By the time I was able to use computers, I had a 28.8k modem, then a 56k. Never used anything that was rated in Bauds… so I was just wondering, how fast would a modern internet connection be if it were still being rated in Bauds? Does this question even make sense? I’m not really sure what a baud is, so I can’t say for sure that a modern DSL line or cable modem or T3 connection could even be measured in bauds… but get creative if you can. If there are any assumptions you could take to make it possible to rate a modern connection in bauds, what would it be? Thanks!
They’re still measured in Baud. If you’re offered a 20MB DSL connection, that’s 20 Megabits per second, which is 20 MegaBaud.
Upload and download speeds aren’t.the same, but they’re still expressed in Baud.
Also, the 28 and 56 k of your modems - those are also in Baud -
When those modems started getting up into the 14400 or 28800 range, the correspondence between Baud and bits broke down, because they achieved those speeds by signalling with more than two states. The actual rate of symbols, the Baud rate, remained at 9600 or whatever (I don’t remember the precise numbers), but the bit rate was higher.
Ximenean: So our modern internet connections still have a relatively low baud rate then, if I am understanding you correctly?
Mangetout: I may stand to be corrected but I am pretty sure baud != bit, except in the good ol’ days.
Baud means ‘bits per second’ like it always has. So if the effective data speed of your connection is 20Mb/s then it’s 20 M Baud. If the underlying framework isn’t signalling using binary, then it’s not meaningful to try to describe that in Baud.
Baud is symbols per second, commonly 0 or 1 (like 99.999%)
Just as a fun educational moment…
There are ways to create signals that can represent more than one bit per symbol, but I don’t think they are in any kind of common use with ISP’s.
Ummmmm . . . no.
Baud refers to line changes per second and IIRC phone lines max out at 1200 baud i.e. 1200 low-to-high or high-to-low changes every second however only 300 and 600 baud was used except in half-duplexing situations and who really wanted to do that.
For 2400 “baud” or higher, phase shifting was used. This was the huge breakthrough in the v2X standards in that it allowed data compression so that a modem could send/receive more bits per second than the limitations of phone line would allow. This means that a full duplex 2400 modem was 2400 bps but only 600 baud.
So to answer the OP, the baud rate is how many signals per second the internet connection transmits which depending on data compression may or may not equal the data transfer rate.
I’m not sure that comms engineers these days really use the term “baud”, anyway. Modern systems such as DSL achieve their speeds by multiplexing many signals in parallel. It’s not clear to me what “symbol” would mean in this context.
Nit: capital B = bytes, lowercase b = bits. They offer 20Mb connections - nobody offers 20MB.
Yeah, Mangetout is in over his head here. Baud means ‘symbols per second’, which, if you’re using a sufficiently complex scheme, could be a fairly small fraction of the total number of bits per second. For example, if you have a quadrature phase-shift keying setup, there are four symbols and, therefore, two bits per symbol, so your baud is one half of your bitrate. If you have 8-PSK, there are eight symbols and, therefore, three bits per symbol, so your baud is one-third your bitrate.
(If that four symbols == two bits isn’t obvious to you, read up on Shannon entropy.)
You can go to higher and higher PSKs (more symbols) and get a higher bitrate given the same baud, but there’s a trade-off involved: Having more distinct symbols means they’re more difficult to distinguish in the presence of noise, either making your system less robust or forcing you to add error-correction, which slows you down.
According to Wikipedia, WiFi uses various PSK schemes depending on the bitrate desired:
BPSK is Binary PSK, so there are two symbols and one bit per symbol. DQPSK is Differential Quadrature PSK, as described above, with four symbols and two bits per symbol.
So what’s the baud of my 12 mbps cable connection? Or is that not even expressable in “baud”?
We can’t say without knowing how it’s modulated, that is, how the signal is encoded to travel down the cable.
And, as a low level concept, it’s more in the realm of physical data transfer and doesn’t generally have a direct relationship anymore to actual connection speeds. It’s like saying “Do you want to ride in a 8 cylinder diesel engined bus or a 10 cylinder gasoline bus? They both get to NY in about 5 hours.” There may be theoretical differences that could matter in disaster situations but most people probably don’t even need to know, let alone care.
Well, ok… but that was the original question. Could you hazard a guess?
Bah. I was hoping someone else would be able to do that for me. 
According to Wikipedia, DOCSIS, the standard for cable Internet access, is varied enough there really is no single answer to that question. Modulation ranges from 64-level to 256-level QAM, giving you from 6 to 8 bits per symbol. Further, your connection is actually a number of channels (commonly 4 or 8, apparently) all bonded together, each of which can be sending a different symbol at the same time. Finally, you lose effective connection speed due to overhead, both from DOCSIS itself and from the Internet protocols required to move your data outside the local network. So you can assign a baud to each channel, based on the modulation, but that won’t give you your actual connection speed.
OK, let’s do a worked example to see if I’m going nuts yet:
According to Wikipedia, a common cable Internet downstream speed is 343.04 Mbit/s, or 304 Mbit/s when you factor out the overhead. Since we want the low-level raw speed, we’ll go with the 343.04 Mbit/s figure. That is split, apparently evenly, into 8 downstream channels, so we divide it by 8 to get 42,880,000, or 42.88 Mbit/s/channel. Now here’s where I make the logical leap to assume anything this juicy is going to be using 256-QAM, the ‘densest’ encoding scheme, so we see that 2**8 (two to the eighth power) is 256, so we have 8 bits per symbol. Thus, we divide 42.88 Mbit/s/channel by 8 to get 5,360,000, or 5.36 Mbaud/channel (that’s 5.36 megabaud per channel).
Many modern digital communications systems do not have a fixed baud rate. For example in OFDM (orthogonal frequency division multiplexing), which is used in both DSL and wireless, the channel is digitally decomposed into a number of subcarriers (using discrete fourier transforms). Each channel is modulated with a bit-to-baud ratio that depends on the measured signal-to-noise ratio on that channel. So for example, there might be 1024 channels. Channel 137 may be using 4 QAM (equivalent to QPSK, 4 symbols per baud, so 2 bits per baud). Simultaneously, channel 795 may have a stronger signal and be using 256 QAM (256 symbols per baud, so 8 bits per baud). Channel 532 may be so weak that it is not being used at all.
If you had proposed such schemes 30 years ago, people would have thought you were nuts, but with the incremental cost of a million transistors being close to zero, the signalling sytems have become highly efficient and incredibly complex.
I stand corrected on my statements above. Ignorance fought.
Thank you all for your posts! I feel like I learned so much. I kinda figured that there wouldn’t be a simple calculation/conversion.