. Air has εr close to 1, so a solid dielectric such as ceramic or polyester gives a much larger capacitance for the same plate size.

Question 5Basic Electricity and Electronics for Telecom

Which instrument connection is correct when measuring the current drawn by a telephone instrument from its line?

  1. Voltmeter in series with the line.
  2. Ohmmeter connected across the line with power on.
  3. Ammeter connected in parallel across the two line wires.
  4. Ammeter in series with the line, with the instrument disconnected from the ammeter's own supply. Correct answer.

Explanation

An ammeter has very low internal resistance and must carry the current, so it is inserted in series with the circuit being measured. Connecting it in parallel would divert almost all the current through it and blow its fuse.

Question 6Basic Electricity and Electronics for Telecom

In a series circuit containing three resistors, which statement about current and voltage is true?

  1. The same current flows through all resistors and the supply voltage divides in proportion to resistance. Correct answer.
  2. The current divides between the resistors and the voltage is the same across each.
  3. The total resistance equals the sum of the reciprocals of the individual resistances.
  4. The current is greatest through the largest resistor.

Explanation

A series path offers only one route for charge, so the current is identical everywhere in the loop. The supply voltage is shared as V = IR across each element, so the largest resistance drops the largest voltage.

Question 7Basic Electricity and Electronics for Telecom

Which semiconductor material and doping combination produces an n-type region?

  1. Silicon doped with boron
  2. Silicon doped with phosphorus Correct answer.
  3. Germanium doped with indium
  4. Silicon doped with aluminium

Explanation

Phosphorus is pentavalent: four electrons bond with silicon and the fifth is loosely held, so it acts as a donor giving excess electrons and an n-type region. Trivalent dopants such as boron, aluminium and indium accept electrons and give p-type material.

Question 8Basic Electricity and Electronics for Telecom

A 12 V supply feeds two resistors of 4 Ω and 6 Ω connected in parallel. What is the total current drawn from the supply?

  1. 5 A Correct answer.
  2. 1.2 A
  3. 12 A
  4. 2 A

Explanation

The parallel equivalent is (4 × 6)/(4 + 6) = 2.4 Ω, so the supply current is 12/2.4 = 5 A. Equivalently, branch currents are 3 A and 2 A, which add to 5 A.

Question 9Basic Electricity and Electronics for Telecom

A transformer used in a telecom power supply has 400 turns on the primary and 100 turns on the secondary. If the primary is fed with 230 V AC, what is the secondary voltage?

  1. 230 V
  2. 920 V
  3. 57.5 V Correct answer.
  4. 92 V

Explanation

For an ideal transformer the turns ratio equals the voltage ratio, so Vs = Vp × Ns/Np = 230 × 100/400 = 57.5 V. The step-down ratio of 4:1 reduces the voltage by the same factor.

Question 10Basic Electricity and Electronics for Telecom

Which statement correctly describes the behaviour of a p-n junction diode under forward bias?

  1. The depletion region narrows and conventional current flows from anode to cathode. Correct answer.
  2. Electrons flow from cathode to anode inside the external circuit.
  3. The depletion region widens and only leakage current flows.
  4. The junction behaves as an open circuit until reverse breakdown.

Explanation

Forward bias reduces the potential barrier, narrowing the depletion layer so that majority carriers cross the junction. Conventional current then flows from the anode to the cathode through the diode and the external circuit.

Question 11Basic Electricity and Electronics for Telecom

A half-wave rectifier feeds a load from a 50 Hz AC supply. What is the ripple frequency of the output?

  1. 50 Hz Correct answer.
  2. 25 Hz
  3. 150 Hz
  4. 100 Hz

Explanation

A half-wave rectifier passes one half-cycle for every input cycle, so the output pulses repeat at the supply frequency of 50 Hz. Full-wave rectification would double this to 100 Hz.

Question 12Basic Electricity and Electronics for Telecom

An amplifier has a voltage gain of 100 and feeds a load with a signal output of 2 V. What input voltage is required?

