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Telecommunications, RF & Antennas (Electronics Engineering) Solved Questions & Notes (2026) - Apex Rankers

Engineering & Technology > Electronics Engineering > Telecommunications, RF & Antennas

35 Total Solved Questions
~53 mins Estimated Reading Time
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Telecommunications, RF & Antennas

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Q. 1 Electronics Engineering
Difficulty: Easy (1 Mark)
In standard Amplitude Modulation (AM), if the carrier power is Pc and the modulation index is m, what is the total transmitted power Pt?
A
Pt = Pc * (1 + m^2 / 2)
✓ Correct
B
Pt = Pc * (1 + m / 2)
C
Pt = Pc * (1 + m^2)
D
Pt = Pc * (1 + 2 * m^2)
💡 Step-by-Step Explanation & Concept Rationale
Total power in DSB-FC AM is Pt = Pc + Pusb + Plsb = Pc + Pc*(m^2/4) + Pc*(m^2/4) = Pc * (1 + m^2 / 2).
Q. 2 Electronics Engineering
Difficulty: Medium (1 Mark)
What is the maximum theoretical transmission efficiency of a standard AM (DSB-FC) broadcast signal with 100% sinusoidal modulation (m = 1)?
A
100%
B
33.33%
✓ Correct
C
50%
D
66.67%
💡 Step-by-Step Explanation & Concept Rationale
Efficiency eta = (Sideband Power / Total Power) = (m^2 / (2 + m^2)). For m = 1, eta = 1 / 3 ≈ 33.33%. Two-thirds of the power is wasted in the carrier which carries no information.
Q. 3 Electronics Engineering
Difficulty: Easy (1 Mark)
According to Carson's Rule, what is the approximate transmission bandwidth (BW) of an FM signal with frequency deviation Delta_f and modulating signal bandwidth fm?
A
BW = 2 * (Delta_f + fm)
✓ Correct
B
BW = Delta_f + fm
C
BW = 2 * Delta_f * fm
D
BW = 2 * fm
💡 Step-by-Step Explanation & Concept Rationale
Carson's Rule states that 98% of FM power is contained within bandwidth BW = 2 * (Delta_f + fm) = 2 * fm * (beta + 1).
Q. 4 Electronics Engineering
Difficulty: Easy (1 Mark)
In a standard superheterodyne AM broadcast receiver, what is the standard intermediate frequency (IF)?
A
455 kHz
✓ Correct
B
10.7 MHz
C
70 MHz
D
1.4 MHz
💡 Step-by-Step Explanation & Concept Rationale
AM broadcast receivers use an IF of 455 kHz. FM broadcast receivers use an IF of 10.7 MHz.
Q. 5 Electronics Engineering
Difficulty: Medium (1 Mark)
In a superheterodyne receiver with local oscillator frequency fo = fs + IF, what is the image frequency f_image?
A
f_image = fs + IF
B
f_image = fs + 2 * IF
✓ Correct
C
f_image = fs - 2 * IF
D
f_image = 2 * fs + IF
💡 Step-by-Step Explanation & Concept Rationale
An unwanted station at f_image = fs + 2*IF mixes with fo = fs + IF to produce |fo - f_image| = IF, entering the IF amplifier unless attenuated by RF preselector filtering.
Q. 6 Electronics Engineering
Difficulty: Easy (1 Mark)
Shannon-Hartley theorem states that the channel capacity C (in bits/sec) for a band-limited AWGN channel is:
A
C = B * log2(1 + S/N)
✓ Correct
B
C = 2 * B * log2(S/N)
C
C = B * ln(1 + S/N)
D
C = B * (1 + S/N)^2
💡 Step-by-Step Explanation & Concept Rationale
Shannon-Hartley theorem: C = B * log2(1 + S/N), where B is channel bandwidth in Hertz and S/N is the signal-to-noise power ratio.
Q. 7 Electronics Engineering
Difficulty: Medium (1 Mark)
In Pulse Code Modulation (PCM), if the number of quantization bits per sample is increased from n to n + 1, by how many dB does the Signal-to-Quantization-Noise Ratio (SQNR) improve?
A
3 dB
B
6.02 dB
✓ Correct
C
10 dB
D
12 dB
💡 Step-by-Step Explanation & Concept Rationale
SQNR (in dB) ≈ 1.76 + 6.02 * n. Adding 1 bit increases the SQNR by approximately 6.02 dB (a 4-fold increase in SNR power ratio).
Q. 8 Electronics Engineering
Difficulty: Medium (1 Mark)
What is the primary purpose of companding (compressing-expanding) using A-law or mu-law in digital telephony PCM systems?
A
To increase transmission data rate by 50%
B
To maintain a relatively constant SNR over a wide dynamic range of weak and strong speech signals
✓ Correct
C
To eliminate the need for an anti-aliasing filter
D
To convert analog voice signals directly into optical pulses
💡 Step-by-Step Explanation & Concept Rationale
Non-uniform quantization (companding) uses smaller step sizes for low amplitude signals and larger step sizes for large signals, keeping SQNR roughly constant across all voice levels.
