Talk:Baud

/N888/B/A/ Telegraph Baud /2006/11/21/

Dear caretaker team of the "Baud" WIKIPEDIA collaborative webpage, I found the following page's "Jean-Maurice-Émile Baudot, and teleprinting" section very useful (from which I quote below): http://www.compkarori.com/dbase/bu07sh.htm

In the (original) telegraph case, a system's Baud is the electro-mechanical maximum signal rate, and since each signal is one bit and about 1/3 are used-up for overhead (see below), the user sees a bits per second transmission rate that is less than the Baud of the transmission system. Shame on the long history of ignorant (or worse) modem speed sales pitches to consumers for the confusion.

If each signal was set and measured more precisely, say at four levels (0,1,2,3) instead of two (0,1) (as was quickly figured-out for modems but not inside computer processors for some reason) the Baud of the system would stay the same while the theoretical transmission speed of the system would increase by eight times, as each signal could represent one of sixteen states instead of two. AVOID CONFUSION, USE BITS PER SECOND, NOT BAUD. NATE88 BETTERDIFFERENT DOT COM N888

Normally, when current is flowing, the line is said to be “marking”, and the machine is sitting there with the motor running, waiting for a character to arrive. To prepare the machine to receive a character, a special zero bit is sent, called the start bit. This bit starts the wheels of the machine turning, and the machine counts and records each of the five bits as they arrive on their fixed schedule. When all five bits have arrived, the bits are used to position a print head or piece of type, which strikes the paper through an inked ribbon, marking the character on the page. After the last bit has arrived, the line is kept in the one state (marking) long enough for the decoding mechanism to reset. In Baudot teleprinters, this was 1.42 times as long as the bit time. It became known as the stop bit, although it was required at the end of each character and really did not stop anything.

With the start bit, five data bits, and 1.42 stop bits, the Baudot code took 7.42 bit times to print each character. Although the speed of different machines varied, a common standard was for a bit time to be .022 second, for a rate (in Baud) of 45.5 bit times per second. At this rate, a teleprinter could print just over six characters every second, so if you are printing five letter words with spaces in between, that is just over 60 words every minute, a very respectable typing speed, and faster than almost all telegraph operators.

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Isn't it true to say that "baud rate" actually refers to the potential number of signal changes/symbols transmitted per second? Since, if one second's worth of 0's are transmitted for some reason, the baud rate of the given channel isn't 0, is it?

done -- guest

"baud rate" is a redundant phrase. "baud" is the commonly accepted word for "symbol rate", in so many symbols per second. a "baud rate", then, is a "symbol rate rate", which can only imply symbol acceleration (equivalently, deceleration).

done -- Tarquin 16:48, 27 Sep 2003 (UTC)

what

what is 'symbol rate rate' ?

correction(?)

'symbol rate rate' == the rate of change of symbol rate. In this context, it was probably used to illustrate why term "baud rate" is, more often than not, nonsensical.

"a 2400-bit/s modem actually transmits at 600 baud, where each quadrature amplitude modulation event carries two bits (four values) of information."

600 events/second * 2 bits/event = 1200 bits/second, right? [drd]

yep 600baud with four values which means log(4)/log(2)=2 bits per value, if there is 600 of this values in one second then it transmits 1200bits/second

modems

I recommend the addition of a statement to the effect of "Modems have been errounously marketed for years and years using the term 'baud' when in fact the measured speed is in bps". Cause I've bought 9600baud, 19.2Kbaud, 33.6Kbaud and 56Kbaud modems. I've learned a couple times that baud=bps. Bleh. thought the wikipedia way was to acknowledge and refute common misconceptions rather than delete them. TomCerul 18:55, 10 October 2005 (UTC)

What about the null state?

The example of a semaphore flag having 8 possible states and thus conveying 3 bits of infomation each second ignores the situation where there is no information in those 8 states to convey and thus the semaphore flag isn't hoisted or is held straight out; e.g., a space between words, no answer yet, etc. The 3-bit-per-second signal rate only holds when a signal is being sent, but the pauses between signals may be equally significant. This is fine where the signals are totally asynchronous, but where the signals are synchronous, there needs to be a null state, which would require 4 bits.

You are quite right. I have modified the text so that only eight states are required now. 194.6.81.93 14:42, 23 January 2006 (UTC)

Excellent work

Excellent analogy with the semaphore, cleared it right up in my head. Have a crisp. --152.78.71.183 17:24, 2 March 2006 (UTC)

Crunch! Thankyou! WLD 23:22, 2 March 2006 (UTC)

Baud for spread spectrum systems

What is the baud for a system such as CDMA, where the channel is modulated at chip rate, with each data symbol comprised of several chips and each data symbol carrying multiple bits, using PSK or QAM to convert bits to symbols? Does it even make sense to ask what the baud is, or is baud only meaningful for a point-to-point channel carrying a single modulated datastream?

Forgetting about the details of the technology for a moment, what is the rate at which symbols are transferred between the sender and receiver? Each symbol may contain multiple bits of information - in the semaphore example, each symbol carries three bits of information. WLD 08:13, 1 September 2006 (UTC)

Common baud values

The “common baud values” section is unnecessary and confusing, and should be deleted. It appears to be a list of common RS-232 DTE rates lumped into arbitrary categories, and has absolutely nothing to do with “baud.”

Agree. WLD 23:00, 25 September 2006 (UTC)

I have removed the section. --Morten LJ 15:40, 19 October 2006 (UTC)

Merge with symbol rate?

Having proposed the merge, it would be a good idea for the merge proposer to put the reason why here on the talk page. In the absence of a good reason, I can see no benefit. Hence:

• Oppose WLD 12:41, 17 November 2006 (UTC)

  Wait a second: you oppose your own merge proposal? Then why exactly are you proposing it? Snoutwood (talk) 19:25, 19 November 2006 (UTC)

  Oh, I apologise, I see now that you didn't propose it. I'm removing the tags since the proposer has not given any rationale. Snoutwood (talk) 20:45, 19 November 2006 (UTC)