  1. 2 V
  2. 200 V
  3. 0.2 V
  4. 0.02 V Correct answer.

Explanation

Voltage gain is the ratio of output to input, so Vin = Vout/Av = 2/100 = 0.02 V, that is 20 mV. This is a typical microphone-level signal for a small-signal audio stage.

Question 13Basic Electricity and Electronics for Telecom

What is the function of the biased Zener diode in a simple shunt voltage regulator?

  1. It conducts in reverse breakdown to hold a nearly constant voltage across the load. Correct answer.
  2. It amplifies the error voltage.
  3. It stores charge to smooth the output.
  4. It rectifies the AC input before regulation.

Explanation

A Zener diode operated in reverse breakdown maintains an almost constant voltage across itself even when the current through it varies. Placed in parallel with the load, it clamps the output at the Zener voltage.

Question 14Basic Electricity and Electronics for Telecom

A coil of 0.5 H carries a current that changes at 200 A per second. What voltage is induced across the coil?

  1. 4 V
  2. 40 V
  3. 0.4 V
  4. 100 V Correct answer.

Explanation

Faraday's law for self-induction gives V = L × di/dt = 0.5 × 200 = 100 V. The induced voltage opposes the change in current, in accordance with Lenz's law.

Question 15Basic Electricity and Electronics for Telecom

Which combination correctly states the frequency bands normally associated with voice-band telephony and with a typical ADSL downstream channel?

  1. Voice band 300–3400 Hz; ADSL downstream typically extends to several hundred kilohertz and above. Correct answer.
  2. Voice band 300–3400 Hz; ADSL downstream is confined to 4–8 kHz.
  3. Voice band 0–300 Hz; ADSL downstream occupies the same 300–3400 Hz band.
  4. Voice band 20–20000 Hz; ADSL downstream occupies 50–60 Hz.

Explanation

Telephony uses the 300–3400 Hz voice band, which is enough for intelligible speech while limiting bandwidth. ADSL reuses the same copper pair above the voice band, with downstream signals occupying frequencies up to hundreds of kilohertz and beyond.

Question 16Basic Electricity and Electronics for Telecom

A 100 Ω resistor dissipates 4 W. What current flows through it?

  1. 0.4 A
  2. 20 A
  3. 2 A
  4. 0.2 A Correct answer.

Explanation

From P = I²R, I = √(P/R) = √(4/100) = 0.2 A. Checking: 0.2 A through 100 Ω gives 20 V and 4 W, which is consistent.

Question 17Basic Electricity and Electronics for Telecom

In a common-emitter transistor amplifier, which phase relationship exists between the base input signal and the collector output signal?

  1. They are exactly 180 degrees out of phase. Correct answer.
  2. They are in phase.
  3. The output leads by 45 degrees.
  4. The output is at half the input frequency.

Explanation

An increase in base current raises the collector current, which increases the drop across the collector resistor and therefore lowers the collector voltage. This inversion makes the output 180 degrees out of phase with the input.

Question 18Basic Electricity and Electronics for Telecom

A capacitor of 10 µF is connected across a 100 V DC supply and fully charged. How much energy does it store?

  1. 1 J
  2. 0.5 J
  3. 10 J
  4. 0.05 J Correct answer.

Explanation

Energy in a capacitor is E = ½CV² = 0.5 × 10 × 10⁻⁶ × 100² = 0.05 J. The microfarad value must be converted to farads before squaring the voltage.

Question 19Basic Electricity and Electronics for Telecom

A telecom rack draws 5 A at 48 V DC. If the supply is fed through a cable whose total loop resistance is 0.4 Ω, what voltage reaches the rack?

  1. 44 V
  2. 48 V
  3. 46 V Correct answer.
  4. 38 V

Explanation

The cable drop is I × R = 5 × 0.4 = 2 V, so the rack receives 48 − 2 = 46 V. This shows why telecom DC distribution uses short, thick conductors to keep loop resistance low.

Question 20Basic Electricity and Electronics for Telecom

Which statement explains why a twisted pair cable reduces interference from external magnetic fields in a telephone line?

