Why do OOK transmissions have bandwidth?Why is channel capacity a factor of bandwidth instead of frequency?Required Bandwidth for ASKBPSK bandwidth for symbols not integer wavelengthsWhy more bandwidth means more bit rate per second?Why does more bandwidth mean higher bit rate in digital transmission?Can't understand Frequency and Bandwidthreceiving transmissions transmitted on different channels sub-1 GHz using one receiverDoes increasing the amplitude of a signal also increase the data rate?Bandwidth-Data rate relationshipWhy does more bandwidth guarantee high bit rate?Bandwidth of triangular wave for desired linearity
Geometric inspiration behind Hal(irutan)'s Wolf(ram Language Logo)
Why can't I share a one use code with anyone else?
History of the Frobenius Endomorphism?
Are there microwaves to heat baby food at Brussels airport?
Why is the Advance Variation considered strong vs the Caro-Kann but not vs the Scandinavian?
What dog breeds survive the apocalypse for generations?
Why did the soldiers of the North disobey Jon?
Did any "washouts" of the Mercury program eventually become astronauts?
Will the volt, ampere, ohm or other electrical units change on May 20th, 2019?
Why is the marginal distribution/marginal probability described as "marginal"?
Understanding Python syntax in lists vs series
My bread in my bread maker rises and then falls down just after cooking starts
Why when I add jam to my tea it stops producing thin "membrane" on top?
Were any of the books mentioned in this scene from the movie Hackers real?
I recently started my machine learning PhD and I have absolutely no idea what I'm doing
What kind of action are dodge and disengage?
Using chord iii in a chord progression (major key)
How to handle professionally if colleagues has referred his relative and asking to take easy while taking interview
Holding rent money for my friend which amounts to over $10k?
Was the dragon prowess intentionally downplayed in S08E04?
Slice a list based on an index and items behind it in python
Promotion comes with unexpected 24/7/365 on-call
Why are goodwill impairments on the statement of cash-flows of GE?
How does a permutation act on a string?
Why do OOK transmissions have bandwidth?
Why is channel capacity a factor of bandwidth instead of frequency?Required Bandwidth for ASKBPSK bandwidth for symbols not integer wavelengthsWhy more bandwidth means more bit rate per second?Why does more bandwidth mean higher bit rate in digital transmission?Can't understand Frequency and Bandwidthreceiving transmissions transmitted on different channels sub-1 GHz using one receiverDoes increasing the amplitude of a signal also increase the data rate?Bandwidth-Data rate relationshipWhy does more bandwidth guarantee high bit rate?Bandwidth of triangular wave for desired linearity
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
With on-off keying, you have a single carrier wave at a constant frequency transmitting its presence or absence. Now I understand that the crystal will not be perfect so you will have some minimal shift over time in the carrier frequency, but assuming the crystal was perfect you would have zero bandwidth because the frequency doesn't change. But datasheets say that the bandwidth is linked to the data rate somehow, what is that equation and what is using up all that extra spectrum if you don't change the frequency?
rf signal bandwidth
asked 5 hours ago
aidanh010aidanh010
11615
$endgroup$
|
show 1 more comment
$begingroup$
With on-off keying, you have a single carrier wave at a constant frequency transmitting its presence or absence. Now I understand that the crystal will not be perfect so you will have some minimal shift over time in the carrier frequency, but assuming the crystal was perfect you would have zero bandwidth because the frequency doesn't change. But datasheets say that the bandwidth is linked to the data rate somehow, what is that equation and what is using up all that extra spectrum if you don't change the frequency?
rf signal bandwidth
asked 5 hours ago
aidanh010aidanh010
11615
$endgroup$
1
$begingroup$
THe on-off data rate has bandwidth, BW. When the carrier is modulated with this the signal bandwidth BW is transferred by this product to add +/- same BW on the carrier,
$endgroup$
– Sunnyskyguy EE75
2 hours ago
1
$begingroup$
During transitions from off to on or vice-verse, the bandwidth is not as narrow as a continuous sine wave. It is a basic idea that an unmodulated carrier conveys no information, and all modulated carriers have non-zero bandwidth. Information theory dictates that you cannot convey information without bandwidth.
