AM Transmitter Modulation Headroom and Receiver Image
A community AM broadcast transmitter runs a 2 kW carrier at 1400 kHz with 7 kHz audio bandwidth into a 50 Ω antenna. The PA is rated 3 kW average and 6 kW peak envelope power. At the studio the modulation monitor shows m = 0.6. The station engineer wants to raise modulation to m = 0.9 to improve loudness at the fringe of the service area; the licence allows up to 100 % modulation, but the manual warns against exceeding the PEP rating. On the receive side, a listener 40 km away uses a simple receiver with a 455 kHz IF and no RF preselector, tuned to 1400 kHz, and suffers interference from a strong short-wave station. The station also wants to know how much of its power actually carries audio.
1. Compute the total power and PEP at m = 0.6, then judge whether raising modulation to m = 0.9 is safe for this PA.
Read the worked answer
At m = 0.6:
P_total = Pc(1 + m²/2) = 2000(1 + 0.18) = 2360 W PEP = Pc(1 + m)² = 2000 × 2.56 = 5120 W
Both are inside the 3 kW average and 6 kW PEP ratings, though PEP is already at 85 % of its limit.
At m = 0.9:
P_total = 2000(1 + 0.405) = 2810 W → 94 % of the 3 kW average rating PEP = 2000 × 3.61 = 7220 W → above the 6 kW PEP rating
The average stays legal but the peak envelope power does not, so the change should not be made as proposed. Continuous high-level peaks stress the modulation transformer, PA devices and output filter, and a limiter that clips peaks would cause splatter. Practical options: keep m near 0.6–0.7, or reduce carrier power and re-license/re-tune if higher modulation is genuinely needed. Average power is not the limiting quantity here; PEP is.
2. State the occupied bandwidth and the sideband power distribution at m = 0.6, and explain why sidebands matter more than carrier power for coverage of audio.
Read the worked answer
For amplitude modulation the bandwidth is twice the highest modulating frequency:
BW = 2 × 7 kHz = 14 kHz
Sideband power per sideband = m²Pc/4 = (0.36 × 2000)/4 = 180 W, so 360 W total in both sidebands, about 15.3 % of the 2360 W transmitted. The remaining 2000 W is carrier, which conveys no information but is needed for envelope detection in simple receivers.
At m = 0.9 each sideband becomes 405 W, more than double the audio power, which is why loudness improves even though total power rises only modestly. Coverage of the audio therefore depends on m and on antenna efficiency, not on carrier watts alone. A 14 kHz signal also needs the antenna and output network to be flat across that width; mis-tuning shows up as sideband asymmetry.
3. The listener's receiver has a 455 kHz IF and no preselector. Determine the image frequency for 1400 kHz tuning and explain the interference.
Read the worked answer
With high-side local oscillator injection:
f_LO = 1400 + 455 = 1855 kHz f_image = f_LO + 455 = 2310 kHz
A strong station at 2.31 MHz is mixed to the same 455 kHz IF as the wanted 1400 kHz signal, with essentially the same conversion gain. Without an RF preselector, the front end cannot reject it, so it appears on top of the wanted programme.
If the receiver instead used low-side injection (f_LO = 945 kHz), the image would move to 490 kHz, which is why LO planning and a tuned front-end bandpass are chosen together. The cure is a 1400 kHz bandpass or tuned ferrite-rod preselector ahead of the mixer, plus good balance in the mixer; simply retuning the IF cannot remove an image. For a station engineer, this is also a reminder that receiver selectivity, not transmitter power, decides fringe reception quality.