_____ __ _____ _____ ___________ _______ ___________ _ _________ _ _______ <^ _____.__\____._____\__ __/______.___ ____\____ /_______.__./____ _\___ ^> |\| |/| | | A BIT about BAUD... | | |-| |-| | | Written by: Jack Deth | | |/|___ _______ ____ _________ ___ _____________ ___ ___ ___________ ______|\| The first thing I would like to say is that there is no such thing as a 1200, 2400, or 9600 BAUD modem. And I know the first thing that you are going to say... 'You don't know what the hell you're talking about. I have one right here'. No, what you have is a 1200, 2400, or 9600 bits per second (bps) modem, not baud. If you have an internal modem you do set a baud rate, but that is for the baud rate of the serial portion of the card, and this is also the baud rate set by terminal programs. The actual rate of transmission is 600 baud; 600 is the optimum rate that can be transmitted over the general switched telephone network for a full duplex FDM (frequency division multiplexing) modem because band limit filters in the central office cut off at about 3000 Hz. Now you may ask 'What exactly is baud?' and 'Why did BAUD and BPS get confused?'. The definition of baud: Baud rate is the reciprocal of the highest frequency component of a transmitted signal. Now, bps is how fast the data is output by the serial interface device of your computer (ie. Super Serial Card). If you are transmitting an entirely digital signal, then the highest frequency is the clock pulse of the serial output device and there is only 1 bit per clock pulse (1 or 0). Here is where the confusion starts. In the case of the output to a printer, for example, the digital signal itself is transmitted so the frequency is the digital signal pulses (1 or 0) and there is 1 bit per pulse or clock period. Therefore the bps will equal the baud rate. When using a modem, the output from the serial card would be, for example, at 2400 bps, transmitted over the cable at 2400 baud, and received by the modem at 2400 bps (it is still all digital). From here the bps and baud will start to differ. The telephone lines are made to transmit audible tones. A digital signal is a series of 1's (+5V) and 0's (zero volts). The telephone line uses pulses of voltage to dial (rotary phones), so if you put a digital signal directly on the phone line you will, in essence, be doing a PULSE dial. This is the whole reason you have a modem; to convert the digital signal to tones. So let's say we convert a '0' to a low tone and a '1' to a high tone. Now we have a signal that can be sent over the phone. But the telephone lines can only efficiently handle a transmission rate of 600 baud and, on phone lines, time is money so we would want to go faster than 600 bps. So, to get a faster bps, then you simply need to encode 2 bits per baud to get a 1200 bps transmission and 4 bits per baud to get a 2400 bps transmission (4 X 600 = 2400) and so on. I'm not going to talk much about 300 bps, because it virtually outdated, or 9600 bps, because it is recent and not much data is available on the actual method used. Four basic types of modulation schemes (that I know of) used to convert digital data into suitable analog signals for the telephone line are: 1) Amplitude Shift Keying (ASK) 2) Frequency Shift Keying (FSK) 3) Differential Phase Shift Keying (DPSK) 4) Quadrature Amplitude Modulation (QAM) Each of these must generate a carrier frequency on which to modulate the data signal. Carrier frequencies used are 1200 Hz and 2400 Hz for the low band and the high band, respectively. The low band is used for transmission by the originating modem and reception by the answering modem, and the high band is used for transmission by the answering modem and reception by the originating modem. So you can see by the diagram that it is as if there were two separate wires, each dedicated for transmitting by opposite ends, and in this way provides FULL duplex. ______________ ____________ | | 2400 Hz | | | Originating | <---------------------------------------- | Answering | | Modem | | Modem | | | ----------------------------------------> | | |______________| 1200 Hz |____________| THIS is frequency division multiplexing (FDM), as you should be able to see the DIVISION of the available telephone frequency range into two separate carriers for data transmission. Some of you might say 'But the phone system has 2 wires. Why not just use one wire for each of the dgdgsgsgdfgsfgsdgsdf ead of all that frequency division stuff'. You can't do that because one wire is the reference (GND) and the other is the signal, so you only have one 'line' on which to transmit. This is the reason all this frequency division 'stuff' was created. And if you only used one transmission 'line' then only one modem (originating or answering) could ever transmit, so at best you could only have a half duplex dumb