$endgroup$
– mkeith
2 hours ago
$begingroup$
While a transmitted signal carrying any information has bandwidth, the larger bandwidth is that of the receiver - which must accommodate both the data bandwidth and the relative frequency error of the transmitter. Common cheap regenerative receivers also have a demodulation bandwidth determined by the quenching rate.
$endgroup$
– Chris Stratton
2 hours ago
$begingroup$
If you attempt to transmit 10Hz 50% duty cycle OOK thru a 1Hz bandwidth, the carrier amplitude will decay only a moderate amount during the OFF time, and the carrier amplitude will increase only a moderate amount during the ON time. You'll need about 5Hz bandwidth to see a large amplitude variation between ON and OFF, because the channel (the filter) has the irksome property of storing energy, and that stored energy causes ISI between a ON and the next OFF, and from that OFF to the next ON. ISI is inter-symbol-interference.
$endgroup$
– analogsystemsrf
2 hours ago
$begingroup$
Related: Why is channel capacity a factor of bandwidth instead of frequency?
$endgroup$
– The Photon
1 hour ago
|
show 1 more comment
$begingroup$
With on-off keying, you have a single carrier wave at a constant frequency transmitting its presence or absence. Now I understand that the crystal will not be perfect so you will have some minimal shift over time in the carrier frequency, but assuming the crystal was perfect you would have zero bandwidth because the frequency doesn't change. But datasheets say that the bandwidth is linked to the data rate somehow, what is that equation and what is using up all that extra spectrum if you don't change the frequency?
rf signal bandwidth
asked 5 hours ago
aidanh010aidanh010
11615
$endgroup$
With on-off keying, you have a single carrier wave at a constant frequency transmitting its presence or absence. Now I understand that the crystal will not be perfect so you will have some minimal shift over time in the carrier frequency, but assuming the crystal was perfect you would have zero bandwidth because the frequency doesn't change. But datasheets say that the bandwidth is linked to the data rate somehow, what is that equation and what is using up all that extra spectrum if you don't change the frequency?
rf signal bandwidth
rf signal bandwidth
asked 5 hours ago
aidanh010aidanh010
11615
asked 5 hours ago
aidanh010aidanh010
11615
asked 5 hours ago
aidanh010aidanh010
11615
asked 5 hours ago
aidanh010aidanh010
11615
asked 5 hours ago
aidanh010aidanh010
11615
11615
1
$begingroup$
THe on-off data rate has bandwidth, BW. When the carrier is modulated with this the signal bandwidth BW is transferred by this product to add +/- same BW on the carrier,
$endgroup$
– Sunnyskyguy EE75
2 hours ago
1
$begingroup$
During transitions from off to on or vice-verse, the bandwidth is not as narrow as a continuous sine wave. It is a basic idea that an unmodulated carrier conveys no information, and all modulated carriers have non-zero bandwidth. Information theory dictates that you cannot convey information without bandwidth.
$endgroup$
– mkeith
2 hours ago
$begingroup$
While a transmitted signal carrying any information has bandwidth, the larger bandwidth is that of the receiver - which must accommodate both the data bandwidth and the relative frequency error of the transmitter. Common cheap regenerative receivers also have a demodulation bandwidth determined by the quenching rate.
$endgroup$
– Chris Stratton
2 hours ago
$begingroup$
If you attempt to transmit 10Hz 50% duty cycle OOK thru a 1Hz bandwidth, the carrier amplitude will decay only a moderate amount during the OFF time, and the carrier amplitude will increase only a moderate amount during the ON time. You'll need about 5Hz bandwidth to see a large amplitude variation between ON and OFF, because the channel (the filter) has the irksome property of storing energy, and that stored energy causes ISI between a ON and the next OFF, and from that OFF to the next ON. ISI is inter-symbol-interference.
