Difficulty: Hard
At room temperature, a 5 MHz receiver has 5 dB noise figure and needs 8 dB SNR. Approximate sensitivity?
A -94 dBm
B -107 dBm
C -102 dBm
D -86 dBm
Answer: A. -94 dBm. Thermal noise is about -107 dBm; add 5 dB noise figure and 8 dB required SNR.
When noise bandwidth doubles, what happens to integrated thermal noise?
A Rises 3 dB
B Falls 3 dB
C Rises 6 dB
D Stays fixed
Answer: A. Rises 3 dB. Doubling bandwidth doubles noise power, a 3 dB increase.
Signal is -80 dBm, interference -90 dBm, and noise -100 dBm. Approximate SINR?
A 20 dB
B 9.6 dB
C 10 dB
D -10 dB
Answer: B. 9.6 dB. Interference plus noise is about -89.6 dBm, giving roughly 9.6 dB SINR.
Two independent receive branches each have 10 dB SNR. Ideal maximal-ratio combining gives about what SNR?
A 10 dB
B 13 dB
C 20 dB
D 23 dB
Answer: B. 13 dB. MRC adds linear branch SNRs: 10 + 10 = 20, or about 13 dB.
Strongly correlated antenna paths usually affect MIMO capacity how?
A Reduce it
B Double it
C Leave it unchanged
D Make it infinite
Answer: A. Reduce it. Correlation reduces independent spatial information and can lower channel capacity.
For a 2 × 2 MIMO channel matrix of rank one, how many independent streams fit?
A Two
B Three
C Four
D One
Answer: D. One. The number of independent spatial streams cannot exceed channel rank.
Why can zero-forcing equalization perform poorly on a weak MIMO channel mode?
A Extra coding gain
B Noise enhancement
C Lower Doppler
D More bandwidth
Answer: B. Noise enhancement. Inverting a small channel singular value amplifies noise.
Compared with zero forcing, what does MMSE equalization explicitly balance?
A Power and wavelength
B Interference and noise
C Delay and range
D Coding and duplex
Answer: B. Interference and noise. MMSE minimizes mean-square error by accounting for both channel interference and noise.
Peak OFDM power is 10 times average power. What is its PAPR?
A 3 dB
B 20 dB
C 10 dB
D 100 dB
Answer: C. 10 dB. PAPR in dB is 10 log10(10), which equals 10 dB.
Why must an OFDM power amplifier allow backoff?
A Increase subcarrier spacing
B Avoid peak clipping
C Remove Doppler
D Create more symbols
Answer: B. Avoid peak clipping. Operating margin keeps high peaks within the amplifier’s linear region.
Useful OFDM time is 66.7 μs and cyclic prefix is 4.7 μs. Useful-time efficiency is about what?
A 93.4%
B 6.6%
C 87.0%
D 98.0%
Answer: A. 93.4%. Efficiency is 66.7/(66.7+4.7), approximately 93.4%.
Carrier offset becomes a noticeable fraction of OFDM subcarrier spacing. What impairment follows?
A Lower path loss
B More guard time
C Higher rank
D Inter-carrier interference
Answer: D. Inter-carrier interference. Frequency error disrupts orthogonality and leaks energy between subcarriers.
Which fading model includes a strong dominant direct path?
A Rician
B Rayleigh
C Uniform
D Gaussian noise
Answer: A. Rician. Rician fading models a dominant component plus scattered multipath.
In a fading channel, outage probability means the chance that what occurs?
A A cell changes colour
B Bandwidth doubles
C SNR falls below target
D A packet is encrypted
Answer: C. SNR falls below target. An outage occurs when instantaneous link quality cannot meet the required threshold.
Two equal-amplitude echoes arrive 180° out of phase. What can occur?
A Power doubling
B No phase change
C Signal cancellation
D Frequency reuse
Answer: C. Signal cancellation. Opposite phases cause destructive addition and a deep fade.
Why is imperfect power control harmful in CDMA uplinks?
A Longer wavelengths
B Near-far interference
C Less coding
D Smaller cells only
Answer: B. Near-far interference. A nearby strong user can overwhelm a weaker distant user’s signal.
Closed-loop power control changes a handset’s transmit level based on what?
A Antenna colour
B Packet length
C Network feedback
D Fibre loss
Answer: C. Network feedback. The network signals adjustments to control received power and interference.
A CDMA signal spreads 1.25 MHz data at 12.5 kb/s. Processing gain is what?
A 100 (20 dB)
B 10 (10 dB)
C 50 (17 dB)
D 1000 (30 dB)
Answer: A. 100 (20 dB). Bandwidth-to-rate ratio is 100; 10 log10(100) equals 20 dB.
For an LNA with 10 dB gain and 3 dB NF, followed by a 6 dB NF stage, approximate cascade NF?
A 2.0 dB
B 4.5 dB
C 3.6 dB
D 9.0 dB
Answer: C. 3.6 dB. Friis’ formula gives Ftotal ≈ 2 + (4 – 1)/10 = 2.3, or about 3.6 dB.
With good SINR, link adaptation can select what for higher throughput?
A Lower coding rate only
B Wider guard band
C More retransmissions
D Higher-order modulation
Answer: D. Higher-order modulation. A cleaner channel can support denser constellations and more bits per symbol.
For a code rate of 3/4 and coded throughput of 12 Mb/s, what is information rate?
A 3 Mb/s
B 12 Mb/s
C 16 Mb/s
D 9 Mb/s
Answer: D. 9 Mb/s. Information rate is 12 × 3/4 = 9 Mb/s.
After a failed block, HARQ stores soft bits before another transmission. What can it do next?
A Discard all prior bits
B Increase path loss
C Combine retransmissions
D Change carrier band
Answer: C. Combine retransmissions. Soft combining pools evidence from repeated transmissions to aid decoding.
In calibrated TDD, when can an uplink channel estimate help downlink beamforming?
A Across any frequency
B After fading changes
C Only in FDD
D Within channel coherence
Answer: D. Within channel coherence. TDD uses the same carrier in both directions, so reciprocity can hold within coherence time after calibration.
Uplink is estimated at 1.8 GHz in FDD. Why cannot it directly represent the 1.9 GHz downlink?
A Same time slot
B No antenna gain
C More coding bits
D Different carrier channels
Answer: D. Different carrier channels. FDD uses separate frequencies, so their channel responses are not inherently reciprocal.
Two strong RF carriers enter a nonlinear amplifier. Which unwanted products may appear?
A Intermodulation products
B Guard intervals
C Pilot symbols
D Independent fades
Answer: A. Intermodulation products. Nonlinear amplification creates new frequencies from combinations of the input tones.