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RBSE Sec Voc Telecom study materials
RBSE Sec Voc Telecom study resources

RBSE Sec Voc Telecom practice questions

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Question 1Basic Electronics and Electrical Fundamentals

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

  1. 1000 Ω with ±2% tolerance
  2. 1000 Ω with ±5% tolerance Correct answer.
  3. 1200 Ω with ±10% tolerance
  4. 100 Ω with ±5% tolerance

Explanation

Brown equals 1, black equals 0, so the significant digits form 10, and the red multiplier adds two zeros to give 1000 Ω (1 kΩ). Gold in the fourth band denotes ±5% tolerance. The value is therefore 1 kΩ ±5% as used for bias and load resistors in telecom circuit boards.

Question 2Basic Electronics and Electrical Fundamentals

Two resistors of 6 Ω and 3 Ω are connected in parallel across a 12 V supply. What current does the supply deliver?

  1. 2 A
  2. 1.33 A
  3. 4 A
  4. 6 A Correct answer.

Explanation

The parallel equivalent is (6×3)/(6+3) = 2 Ω. Supply current equals voltage divided by equivalent resistance, so 12 V ÷ 2 Ω = 6 A. The branch currents of 2 A and 4 A add to this total, confirming the result.

Question 3Basic Electronics and Electrical Fundamentals

A capacitor of 10 µF is charged to 50 V. How much energy is stored in its electric field?

  1. 0.25 J
  2. 0.5 J
  3. 0.025 J
  4. 0.0125 J Correct answer.

Explanation

Energy stored in a capacitor is ½CV². Substituting gives ½ × 10×10⁻⁶ F × (50)² V² = ½ × 10×10⁻⁶ × 2500 = 0.0125 J. This stored energy matters in telecom power supplies, where large capacitors can retain charge after switch-off.

Question 4Basic Electronics and Electrical Fundamentals

Which property of an inductor opposes any change in the current flowing through it?

  1. Self-inductance Correct answer.
  2. Inductive reactance
  3. Reluctance
  4. Permeance

Explanation

Self-inductance is the property by which a changing current induces a back EMF that opposes the change, in line with Lenz's law. Reactance, by contrast, is the opposition to AC current at a given frequency and depends on inductance and frequency. Reluctance and permeance describe the magnetic circuit, not current opposition.

Question 5Basic Electronics and Electrical Fundamentals

A transformer with 400 primary turns and 100 secondary turns is fed from a 230 V AC mains. What is the secondary voltage?

  1. 92 V
  2. 575 V
  3. 57.5 V Correct answer.
  4. 920 V

Explanation

The turns ratio is 400:100, or 4:1, and a step-down transformer reduces voltage in the same ratio. Dividing 230 V by 4 gives 57.5 V on the secondary. This principle underlies the mains transformers used in telecom power shelves.

Question 6Basic Electronics and Electrical Fundamentals

In a semiconductor diode, what is the typical forward voltage drop of a silicon PN junction under normal conduction?

  1. 0.7 V Correct answer.
  2. 0.2 V
  3. 5 V
  4. 1.5 V

Explanation

A forward-biased silicon junction needs roughly 0.7 V to overcome the barrier before significant conduction begins. Germanium junctions are lower, about 0.3 V, while 0.2 V is typical of a Schottky diode. Designers allow for this drop when calculating rectifier and clamp circuits.

Question 7Basic Electronics and Electrical Fundamentals

What is the main function of a bridge rectifier in a telecom DC power supply?

  1. To regulate the output against load changes
  2. To smooth ripple on the output
  3. To convert AC into pulsating DC Correct answer.
  4. To step the mains voltage down

Explanation

A bridge rectifier uses four diodes so that both half-cycles of the AC input produce current in the same direction, giving pulsating DC. Smoothing is then done by filter capacitors and regulation by a regulator stage. Confusing these roles leads to poor power-supply design.

