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.
You may be right that the 1.2 is just an empirical factor. But it could also be motivated by the ratio of the -6 dB bandwidth and the peak-to-zero bandwidth of a rectangular pulse which is 1.21 (from the properties of a sinc-shaped spectrum)
But anyway, it’s more straight-forward with the 468/f and it becomes easy to see if one thinks in metric units. A dipole is half the wavelength and that is 300/2f meters, with 300 coming from the speed of light. Assume then a velocity factor of 95% (empirical value) and you get 142.5/f. Then convert from meters to feet and you have 467.5/f which when rounded off gives you the 468/f.
Actually, some people think that it’s capacitive loading, not velocity factor that is the reason a half-wave dipole is actually shorter than a half wavelength. See my post, Velocity factor? No, capacitive loading.
The 468/f number does have real science behind it, but it is empirical rather than theoretical. As people will find out if they click the “rule of thumb” link above, the experimentation was done in 1926 and published in QST. Ward Silver, N0AX, repeated the experiments with modeling software and duplicated the original results for the same heights and wavelengths (they didn’t use “frequency” in 1926).
http://www.eham.net/articles/23802
Well, perhaps I should have said that theory rather than science. Even so, 468 is still not a hard and fast number. The simulations gave a range of values. That smells like a rule of thumb to me.
Hi Dan, I follow your blog thorough my Kindle and I really enjoy it, thank you very much!
I thought I might help with this FM bandwidth issue.
In theory, any FM signal would have infinite bandwidth, so there’s always some level of distortion in a band-limited system such as transmitter-RF channel-receiver. What we need to consider is how much distortion is acceptable, be it speech noise or BER (bit error rate). In other words, how critical is the application, how susceptible it is to further distortion (multi-path, atmospheric noise etc.) and how good are the filters/circuit/design of the equipment/software in question.
The FM bandwidth spectrum and the actual distribution of the sidebands (levels and spacing) depends on the modulation index, which is determined by frequency shift AND maximum modulation frequency – again, wether it’s voice audio band or digital baud rate. The significant bandwidth for a typical ham analog voice FM (or any similar system) is generally accepted to be
2x (MaxDeviation + MaxModulationFrequency) = 2 x (5kHz + 3kHz) = around 16kHz per channel.
That’s why we find 15, 20 and 30kHz VHF channels around the world, depending on how “hi-fi” is the channel potential and how crowded is the band.
In the case of digital communications, further baseband shaping (pulse filtering) is applied to the square/step digital signal before modulation as to MINIMIZE BER. The kind of pulse shaping chosen will also determine the “significant” bandwidth, where 98% of the energy lies. Needless to say, the less BER is desired, the more bandwidth is required.
A modulation scheme (C4FM, BPSK, GMSK etc.) will prioritize one aspect in detriment of others. Some types of modulation are very spectrum efficient (for a given BER) but are more susceptible to multi-path distortion and/or require much more sophisticated/expensive filters/circuitry. It’s all about design compromises, as always!
Having said that, I believe there’s no simple formula to determine the required bandwidth for a digital channel. But then again, we can estimate it with good confidence since most ham technology will use low modulation indexes (0.5 to 2) in non-critical mission applications.
We can use the formula in your post
BW = f shift + baud rate
IF we notice that:
1) “frequency shift” here refers to 2x MaxDev. In the case of analog voice: frequency shift = 10kHz = +- 5kHz around center carrier frequency.
2) “baud rate” is actually a misnomer, the correct would be “bits per second” (BPS) and it equals two times the required audio bandwidth for 2-bits/symbol modulation types only, such as BPSK used for 9600 bps packet radio. Other types of modulation will have different bits/symbol rates and thus bps/frequency relationships. In the end, the actual modulating FREQUENCY (baud rate) is the bandwidth factor, not the bit rate.
Then your formula is exactly the same as the previous:
BW = 2x (MaxDeviation + MaxModulationFrequency)
(aka. Carlson’s Rule)
As far as I know, most digital designs will target/assume a +-3kHz deviation – which is 60% of the typical 5kHz max of all ham gear. So it’s only a matter of knowing the required audio bandwidth (BAUD RATE) to estimate the channel BW!
By the way, most ham equipment will present an audio BW in excess of 3kHz, sometimes 6 or 7kHz – which means we can push the modulating frequency and achieve higher bit-rates in the typical 20kHz channel (I actually encourage analog FM repeaters to be adjusted for 5kHz audio bandwidth, where channel spacing and band usage permit. Most users will notice a great improvement in audio quality. I’ve done that with our local repeater.)
Going back to digital, 3kHz deviation + 5kHz audio would require a bandwidth of 2x (3 + 5) = 16kHz, similar to analog FM. 5000 baud rate (4800 much more common) can deliver very reliable 9600bps and somewhat good performance with 20kbps or even higher, depending on the modulation technology used.
As an example, Yaesu’s System Fusion uses C4FM (a particular case of 4FSK) to achieve 9600bps with 4800baud rate, using a 5kHz flat (level and phase) audio spectrum and 3kHz deviation in a 12.5kHz channel. According to the formula above, the required BW would be around 16kHz. But the choice of modulation type, reliability level (BER, error correction etc.) and baseband shaping further improves BW efficiency from 16 to 12.5kHz, so it can fit perfectly in a 15kHz channel or even operate half-speed in a narrowband 6.25kHz. (I have also modified our local repeater so that the GM300’s are able to pass flat 5kHz audio, possibly creating the first analog+digital System Fusion compatible homebrew repeater in the world.)
Hope that helps and that I did not make any mistakes.
My best!
Dennis Zasnicoff
PY2DZA
As for the 1.2….
I guess it came from a particular modulation/protocol and stayed there as a “safe margin”. I don’t particularly like it and it does not make any sense as a general rule.
An equation for a rule of thumb shouldn’t use the equal sign. Instead, the approximate sign (≈) should be used to show that it’s an approximation.
I was not convinced by my own explanation… so I went out and found this doc which hopefully explains everything:
http://web.stanford.edu/class/ee384e/cgi-bin/readings/r03-radiosintro.pdf
Good read!
1.2 is actually 1+(0.2) and 0.2 is the common filter alpha (excess-bandwidth or “splatter” factor) of most modern digital radios.