  1. Twisting converts the interference into a DC offset that the receiver ignores.
  2. The twisting cancels the induced voltages because successive loops face opposite directions. Correct answer.
  3. The twisting produces a magnetic shield of copper around each conductor.
  4. The twists increase the total loop resistance.

Explanation

In each successive twist the two conductors swap positions relative to the interfering field, so the voltages induced in adjacent sections have opposite polarity and largely cancel. This is why balanced twisted pairs reject common-mode interference.

Question 21Basic Electricity and Electronics for Telecom

A bridge rectifier with a capacitor-input filter feeds a light load. Which change will reduce the peak-to-peak ripple voltage at the output?

  1. Increasing the value of the filter capacitor Correct answer.
  2. Increasing the load current
  3. Removing the filter capacitor
  4. Reducing the transformer secondary voltage

Explanation

Ripple is approximately Vripple = I/(f × C), so a larger filter capacitance holds more charge between charging peaks and reduces the ripple. With light loads the discharge is slower, so the effect is pronounced.

Question 22Basic Electricity and Electronics for Telecom

A transistor in common-emitter configuration has base current 50 µA and collector current 4 mA. What is its current gain, and what does this imply for the emitter current?

  1. β = 0.0125, and emitter current is 4.05 mA
  2. β = 8, and emitter current is 32 mA
  3. β = 80, and emitter current is about 4.05 mA Correct answer.
  4. β = 80, and emitter current is about 3.95 mA

Explanation

Current gain β = Ic/Ib = 4 mA / 50 µA = 80. The emitter current is the sum of collector and base currents, 4 mA + 0.05 mA = 4.05 mA, so it is only slightly larger than the collector current.

Question 23Basic Electricity and Electronics for Telecom

Two 8 Ω loudspeakers are connected in parallel to an amplifier output. Which load does the amplifier see, and what happens to the power drawn for the same output voltage compared with a single speaker?

  1. 4 Ω, and the power is doubled Correct answer.
  2. 16 Ω, and the power is halved
  3. 4 Ω, and the power is halved
  4. 8 Ω, and the power is unchanged

Explanation

Two equal resistors in parallel give 8/2 = 4 Ω. Since P = V²/R, halving the load resistance at the same output voltage doubles the power drawn, which is why amplifier load ratings must be respected.

Question 24Basic Electricity and Electronics for Telecom

A Zener shunt regulator uses a 12 V Zener with a 470 Ω series resistor from an 18 V supply. If the load draws 5 mA, what current flows through the Zener?

  1. About 10.2 mA
  2. About 7.8 mA Correct answer.
  3. About 2.2 mA
  4. About 12.8 mA

Explanation

The series resistor carries (18 − 12)/470 ≈ 12.8 mA. Of this, 5 mA goes to the load, leaving about 7.8 mA through the Zener, which must exceed the minimum knee current for regulation.

Question 25Telecommunication Fundamentals and Signals

In a telecommunication system, which block converts the original voice message into an electrical signal suitable for transmission?

  1. The receiver
  2. The channel
  3. The transmitter Correct answer.
  4. The noise source

Explanation

A communication system has three essential parts: transmitter, channel and receiver. The transmitter performs the conversion of the message into a form, such as an electrical or modulated signal, that the channel can carry. The receiver reverses this process at the far end.

Question 26Telecommunication Fundamentals and Signals

A sinusoidal signal completes 500 cycles in 10 milliseconds. What is its frequency?

  1. 5 kHz
  2. 500 Hz
  3. 50 kHz Correct answer.
  4. 20 kHz

Explanation

Frequency equals the number of cycles divided by the time taken, so f = 500 / 0.01 s = 50,000 Hz. That is 50 kHz. Always convert milliseconds to seconds before dividing.

Question 27Telecommunication Fundamentals and Signals

Which property of a signal is measured in hertz?

