Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 6/24/83; site redwood.UUCP Path: utzoo!linus!philabs!cmcl2!seismo!hao!hplabs!hpda!fortune!redwood!rpw3 From: rpw3@redwood.UUCP (Rob Warnock) Newsgroups: net.music.synth Subject: Re: MIDI and RS232 Message-ID: <190@redwood.UUCP> Date: Thu, 14-Mar-85 03:27:37 EST Article-I.D.: redwood.190 Posted: Thu Mar 14 03:27:37 1985 Date-Received: Sun, 17-Mar-85 20:59:48 EST References: <1854@pucc-h> <65@daisy.UUCP> <426@x.UUCP> Organization: [Consultant], Foster City, CA Lines: 90 +------edited for brevity--------- | : ...the following quote is so confused that I couldn't resist. | > is actually a 1200 bps modem. The Baud rate is 600 and Vadic puts two bits | > (called a dibit) into each signal change (Baud). 9600 Baud modems... | > ...typically, they pack three or four bits into each signal change. | This doen't make any sense. ...[NRZ/RS-232/bit=baud]... All other schemes | I know of (FM, MFM, etc.) use more than one signal change to encode | each bit... I belive that it is theoreticaly impossible to encode more | than one bit into one signal change. | William J. Richard @ Charles River Data Systems | uucp: ...!decvax!frog!wjr +--------------- I'm sorry Mr. Richard, but WAS correct. The ONLY way to get 9600 through dial-up lines is to have more than one bit per baud, since dialup lines are bandlimited at about 3 kilohertz. There are many methods for encoding several bits/baud, but all of the ones you mentioned -- RS-232, NRZ, FM, MFM, etc. -- use BINARY signalling (I assume you are speaking of the FM and MFM of disk storage) and of course binary signalling cannot exceed one bit/baud. But there are other methods than binary. As an extreme example (but still feasible and quite practical over "clean" wires), you can use AM or PAM (Pulse Amplitude Modulation), using 256 voltage levels from 0 to 25.5 volts, with each of the 1/10 volt values representing a different 8-bit number. (If this implies to you that I am "transmitting" with a D/A and "receiving" with an A/D, you are quite correct.) Using raised-cosine filtering, over a low-noise bandlimited channel, with good equalization at the receiver, one can send very nearly two baud per Hertz of bandwidth. (Loosely speaking, you get an "up" symbol and a "down" symbol per cycle, hence two baud/Hertz.) Each baud (symbols or transmit states per second) conveys 8 bits of information, so the system is capable of nearly 16 bits/sec per Hertz of bandwidth. Over a noise-free channel of infinite bandwidth, there is no theoretical limit to the number of bits per symbol, since you can pack as many bits into a symbol (physical state) as you can measure. Practical systems have noise, and AM is not ideal under those conditions. A popular modulation scheme is QPSK (Quadrature Phase-shift Keying), in which one of 4 phases of a carrier signal is sent during each symbol interval (normally evenly spaced at 0, 90 180, and 270 degrees, hence "quadrature). Since there are 4 possible states for each signalling interval, there are log2(4) = 2 bits/sec per baud. (This is the "di-bit" mentioned.) [Aside: Remember that "baud" is symbols per second, and "bits" has no time dimension, so we should compare "bits/sec" with "baud". Though normally it does no harm to get a little sloppy and talk about "bits/baud", to be precise one should say "bits/symbol" (no time implied) or "bits/sec per baud".] Anyway, as I recall, the Vadic and Bell 212 modems both use QPSK, and thus are 600 baud ON THE ANALOG OR TELEPHONE SIDE OF THE MODEM. (They are of course 1200 baud on the digital or RS-232 side, where binary signalling is being used.) The next higher-density popular coding scheme is QAM, or Quadrature Amplitude Modulation, which is a mixture of both AM and QPSK. Each symbol can be any on of the 4 phases (0/90/180/270) AND can be any one of several amplitudes, say (for 4-level) +3/+1/-1/-3 volts. This produces 16 possible states for each symbol, so you can send log2(16) = 4 bits/symbol, or 4 bits/sec per baud. In practice, in order to separate the symbol states so as to maximize noise immunity and signal detection ability, the same set of phases is not used at each amplitude, but the symbol states are spaced out in amplitude/phase space. So for 16 state coding, you might have the 4 phases 0/90/180/270 degrees used at the lowest amplitude, the 4 phases 45/135/225/315 degrees at the middle amplitude, and the 8 phases 22.5/67.5/112.5/.../337.5 at the highest amplitude. (I may be off a few degrees, but you can see the pattern very clearly on an oscilloscope displaying what is called an "eye" diagram.) In any case, there are 16 discrete states, so you still get 4 bits/sec per baud. (Most installed 9600 b/s modems use QAM, and thus run at 2400 baud ON THE ANALOG SIDE -- the telephone line. That SHOULD only need 1200 Hertz bandwidth, but the phone line is noisy, and has amplitude and phase errors with time and frequency -- so it's rough squeezing it into 3000 Hertz.) I hope this settles the controversy -- if you are using BINARY signalling, you can only get 1 bit/sec per baud (or less if some of you symbols are used for clock). With ternary, quaternary, or higher modulations, you can get many bits/sec per baud. Rob Warnock Systems Architecture Consultant UUCP: {ihnp4,ucbvax!dual}!fortune!redwood!rpw3 DDD: (415)572-2607 USPS: 510 Trinidad Lane, Foster City, CA 94404