terminal. When you select Half duplex in your terminal program, you are just telling the computer to send what you type to your screen as well as to the other modem because you know the other modem is not going to echo your signal back. You are still in a Full duplex system. ASK is an abandoned method probably used by the old teletype systems that ran at 110 bps. Because FSK is relatively simple in its operation and thus inexpensive to implement, it is used for 300 bps modems. FSK is not efficient, however, in its use of spectrum space, so a different method must be used for higher bit rates. If you have a 300/1200 bps modem, then it will use FSK for 300 bps and 4-phase DPSK for 1200 bps. In 4-phase DPSK the data stream to be transmitted is divided into groups of two consecutive bits (dibits) per baud. Each dibit is encoded as a phase change relative to the phase of the preceding signal dibit element. I don't know any other way to explain this without getting technical, and I can't draw good diagrams in text, so I will do the best I can. o o 135 | 45 _____________________________ \ | / | Dibit | Phase Shift | \ | / |-----------|-----------------| \ | / | 00 | +90 degrees | \ | / | 01 | 0 degrees | \|/ | 10 | 180 degrees | ---------/---------- | 11 | -90 degrees | /|\ |___________|_________________| / | \ / | \ / | \ /o | \ o 225 | 315 Now you should see where the '4-phase' comes from; two bits give you four different combinations which have to be this case, by different phases. For example, say it starts on the 45 degrees axis and the first dibit to be sent is '10'. Then the phase would shift to the 225 degrees axis. And let's say the next dibit is '01' then the phase would remain at the 225 degrees axis. Then the next dibit is '11' so the phase would shift to the 135 degrees axis and so on. If you do not understand how the phase shift is generated by changing the degrees and really want to know, then you'll have to get a book on it because I'm not going to explain that here. Now let's get on to the 2400/1200/300 bps modems. They still use FSK for the 300 bps mode, but 2400/1200 is done with QAM. Since they have to use QAM for 2400 bps they simply use that same method to do 1200 bps so they won't have to add more hardware to do the DPSK as in 1200/300 bps modems. At the 2400 bps data rate, the data to be transmitted is divided into groups of four consecutive bits (quadbits). Now, with four bits to send, that gives us 16 different combinations to be represented. If they tried to use 16 different phases to do this, the phases would be much too close together to give a decent error margin. This is where the Amplitude part of the name comes in. The earlier method of doing this was to divide it into 12 different phases with 4 of the phases having a possibility of 2 different amplitudes. So that is 8 phases with one set amplitude and 4 with 2 different amplitudes (8 + 4 X 2 = 16). This is almost impossible to diagram in text, but I believe a diagram would help you to understand this much better, so I'm going to try. * 0001 0110 0010 / * * / \ | / \ | * 0011 0111 *--..__ 0101 \ | / _* 0000 ~~---*__ \ | / __..--~ ~~--..\|/..--~~ 0100 *----------------|----------------* 1000 __..--~~/|\~~--..__ _.--~~ / | \ *---..__ 1100 * / | \ 1001 ~~--* 1011 1111 * | \ / | \ / * * / 1110 1010 * 1101 The longer the line, the higher the amplitude. The '*' marks the amplitude point for each particular quadbit. So '1111' is at the same phase as '1101' but half the amplitude. As I said, that was an earlier method and is not in use in today's 2400 bps modems. Today they encode the first two bits of the quadbit as a phase quadrant change relative to the quadrant occupied by the preceding signal (as in the DPSK with 4 possibilities). The last two bits define one of four amplitudes associated with the new quadrant. This makes it easier to do both 2400 and 1200 bps with the same hardware. In 1200 bps mode, they simply use the dibit to determine the phase quadrant and keep a constant amplitude. To get 9600 bps, they will have to encode 16 bits per baud which gives 256 different combinations to be represented. I have no idea how this is done and haven't looked for any information on it. All of this may help explain why modems take so long to go up in bps rate and those of you looking for a 19.2K bps modem may be waiting a long time. Because 19.2K bps will require encoding 32 bits per baud which gives 1024 different combinations to be represented. In closing, most of this may be too technical and go over the heads of the average modem user, but, hopefully, it gives you a better understanding of how the modem works. And I'm like you, I still call my modem by BAUD rate because everyone else does. ___________________ ___________________ <___________________| Last Dimension AE (214) 827-5249 |___________________>