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WGU D413 – Telecommunications & Wireless Communications Objective Assessment | OA V1 and V2 | Questions and Answers | 2026 Update | 100% Correct

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WGU D413 – Telecommunications & Wireless Communications Objective Assessment | OA V1 and V2 | Questions and Answers | 2026 Update | 100% Correct

Institution
WGU D413
Course
WGU D413

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WGU D413 – Telecommunications & Wireless
Communications Objective Assessment | OA V1 and V2 |
Questions and Answers | 2026 Update | 100% Correct


Q1. What is the primary cause of free space path loss (FSPL) in a wireless link?
A) Absorption by atmospheric gases
B) Spreading of electromagnetic energy as distance increases
C) Multipath cancellation at the receiver
D) Interference from other transmitters

Answer: B
Rationale: FSPL occurs because RF energy spreads out over a larger area as distance
increases, reducing power density at the receiver. It is not primarily caused by absorption
(A), multipath (C), or interference (D).

Q2. A signal at 900 MHz has a wavelength of approximately:
A) 0.33 meters
B) 0.33 centimeters
C) 3.3 meters
D) 33 meters

Answer: A
*Rationale: Wavelength (λ) = c / f = 3×10⁸ / 900×10⁶ ≈ 0.33 meters. Lower frequencies
have longer wavelengths.*

Q3. Which fading type is caused by reflections from large obstacles like buildings,
resulting in multiple delayed copies of the same signal?
A) Rayleigh fading
B) Rician fading
C) Multipath fading
D) Shadow fading

Answer: C
Rationale: Multipath fading occurs when a signal arrives via multiple paths with different
delays. Rayleigh (A) is a specific statistical model for multipath with no LOS. Rician (B)
includes a dominant LOS component. Shadow fading (D) is due to large obstructions.

Q4. Increasing transmitter power from 1W to 2W is an increase of approximately:
A) 1 dB

,B) 3 dB
C) 6 dB
D) 10 dB

Answer: B
*Rationale: dB = 10 × log₁₀(P2/P1) = 10 × log₁₀(2) = 10 × 0.301 = 3.01 dB. Doubling
power = +3 dB.*

Q5. What does SNR stand for, and why is it important?
A) Signal Noise Ratio – higher is worse
B) Signal-to-Noise Ratio – higher is better for data integrity
C) Serial Noise Reduction – irrelevant to wireless
D) Signal Node Routing – used in mesh networks

Answer: B
Rationale: SNR measures signal power relative to noise power. Higher SNR means cleaner
signal, lower bit error rate, and higher possible data rate.




Competency 2: Modulation & Encoding
Q6. Which digital modulation scheme changes both amplitude and phase to represent
multiple bits per symbol?
A) BPSK
B) QPSK
C) 16-QAM
D) FSK

Answer: C
*Rationale: QAM (Quadrature Amplitude Modulation) varies both amplitude and phase.
16-QAM carries 4 bits per symbol. BPSK (A) and QPSK (B) are phase-only. FSK (D) varies
frequency.*

Q7. What is the primary advantage of OFDM over single-carrier modulation?
A) Simpler transceiver design
B) Lower peak-to-average power ratio
C) Resilience to multipath fading via guard intervals
D) Higher power efficiency

, Answer: C
Rationale: OFDM uses multiple subcarriers with cyclic prefixes (guard intervals) to resist
inter-symbol interference from multipath. It is not simpler (A) nor more power-efficient
(D).

Q8. A modulation scheme uses 8 different phase angles. How many bits per symbol
does it carry?
A) 2
B) 3
C) 4
D) 8

Answer: B
*Rationale: Bits per symbol = log₂(M) where M = number of symbols. log₂(8) = 3 bits.*

Q9. In QPSK, the two carriers used are:
A) Same frequency, same phase
B) Same frequency, 90° out of phase
C) Different frequencies, same phase
D) Same frequency, 180° out of phase

Answer: B
*Rationale: QPSK uses in-phase (I) and quadrature (Q) carriers at the same frequency but
shifted by 90°. This allows two bits per symbol.*

Q10. Which factor most directly limits the maximum data rate of a wireless channel (per
Shannon-Hartley)?
A) Transmitter antenna gain
B) Bandwidth and SNR
C) Number of users
D) Encryption type

Answer: B
*Rationale: Shannon-Hartley theorem: C = B × log₂(1 + SNR). Capacity increases with
bandwidth (B) and SNR.*




Competency 3: Antennas & Propagation

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