Lyle:
As I understand modulation/demodulation over power-limited and bandwidth-limited channels, the basis for the Shannon Limit is not too difficult to understand, even for non-technical people, although it is very difficult for me to come up with an explanation which works equally well for ALL people. I'd like to try though....and get your reaction to it.
Computer data, that is, "ones and zeros", can't be sent directly over the phone lines at more than a few bits per second. Data is sent over phone lines at high speed by converting the data to "tones" which the phone system is capable of transmitting. This is very similar to something we are all familiar with: the touch tone telephone. When you depress a button to dial a phone number, you hear a distinct set of tones in the earpiece; each number has a unique set of tones which is formed much as a chord is formed on a piano, by summing separate "notes", played at the same time, to produce a "chord". These chords are unique enough that the phone system "knows" they are numbers, and not just part of the transmission. The system "demodulates" these tones and is able to dial the correct number for you.
A modem does essentially the same thing, only at a much higher level of sophistication. (Early modems, running at 300 buad, simply sent one tone for a one, and another for a zero, directly mimicking the touch tone telephone!) The "chords" are much more complex, there are many more of them, and they change very rapidly. The amount of data which may be sent by a modem is determined by how many "keys" (bits) a chord may contain, and by how many "chords" (symbols) may be sent in a period of time, which is determined by how quickly the phone system can respond to a change in chord. If one attempts to send the chords too quickly, they tend to "pile up" on top of each other (in a manner of speaking), and the modem cannot accurately determine which of the "piled up" chords was sent, and at which time. Sort of like putting your hands on the piano and pushing all the keys you can reach at once; there may be a lot of valid chords in there, but the result is usually not harmonious, and conveys little emotion, except perhaps chaos!
The Shannon-Hartley Limit for a channel like the POTS, simply puts all this into precise mathematical terms, to allow us to determine exactly how many of the chords we may send per second, and, if properly formatted, how many keys we may place in a chord. There are also limits to the contents of the chords, but as long as we use only the permitted "keys" and do not transmit the chords too fast, the receiving modem will be able to "understand" each distinct chord. Exceeding this limit is like playing many chords at once on a piano; the chords are all there, but we cannot distinguish one from another, especially since some of them share the use of some of the keys. The information about all the chords is there, unfortunately, we don't know how to separate the composite chords into separate keys.
The chords can be thought of as "symbols", that is, groups of individual keys played in unison. A trained musician can often hear a chord and tell you which individual keys were pressed. Modems also have "symbols" (groups of data bits), which can be broken down into the individual data bits used to formulate the symbol. If a piano chord is struck using too many keys, the musician cannot separate the combined sound into individual chords and keys without error; he will get some right but not all of them, and eventually, will have a 50-50 rate of success by merely guessing! A modem also has a limit to the number of "keys" (bits) which it can decipher from a "chord" (symbol) WITHOUT ERROR. And it also can acheive a long-term error rate of 50% by guessing; thus, a modem ought to have a much lower error rate or it would be just as well to let the reciever guess the bits as to even transmit them!
Shannon's Channel Capacity Formula (combined with formulas from Nyquist, etc.) simply tells us how many chords (symbols) we may send in period of time, and, how many keys (bits) we may put into a single chord. Sophisticated modulation systems are able to place many keys into a chord and to send many chords per second. In the case of V.34, if you are operating at 28800 bps while connecting to the internet, your modem has the option of selecting a number of different combinations of keys and chords. For example, you may be sending 2400 "chords per second" (symbols per second), and each cord may be composed of 12 individual "keys" (bits per symbol), resulting in a tramsmission speed of 28,800 "keys per second" (bits per second or bps or "baud"). If you attempt to send more chords/symbols per second, or to put more than twelve keys/bits in those chords, you will certainly be able to compose the "tune", but the phone system will only "hear" a confused collection of noises, and will not be capable of receiving it without error. The physical laws which control how the system operates simply won't allow more chords per second or more keys per chord.
What limits the speed of such a system? The number of symbols which may be send in a second is mostly limited by the bandwidth, while the number of bits which may be sent is mostly limited by the noise of the system (all real world systems have some noise). The noise problem is rather easy to visualize: the difference between two "keys" (bits) in a symbol must be great enough to allow the receiver to separate a chord into its individual keys. In the case of the modem example, where a symbol (or chord) carries twelve bits (or keys) of information, the symbol must have 4096 distinctly different characteristics in order for the receiver to extract the correct bits from the symbol!!! (V.34 uses several very complex systems which, together, produce this level of resolution.) Thus, if we take the ratio of the total power available in a "chord" (symbol) to the total noise power present in the symbol, we get the "signal to noise ratio", which controls how many bits we may put in a symbol. If we attempt to put more bits into the symbol, we will incur errors in the receiver, because the noise which the system adds to our transmitted symbol will prevent the receiver from distinguishing between the individual bits in the symbol; the noise adds to the bit and changes its characteristics so that the receiver thinks it sees a different bit than the correct one. If the correct bit signal is stronger than the noise, the receiver will "see through" the noise and select the correct bit. Again, the piano analogy works here: if you hear a piano chord in a noisy room, you can still correctly perceive it even if there is noise in the room, as long as the chord is louder than the noise.
Modem signal to noise ratio is intuitively easy to understand by analogy. If you are at a party, and many people are talking, there is a constant background "noise" in the room. If you are going to talk to someone else in the room, you must speak louder than the noise level at that person's location in order for them to correctly understand what you say. If you speak at a level which is less than the noise level at the time you speak, the other person will not understand you; the noise will overwhelm their perception of your speech. Oh, they will hear SOMETHING, and may know you are talking to them, but they will not be able to distinguish between your conversation and the mixed conversations of others in the room. In other words, the ratio of your "signal" to the level of the noise in the room must be larger than the level of the noise.
The identical condition exists for modems. The ratio of the received signal to the system noise must be greater than the level of noise itself. If it is not, at least some of the bits extracted from the symbols will be wrong; the worse the "signal to noise ratio", the greater the errors will be. Once the error rate reaches about 50%, you will have about as much success "guessing" the bits as you will demodulating them!
The power of Shannon's Theorem is that it allows a designer to measure certain parameters of the system he is designing for, and using those parameters, determine whether or not the proposed design is going to demand more symbols (chords) per second and/or more keys (bits) per chord (symbol) than the channel can support. The point to measure these things is at the receiver, where the cumulative effects of the channel show up. Regardless of how you arrange the keys and notes that you are sending, Shannon's Channel Capacity Formula defines how many chords and how many keys you may send for a given time frame. "Play" the notes faster than that and they "run together" or are simply not heard at all; play at or below the Limit, and your "tune" may be correctly heard at the receiver.
Obviously, if anyone could figure out how to remove or greatly reduce the noise added by the system, without affecting the signal, one could place more bits in a symbol and could improve upon the transmission speed. However, there are other, complex reasons why random noise cannot be completely removed from the signal without altering the signal proportionally. Until someone finds a way to remove noise without removing signal, Shannon's Limit will stand. Actually, even such a noise reduction method would not invalidate the Limit, it would only change the way it is calculated. And, in the limit, it turns out that S/N ratio, and not bandwidth, is the factor which could be changed; even at infinite bandwidth there is a finite limit to capacity and to S/N ratio. "Breaking" Shannon's Limit would require that this S/N problem be addressed, or actually, eliminated entirely.
regards,
Larry
p.s. critique welcomed |