$endgroup$
– analogsystemsrf
2 hours ago
$begingroup$
Related: Why is channel capacity a factor of bandwidth instead of frequency?
$endgroup$
– The Photon
1 hour ago
|
show 1 more comment
1
$begingroup$
THe on-off data rate has bandwidth, BW. When the carrier is modulated with this the signal bandwidth BW is transferred by this product to add +/- same BW on the carrier,
$endgroup$
– Sunnyskyguy EE75
2 hours ago
1
$begingroup$
During transitions from off to on or vice-verse, the bandwidth is not as narrow as a continuous sine wave. It is a basic idea that an unmodulated carrier conveys no information, and all modulated carriers have non-zero bandwidth. Information theory dictates that you cannot convey information without bandwidth.
$endgroup$
– mkeith
2 hours ago
$begingroup$
While a transmitted signal carrying any information has bandwidth, the larger bandwidth is that of the receiver - which must accommodate both the data bandwidth and the relative frequency error of the transmitter. Common cheap regenerative receivers also have a demodulation bandwidth determined by the quenching rate.
$endgroup$
– Chris Stratton
2 hours ago
$begingroup$
If you attempt to transmit 10Hz 50% duty cycle OOK thru a 1Hz bandwidth, the carrier amplitude will decay only a moderate amount during the OFF time, and the carrier amplitude will increase only a moderate amount during the ON time. You'll need about 5Hz bandwidth to see a large amplitude variation between ON and OFF, because the channel (the filter) has the irksome property of storing energy, and that stored energy causes ISI between a ON and the next OFF, and from that OFF to the next ON. ISI is inter-symbol-interference.
$endgroup$
– analogsystemsrf
2 hours ago
$begingroup$
Related: Why is channel capacity a factor of bandwidth instead of frequency?
$endgroup$
– The Photon
1 hour ago
1
1
$begingroup$
THe on-off data rate has bandwidth, BW. When the carrier is modulated with this the signal bandwidth BW is transferred by this product to add +/- same BW on the carrier,
$endgroup$
– Sunnyskyguy EE75
2 hours ago
$begingroup$
THe on-off data rate has bandwidth, BW. When the carrier is modulated with this the signal bandwidth BW is transferred by this product to add +/- same BW on the carrier,
$endgroup$
– Sunnyskyguy EE75
2 hours ago
1
1
$begingroup$
During transitions from off to on or vice-verse, the bandwidth is not as narrow as a continuous sine wave. It is a basic idea that an unmodulated carrier conveys no information, and all modulated carriers have non-zero bandwidth. Information theory dictates that you cannot convey information without bandwidth.
$endgroup$
– mkeith
2 hours ago
$begingroup$
During transitions from off to on or vice-verse, the bandwidth is not as narrow as a continuous sine wave. It is a basic idea that an unmodulated carrier conveys no information, and all modulated carriers have non-zero bandwidth. Information theory dictates that you cannot convey information without bandwidth.
$endgroup$
– mkeith
2 hours ago
$begingroup$
While a transmitted signal carrying any information has bandwidth, the larger bandwidth is that of the receiver - which must accommodate both the data bandwidth and the relative frequency error of the transmitter. Common cheap regenerative receivers also have a demodulation bandwidth determined by the quenching rate.
$endgroup$
– Chris Stratton
2 hours ago
$begingroup$
While a transmitted signal carrying any information has bandwidth, the larger bandwidth is that of the receiver - which must accommodate both the data bandwidth and the relative frequency error of the transmitter. Common cheap regenerative receivers also have a demodulation bandwidth determined by the quenching rate.
$endgroup$
– Chris Stratton
2 hours ago
$begingroup$
If you attempt to transmit 10Hz 50% duty cycle OOK thru a 1Hz bandwidth, the carrier amplitude will decay only a moderate amount during the OFF time, and the carrier amplitude will increase only a moderate amount during the ON time. You'll need about 5Hz bandwidth to see a large amplitude variation between ON and OFF, because the channel (the filter) has the irksome property of storing energy, and that stored energy causes ISI between a ON and the next OFF, and from that OFF to the next ON. ISI is inter-symbol-interference.