Q. 9 Electronics Engineering
Difficulty: Easy (1 Mark)
What is the radiation resistance of an ideal center-fed half-wave dipole antenna in free space?
A
36.5 ohms
B
73.13 ohms
✓ Correct
C
50 ohms
D
377 ohms
💡 Step-by-Step Explanation & Concept Rationale
The terminal impedance of a resonant half-wave thin dipole in free space is approximately 73.13 + j42.5 ohms (purely resistive ~73 ohms when cut slightly shorter by ~5%).
Q. 10 Electronics Engineering
Difficulty: Easy (1 Mark)
What is the radiation resistance of an ideal quarter-wave monopole antenna over an infinite, perfectly conducting ground plane?
A
73.13 ohms
B
36.56 ohms
✓ Correct
C
120 pi ohms
D
300 ohms
💡 Step-by-Step Explanation & Concept Rationale
By image theory, a quarter-wave monopole over a ground plane radiates half the total power of a half-wave dipole for the same current. Thus, its radiation resistance is half: 73.13 / 2 ≈ 36.56 ohms.
Q. 11 Electronics Engineering
Difficulty: Easy (1 Mark)
What is the characteristic intrinsic impedance of free space (vacuum) for electromagnetic waves?
A
50 ohms
B
75 ohms
C
120 * pi ≈ 377 ohms
✓ Correct
D
600 ohms
💡 Step-by-Step Explanation & Concept Rationale
The wave impedance of free space is eta_0 = sqrt(mu_0 / epsilon_0) ≈ 120 * pi ≈ 376.73 ≈ 377 ohms.
Q. 12 Electronics Engineering
Difficulty: Easy (1 Mark)
In a Yagi-Uda antenna array, which element is connected directly to the transmission feedline?
A
Reflector
B
Driven element (usually a folded dipole)
✓ Correct
C
Directors
D
Balun shield only
💡 Step-by-Step Explanation & Concept Rationale
Only the driven element (typically a folded dipole for ~300 ohm impedance matching) is actively fed. The reflector and directors are parasitic elements excited via reradiation.
Q. 13 Electronics Engineering
Difficulty: Medium (1 Mark)
In a Yagi-Uda antenna, how do the physical lengths of the reflector and directors compare to the driven element?
A
Reflector is ~5% longer, and directors are ~5% shorter than the driven element
✓ Correct
B
Reflector is shorter, and directors are longer
C
All elements have identical lengths
D
Directors are twice as long as the reflector
💡 Step-by-Step Explanation & Concept Rationale
The reflector is slightly longer (~5% longer) to make it inductive, while the directors are slightly shorter (~5% shorter) to make them capacitive, shaping a highly directive beam toward the directors.
Q. 14 Electronics Engineering
Difficulty: Medium (1 Mark)
In ionospheric radio propagation, the highest frequency that can be reflected back to Earth for vertical incidence is known as the:
A
Maximum Usable Frequency (MUF)
B
Critical Frequency (fc)
✓ Correct
C
Optimum Working Frequency (OWF)
D
Cutoff Frequency (fco)
💡 Step-by-Step Explanation & Concept Rationale
The critical frequency fc is the highest frequency returned to Earth when transmitted vertically: fc ≈ 9 * sqrt(Nmax), where Nmax is maximum electron density per m^3.
Q. 15 Electronics Engineering
Difficulty: Medium (1 Mark)
The relationship between Maximum Usable Frequency (MUF) and Critical Frequency (fc) for oblique incidence at angle of incidence theta is given by the Secant Law as:
A
MUF = fc * sec(theta) = fc / cos(theta)
✓ Correct
B
MUF = fc * cos(theta)
C
MUF = fc * sin(theta)
D
MUF = fc / sin(theta)
💡 Step-by-Step Explanation & Concept Rationale
According to the secant law, MUF = fc * sec(theta) = fc / cos(theta). For oblique angles (theta > 0), MUF is always greater than the vertical critical frequency fc.
Q. 16 Electronics Engineering
Difficulty: Easy (1 Mark)
The zone between the end of ground-wave coverage and the first point where the sky wave returns to Earth is called the:
A
Fresnel zone
B
Skip zone (or dead zone)
✓ Correct
C
Near-field region
D
Shadow boundary
💡 Step-by-Step Explanation & Concept Rationale
The skip zone is the silent gap where the ground wave has completely attenuated, but the refracted sky wave has not yet reached the ground.
Q. 17 Electronics Engineering
Difficulty: Easy (1 Mark)
In digital modulation, Quadrature Phase Shift Keying (QPSK) transmits how many bits per symbol?
A
1 bit/symbol
B
2 bits/symbol
✓ Correct
C
3 bits/symbol
D
4 bits/symbol
💡 Step-by-Step Explanation & Concept Rationale
QPSK uses 4 phase states (0, 90, 180, 270 degrees). Since M = 4, the number of bits per symbol is log2(4) = 2 bits/symbol.
