Question T8A11
From subelement T8 - T8A
What is the approximate bandwidth required to transmit a CW signal?
Why is this correct?
CW (Morse code) signals require approximately 150 Hz of bandwidth, making them the narrowest bandwidth mode in amateur radio. This extremely narrow bandwidth occurs because CW is simply an on/off keying of a carrier wave to create dots and dashes, requiring minimal frequency space. In contrast, voice modes need much wider bandwidths: SSB voice needs about 3 kHz, FM voice needs 10-15 kHz, and AM fast-scan TV needs about 6 MHz.
Memory tip
Remember the bandwidth hierarchy: CW is always the narrowest at 150 Hz, then SSB voice at 3 kHz, then FM voice at 10-15 kHz, then TV at 6 MHz. This pattern appears across multiple exam questions—memorizing this ascending order helps you quickly eliminate wrong answers.
Learn more
CW's narrow 150 Hz bandwidth makes it ideal for weak signal work and crowded band conditions. This efficiency allows many CW operators to fit in the same frequency space that would accommodate just one SSB contact. Under Part 97 emission standards, CW's spectral efficiency enables its authorization across all amateur frequency privileges, from HF through microwave bands, making it valuable for emergency communications when maximum range with minimum power is critical.
Think about it
Why do you think CW's extremely narrow bandwidth makes it particularly valuable during emergency communications or when atmospheric conditions are poor for radio propagation?