$endgroup$
– analogsystemsrf
2 hours ago
$begingroup$
If you attempt to transmit 10Hz 50% duty cycle OOK thru a 1Hz bandwidth, the carrier amplitude will decay only a moderate amount during the OFF time, and the carrier amplitude will increase only a moderate amount during the ON time. You'll need about 5Hz bandwidth to see a large amplitude variation between ON and OFF, because the channel (the filter) has the irksome property of storing energy, and that stored energy causes ISI between a ON and the next OFF, and from that OFF to the next ON. ISI is inter-symbol-interference.
$endgroup$
– analogsystemsrf
2 hours ago
$begingroup$
Related: Why is channel capacity a factor of bandwidth instead of frequency?
$endgroup$
– The Photon
1 hour ago
$begingroup$
Related: Why is channel capacity a factor of bandwidth instead of frequency?
$endgroup$
– The Photon
1 hour ago
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
On-Off Keying (OOK) is equivalent to AM modulation of the carrier with the data bit-stream with a 100 % modulation index.
As a result the entire spectrum of the bit stream is up-converted to the carrier frequency of the OOK modulator. Additionally, due to AM modulation an image of the data-stream exists below the carrier frequency in addition to the translated spectrum above the carrier.
answered 2 hours ago
sstobbesstobbe
2,278159
$endgroup$
add a comment |
$begingroup$
Only an ideal sine wave with constant amplitude and single frequency that runs for infinitely long in time has extremely narrow bandwidth of a single peak. Having any changes to the ideal sine wave such as changing the amplitude means that since it cannot be represented as a single sine wave any more but a sum of many sine waves of different amplitudes and different frequencies, in frequency domain it means it has a range of frequencies. The faster the changes, the wider the bandwidth.
answered 1 hour ago
JustmeJustme
2,9481413
$endgroup$
add a comment |
Your Answer
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%2f438512%2fwhy-do-ook-transmissions-have-bandwidth%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
On-Off Keying (OOK) is equivalent to AM modulation of the carrier with the data bit-stream with a 100 % modulation index.
As a result the entire spectrum of the bit stream is up-converted to the carrier frequency of the OOK modulator. Additionally, due to AM modulation an image of the data-stream exists below the carrier frequency in addition to the translated spectrum above the carrier.
answered 2 hours ago
sstobbesstobbe
2,278159
$endgroup$
add a comment |
$begingroup$
On-Off Keying (OOK) is equivalent to AM modulation of the carrier with the data bit-stream with a 100 % modulation index.
As a result the entire spectrum of the bit stream is up-converted to the carrier frequency of the OOK modulator. Additionally, due to AM modulation an image of the data-stream exists below the carrier frequency in addition to the translated spectrum above the carrier.
answered 2 hours ago
sstobbesstobbe
2,278159
$endgroup$
add a comment |
$begingroup$
On-Off Keying (OOK) is equivalent to AM modulation of the carrier with the data bit-stream with a 100 % modulation index.
As a result the entire spectrum of the bit stream is up-converted to the carrier frequency of the OOK modulator. Additionally, due to AM modulation an image of the data-stream exists below the carrier frequency in addition to the translated spectrum above the carrier.
answered 2 hours ago
sstobbesstobbe
2,278159
$endgroup$
On-Off Keying (OOK) is equivalent to AM modulation of the carrier with the data bit-stream with a 100 % modulation index.
