Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!mailrus!accuvax.nwu.edu!nucsrl!telecom-request From: rpw3%rigden.wpd@sgi.com (Rob Warnock) Newsgroups: comp.dcom.telecom Subject: Re: Baud per Hertz Message-ID: <8806@accuvax.nwu.edu> Date: 8 Jun 90 11:30:08 GMT Sender: news@accuvax.nwu.edu Reply-To: Rob Warnock Organization: Silicon Graphics Inc., Mountain View, CA Lines: 137 Approved: Telecom@eecs.nwu.edu X-Submissions-To: telecom@eecs.nwu.edu X-Administrivia-To: telecom-request@eecs.nwu.edu X-Telecom-Digest: Volume 10, Issue 422, Message 1 of 8 In article <8731@accuvax.nwu.edu> Henry Troup writes: | In article <8683@accuvax.nwu.edu> Rob Warnock 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