  1. Wavelength
  2. Frequency Correct answer.
  3. Phase
  4. Amplitude

Explanation

Frequency is the number of complete cycles per second and its unit is the hertz, one cycle per second. Amplitude is measured in volts or amperes, phase in degrees or radians, and wavelength in metres.

Question 28Telecommunication Fundamentals and Signals

A signal whose value changes continuously over time, taking every possible value within a range, is best described as:

  1. Digital
  2. Analog Correct answer.
  3. Quantised
  4. Discrete-time

Explanation

An analog signal is continuous in both time and amplitude, so it can take any value within its range. Digital signals, by contrast, exist only at specified instants and at a finite set of levels after sampling and quantisation.

Question 29Telecommunication Fundamentals and Signals

A signal has a peak voltage of 10 V. What is its RMS value if the waveform is a pure sine wave?

  1. 7.07 V Correct answer.
  2. 10 V
  3. 14.14 V
  4. 6.37 V

Explanation

For a sine wave the RMS value equals the peak value divided by the square root of two, so 10 / 1.414 gives about 7.07 V. The RMS value is the effective value that would produce the same heating effect in a resistor.

Question 30Telecommunication Fundamentals and Signals

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

  1. 8 kbps
  2. 16 kbps Correct answer.
  3. 4 kbps
  4. 60 kbps

Explanation

Shannon's formula is C = B log2(1 + S/N). With B = 4000 Hz and S/N = 15, the term log2(16) equals 4, so C = 4000 x 4 = 16,000 bits per second. This is a theoretical upper bound, not a guaranteed practical rate.

Question 31Telecommunication Fundamentals and Signals

According to the Nyquist sampling theorem, a band-limited signal containing frequencies up to 3.4 kHz must be sampled at a minimum rate of:

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

Explanation

The Nyquist rate is twice the highest frequency present in the band-limited signal, so 2 x 3.4 kHz = 6.8 kHz. Sampling below this rate causes aliasing, in which high-frequency components fold back and corrupt the recovered signal.

Question 32Telecommunication Fundamentals and Signals

In an 8-bit PCM system, how many quantisation levels can be represented?

  1. 8
  2. 256 Correct answer.
  3. 128
  4. 16

Explanation

The number of quantisation levels is 2 raised to the number of bits, so 2^8 equals 256. Each sample is then coded as one of these 256 discrete levels, giving a quantisation step of the input range divided by 256.

Question 33Telecommunication Fundamentals and Signals

Quantisation error in a PCM system is reduced most directly by:

  1. Reducing the sampling rate
  2. Lowering the transmitter power
  3. Increasing the channel noise
  4. Increasing the number of quantisation levels Correct answer.

Explanation

Quantisation error is the difference between the true sample value and the nearest allowed level. Adding more levels makes the step size smaller, so the maximum error falls. The trade-off is a higher bit rate, since more bits per sample are needed.

Question 34Telecommunication Fundamentals and Signals

In amplitude modulation, the modulation index is defined as the ratio of:

  1. Carrier amplitude to modulating amplitude
  2. Sideband frequency to carrier frequency
  3. Modulating amplitude to carrier amplitude Correct answer.
  4. Modulating frequency to carrier frequency

Explanation

The AM modulation index m equals the peak modulating voltage divided by the carrier voltage, Vm/Vc. Values above one cause overmodulation and envelope distortion. It is a voltage ratio and is dimensionless.

Question 35Telecommunication Fundamentals and Signals

A carrier of 1000 kHz is amplitude modulated by a 5 kHz audio tone. What are the frequencies of the two sidebands?

  1. 995 kHz and 990 kHz
  2. 1005 kHz and 1005 kHz
  3. 1005 kHz and 995 kHz Correct answer.
  4. 1000 kHz and 1005 kHz

Explanation

AM produces an upper sideband at fc + fm and a lower sideband at fc - fm. Adding and subtracting 5 kHz from 1000 kHz gives 1005 kHz and 995 kHz. The total bandwidth occupied is therefore 10 kHz.

