hwt@uunet.uu.net> (06/06/90)
In article <8683@accuvax.nwu.edu> Rob Warnock <rpw3@sgi.com> writes: >...(Pushing a 7 baud signal through a 5 Hz pipe is >quite good! The theoretical maximum is 2 baud/Hz: one state for each >half-cycle of bandwidth.)... I don't see a theoretical limit, not if you allow phase modulation. For real phase discriminators and real lines there certainly are limits, but in theory you could shift each half cycle by as fine an increment as you could measure ... I guess Heisenberg limits that somewhere, but not for a long time. Still, in the real world 7 baud on 5 Hz is very good! Henry Troup - BNR owns but does not share my opinions ..uunet!bnrgate!hwt%bwdlh490 or HWT@BNR.CA
cdc@uafhp.uark.edu (C. D. Covington) (06/07/90)
In article <8731@accuvax.nwu.edu>, bnrgate!.bnr.ca!hwt@uunet.uu.net (Henry Troup) writes: > In article <8683@accuvax.nwu.edu> Rob Warnock <rpw3@sgi.com> writes: > >...(Pushing a 7 baud signal through a 5 Hz pipe is > >quite good! The theoretical maximum is 2 baud/Hz: one state for each > >half-cycle of bandwidth.)... > I don't see a theoretical limit, not if you allow phase modulation. Unlimited bandwidth if there is no noise. We do not have that privilege on analog lines. Theoretical maximum rate of error free information interchange is limited by signal to noise ratio of the connecting channel. 7 baud in 5 Hz is a conclusion based on assumed SNR. C. David Covington (WA5TGF) cdc@uafhcx.uark.edu (501) 575-6583 Asst Prof, Elec Eng Univ of Arkansas Fayetteville, AR 72701
peterd@uunet.uu.net> (06/07/90)
bnrgate!.bnr.ca!hwt@uunet.uu.net (Henry Troup) writes: >In article <8683@accuvax.nwu.edu> Rob Warnock <rpw3@sgi.com> writes: >>...(Pushing a 7 baud signal through a 5 Hz pipe is >>quite good! The theoretical maximum is 2 baud/Hz: one state for each >>half-cycle of bandwidth.)... >I don't see a theoretical limit, not if you allow phase modulation. >For real phase discriminators and real lines there certainly are >limits, but in theory you could shift each half cycle by as fine an >increment as you could measure ... I guess Heisenberg limits that >somewhere, but not for a long time. You can encode multiple bits per baud with amplitude and phase shift keying, and in fact every modem above 300 bits per second does so. The only limit here is Shannon's limit - bits/sec < 2*f*log2(S/N) where S is the signal power, N is the noise power, and f is the bandwidth. However, a baud is not a bit. By the Nyquist theorem, you can only get 2f bauds per second. In practice high-speed modems such as V.32 run at about 2500-3000 bauds/sec over lines with a 3000Hz bandwidth. >Still, in the real world 7 baud on 5 Hz is very good! 9600bps over 3000 Hz is a good deal better, and is quite common. Peter Desnoyers
Norman Yarvin <yarvin-norman@cs.yale.edu> (06/08/90)
In article <8731@accuvax.nwu.edu> Henry Troup <bnrgate!bwdlh490. bnr.ca!hwt@uunet.uu.net> writes: >I don't see a theoretical limit, not if you allow phase modulation. >For real phase discriminators and real lines there certainly are >limits, but in theory you could shift each half cycle by as fine an >increment as you could measure. That would expand the frequency range of your signal. Actually staying within a bandwidth limit is not so easy; a series of pieces of 2000 Hz tones do not make a 2000 Hz tone.
