(Textbook error?) Nyquist's relation: telegraph speed of transmission
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.
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.
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.
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:
- By going from two-current telegraphy to three-current telegraphy, we can increase the speed of sending letters or other symbols by 60%.
- By going from two-current telegraphy to four-current telegraphy, we can double the speed of sending letters or other symbols.
- 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.
- 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.
- $$log 4 = 2$$
This is double the value from 2-current telegraphy. Again, so far as good.
- $$log 8 = 3$$
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.
current telecommunications
|
show 1 more comment
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.
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.
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.
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:
- By going from two-current telegraphy to three-current telegraphy, we can increase the speed of sending letters or other symbols by 60%.
- By going from two-current telegraphy to four-current telegraphy, we can double the speed of sending letters or other symbols.
- 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.
- 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.
- $$log 4 = 2$$
This is double the value from 2-current telegraphy. Again, so far as good.
- $$log 8 = 3$$
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.
current telecommunications
I would note that "the author" in this context is not Mr. Nyquist.
– Brian Drummond
Dec 30 '18 at 14:25
@BrianDrummond of course, I never claimed it was.
– The Pointer
Dec 30 '18 at 14:26
To spell it out, I think you have found a mistake in the text.
– Brian Drummond
Dec 30 '18 at 14:30
@BrianDrummond ok, thanks for that. I just wanted to make sure that I wasn’t misunderstanding something.
– The Pointer
Dec 30 '18 at 14:34
No textbook typed in MS Word is reliable imo.
– Chase Ryan Taylor
Dec 30 '18 at 18:00
|
show 1 more comment
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.
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.
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.
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:
- By going from two-current telegraphy to three-current telegraphy, we can increase the speed of sending letters or other symbols by 60%.
- By going from two-current telegraphy to four-current telegraphy, we can double the speed of sending letters or other symbols.
- 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.
- 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.
- $$log 4 = 2$$
This is double the value from 2-current telegraphy. Again, so far as good.
- $$log 8 = 3$$
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.
current telecommunications
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.
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.
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.
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:
- By going from two-current telegraphy to three-current telegraphy, we can increase the speed of sending letters or other symbols by 60%.
- By going from two-current telegraphy to four-current telegraphy, we can double the speed of sending letters or other symbols.
- 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.
- 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.
- $$log 4 = 2$$
This is double the value from 2-current telegraphy. Again, so far as good.
- $$log 8 = 3$$
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.
current telecommunications
current telecommunications
asked Dec 30 '18 at 14:06
The PointerThe Pointer
30919
30919
I would note that "the author" in this context is not Mr. Nyquist.
– Brian Drummond
Dec 30 '18 at 14:25
@BrianDrummond of course, I never claimed it was.
– The Pointer
Dec 30 '18 at 14:26
To spell it out, I think you have found a mistake in the text.
– Brian Drummond
Dec 30 '18 at 14:30
@BrianDrummond ok, thanks for that. I just wanted to make sure that I wasn’t misunderstanding something.
– The Pointer
Dec 30 '18 at 14:34
No textbook typed in MS Word is reliable imo.
– Chase Ryan Taylor
Dec 30 '18 at 18:00
|
show 1 more comment
I would note that "the author" in this context is not Mr. Nyquist.
– Brian Drummond
Dec 30 '18 at 14:25
@BrianDrummond of course, I never claimed it was.
– The Pointer
Dec 30 '18 at 14:26
To spell it out, I think you have found a mistake in the text.
– Brian Drummond
Dec 30 '18 at 14:30
@BrianDrummond ok, thanks for that. I just wanted to make sure that I wasn’t misunderstanding something.
– The Pointer
Dec 30 '18 at 14:34
No textbook typed in MS Word is reliable imo.
– Chase Ryan Taylor
Dec 30 '18 at 18:00
I would note that "the author" in this context is not Mr. Nyquist.
– Brian Drummond
Dec 30 '18 at 14:25
I would note that "the author" in this context is not Mr. Nyquist.
– Brian Drummond
Dec 30 '18 at 14:25
@BrianDrummond of course, I never claimed it was.
– The Pointer
Dec 30 '18 at 14:26
@BrianDrummond of course, I never claimed it was.
– The Pointer
Dec 30 '18 at 14:26
To spell it out, I think you have found a mistake in the text.
– Brian Drummond
Dec 30 '18 at 14:30
To spell it out, I think you have found a mistake in the text.
– Brian Drummond
Dec 30 '18 at 14:30
@BrianDrummond ok, thanks for that. I just wanted to make sure that I wasn’t misunderstanding something.
– The Pointer
Dec 30 '18 at 14:34
@BrianDrummond ok, thanks for that. I just wanted to make sure that I wasn’t misunderstanding something.
– The Pointer
Dec 30 '18 at 14:34
No textbook typed in MS Word is reliable imo.
– Chase Ryan Taylor
Dec 30 '18 at 18:00
No textbook typed in MS Word is reliable imo.
– Chase Ryan Taylor
Dec 30 '18 at 18:00
|
show 1 more comment
1 Answer
1
active
oldest
votes
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.
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
– Sunnyskyguy EE75
Dec 30 '18 at 17:03
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
– Sunnyskyguy EE75
Dec 30 '18 at 17:08
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");
StackExchange.ifUsing("editor", function () {
return StackExchange.using("schematics", function () {
StackExchange.schematics.init();
});
}, "cicuitlab");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "135"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f414452%2ftextbook-error-nyquists-relation-telegraph-speed-of-transmission%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
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
– Sunnyskyguy EE75
Dec 30 '18 at 17:03
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
– Sunnyskyguy EE75
Dec 30 '18 at 17:08
add a comment |
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.
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
– Sunnyskyguy EE75
Dec 30 '18 at 17:03
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
– Sunnyskyguy EE75
Dec 30 '18 at 17:08
add a comment |
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.
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.
answered Dec 30 '18 at 14:49
Brian DrummondBrian Drummond
47.3k137104
47.3k137104
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
– Sunnyskyguy EE75
Dec 30 '18 at 17:03
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
– Sunnyskyguy EE75
Dec 30 '18 at 17:08
add a comment |
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
– Sunnyskyguy EE75
Dec 30 '18 at 17:03
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
– Sunnyskyguy EE75
Dec 30 '18 at 17:08
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
– Sunnyskyguy EE75
Dec 30 '18 at 17:03
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
– Sunnyskyguy EE75
Dec 30 '18 at 17:03
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
– Sunnyskyguy EE75
Dec 30 '18 at 17:08
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
– Sunnyskyguy EE75
Dec 30 '18 at 17:08
add a comment |
Thanks for contributing an answer to Electrical Engineering Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2felectronics.stackexchange.com%2fquestions%2f414452%2ftextbook-error-nyquists-relation-telegraph-speed-of-transmission%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
I would note that "the author" in this context is not Mr. Nyquist.
– Brian Drummond
Dec 30 '18 at 14:25
@BrianDrummond of course, I never claimed it was.
– The Pointer
Dec 30 '18 at 14:26
To spell it out, I think you have found a mistake in the text.
– Brian Drummond
Dec 30 '18 at 14:30
@BrianDrummond ok, thanks for that. I just wanted to make sure that I wasn’t misunderstanding something.
– The Pointer
Dec 30 '18 at 14:34
No textbook typed in MS Word is reliable imo.
– Chase Ryan Taylor
Dec 30 '18 at 18:00