Question 36Telecommunication Fundamentals and Signals

The main advantage of single sideband suppressed carrier transmission over conventional AM is that it:

  1. Occupies half the bandwidth and saves power Correct answer.
  2. Works only for low audio frequencies
  3. Requires no carrier recovery at the receiver
  4. Produces a simpler envelope detector output

Explanation

SSB removes the carrier and one sideband, so it needs only the bandwidth of one sideband and transmits no wasteful carrier power. The cost is that the receiver must reinsert a carrier for coherent detection, which makes the circuit more complex.

Question 37Telecommunication Fundamentals and Signals

In frequency modulation, increasing the amplitude of the modulating signal while keeping its frequency constant will:

  1. Increase the frequency deviation Correct answer.
  2. Reduce the modulation index to zero
  3. Increase the carrier frequency
  4. Change the carrier amplitude

Explanation

In FM the instantaneous frequency shifts away from the carrier in proportion to the modulating signal amplitude. A larger modulating amplitude therefore produces a larger peak deviation. The rate at which the frequency shifts is set by the modulating frequency, not its amplitude.

Question 38Telecommunication Fundamentals and Signals

FM broadcasting is more resistant to noise than AM broadcasting mainly because:

  1. Information is carried in frequency variations while amplitude is held constant Correct answer.
  2. It occupies a narrower bandwidth
  3. It transmits only one sideband
  4. It uses a lower carrier frequency

Explanation

Most noise, such as lightning and spark interference, appears as amplitude variations. An FM receiver limits and clips the signal amplitude before demodulation, so these variations are largely removed. The message survives because it is encoded in frequency changes.

Question 39Telecommunication Fundamentals and Signals

A radio receiver that mixes the incoming signal with a local oscillator to produce a fixed intermediate frequency is called a:

  1. Regenerative detector
  2. Crystal video receiver
  3. Tuned radio frequency receiver
  4. Superheterodyne receiver Correct answer.

Explanation

The superheterodyne receiver converts every incoming frequency to a fixed intermediate frequency by mixing with a local oscillator. Because the IF stages always work at the same frequency, selectivity and gain are easier to design and remain uniform across the band.

Question 40Telecommunication Fundamentals and Signals

In a superheterodyne receiver with an intermediate frequency of 455 kHz, a station is tuned at 1000 kHz. What is the local oscillator frequency if it is set above the signal?

  1. 545 kHz
  2. 455 kHz
  3. 1455 kHz Correct answer.
  4. 1000 kHz

Explanation

The mixer output must equal the intermediate frequency, so the oscillator runs at fs + IF when placed above the signal. Adding 1000 kHz and 455 kHz gives 1455 kHz. The difference between oscillator and signal is then exactly 455 kHz.

Question 41Telecommunication Fundamentals and Signals

The image frequency problem in a superheterodyne receiver arises because:

  1. The antenna is too short for the wavelength
  2. Two input frequencies can produce the same intermediate frequency Correct answer.
  3. The local oscillator drifts with temperature
  4. The IF amplifier has too much gain

Explanation

Any input at the oscillator frequency plus or minus the IF will mix down to the same IF, so an unwanted image station can pass through. A higher IF or a preselector filter ahead of the mixer reduces this effect. The image lies twice the IF away from the wanted signal.

Question 42Telecommunication Fundamentals and Signals

Which type of transmission line carries the signal as an electric field between a centre conductor and an outer shield, with the shield also preventing radiation?

  1. Rectangular waveguide
  2. Coaxial cable Correct answer.
  3. Two-wire parallel line
  4. Optical fibre

Explanation

In a coaxial cable the inner conductor and the surrounding braid form the two electrodes, and the dielectric between them holds the field. The outer conductor confines the field and acts as a shield against interference. This makes coax suitable for carrying high-frequency signals with low leakage.

Question 43Telecommunication Fundamentals and Signals

A transmission line of characteristic impedance 50 ohm is terminated by a purely resistive load of 50 ohm. What is the voltage standing wave ratio on the line?

