The 2 m Repeater and the Shared Antenna
A newly licensed Technician, Maya, lives in a town where the local 2 m repeater output is 146.940 MHz and its input is 146.340 MHz. She has a single dual-band vertical on the roof, a 50 W mobile transceiver, and a duplexer rated for 100 W. She also wants to work the local 70 cm repeater, whose output is 442.100 MHz and input is 447.100 MHz. She plans to use one feedline for both bands and to run the 2 m radio at full power. A neighbor reports hearing her voice on a television set in the house. Maya asks whether she can simply turn up the power to reach a distant repeater 40 km away, and whether her license class permits operating on the 70 cm repeater at all.
1. Identify the offset and direction for each repeater, and state whether Maya's Technician privileges allow her to transmit on both.
Read the worked answer
For the 2 m repeater, the output is 146.940 MHz and the input is 146.340 MHz. The difference is 146.940 − 146.340 = 0.600 MHz, a standard 600 kHz offset. Because the input frequency is below the output, the repeater uses a negative offset: Maya's radio must transmit 600 kHz below the repeater output. For the 70 cm repeater, the output is 442.100 MHz and the input is 447.100 MHz. The difference is 5.000 MHz, and the input is above the output, so this is a positive offset of 5 MHz: her radio transmits 5 MHz above the output. Technician class licensees hold full privileges on the 2 m and 70 cm bands, so she may transmit on both repeaters. The key judgement is not memorizing offsets but computing them from the listed input and output frequencies and then setting the radio's offset direction and magnitude correctly.
2. Explain whether increasing power is likely to solve the 40 km coverage problem, and what she should check first.
Read the worked answer
Power increases do not reliably extend repeater coverage, because a repeater is limited by its receive path. If Maya can hear the repeater but cannot key it, the weak link is her signal arriving at the repeater input, and doubling power adds only about 3 dB - roughly a 1.4× range improvement in free space, and far less in terrain with obstructions. Before raising power she should verify: (1) the antenna is resonant and well matched on 2 m (check SWR); (2) the feedline loss at 146 MHz; (3) that she is actually transmitting on the correct offset and CTCSS tone; (4) antenna height and any nearby obstructions. Raising power to 50 W may also worsen the TV interference report, since strong nearby fields can overload consumer electronics. The disciplined answer is to diagnose the link budget and interference before adding power.
3. The neighbor's TV interference appears when Maya transmits on 2 m. Describe the likely mechanism and a reasonable first response consistent with good amateur practice.
Read the worked answer
The most likely mechanism is fundamental overload or harmonic/out-of-band energy coupling into the TV's front end, not necessarily a fault in Maya's station. A 50 W transmitter near a consumer receiver can produce interference through direct pickup, poor shielding, or a nonlinear junction. Good practice: confirm the interference is caused by her transmissions (key down briefly while the neighbor observes), reduce power, move the antenna farther from the TV, check connectors and feedline for common-mode current (a choke or balun may help), and ensure the transmitter is not producing spurious emissions. She should cooperate with the neighbor, document what she finds, and address station-caused problems. She should not simply dismiss the complaint; the operating standards expect courtesy and technical effort to resolve interference.