bbs@NCoast.ORG (XBBS System) (01/02/91)
Could someone out there explain to me why a signal can physically travel further on a metal media at a slower baud rate ? I see this often when looking at specifications for line drivers, etc. And usually the slower the baud rate is, the further the distance can be. Someone says it has something to do with the waves and frequency of the signal. Another person suggested that the more powerful the signal, the attenuation it will have. I didn't quite understand until he compared it to how FM radio stations usually have more power but the signal doesn't travel as far to AM stations that may be heard by half the country. Yet, I have not got a definate answer from anyone. -- Rich Banks ah335@thor.ins.cwru.edu richb@railnet.uucp
mcovingt@athena.cs.uga.edu (Michael A. Covington) (01/02/91)
In article <1991Jan2.055516.14616@NCoast.ORG> richb@railnet.uucp writes: > >Could someone out there explain to me why a signal can physically travel >further on a metal media at a slower baud rate ? > Simple. A serial signal is simply a voltage that switches from one value to another (binary one and binary zero) very fast. The higher the baud rate, the faster it has to be able to switch. A long cable has both capacitance and inductance. Both of these tend to slow down the switching, so that a sudden change in voltage at one end is not so sudden when it gets to the other end because of the time taken to alter the magnetic and electric field of the cable. So a signal traveling through a long cable can't switch very fast. Hence a lower baud rate is necessary. (There are impedance-matched coaxial cables that don't have this problem. I'm speaking about a typical RS-232 signal in a typical wire cable.)
don@zl2tnm.gp.co.nz (Don Stokes) (01/02/91)
bbs@NCoast.ORG (XBBS System) writes: > Could someone out there explain to me why a signal can physically travel > further on a metal media at a slower baud rate ? Ok. First, computer signals are a square wave, typically a negative voltage for a one, positive voltage for a zero. The faster the baud rate, the shorter the "wavelength". Now a square wave isn't really a wave at all, but can be thought of as the sum of a wave and all the odd harmonics, ie third, fifth, seventh and so-on (the second harmonic is twice the frequency of the primary, he third harmonic is three times the primary etc). The interesting thing about this model is that the "corners" of the square wave are made up of the higher frequency harmonics; the cleaner the corner, the higher the frequency of the harmonics that make it up. Conversely, the lower the frequency of the highest harmonics, the rounder the corner. The power of a harmonic is proportional to its order, ie the third harmonic is a third of the power (???, help me out guys!) of the primary. As the order of the harmonic goes up, the power goes down, and makes the wave more likely to be squashed by the attenuation of the line. Now, by the time the signal gets from one end of the wire to the other, it can be quite rounded. A single bit at 9600 baud might look like: TX end RX end +6V - .......... .... . . .. .. 0V - . . . . . . .. .. -6V -....... ....... .... .... | 1/9600s| | 1/9600s| That contains a recognisable bit, but at 19200, with the same level of attenuation, our bit is going to look a bit like: TX end RX end +6V - ...... . . .... 0V - . . . . . . .. .. -6V - ....... ....... .... .... | | | | 1/19200s 1/19200s Now, the signal hasn't finished rising to the +6V required to register as a space (RS232 idles in the mark or 1 state, between -6V and -12V(?), a space or 0 being a +6V to +12V signal) in the UART (although most trigger well below the +/- 6V mark) before it starts diving back toward the mark state. Of course two zero bits will be recognisable (as at least one) as they'll look just like a single bit at 9600 baud. > I see this often when looking at specifications for line drivers, etc. > And usually the slower the baud rate is, the further the distance can be. Yes, although it always amazes me what you can get away with if you know what you are doing ... I can show you cables that go well over 100m at 19200 baud through a factory, in violation of all the rules.... (the cables are shielded and grounded, each data line is twisted with a signal ground line to provide that extra bit of shielding, but otherwise it's pure RS232). It works, too. (I've also seen cables/equipment in a room twenty yards from the host computer that couldn't hack it at 19200 though.) Specifications always tell you what is guaranteed to work or your money back. > Someone says it has something to do with the waves and frequency of the signa > Another person suggested that the more powerful the signal, the attenuation > it will have. I didn't quite understand until he compared it to how FM radio > stations usually have more power but the signal doesn't travel as far to > AM stations that may be heard by half the country. Ah, now this is quite different territory. It has to do with the fact that radio wave propagation in the AM band (500 or so KHz to 1800 or so KHz) is utterly and totally different to the FM band (~80MHz to ~110 MHz). FM is mostly line-of-sight, whereas AM is mostly ground wave. Borrow an ARRL handbook or something -- this is hard to explain without a lot of hand-waving and glossing over important bits. Hope that helps. Don Stokes, ZL2TNM / / don@zl2tnm.gp.co.nz (home) Systems Programmer /GP/ GP PRINT LIMITED Wellington, don@gp.co.nz (work) __________________/ / ---------------- New_Zealand__________________________