As a result the entire spectrum of the bit stream is up-converted to the carrier frequency of the OOK modulator. Additionally, due to AM modulation an image of the data-stream exists below the carrier frequency in addition to the translated spectrum above the carrier.
answered 2 hours ago
sstobbesstobbe
2,278159
answered 2 hours ago
sstobbesstobbe
2,278159
answered 2 hours ago
sstobbesstobbe
2,278159
answered 2 hours ago
sstobbesstobbe
2,278159
2,278159
add a comment |
add a comment |
$begingroup$
Only an ideal sine wave with constant amplitude and single frequency that runs for infinitely long in time has extremely narrow bandwidth of a single peak. Having any changes to the ideal sine wave such as changing the amplitude means that since it cannot be represented as a single sine wave any more but a sum of many sine waves of different amplitudes and different frequencies, in frequency domain it means it has a range of frequencies. The faster the changes, the wider the bandwidth.
answered 1 hour ago
JustmeJustme
2,9481413
$endgroup$
add a comment |
$begingroup$
Only an ideal sine wave with constant amplitude and single frequency that runs for infinitely long in time has extremely narrow bandwidth of a single peak. Having any changes to the ideal sine wave such as changing the amplitude means that since it cannot be represented as a single sine wave any more but a sum of many sine waves of different amplitudes and different frequencies, in frequency domain it means it has a range of frequencies. The faster the changes, the wider the bandwidth.
answered 1 hour ago
JustmeJustme
2,9481413
$endgroup$
add a comment |
$begingroup$
Only an ideal sine wave with constant amplitude and single frequency that runs for infinitely long in time has extremely narrow bandwidth of a single peak. Having any changes to the ideal sine wave such as changing the amplitude means that since it cannot be represented as a single sine wave any more but a sum of many sine waves of different amplitudes and different frequencies, in frequency domain it means it has a range of frequencies. The faster the changes, the wider the bandwidth.
answered 1 hour ago
JustmeJustme
2,9481413
$endgroup$
Only an ideal sine wave with constant amplitude and single frequency that runs for infinitely long in time has extremely narrow bandwidth of a single peak. Having any changes to the ideal sine wave such as changing the amplitude means that since it cannot be represented as a single sine wave any more but a sum of many sine waves of different amplitudes and different frequencies, in frequency domain it means it has a range of frequencies. The faster the changes, the wider the bandwidth.
answered 1 hour ago
JustmeJustme
2,9481413
answered 1 hour ago
JustmeJustme
2,9481413
answered 1 hour ago
JustmeJustme
2,9481413
answered 1 hour ago
JustmeJustme
2,9481413
2,9481413
add a comment |
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.
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%2f438512%2fwhy-do-ook-transmissions-have-bandwidth%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
1
$begingroup$
THe on-off data rate has bandwidth, BW. When the carrier is modulated with this the signal bandwidth BW is transferred by this product to add +/- same BW on the carrier,
$endgroup$
– Sunnyskyguy EE75
2 hours ago
1
$begingroup$
During transitions from off to on or vice-verse, the bandwidth is not as narrow as a continuous sine wave. It is a basic idea that an unmodulated carrier conveys no information, and all modulated carriers have non-zero bandwidth. Information theory dictates that you cannot convey information without bandwidth.
$endgroup$
– mkeith
2 hours ago
$begingroup$
While a transmitted signal carrying any information has bandwidth, the larger bandwidth is that of the receiver - which must accommodate both the data bandwidth and the relative frequency error of the transmitter. Common cheap regenerative receivers also have a demodulation bandwidth determined by the quenching rate.
$endgroup$
– Chris Stratton
2 hours ago
$begingroup$
If you attempt to transmit 10Hz 50% duty cycle OOK thru a 1Hz bandwidth, the carrier amplitude will decay only a moderate amount during the OFF time, and the carrier amplitude will increase only a moderate amount during the ON time. You'll need about 5Hz bandwidth to see a large amplitude variation between ON and OFF, because the channel (the filter) has the irksome property of storing energy, and that stored energy causes ISI between a ON and the next OFF, and from that OFF to the next ON. ISI is inter-symbol-interference.
$endgroup$
– analogsystemsrf
2 hours ago
$begingroup$
Related: Why is channel capacity a factor of bandwidth instead of frequency?
$endgroup$
– The Photon
1 hour ago