Question 8Basic Electronics and Electrical Fundamentals

A zener diode with a rated zener voltage of 5.1 V is used as a shunt regulator. What happens if the load draws more current while the input stays constant?

  1. The zener current rises and the output rises sharply
  2. The zener stops conducting and the output falls to zero
  3. The zener current falls and the output stays near 5.1 V Correct answer.
  4. The zener voltage doubles to about 10.2 V

Explanation

In a shunt regulator the series resistor carries the sum of load and zener current. If load current increases, the zener current must fall to keep the total roughly constant, and the output stays near the zener voltage. Regulation holds as long as the zener remains in reverse breakdown.

Question 9Basic Electronics and Electrical Fundamentals

Which statement correctly describes the base-emitter junction condition for a silicon NPN bipolar transistor to operate in the active region?

  1. Both junctions reverse biased
  2. Base-emitter forward biased, base-collector reverse biased Correct answer.
  3. Both junctions forward biased
  4. Base-emitter reverse biased, base-collector forward biased

Explanation

In the active or linear region the emitter junction is forward biased so carriers are injected, while the collector junction is reverse biased to sweep them across. Both junctions forward biased is saturation, and both reverse biased is cut-off. Amplifier stages in telecom equipment rely on the active region.

Question 10Basic Electronics and Electrical Fundamentals

A transistor has a collector current of 4 mA and a base current of 40 µA. What is its DC current gain?

  1. 1000
  2. 100 Correct answer.
  3. 160
  4. 10

Explanation

DC current gain hFE is collector current divided by base current. Converting 40 µA to 0.04 mA gives 4 mA ÷ 0.04 mA = 100. This gain figure is used when calculating bias resistors and stage loading in amplifier design.

Question 11Basic Electronics and Electrical Fundamentals

Why is a field-effect transistor often preferred over a bipolar transistor as the front-end device in a low-noise receiver amplifier?

  1. It requires no bias supply
  2. It has a lower input impedance
  3. It can handle much larger currents
  4. It has a very high input impedance and low noise figure Correct answer.

Explanation

A FET controls current by an electric field on a high-impedance gate, so it draws almost no input current and generates little shot noise, giving a low noise figure. This suits weak-signal front ends such as receiver preamplifiers. High current capability is not the reason, and a FET still needs bias.

Question 12Basic Electronics and Electrical Fundamentals

Which logic gate outputs HIGH only when all its inputs are HIGH?

  1. AND gate Correct answer.
  2. XOR gate
  3. OR gate
  4. NOR gate

Explanation

An AND gate produces a logic 1 output only when every input is at logic 1; any low input forces the output low. This behaviour is used for enable and interlock conditions in digital control circuits. OR, NOR and XOR all give a high output for some single-input-high case.

Question 13Basic Electronics and Electrical Fundamentals

A two-input XOR gate has inputs A = 1 and B = 1. What is the output?

  1. 0 Correct answer.
  2. Indeterminate
  3. Equal to A only
  4. 1

Explanation

The exclusive-OR output is HIGH only when the inputs are different. With both inputs at logic 1 they are equal, so the output is 0. This property makes XOR useful for parity generation and comparison in digital transmission systems.

Question 14Basic Electronics and Electrical Fundamentals

What does the binary number 1011 equal in decimal?

  1. 15
  2. 9
  3. 13
  4. 11 Correct answer.

Explanation

Reading place values from the left, 1011 equals 1×8 + 0×4 + 1×2 + 1×1 = 11. Binary counting underpins addressing and data representation in digital telecom systems. Each position doubles in weight from right to left.

Question 15Basic Electronics and Electrical Fundamentals

A signal of 1 kHz is sampled at 8 kHz. Which statement about the sampled signal is correct?