rpw3%rigden.wpd@sgi.com (Rob Warnock) (06/08/90)
In article <8731@accuvax.nwu.edu> Henry Troup <bnrgate!bwdlh490. bnr.ca!hwt@uunet.uu.net> writes: | In article <8683@accuvax.nwu.edu> Rob Warnock <rpw3@sgi.com> writes: | >...(Pushing a 7 baud signal through a 5 Hz pipe is | >quite good! The theoretical maximum is 2 baud/Hz: one state for each | >half-cycle of bandwidth.)... | I don't see a theoretical limit, not if you allow phase modulation. | For real phase discriminators and real lines there certainly are | limits, but in theory you could shift each half cycle by as fine an | increment as you could measure ... I guess Heisenberg limits that | somewhere, but not for a long time. Please read what I said again: the theoretical limit is 2 *BAUD*/Hz (the Nyquist limit), not two bits/Hz. And since Hertz == cycles per second, your next statement, "you could shift each half cycle", confirms this. "Baud" == "symbols/second" == "state changes/second". So if one changes something on each half-cycle, that *is* 2 Baud/Hz, just as I said. Now what you really seem to be talking about when you say "by as fine an increment as you could measure" is how many states you can differentiate from each other, or how big your symbol alphabet is. And the limit there is not Heisenberg, particularly, but Shannon. The channel (or system) noise which is added to the signal puts an upper limit on how many states you can distinguish. Whatever the method of modulation -- phase, amplitude, frequency -- at some point in the receiver you will eventually need to decide which of your finite set of states (symbols) to assign to the actual current analog value of the received signal. The circuit for this is an analog-to-digital converter (A/D), also called a quantizer or "slicer". For each signaling interval (the length of which is 1/Baud_rate) you will get one digitized sample which purports to name which symbol was sent during that interval. How many bits it takes to name all the states is how many bits per symbol you have. And bits per symbol times Baud rate is bits per second. So *of course* if your phase discriminators (A/D's) are perfect, and there is zero noise, you can discriminate as many distinct states as you like, and send as many bits per symbol as you like, and therefore as many bits-per-second per Hertz of bandwidth. But real channels have noise -- they are *not* perfect, thus your measurements will not be, either. And the theoretical limit to how many different phases (or whatever) you can use effectively is just the Shannon limit: C = W * log2(SNR + 1) [Page 26 in reference given below] The fundamental "channel capacity" C -- the upper limit on the number of bits per second you can push through a channel with an error rate "as low as you like" assuming you use a coding scheme that is "good enough" -- is the bandwidth "W" in Hertz times the logarithm base-2 of the signal-to-noise ratio SNR -- the *power* ratio (energy per time or energy per bit) -- plus one. The Nyquist theorem says that with perfect A/D's [which we actually can come quite close to these days, at least as far as voice-grade modems care] we can get all the information from a band-limited signal by sampling at a Baud rate B = 2 * W (which is the same as your "each half cycle", above), so if you send R bits/symbol we can equate C = R * B, which gives: C = R * B = (B/2) * log2 (SNR + 1) Simplifying, the maximum useful bits/symbol is: R = log2(SNR + 1) / 2 [Page 40 in reference given below] This says if you have a signal/noise power ratio of 255 = 24.1dB (which is a *voltage* ratio of just under 16), and you have a perfect modulation/demodulation method and a perfect [actually, good enough] error-correcting coding scheme, you can send 4 bits/symbol or 8 bps/Hz of bandwidth. [For those who care about such details, this is ~15 dB "Eb/N0", that is, energy per bit divided by noise power density per Hertz.] But in practice, modulation methods such as phase-shift modulation (PSK) are *not* ideal, they "waste information". I don't have tables for 16 bps/Hz, but I have a chart for coherent PSK that goes up to 10 bps/Hz [Ref: Fig 1.7]. In order to achieve a bit error rate of 1 in 10**5 (no error-correction code), you need the following SNRs (plus or minus a few tenths of a dB for my chart-reading error): Overall Shannon # of phases bits/symbol bps/Hz Eb/N0 (dB) SNR (dB) Limit (dB) 2 (0/180 deg) 1 2 9.5 9.5 0.0 4 (0/90/180/270) 2 4 10.0 13.0 4.8 8 (0/45/90/...) 3 6 14.0 18.8 8.5 16 (0/22.5/45...) 4 8 18.8 24.8 11.8 32 (0/11.25/...) 