2
\$\begingroup\$

An Introduction to Information Theory: Symbols, Signals and Noise, by John R. Pierce, says the following:

In 1917, Harry Nyquist came to the American Telephone and Telegraph Company immediately after receiving his Ph.D. at Yale (Ph.D.’s were considerably rarer in those days). Nyquist was a much better mathematician than most men who tackled the problems of telegraphy, and he always was a clear, original, and philosophical thinker concerning communication. He tackled the problems of telegraphy with powerful methods and with clear insight. In 1924, he published his results in an important paper, “Certain Factors Affecting Telegraph Speed.”

This paper deals with a number of problems of telegraphy. Among other things, it clarifies the relation between the speed of telegraphy and the number of current values such as +1, – 1 (two current values) or +3, +1, – 1, – 3 (four current values). Nyquist says that if we send symbols (successive current values) at a constant rate, the speed of transmission, W, is related to m, the number of different symbols or current values available, by

$$W = K \log m$$

Here K is a constant whose value depends on how many successive current values are sent each second. The quantity log m means logarithm of m. There are different bases for taking logarithms. If we choose 2 as a base, then the values of log m for various values of m are given in Table II.

enter image description here

We can easily see by means of an example why the logarithm is the appropriate function in Nyquist’s relation. Suppose that we wish to specify two independent choices of off-or-on, 0-or-1, simultaneously. There are four possible combinations of two independent 0-or-1 choices, as shown in Table III.

enter image description here

Further, if we wish to specify three independent choices of 0-or-1 at the same time, we find eight combinations, as shown in Table IV.

enter image description here

Similarly, if we wish to specify four independent 0-or-1 choices, we find sixteen different combinations, and, if we wish to specify M different independent 0-or-1 choices, we find 2^M different combinations.

If we can specify M independent 0-or-1 combinations at once, we can in effect send M independent messages at once, so surely the speed should be proportional to M. But, in sending M messages at once we have 2^M possible combinations of the M independent 0-or-1 choices. Thus, to send M messages at once, we need to be able to send 2^M different symbols or current values. Suppose that we can choose among 2^M different symbols. Nyquist tells us that we should take the logarithm of the number of symbols in order to get the line speed, and

$$\log 2^M = M$$

Thus, the logarithm of the number of symbols is just the number of independent 0-or-1 choices that can be represented simultaneously, the number of independent messages we can send at once, so to speak. Nyquist’s relation says that by going from off-on telegraphy to three-current ( + 1, 0, – 1) telegraphy we can increase the speed of sending letters or other symbols by 60 per cent, and if we use four current values (+3, +1, – 1, – 3) we can double the speed. This is, of course, just what Edison did with his quadruplex telegraph, for he sent two messages instead of one. Further, Nyquist showed that the use of eight current values (0, 1, 2, 3, 4, 5, 6, 7, or +7, +5, +3, +1, – 1, – 3, —5, – 7) should enable us to send four times as fast as with two current values. However, he clearly realized that fluctuations in the attenuation of the circuit, interference or noise, and limitations on the power which can be used, make the use of many current values difficult.

In the last paragraph, the following claims are made by the author:

  1. By going from two-current telegraphy to three-current telegraphy, we can increase the speed of sending letters or other symbols by 60%.
  2. By going from two-current telegraphy to four-current telegraphy, we can double the speed of sending letters or other symbols.
  3. By going from two-current telegraphy to eight-current telegraphy, we can get 4 times the speed of sending letters or other symbols.

Let's start off by defining our variables:

m is the number of different symbols or current values available.

M is the number of independent 0-or-1 choices.

There are

$$2^M$$

possible combinations of the M independent 0-or-1 choices.

So if we have M = 2 independent 0-or-1 choices, then we have 4 possible combinations. If we have M = 3 independent 0-or-1 choices, then we have 8 possible combinations. This is all consistent with tables III and IV.

Now let's address the author's claims.

  1. Three current telegraphy is no longer 0-or-1 -- it is, for instance, +1, 0, or -1. So we have m = 3 (m -- NOT M), which gives us

$$\log 3 = 1.6,$$

which is consistent with table II.

1.6 is 60% more than 1, which is table II's value for 2-current telegraphy.

So far so good.

  1. $$\log 4 = 2$$

This is double the value from 2-current telegraphy. Again, so far as good.

  1. $$\log 8 = 3$$

This is not four-times the speed of 2-current telegraphy, it is three-times the speed.

I would appreciate it if people could please check my work to see whether the author has made an error. If not, please explain where I went wrong.

\$\endgroup\$
6
  • \$\begingroup\$ I would note that "the author" in this context is not Mr. Nyquist. \$\endgroup\$
    – user16324
    Dec 30, 2018 at 14:25
  • \$\begingroup\$ @BrianDrummond of course, I never claimed it was. \$\endgroup\$ Dec 30, 2018 at 14:26
  • \$\begingroup\$ To spell it out, I think you have found a mistake in the text. \$\endgroup\$
    – user16324
    Dec 30, 2018 at 14:30
  • \$\begingroup\$ @BrianDrummond ok, thanks for that. I just wanted to make sure that I wasn’t misunderstanding something. \$\endgroup\$ Dec 30, 2018 at 14:34
  • \$\begingroup\$ No textbook typed in MS Word is reliable imo. \$\endgroup\$ Dec 30, 2018 at 18:00

1 Answer 1

7
\$\begingroup\$

Funnily enough if you search "Certain Factors Affecting Telegraph Speed." in the usual online sources, you will find the original text by Nyquist.

It includes a table:

Number of current values employed    Relative amount of intelligence ... transmitted
2                                    100
3                                    158
4                                    200
5                                    230
8                                    300
16                                   400

so the mistake is, as expected, not Nyquist's.

\$\endgroup\$
2
  • \$\begingroup\$ Surprisingly even with editors like Mandelbrot, Chapter II did not change in the 1980 Version II when the author was age 70 after the initial release in 1961. It was intentionally written NOT to dwell too much on the math so a non-engineer could appreciate the beauty in science. But this example proves many readers glossed over non-linear example and the fact that there are 4 times as many levels but only 3 times as much information. with 2^3 vs 2^1 archive.org/details/introductiontoin00john/page/37 \$\endgroup\$ Dec 30, 2018 at 17:03
  • \$\begingroup\$ Then Shannon showed the rise of SNR needed at the same error rate to support the ratio of Bits/Baud symbol with the rise of m . notice no change on p38 archive.org/details/introductiontoin00john/page/37 WTG @The Pointer \$\endgroup\$ Dec 30, 2018 at 17:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.