  1. It must be sampled at exactly 1 kHz
  2. It satisfies the Nyquist criterion and can be reconstructed Correct answer.
  3. It will show aliasing because the rate is too low
  4. It cannot be digitised without a low-pass filter at 8 kHz

Explanation

The Nyquist criterion requires a sampling rate of at least twice the highest signal frequency. Here 8 kHz is well above 2 × 1 kHz, so the signal can be reconstructed without aliasing. Anti-alias filtering is still good practice, but the rate itself is adequate.

Question 16Basic Electronics and Electrical Fundamentals

In a series RLC circuit at resonance, which condition holds?

  1. Inductive reactance equals capacitive reactance Correct answer.
  2. Resistance equals reactance
  3. The current is at its minimum
  4. The impedance is at its maximum

Explanation

At series resonance the inductive and capacitive reactances are equal in magnitude and cancel, leaving only resistance. Impedance is then at its minimum and current at its maximum. This principle is used in tuned circuits and filters in radio equipment.

Question 17Basic Electronics and Electrical Fundamentals

A coil has an inductive reactance of 314 Ω at 50 Hz. What is its inductance?

  1. 0.5 H
  2. 3.14 H
  3. 2 H
  4. 1 H Correct answer.

Explanation

Inductive reactance is XL = 2πfL. Rearranging gives L = 314 ÷ (2π × 50) = 314 ÷ 314 ≈ 1 H. Reactance rises with frequency, so the same coil presents more opposition at higher frequencies.

Question 18Basic Electronics and Electrical Fundamentals

What is the RMS value of a sinusoidal voltage whose peak value is 325 V?

  1. 230 V Correct answer.
  2. 460 V
  3. 325 V
  4. 260 V

Explanation

For a sine wave the RMS value is the peak divided by √2, so 325 ÷ 1.414 ≈ 230 V. This is why a 230 V mains supply has a peak of about 325 V. RMS values are used because they give the equivalent heating effect.

Question 19Basic Electronics and Electrical Fundamentals

A 12 V battery delivers 2 A to a load. What power is dissipated in the load, assuming negligible losses?

  1. 48 W
  2. 6 W
  3. 14 W
  4. 24 W Correct answer.

Explanation

Electrical power equals voltage multiplied by current, so 12 V × 2 A = 24 W. This figure determines heat dissipation and component ratings in DC-powered telecom equipment. It can also be found as I²R once the load resistance is known.

Question 20Basic Electronics and Electrical Fundamentals

Which material is classed as a semiconductor because its conductivity lies between that of conductors and insulators?

  1. Rubber
  2. Glass
  3. Copper
  4. Silicon Correct answer.

Explanation

Silicon has four valence electrons and a band gap of about 1.1 eV, giving conductivity between metals and insulators, and it can be doped to control that conductivity. Copper is a good conductor, while glass and rubber are insulators. This controlled conductivity is the basis of all semiconductor devices.

Question 21Basic Electronics and Electrical Fundamentals

A PN junction is described as reverse biased. What is the dominant current mechanism then?

  1. Large majority-carrier current
  2. Small leakage current of minority carriers Correct answer.
  3. Current limited only by forward resistance
  4. Zero current under all conditions

Explanation

Reverse bias widens the depletion region and blocks majority carriers, leaving only a small leakage current from thermally generated minority carriers. This leakage grows with temperature and is the basis of the diode's reverse blocking behaviour in rectifiers and protection circuits.

Question 22Basic Electronics and Electrical Fundamentals

An ideal operational amplifier used in a linear negative-feedback circuit has which input condition?

  1. Infinite differential voltage
  2. Zero output impedance with no feedback
  3. Zero differential voltage between its inputs Correct answer.
  4. Equal currents at both inputs

Explanation

With negative feedback and very high open-loop gain, the amplifier drives its output so that the differential input voltage is almost zero, the virtual-short condition. Input currents are also nearly zero, but they are not equal in general. This assumption simplifies gain calculations for inverting and non-inverting stages.

Question 23Basic Electronics and Electrical Fundamentals

A coil and a capacitor are connected in parallel and the combination is fed at its resonant frequency. What is the approximate impedance of the parallel network?

