Communications practice questions

Answer: A) The probability of bit error decreases

With a fixed bit rate, a higher Eb/N0 means more signal energy per bit against the same noise density. For BPSK, Pb = Q(√(2Eb/N0)) falls as Eb/N0 rises.

• ACorrect. For BPSK, Pb = Q(√(2Eb/N0)). The Q-function falls as its argument grows, so a higher Eb/N0 gives a lower bit error probability.
• BIncorrect. Channel noise (N0) does not depend on transmit power. More energy per bit against the same noise lowers the error rate.
• CIncorrect. BPSK bandwidth depends on the symbol rate and pulse shaping, not on signal power. The bit rate is fixed, so the bandwidth stays the same.
• DIncorrect. The bit rate is set by the modulator and is fixed here. Raising Eb/N0 improves reliability, not throughput.

Answer: A) 3 m

• ACorrect. \(\lambda = c/f = 3\times10^8/10^8\).
• BIncorrect. Recheck the powers of ten.
• CIncorrect. That's for 1 GHz.
• DIncorrect. That's for 1 MHz.

Answer: C) 20 dB

• AIncorrect. Take the logarithm.
• BIncorrect. That uses 20 log for power.
• CCorrect. \(10\log_{10} 100\).
• DIncorrect. Multiply the log by 10.

Answer: D) 180 kHz

Carson's rule estimates FM bandwidth as B ≈ 2(Δf + fm). Here that is 2(75 + 15) = 180 kHz.

• AIncorrect. This is 2Δf only. Carson's rule also adds the message frequency: B = 2(Δf + fm).
• BIncorrect. This is Δf + fm without the factor of 2. The rule counts both sidebands, so the sum is doubled.
• CIncorrect. 2fm is the bandwidth of a DSB AM signal. It ignores the 75 kHz frequency deviation.
• DCorrect. B ≈ 2(Δf + fm) = 2(75 kHz + 15 kHz) = 180 kHz.

Answer: B) Output signal-to-noise ratio

• AIncorrect. It maximizes SNR.
• BCorrect. At the sampling instant.
• CIncorrect. It maximizes SNR.
• DIncorrect. It maximizes SNR.

Answer: D) 0.25 m

• AIncorrect. That multiplies by 4.
• BIncorrect. A quarter wave is shorter.
• CIncorrect. That's a half wave.
• DCorrect. λ = c/f = 1 m, divided by 4.

Answer: D) 6 dB

• AIncorrect. That's too little.
• BIncorrect. That's a doubling of power.
• CIncorrect. That's too much.
• DCorrect. An 8-bit converter gives about 50 dB.

Answer: B) 4,000 bps

• AIncorrect. Use \(\log_2 16 = 4\) bits per symbol.
• BCorrect. 16-QAM carries 4 bits per symbol.
• CIncorrect. Multiply by bits per symbol.
• DIncorrect. Multiply, don't divide.

Answer: A) It splits the stream across many closely spaced subcarriers whose spectra overlap without mutual interference

OFDM divides a high-rate stream into many low-rate streams on closely spaced orthogonal subcarriers. Their spectra overlap without interfering, which uses bandwidth efficiently and resists multipath.

• ACorrect. OFDM sends parallel low-rate streams on many subcarriers spaced at 1/T. That spacing makes them orthogonal, so their spectra can overlap without interfering, and the long symbols resist multipath.
• BIncorrect. This describes conventional single-carrier transmission. OFDM avoids very short symbols by dividing the data among many subcarriers.
• CIncorrect. This describes time-division multiple access (TDMA), not OFDM.
• DIncorrect. This describes direct-sequence spread spectrum and CDMA, not OFDM.

Answer: A) Gain increases and the half-power beamwidth becomes narrower

Parabolic dish gain is G = η(πD/λ)², and beamwidth is roughly proportional to λ/D. A larger diameter at the same frequency therefore raises the gain and narrows the beam.

• ACorrect. Gain G = η(πD/λ)² grows with the square of the diameter (doubling D adds about 6 dB). Beamwidth is roughly proportional to λ/D, so it narrows.
• BIncorrect. Higher gain means the energy is concentrated in a narrower beam. Beamwidth goes roughly as λ/D, so a bigger D gives a narrower beam, not a wider one.
• CIncorrect. Gain rises with aperture area (proportional to D²) at a fixed wavelength, so a larger dish cannot lower the gain.
• DIncorrect. Gain is directly proportional to effective aperture (G = 4πA_e/λ²), so a larger effective aperture always means higher gain at the same frequency.