Compared with DSB-SC, SSB saves which resource?
A Carrier power
B Audio frequency
C Transmission bandwidth
D Detector timing
Answer: C. Transmission bandwidth. SSB sends one sideband, reducing ideal bandwidth from 2B to B.
Why does DSB-SC detection use a local oscillator?
A The carrier is absent
B Both sidebands are absent
C The message is digital
D Noise is eliminated
Answer: A. The carrier is absent. A local synchronized carrier replaces the suppressed carrier for coherent recovery.
When a product detector has a 90° phase error, how does its output change?
A Output doubles
B Recovered output nulls
C Carrier frequency shifts
D Noise becomes zero
Answer: B. Recovered output nulls. The recovered message is scaled by cos(phi), which is zero at 90 degrees.
In the phasing method, which transform creates quadrature?
A Fourier series
B Laplace transform
C Z-transform
D Hilbert transform
Answer: D. Hilbert transform. A Hilbert transform supplies a 90-degree phase-shifted signal for SSB generation.
Which component remains in ideal SSB-SC spectrum?
A One sideband
B Carrier only
C Both sidebands
D Message DC only
Answer: A. One sideband. SSB-SC suppresses the carrier and transmits only one sideband.
Which scheme pairs a full sideband with a small remainder?
A DSB-SC
B FM
C VSB
D PM
Answer: C. VSB. VSB retains one sideband plus a vestige of the other.
For an envelope detector, how should RC compare with the carrier period?
A Much shorter
B Much longer
C Exactly equal
D Zero
Answer: B. Much longer. The RC constant should smooth carrier cycles while still following the message envelope.
Excessive detector RC mainly causes what?
A Carrier suppression
B Frequency doubling
C Phase reversal
D Diagonal clipping
Answer: D. Diagonal clipping. A slow discharge cannot follow rapid envelope decreases, causing diagonal clipping.
Very small detector RC produces what?
A Diagonal clipping
B Sideband cancellation
C Carrier ripple
D Overdeviation
Answer: C. Carrier ripple. A short RC constant fails to smooth carrier-frequency variations.
In AM power calculations, Pc represents what?
A Unmodulated carrier power
B Total sideband power
C Message signal power
D Noise power
Answer: A. Unmodulated carrier power. Pc is the power in the carrier before modulation.
With m=1 and Pc=60 W, total AM power equals?
A 60 W
B 75 W
C 120 W
D 90 W
Answer: D. 90 W. Ptotal = Pc(1 + m²/2) = 60 × 1.5 = 90 W.
What is AM efficiency when m=0.5?
A 6.25%
B 11.1%
C 20%
D 33.3%
Answer: B. 11.1%. Efficiency is m²/(2 + m²), giving 0.25/2.25 ≈ 11.1%.
Frequency deviation in FM is proportional to what?
A Carrier amplitude
B Message phase only
C Message amplitude
D Carrier power
Answer: C. Message amplitude. In FM, instantaneous frequency deviation follows the modulating signal amplitude.
For fixed Δf, doubling fm changes β how?
A It halves
B It doubles
C It stays constant
D It becomes zero
Answer: A. It halves. FM index β = Δf/fm, so doubling fm halves β.
Given Δf=30 kHz and fm=10 kHz, find β.
A 0.3
B 2
C 40
D 3
Answer: D. 3. The modulation index is 30/10 = 3.
Carson’s rule estimates FM bandwidth using which expression?
A Δf − fm
B 2(Δf + fm)
C 2Δf − fm
D Δf + 2fm
Answer: B. 2(Δf + fm). Carson’s rule approximates bandwidth as twice peak deviation plus highest modulating frequency.
For Δf=20 kHz and fm,max=5 kHz, estimate Carson bandwidth.
A 50 kHz
B 25 kHz
C 40 kHz
D 60 kHz
Answer: A. 50 kHz. 2(20 + 5) kHz equals 50 kHz.
What distinguishes ideal FM from AM envelope amplitude?
A Envelope follows audio
B Carrier is absent
C Constant envelope
D Two fixed sidebands
Answer: C. Constant envelope. Ideal FM varies instantaneous frequency while maintaining constant amplitude.
What quantity does a frequency discriminator output track?
A Carrier power
B Instantaneous frequency deviation
C Message phase offset
D Sideband count
Answer: B. Instantaneous frequency deviation. An FM discriminator converts instantaneous frequency changes into an output voltage.
Why use an amplitude limiter before FM detection?
A Increase deviation
B Recover the carrier
C Create sidebands
D Reject amplitude noise
Answer: D. Reject amplitude noise. An amplitude limiter removes unwanted amplitude variations before frequency detection.
Narrowband FM generally has what modulation index?
A Exactly one
B Much greater than one
C Much less than one
D No defined index
Answer: C. Much less than one. Narrowband FM is characterized by a small modulation index.
When β is much greater than one, FM is called what?
A Wideband FM
B Narrowband FM
C DSB-SC
D Vestigial FM
Answer: A. Wideband FM. A large modulation index corresponds to wideband FM.
In phase modulation, instantaneous phase follows what?
A Carrier power
B Message signal
C Noise variance
D Envelope minimum
Answer: B. Message signal. PM varies carrier phase in proportion to the message signal.
Which modulation’s instantaneous frequency follows the message directly?
A AM
B SSB
C PAM
D FM
Answer: D. FM. FM frequency deviation is directly proportional to the modulating signal.
In single-tone FM, β divides peak deviation by what?
A Carrier frequency
B Message amplitude
C Modulating frequency
D Channel bandwidth
Answer: C. Modulating frequency. For a single tone, β = Δf/fm.