  1. Equal to the capacitive reactance
  2. Equal to the inductive reactance
  3. Very high, close to the dynamic resistance Correct answer.
  4. Very low, close to zero

Explanation

At parallel resonance the circulating reactive currents cancel within the tank, so the terminal impedance rises to a high value set mainly by the coil's losses, often called the dynamic resistance. This high impedance is exploited in tuned amplifier and oscillator tank circuits.

Question 24Basic Electronics and Electrical Fundamentals

A conductor carries 5 A for 2 minutes. How much charge has passed through it?

  1. 10 C
  2. 600 C Correct answer.
  3. 60 C
  4. 1200 C

Explanation

Charge equals current multiplied by time in seconds, so 5 A × 120 s = 600 C. Converting minutes to seconds is essential before applying Q = It. This relationship underlies battery capacity ratings and fuse selection in DC systems.

Question 25Communication Systems and Modulation Techniques

In an amplitude modulation system, the modulation index is defined as the ratio of the modulating signal amplitude to which quantity?

  1. Carrier amplitude Correct answer.
  2. Sideband amplitude
  3. Modulated wave amplitude
  4. Noise amplitude

Explanation

The AM modulation index m equals the peak modulating voltage divided by the peak carrier voltage. It describes how deeply the carrier envelope is varied by the intelligence signal. Values above unity indicate overmodulation and cause envelope distortion.

Question 26Communication Systems and Modulation Techniques

A carrier of 1 MHz is amplitude modulated by a 5 kHz sine wave. What are the upper and lower sideband frequencies?

  1. 1.050 MHz and 0.950 MHz
  2. 1.005 MHz and 0.995 MHz Correct answer.
  3. 2.000 MHz and 1.000 MHz
  4. 1.005 kHz and 0.995 kHz

Explanation

AM produces sidebands at the sum and difference of carrier and modulating frequencies. Adding and subtracting 5 kHz from 1 MHz gives 1.005 MHz and 0.995 MHz. The sidebands sit symmetrically about the carrier, spaced by the modulating frequency.

Question 27Communication Systems and Modulation Techniques

Which property of a carrier is varied in frequency modulation?

  1. Instantaneous frequency Correct answer.
  2. Amplitude
  3. Phase only
  4. Polarization

Explanation

In FM the instantaneous frequency of the carrier deviates above and below its resting value in step with the modulating signal. The amplitude stays essentially constant. This constant-envelope property is why FM is more resistant to amplitude noise.

Question 28Communication Systems and Modulation Techniques

Carson's rule estimates the bandwidth of an FM signal. Which expression is correct?

  1. BW = 2fm
  2. BW = Δf / fm
  3. BW = Δf + fm
  4. BW = 2(Δf + fm) Correct answer.

Explanation

Carson's rule gives bandwidth as twice the sum of peak deviation and the highest modulating frequency. It captures both the deviation spread and the sideband width. The result is an approximation that contains most of the signal power.

Question 29Communication Systems and Modulation Techniques

Which modulation scheme transmits information by shifting the carrier between two discrete frequencies?

  1. QAM
  2. PSK
  3. FSK Correct answer.
  4. ASK

Explanation

Frequency-shift keying represents binary symbols by switching the carrier between distinct frequencies. ASK varies amplitude instead, and PSK varies phase. FSK is common in low-cost data links and telemetry.

Question 30Communication Systems and Modulation Techniques

What is the primary purpose of a mixer in a superheterodyne receiver?

  1. To demodulate the baseband signal directly
  2. To amplify the audio output
  3. To radiate the local oscillator signal
  4. To translate the received signal to a fixed intermediate frequency Correct answer.

Explanation

A mixer multiplies the incoming RF with the local oscillator to produce sum and difference frequencies. The difference product becomes the intermediate frequency, which allows fixed-tuned, highly selective IF filtering. Tuning is then done by changing the local oscillator.

