Question E1A03:
[97.305, 97.307(b)] What is the highest legal carrier frequency on the 20-meter band for transmitting a 2.8 kHz wide USB data signal?  

Choices:  
A: 14.0708 MHz  
B: 14.1002 MHz  
C: 14.1472 MHz  
D: 14.3490 MHz    

Correct Answer is  C: 14.1472 MHz  

To keep an SSB signal inside the authorized band segment, all transmitted energy must stay within the edge. For USB, the sideband energy lies above the suppressed carrier (dial) frequency.

Rule:
- Near an upper band edge (USB):  
  $$f_\text{carrier,max} = f_\text{upper edge} - BW$$
- Near a lower band edge (LSB):  
  $$f_\text{carrier,min} = f_\text{lower edge} + BW$$

On 20 meters, RTTY/data is authorized only up to 14.150 MHz [97.305], and emissions must stay within the band edges [97.307(b)]. For a 2.8 kHz wide USB data signal:
$$f_\text{carrier,max} = 14.1500\ \text{MHz} - 0.0028\ \text{MHz} = 14.1472\ \text{MHz}$$
So, 14.1472 MHz is the highest legal carrier frequency.

Why the correct answer (C) is correct:
- USB energy extends upward from the carrier; subtracting the 2.8 kHz bandwidth from the 14.150 MHz data sub-band edge gives 14.1472 MHz.

Why the other choices are incorrect:
- A: 14.0708 MHz — Well inside the data band but not the highest permissible frequency.
- B: 14.1002 MHz — Also inside the data band but not the highest permissible frequency.
- D: 14.3490 MHz — Near the top of the 20 m band; a USB signal here would extend above 14.350 MHz, violating 97.307(b). Additionally, the segment above 14.150 MHz is not authorized for RTTY/data per 97.305.

Useful exam tips:
- For USB at an upper edge: keep your dial at least the occupied bandwidth below the edge.
- For LSB at a lower edge: keep your dial at least the occupied bandwidth above the edge.