morrison@cs.uiuc.edu (Vance Morrison) (01/03/91)
In <1991Jan2.055516.14616@NCoast.ORG> bbs@NCoast.ORG (XBBS System) writes: >Could someone out there explain to me why a signal can physically travel >further on a metal media at a slower baud rate ? There are two reasons main reasons for a length restriction for cableing, attenuation and dispersion. Attenuation is relatively easy to deal with and is independent of baud rate, it can also be compensated for relatively easily by simply boosting the signal at the source. Dispersion is the limiting factor in your case. So what is dispersion? The key is to realize that a cable is NOT a perfect transmitting media, it also distorts the signal passing through it. To actually calculate the distortion of a cable involves essentially solving some partial differntial equations (maxwells equations), given the geometry of the cable. This sounds nasty, and doing it from first principles is nasty, but EE's have developed some good 'tricks' that simply the problem greatly. The first trick is to realise that the cable is simply a linear system (that is a system with only resistors, capacitors and inductors). For such systems, we know if they are fed a sinusoidal waveform, the output MUST be a sinusoidal waveform of the same frequency. Thus the only thing a linear system can do is to change the phase and amplitude of of the sinusoid. The second trick is to realise that ANY particular waveform can be expressed as a sum of sinusoids (fourier's theorem). The final trick is to realize that since the system linear, superposition holds, that is, the responce of a sum of two signals is simply the sum of the responce of each of the signals applied seperately. Combining these three tricks we now have a procedure for caclulating the output of a cable (or any other linear system), given the input. We simply decompose the input into a sum of sinusoids (find the fourier series or transform), compute how the cable distorts each of the sinusoids (remember only phase and amplitude can change), and then add the responces back up (compute the inverse fourier series or transform). Thus the key to all of this is determining how the cable changes the phase and amplitude of a sinusoid (this is a much easier problem then the general case. As it turns out to a very good approximation, cables distort sinusoids in a very predictable way. Basically all that happens is that the high frequencies travel faster then the low frequencies (or the other way around, I can never keep that straight). This dependance of signal velocity on frequency is called dispersion. Now it is relatively easy to see what is going on. If we sent a square waveform down a cable, the high frequency components of the waveform will arrive before the lower frequency components. Thus the signal does not 'add up' right and the signal is distorted. This causes a rounding of the edges of the waveform. Notice, however, that the longer the cable the more the high frequencies will be out of phase the the lower frequency components and the worse the distortion will be. Now by lowering the baud rate, the reciever will not care so much about the edges of the waveform and so more phase distortion can be tolerated, so longer cables can be used. Note that to a large degree, this distortion is very predictable. Because of this it is possible to 'predistort' the signal so that the the cable distortion is canceled (it can also be done at the receiving end). Modern modems do this kind of fancy signal processing. One final note, RS232 cables have a common ground for all signals, this type of cable has VERY bad dispersion characteristics and this is why RS232 cables can only go limited distances. On the other hand if each signal wire has its own return (differential mode), (like RS422), the characteristics are MUCH better and much longer distances (4000 ft), are quite reasonable. Hope this helps Vance
bhoughto@cmdnfs.intel.com (Blair P. Houghton) (01/03/91)
In article <1991Jan2.164234.21958@ux1.cso.uiuc.edu> morrison@cs.uiuc.edu (Vance Morrison) writes: >In <1991Jan2.055516.14616@NCoast.ORG> bbs@NCoast.ORG (XBBS System) writes: >>Could someone out there explain to me why a signal can physically travel >>further on a metal media at a slower baud rate ? >There are two main reasons for a length restriction for cableing, >attenuation and dispersion. [...most of user-friendly description of the horribly hairy math involved in dispersion calculations deleted...and yes, we (current/former) EE students actually (do/did) have to do these in homework problems in required undergraduate courses, occasionally in time domain only... (:-(/:-))] >Now it is relatively easy to see what is going on. If we sent a square >waveform down a cable, the high frequency components of the waveform will >arrive before the lower frequency components. Thus the signal does not >'add up' right and the signal is distorted. This causes a rounding of >the edges of the waveform. Rounding and ringing, actually. --Blair "One nit, no charge."