Question 31Communication Systems and Modulation Techniques

In a PCM system, what does the quantizer do?

  1. Converts serial bits into parallel words
  2. Removes high-frequency noise before sampling
  3. Rounds each sampled value to the nearest discrete level Correct answer.
  4. Recovers the carrier at the receiver

Explanation

Quantization maps the continuous range of sample amplitudes onto a finite set of levels. The rounding error introduced is quantization noise. More levels, or more bits per sample, reduce that error.

Question 32Communication Systems and Modulation Techniques

A signal band-limited to 4 kHz is sampled at exactly the Nyquist rate. What is the sampling frequency?

  1. 2000 Hz
  2. 16000 Hz
  3. 4000 Hz
  4. 8000 Hz Correct answer.

Explanation

The Nyquist rate equals twice the highest frequency present in the signal. For a 4 kHz maximum frequency the minimum rate is 8 kHz. Sampling below this causes aliasing, where high-frequency content folds into the baseband.

Question 33Communication Systems and Modulation Techniques

A 100 MHz carrier is frequency modulated with a peak deviation of 25 kHz by a 5 kHz tone. Using Carson's rule, what is the approximate bandwidth?

  1. 125 kHz
  2. 50 kHz
  3. 30 kHz
  4. 60 kHz Correct answer.

Explanation

Carson's rule gives BW = 2(Δf + fm). Substituting 25 kHz deviation and 5 kHz modulating frequency yields 2(30 kHz) = 60 kHz. This approximation contains the significant sidebands of the FM spectrum.

Question 34Communication Systems and Modulation Techniques

An AM wave has a modulation index of 1.0 with a 50 kW unmodulated carrier. What is the total transmitted power?

  1. 100 kW
  2. 75 kW Correct answer.
  3. 50 kW
  4. 62.5 kW

Explanation

Total AM power equals carrier power times (1 + m²/2). With m = 1 the factor is 1.5, so 50 kW becomes 75 kW. The extra 25 kW resides in the two sidebands, each carrying one quarter of the carrier power.

Question 35Communication Systems and Modulation Techniques

Why is a high-side injection local oscillator used in a superheterodyne receiver tuned to 1 MHz with a 455 kHz IF?

  1. The local oscillator runs at 1 MHz and the IF equals zero
  2. The local oscillator runs at 910 kHz and doubles the IF
  3. The local oscillator runs at 545 kHz and the sum product falls at the IF
  4. The local oscillator runs at 1.455 MHz and the difference product falls at the IF Correct answer.

Explanation

With high-side injection the oscillator sits above the signal by the IF value. Mixing 1.455 MHz with 1 MHz produces a 455 kHz difference that the IF filter selects. The oscillator frequency is therefore always signal plus IF.

Question 36Communication Systems and Modulation Techniques

What is the image frequency for a receiver with a 10 MHz signal frequency and a 2 MHz IF using high-side injection?

  1. 8 MHz
  2. 22 MHz
  3. 12 MHz
  4. 14 MHz Correct answer.

Explanation

The image lies on the opposite side of the local oscillator from the wanted signal. The oscillator runs at 12 MHz, so the image is 12 MHz plus 2 MHz, namely 14 MHz. Image rejection depends on the RF filter ahead of the mixer.

Question 37Communication Systems and Modulation Techniques

In FM, what happens to the modulation index and the bandwidth as the modulating frequency is reduced while peak deviation is held constant?

  1. The modulation index rises and the bandwidth approaches twice the deviation Correct answer.
  2. Both remain unchanged because deviation is constant
  3. The modulation index rises but the bandwidth collapses to the modulating frequency
  4. The modulation index falls and the bandwidth shrinks toward zero

Explanation

The FM modulation index is the ratio of peak deviation to modulating frequency. Lowering the modulating frequency raises the index and spreads energy into more sidebands. Carson's rule then shows the bandwidth tending toward twice the deviation.

