Question T8A10
From subelement T8 - T8A
What is the approximate bandwidth of AM fast-scan TV transmissions?
Why is this correct?
AM fast-scan TV transmissions require about 6 MHz of bandwidth because they must carry complete video information including luminance, chrominance, and synchronization signals in real-time. This is enormously wide compared to voice modes - 2000 times wider than CW (150 Hz) and 400 times wider than FM voice (15 kHz). The other options are incorrect: 10+ MHz exceeds what's needed, while 3 MHz and 1 MHz are insufficient for full-motion color television transmission.
Memory tip
Remember the bandwidth hierarchy: CW uses Hz, voice uses kHz, but video requires MHz. Fast-scan TV needs such wide bandwidth because it transmits 30 complete picture frames per second with full color information - think of it as sending thousands of photographs every second through the air.
Learn more
Fast-scan television requires approximately 6 MHz because it transmits complete analog video frames at broadcast rates. This enormous bandwidth requirement explains why amateur television is restricted to frequency privileges on 70 cm and above - lower frequency amateur allocations simply don't have enough spectrum width to accommodate such wide emission standards. The 6 MHz requirement matches commercial broadcast television channel spacing.
Think about it
Why do you think amateur fast-scan TV is only permitted on 70 cm and higher frequency bands, but not on HF or lower VHF frequencies?