  1. Infinite
  2. 1.0 Correct answer.
  3. 0.5
  4. 2.0

Explanation

When the load equals the characteristic impedance the line is matched, so no power is reflected. With no reflected wave the maximum and minimum voltages are equal and the VSWR is 1. A VSWR of 1 therefore indicates a perfectly matched line.

Question 44Telecommunication Fundamentals and Signals

In free space, a radio wave with a frequency of 100 MHz has a wavelength of approximately:

  1. 30 m
  2. 3 m Correct answer.
  3. 300 m
  4. 0.3 m

Explanation

Wavelength equals the speed of light divided by frequency, so 3 x 10^8 m/s divided by 1 x 10^8 Hz gives 3 m. This relation is useful when choosing antenna dimensions, since a half-wave dipole at this frequency would be about 1.5 m long.

Question 45Telecommunication Fundamentals and Signals

A half-wave dipole antenna is designed for a frequency of 150 MHz. Ignoring the shortening effect, its approximate physical length is:

  1. 0.5 m
  2. 2 m
  3. 1 m Correct answer.
  4. 4 m

Explanation

The wavelength at 150 MHz is 300/150 = 2 m, so a half-wave dipole is about 1 m long. In practice the element is cut a few percent shorter because of end effects and the thickness of the conductor. This makes the resonant length slightly less than the free-space half wavelength.

Question 46Telecommunication Fundamentals and Signals

An antenna has a gain of 10 dB over a reference. Expressed as a power ratio, this gain is:

  1. 10 Correct answer.
  2. 100
  3. 1000
  4. 2

Explanation

A power ratio in decibels is ten times the base-10 logarithm of the ratio. For 10 dB, log10(P/Pref) equals 1, so the ratio is 10. Each additional 10 dB multiplies the power ratio by ten again.

Question 47Telecommunication Fundamentals and Signals

Compared with a low-frequency signal, a high-frequency signal propagating along a practical transmission line shows greater attenuation mainly because:

  1. The line becomes shorter
  2. Skin effect and dielectric loss increase with frequency Correct answer.
  3. The signal voltage is amplified by the line
  4. The characteristic impedance rises to infinity

Explanation

At higher frequencies current crowds toward the surface of the conductor, raising the effective resistance through the skin effect. Dielectric losses in the insulation also grow with frequency. Together these make attenuation rise as frequency increases, which is why long cables are equalised or repeated.

Question 48Telecommunication Fundamentals and Signals

In a cellular mobile system, frequency reuse is possible because:

  1. Co-channel cells are separated by enough distance to keep interference acceptable Correct answer.
  2. The same channel is assigned to every cell
  3. All base stations share one antenna
  4. The mobile unit transmits at very high power

Explanation

Cells using the same frequencies are grouped into clusters and repeated across the service area. Reuse distance is chosen so that the signal from the wanted base station dominates the co-channel interference from distant cells. This lets a limited spectrum serve many subscribers.

Question 49Telecom Transmission Media and Cabling

Which type of optical fibre has a core diameter of about 9 µm and carries a single propagation mode at 1310 nm?

  1. Step-index multimode fibre
  2. Graded-index multimode fibre
  3. Single-mode fibre Correct answer.
  4. Plastic optical fibre

Explanation

Single-mode fibre has a very small core, around 8 to 10 µm, so only one mode propagates at the usual 1310 nm and 1550 nm windows. This low mode count removes intermodal dispersion and permits long-haul transmission. Multimode and plastic fibres have much larger cores and support many modes.

Question 50Telecom Transmission Media and Cabling

In a twisted-pair cable, what is the main purpose of twisting the two conductors of a pair?

  1. To lower the characteristic impedance to 50 ohms
  2. To increase the tensile strength of the copper
  3. To allow the pair to carry a DC power feed
  4. To cancel the magnetic fields and reduce crosstalk and noise pickup Correct answer.

Explanation

The two wires of a pair carry equal and opposite signal currents, so their magnetic fields largely cancel and external interference couples equally into both conductors. Twisting keeps this balance along the whole length, which suppresses crosstalk and common-mode noise. Twisting does not set impedance to 50 ohms or provide power feed.