5 10 23.8 30.8 14.9 As you can see, multi-phase coherent PSK is roughly 10dB worse than the Shannon limit, and is worse off at the higher bits/symbol end (although asymptotically is within a constant factor of the Shannon limit [Ref: p.41]). The best way to use PSK seems to be with four phases, where the excess loss is "only" 8.2dB, which is probably why this version (a.k.a. QPSK) is quite popular in modems. Most higher bit-rate modems use a combination of amplitude and phase modulation (often called "QAM" whether or not the phase-modulation is really "quadrature"), which gives better performance than either AM or PSK alone. [Reference: Michelson & Levesque, "Error-Control Techniques For Digital Communication" (Wiley-Interscience 1985), pp.26-41. Since Figure 1.7 actually graphed Eb/N0 versus PsubM, the *symbol* error rate, I had to extrapolate all the curves (except M=2) down below the chart to get an Eb/N0 corresponding to a 1.0e-5 *bit* error rate. This is the source of much of the "chart-reading error" mentioned above.] In article <8772@accuvax.nwu.edu> codex!peterd@uunet.uu.net (Peter Desnoyers) writes: | However, a baud is not a bit. By the Nyquist theorem, you can only get | 2f bauds per second. In practice high-speed modems such as V.32 run at | about 2500-3000 bauds/sec over lines with a 3000Hz bandwidth. | >Still, in the real world 7 baud on 5 Hz is very good! | 9600bps over 3000 Hz is a good deal better, and is quite common. Oops! You fell in the trap, too! Those 9600 b/s modems use 4 bit/symbol modulation, and so actually run at 2400 baud. And 2400 baud on 2700 Hz (3000 - 300) is not as good as 7 baud on 5 Hz. On clean lines the Telebit will run 6 bits/symbol, so at 7.35 baud that's 44.1 bits/sec in 5 Hz, or 8.8 bps/Hz. 9600/2700 is a mere 3.6 bps/Hz. Rob Warnock, MS-9U/510 rpw3@sgi.com rpw3@pei.com Silicon Graphics, Inc. (415)335-1673 Protocol Engines, Inc. 2011 N. Shoreline Blvd. Mountain View, CA 94039-7311
peter@ficc.ferranti.com (Peter da Silva) (06/08/90)
In article <8772@accuvax.nwu.edu> codex!peterd@uunet.uu.net (Peter Desnoyers) writes: > >Still, in the real world 7 baud on 5 Hz is very good! > 9600bps over 3000 Hz is a good deal better, and is quite common. Not really. Those 7 bauds at 6 bits per baud come to 42 bps over 5 Hz. The highest rate I've heard of over the 3000 Hz band using PEP is something like 24000 bps, or about 2.5 times as much as V.32. 18000 bps is not terribly uncommon. `-_-' Peter da Silva. +1 713 274 5180. <peter@ficc.ferranti.com> 'U` Have you hugged your wolf today? <peter@sugar.hackercorp.com> @FIN Dirty words: Zhghnyyl erphefvir vayvar shapgvbaf.
meier@uunet.uu.net (Rolf Meier) (06/08/90)
In article <8731@accuvax.nwu.edu> Henry Troup <bnrgate!bwdlh490.bnr. ca!hwt@uunet.uu.net> writes: >In article <8683@accuvax.nwu.edu> Rob Warnock <rpw3@sgi.com> writes: >>...(Pushing a 7 baud signal through a 5 Hz pipe is >>quite good! The theoretical maximum is 2 baud/Hz: one state for each >>half-cycle of bandwidth.)... >I don't see a theoretical limit, not if you allow phase modulation. >For real phase discriminators and real lines there certainly are >limits, but in theory you could shift each half cycle by as fine an >increment as you could measure ... I guess Heisenberg limits that >somewhere, but not for a long time. Not Heisenberg, but Shannon sets the limit. The theoretical maximum is: max bit rate = bandwidth x log(2)(1 + S/N) where S/N is the signal to noise ratio log(2) is log base 2 (not 0.30103 :-)) For example, using a normal telephone line with a 3 kHz bandwidth and a 60 dB (1000:1 for the formula) S/N ratio, you can in theory transmit 30,000 bits/sec. You can use phase, frequency, or amplitude modulation. The maximum bit rate is reached when you can no longer resolve the signal variation due to noise. Rolf Meier Mitel Corporation
segal@uunet.uu.net (Gary Segal) (06/11/90)
rpw3%rigden.wpd@sgi.com (Rob Warnock) writes: [Very in depth and useful analysis of noise and it's effects on analog signals deleted] >In article <8772@accuvax.nwu.edu> codex!peterd@uunet.uu.net (Peter >Desnoyers) writes: >| However, a baud is not a bit. By the Nyquist theorem, you can only get >| 2f bauds per second. In practice high-speed modems such as V.32 run at >| about 2500-3000 bauds/sec over lines with a 3000Hz bandwidth. >| >Still, in the real world 7 baud on 5 Hz is very good! >| 9600bps over 3000 Hz is a good deal better, and is quite common. >Oops! You fell in the trap, too! Those 9600 b/s modems use 4 >bit/symbol modulation, and so actually run at 2400 baud. And 2400 baud >on 2700 Hz (3000 - 300) is not as good as 7 baud on 5 Hz. True, V.32 modems can run at 4 bits/baud (16 symbols), but most are usually run at 5 bits/buad (32 symbols) with trellis coding. Trellis coding provides a type of forward error correction (one extra bit for every four data bits) at a very low level of data transfer. (Is this OSI layer 0.5??? :-). In general, trellis coding gives the modem a performace gain of about one to two db over the uncoded signal. Note that both run at 2400 baud, so that trellis coding sends 12,000 bps, of which 20% is error correction. Gary Segal ...!uunet!motcid!segal +1-708-632-2354 Motorola INC., 1501 W. Shure Drive, Arlington Heights IL, 60004 The opinions expressed above are those of the author, and do not consititue the opinions of Motorola INC.