Q. 18 Electronics Engineering
Difficulty: Medium (1 Mark)
What is the spectral efficiency of 16-QAM (Quadrature Amplitude Modulation) in theoretical bits/s/Hz?
A
2 bits/s/Hz
B
4 bits/s/Hz
✓ Correct
C
8 bits/s/Hz
D
16 bits/s/Hz
💡 Step-by-Step Explanation & Concept Rationale
In 16-QAM, each symbol represents log2(16) = 4 bits. With ideal Nyquist pulse shaping, the spectral efficiency is 4 bits/s/Hz.
Q. 19 Electronics Engineering
Difficulty: Hard (1 Mark)
The Friis transmission equation expresses received power Pr in free space as:
A
Pr = Pt * Gt * Gr * (lambda / (4 * pi * d))^2
✓ Correct
B
Pr = Pt * Gt * Gr * (4 * pi * d / lambda)^2
C
Pr = Pt * Gt * Gr / (d^2)
D
Pr = Pt * (lambda / d)^2
💡 Step-by-Step Explanation & Concept Rationale
Friis formula: Pr = Pt * Gt * Gr * (lambda / (4 * pi * d))^2. Received power decreases with the square of distance and square of frequency (since lambda = c / f).
Q. 20 Electronics Engineering
Difficulty: Easy (1 Mark)
What does a Voltage Standing Wave Ratio (VSWR) of 1.0 indicate on an RF transmission line?
A
Total reflection with an open circuit
B
A perfect impedance match with zero reflected power (reflection coefficient Gamma = 0)
✓ Correct
C
A direct short circuit to ground
D
Equal forward and reverse power flows
💡 Step-by-Step Explanation & Concept Rationale
VSWR = (1 + |Gamma|) / (1 - |Gamma|). When |Gamma| = 0 (perfect impedance match, Zload = Z0), VSWR = 1.0 (1:1), meaning 100% of forward power is absorbed by the load.
Q. 21 Electronics Engineering
Difficulty: Medium (1 Mark)
If an RF load has a reflection coefficient magnitude |Gamma| = 0.333 (or 1/3), what is the VSWR?
A
1.5 : 1
B
2.0 : 1
✓ Correct
C
3.0 : 1
D
1.33 : 1
💡 Step-by-Step Explanation & Concept Rationale
VSWR = (1 + 1/3) / (1 - 1/3) = (4/3) / (2/3) = 2.0 : 1.
Q. 22 Electronics Engineering
Difficulty: Easy (1 Mark)
In fiber optic communications, what optical principle confines light within the core?
A
Diffuse reflection
B
Total Internal Reflection (TIR) with core refractive index n1 > cladding index n2
✓ Correct
C
Polarization scattering
D
Fresnel diffraction
💡 Step-by-Step Explanation & Concept Rationale
Light propagates inside the fiber core via total internal reflection, which requires the core index of refraction (n1) to be slightly higher than the cladding index (n2), and incident angles exceeding the critical angle.
Q. 23 Electronics Engineering
Difficulty: Medium (1 Mark)
In a quarter-wave transmission line impedance transformer, to match a line of characteristic impedance Z0 to a load RL, the transformer impedance Zmatch must be:
A
Zmatch = sqrt(Z0 * RL)
✓ Correct
B
Zmatch = (Z0 + RL) / 2
C
Zmatch = Z0 * RL
D
Zmatch = Z0^2 / RL
💡 Step-by-Step Explanation & Concept Rationale
A quarter-wave transformer transforms input impedance according to Zin = Zmatch^2 / RL. For Zin = Z0, Zmatch = sqrt(Z0 * RL).
Q. 24 Electronics Engineering
Difficulty: Medium (1 Mark)
What type of optical fiber dispersion is caused by different spectral wavelengths of light traveling at slightly different velocities in the glass core?
A
Modal dispersion
B
Chromatic (material) dispersion
✓ Correct
C
Polarization mode dispersion
D
Rayleigh scattering
💡 Step-by-Step Explanation & Concept Rationale
Chromatic dispersion arises because the glass refractive index varies with optical wavelength, causing different spectral components from a laser/LED pulse to travel at different group velocities, spreading the pulse.
Q. 25 Electronics Engineering
Difficulty: Medium (1 Mark)
Which multiple access technique divides the available frequency spectrum into orthogonal, overlapping sub-carriers to maximize spectral efficiency and resist multipath fading?
A
TDMA (Time Division Multiple Access)
B
OFDM (Orthogonal Frequency Division Multiplexing)
✓ Correct
C
CDMA using Barker codes
D
FHSS (Frequency Hopping Spread Spectrum)
💡 Step-by-Step Explanation & Concept Rationale
OFDM divides high-rate data streams into many closely spaced orthogonal narrowband subcarriers, mitigating intersymbol interference (ISI) and multipath fading in 4G LTE, 5G NR, and Wi-Fi.
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