Recently, a reader asked:
On question E8C06 and E8C07 the formula uses 1.2. Where did the 1.2 come from and what does it represent?
I wasn’t sure what he was referring to since my study doesn’t mention how to calculate that value at all. Instead, it reads:
The bandwidth needed for ASCII digital transmissions increases as the data rate increases. The bandwidth necessary for a 170-hertz shift, 300-baud ASCII transmission is 0.5 kHz. (E8C06) The bandwidth necessary for a 4800-Hz frequency shift, 9600-baud ASCII FM transmission is 15.36 kHz. (E8C07)
I e-mailed him, asking him, “”Is the formula you’re referring to perhaps in another license manual? If so, and if you can send that to me, perhaps I can explain it to you.”
He replied, “This was out of the Gordon West Extra Class book page 81.” He attached a copy of the page, and it did indeed refer to the formula:
BW = baud rate + (1.2 x f shift)
Now, I had never run across this particular formula, but I decided to do a little Googling. What I turned up was interesting. It appears that the 1.2 number basically comes from some version of the ARRL Extra Class License Study Guide. Where they got it from I don’t really know.
In my Google search, I turned up one source that simply says that:
BW = baud rate + the frequency shift
Perhaps someone along the line said, “Well, that’s the theoretical value. Practically, if we increase that by 20% to 1.2 times the frequency shift then the signal will definitely fit in that bandwidth.” I’m just guessing here. I’m not really sure.
I told my reader that, for what it’s worth, there’s a lot of this in amateur radio. The formula used to calculate the length of a half-wave dipole antenna is perhaps the biggest example of this. There’s no real science behind the formula length in feet = 468 / frequency in MHz. It’s just a rule of thumb.
While it may be disappointing that the science behind this is perhaps a bit shaky, the good news is that using these rules of thumbs produce circuits and systems that generally work.