Question 38Communication Systems and Modulation Techniques

Which statement correctly describes single-sideband suppressed-carrier transmission compared with full-carrier AM?

  1. It uses roughly half the bandwidth and transmits no carrier power Correct answer.
  2. It transmits the carrier plus one sideband for simpler detection
  3. It uses twice the bandwidth but improves signal-to-noise ratio
  4. It occupies the same bandwidth but with a higher carrier level

Explanation

SSB removes the carrier and one redundant sideband, halving occupied bandwidth and concentrating power in the remaining sideband. Efficiency improves because no power is wasted on the carrier. The trade-off is the need for accurate carrier reinsertion at the receiver.

Question 39Communication Systems and Modulation Techniques

A binary channel transmits at 2400 baud using 16-level QAM. Assuming each symbol carries the full number of bits, what is the bit rate?

  1. 9600 bit/s Correct answer.
  2. 38400 bit/s
  3. 2400 bit/s
  4. 4800 bit/s

Explanation

Each 16-QAM symbol carries log2(16) = 4 bits. Multiplying 4 bits per symbol by 2400 symbols per second gives 9600 bit/s. Higher-order QAM raises bits per symbol at the cost of tighter noise margins.

Question 40Communication Systems and Modulation Techniques

In a PCM link the signal is sampled at 8 kHz and each sample is encoded into 8 bits. What is the resulting bit rate?

  1. 8 kbit/s
  2. 16 kbit/s
  3. 64 kbit/s Correct answer.
  4. 128 kbit/s

Explanation

Bit rate equals sampling rate multiplied by bits per sample. Here 8000 samples per second times 8 bits gives 64 000 bit/s. This is the standard figure for a single voice channel in telephony PCM.

Question 41Communication Systems and Modulation Techniques

What is the main advantage of delta modulation over conventional PCM?

  1. Simpler circuitry because only one bit per sample is sent Correct answer.
  2. Wider dynamic range for large-amplitude signals
  3. Higher signal-to-noise ratio for the same bit rate
  4. Immunity to slope overload without any added complexity

Explanation

Delta modulation encodes only the direction of change in the signal, so it needs a single bit per sample and very simple hardware. The price is granular noise and slope overload when the signal changes too quickly. Companding or adaptive step size is used to widen its dynamic range.

Question 42Communication Systems and Modulation Techniques

Which effect causes interference when a strong signal near the receiver's tuned frequency is amplified nonlinearly by the front-end stage?

  1. Image response
  2. Doppler shift
  3. Cross-modulation Correct answer.
  4. Selective fading

Explanation

Cross-modulation occurs when a strong unwanted signal modulates the gain of a nonlinear stage, transferring its modulation onto the wanted signal. It is a front-end linearity problem. Improving dynamic range and preselection reduces it.

Question 43Communication Systems and Modulation Techniques

Why is pre-emphasis applied to the modulating signal in FM broadcast transmitters?

  1. To convert frequency modulation into amplitude modulation
  2. To increase the carrier deviation beyond the licensed limit
  3. To widen the audio bandwidth to 15 kHz
  4. To boost high frequencies so that de-emphasis at the receiver reduces high-frequency noise Correct answer.

Explanation

FM demodulators produce noise that rises with frequency, so high audio frequencies suffer worse signal-to-noise ratio. Pre-emphasis raises them before transmission, and a matching de-emphasis network at the receiver restores flat response while attenuating the boosted noise. The two curves must complement each other.

Question 44Communication Systems and Modulation Techniques

A superheterodyne receiver uses an IF of 455 kHz. Which intermediate frequency would most likely be chosen to improve image rejection in a VHF receiver?

  1. A higher IF, such as 10.7 MHz Correct answer.
  2. A lower IF, such as 100 kHz
  3. An IF equal to the received frequency
  4. An IF of zero with no local oscillator

Explanation

Image rejection improves when the IF is large relative to the signal, because the image then lies far from the tuned frequency and can be filtered. VHF receivers therefore use higher IFs such as 10.7 MHz. The trade-off is that wider IF filters make channel selectivity harder.

