Carrier power Pc=80 W with m=0.5 produces what total sideband power?
A 5 W
B 10 W
C 20 W
D 40 W
Answer: B. 10 W. Psb,total = m²Pc/2 = 0.25 × 80/2 = 10 W.
When an AM envelope reaches 18 V and 6 V, find m.
A 0.5
B 0.25
C 0.6
D 0.75
Answer: A. 0.5. m = (18 − 6)/(18 + 6) = 0.5.
At m=0.8 and Pc=200 W, total AM power is?
A 216 W
B 232 W
C 280 W
D 264 W
Answer: D. 264 W. Ptotal = 200(1 + 0.64/2) = 264 W.
At fc=1.25 MHz and fm=7 kHz, find the upper sideband.
A 1.243 MHz
B 1.250 MHz
C 1.257 MHz
D 1.264 MHz
Answer: C. 1.257 MHz. The upper sideband is fc + fm = 1.257 MHz.
For conventional DSB AM, a 12 kHz channel allows what highest message frequency?
A 3 kHz
B 8 kHz
C 12 kHz
D 6 kHz
Answer: D. 6 kHz. For conventional DSB AM, Bchannel = 2Bmessage.
At 100% AM and Pc=120 W, what is power in each sideband?
A 15 W
B 30 W
C 40 W
D 60 W
Answer: B. 30 W. Each sideband carries m²Pc/4 = 120/4 = 30 W.
SSB carrying 0–4 kHz speech has what ideal RF bandwidth?
A 2 kHz
B 8 kHz
C 4 kHz
D 12 kHz
Answer: C. 4 kHz. An ideal SSB channel occupies the message bandwidth B.
When the local carrier is 60° out of phase, what amplitude scale remains?
A 0.5
B 0
C 0.25
D 0.866
Answer: A. 0.5. The output scales by cos(60°) = 0.5.
Carson’s rule for Δf=45 kHz and fm,max=15 kHz gives what bandwidth?
A 60 kHz
B 120 kHz
C 90 kHz
D 150 kHz
Answer: B. 120 kHz. Carson’s estimate is 2(45 + 15) = 120 kHz.
Use β=4 and fm=5 kHz to determine peak deviation.
A 1.25 kHz
B 9 kHz
C 25 kHz
D 20 kHz
Answer: D. 20 kHz. β = Δf/fm, so Δf = 4 × 5 = 20 kHz.
As fm rises from 10 to 15 kHz at fixed Δf=30 kHz, how does β change?
A 3 to 2
B 2 to 3
C 3 to 4.5
D It stays 30
Answer: A. 3 to 2. β falls from 30/10 = 3 to 30/15 = 2.
In PM, doubling message frequency at fixed amplitude does what to peak deviation?
A Halves frequency deviation
B Leaves it unchanged
C Doubles frequency deviation
D Sets it to zero
Answer: C. Doubles frequency deviation. PM instantaneous frequency deviation depends on the time derivative of message phase.
To prevent envelope crossing in sinusoidal AM, require what?
A m > 1
B m ≤ 1
C m = 2
D m < 0
Answer: B. m ≤ 1. The envelope stays nonnegative when the modulation index does not exceed one.
Which adjustment helps a detector with diagonal clipping?
A Increase RC greatly
B Remove the diode
C Raise carrier frequency
D Reduce RC
Answer: D. Reduce RC. Reducing RC lets the detector follow faster envelope decreases.
Which adjustment helps when detector output has carrier ripple?
A Reduce RC greatly
B Suppress both sidebands
C Increase RC
D Lower message amplitude
Answer: C. Increase RC. A larger RC smooths the carrier cycles, while remaining short enough to follow the envelope.
Which phase error inverts a DSB-SC output?
A 180°
B 45°
C 90°
D 270°
Answer: A. 180°. A 180-degree local-carrier error changes the sign of the recovered message.
Double-sideband transmission of 15 kHz baseband requires what bandwidth?
A 7.5 kHz
B 15 kHz
C 45 kHz
D 30 kHz
Answer: D. 30 kHz. Double-sideband transmission requires 2B = 30 kHz.
VSB with an 8 kHz message uses what bandwidth range?
A Exactly 4 kHz
B Between 8 and 16 kHz
C Exactly 8 kHz
D Above 16 kHz
Answer: B. Between 8 and 16 kHz. VSB sends one full sideband and a vestige, so its bandwidth lies between SSB and DSB.
AM total power is 25% above Pc. Find m.
A 0.707
B 0.25
C 0.5
D 1.0
Answer: A. 0.707. 1 + m²/2 = 1.25 gives m = √0.5 ≈ 0.707.
For m=0.6, each AM sideband carries what fraction of Pc?
A 4.5%
B 18%
C 9%
D 36%
Answer: C. 9%. Each sideband power is m²Pc/4 = 0.09Pc.
Carson’s rule gives what bandwidth for Δf=1 kHz and fm=5 kHz?
A 5 kHz
B 10 kHz
C 15 kHz
D 12 kHz
Answer: D. 12 kHz. Carson’s estimate is 2(1 + 5) = 12 kHz.
Given a 200 W carrier under 100% AM, what is total power?
A 200 W
B 250 W
C 300 W
D 400 W
Answer: C. 300 W. Ptotal = Pc(1 + 1/2) = 300 W.
When TV video uses a full sideband plus part of another, what portion remains?
A Vestige sideband portion
B Full carrier sideband
C Message harmonic only
D Noise guard band
Answer: A. Vestige sideband portion. VSB retains a vestige of the suppressed sideband to preserve low-frequency response.
When the DSB-SC local carrier is 180° out of phase, what is recovered?
A Unchanged message
B Inverted message
C Zero carrier only
D Doubled bandwidth
Answer: B. Inverted message. A 180-degree phase reversal changes the sign of the synchronously detected message.
Why place an amplitude limiter before an FM discriminator?
A Widen the message band
B Restore the carrier
C Suppress amplitude noise
D Increase phase deviation
Answer: C. Suppress amplitude noise. An ideal FM signal has constant envelope, so limiting removes amplitude noise before frequency detection.