Question 45Communication Systems and Modulation Techniques

An FM signal has a deviation ratio of 5 and a maximum modulating frequency of 3 kHz. What is the peak frequency deviation?

  1. 30 kHz
  2. 15 kHz Correct answer.
  3. 0.6 kHz
  4. 8 kHz

Explanation

The deviation ratio D equals peak deviation divided by the highest modulating frequency. Rearranging gives Δf = D × fm, so 5 × 3 kHz = 15 kHz. This ratio is used to read the required bandwidth from Bessel-function tables.

Question 46Communication Systems and Modulation Techniques

In 16-QAM, how do the required signal-to-noise ratio and spectral efficiency compare with QPSK for the same bit error rate?

  1. 16-QAM needs a higher signal-to-noise ratio but carries twice the bits per symbol Correct answer.
  2. 16-QAM needs the same signal-to-noise ratio and carries half the bits per symbol
  3. 16-QAM needs a higher signal-to-noise ratio and carries half the bits per symbol
  4. 16-QAM needs a lower signal-to-noise ratio and carries the same bits per symbol

Explanation

16-QAM places four bits per symbol, double QPSK's two, so it doubles spectral efficiency. The constellation points are closer together, so more energy per bit is needed to keep the same error rate. That trade-off between efficiency and robustness governs modulation choice in a link budget.

Question 47Communication Systems and Modulation Techniques

A signal is sampled at 10 kHz without an anti-alias filter. A 7 kHz tone is present in the input. At what frequency does it appear after sampling?

  1. 7 kHz
  2. 17 kHz
  3. 2 kHz
  4. 3 kHz Correct answer.

Explanation

Sampling at 10 kHz folds frequencies above the 5 kHz Nyquist limit back into the baseband. The 7 kHz component appears at 10 kHz minus 7 kHz, which is 3 kHz. An anti-alias filter ahead of the sampler is the standard remedy.

Question 48Communication Systems and Modulation Techniques

Why does a coherent BPSK receiver typically outperform a non-coherent ASK receiver at the same bit rate?

  1. Coherent detection avoids the need for any synchronization at the receiver
  2. Non-coherent ASK has a larger constellation and therefore more noise margin
  3. Coherent detection uses carrier phase reference and requires less energy per bit for a given error rate Correct answer.
  4. BPSK occupies a wider bandwidth so errors are spread over more frequencies

Explanation

Coherent detection compares the received phase against a locally regenerated carrier, which is the most efficient use of signal energy. Non-coherent ASK compares envelope levels and pays a penalty of roughly 3 dB. The cost of coherent operation is the carrier recovery circuitry it demands.

Question 49Communication Systems and Modulation Techniques

A receiver has a noise figure of 6 dB and a bandwidth of 1 MHz. Referred to a reference temperature of 290 K, what is the approximate input noise power in dBm?

  1. -120 dBm
  2. -108 dBm Correct answer.
  3. -114 dBm
  4. -102 dBm

Explanation

Thermal noise power is kTB, which at 290 K and 1 MHz is about -114 dBm. Adding the 6 dB noise figure gives roughly -108 dBm at the receiver input. Noise figure therefore directly raises the effective noise floor.

Question 50Communication Systems and Modulation Techniques

In an amplitude modulation system, the carrier is 200 V peak and the modulating tone produces a peak envelope variation of 80 V above and below the unmodulated carrier level. What is the modulation index?

  1. 1.25
  2. 0.8
  3. 0.4 Correct answer.
  4. 0.2

Explanation

The modulation index equals the peak envelope deviation divided by the unmodulated carrier amplitude, so m = 80/200 = 0.4. This value lies between 0 and 1, meaning the envelope never reaches zero and no overmodulation distortion occurs. Expressing it as a percentage gives 40 percent modulation depth.