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
My bread in my bread maker rises and then falls down just after cooking starts
Do high-wing aircraft represent more difficult engineering challenges than low-wing aircraft?
Does this "yield your space to an ally" rule my 3.5 group uses appear anywhere in the official rules?
Can my American children re-enter the USA by International flight with a passport card? Being that their passport book has expired
What kind of action are dodge and disengage?
Polynomial division: Is this trick obvious?
Could a space colony 1g from the sun work?
How does a permutation act on a string?
1970s short story about a famous hunter who is cloned and will die after three days?
Understanding Deutch's Algorithm
Promotion comes with unexpected 24/7/365 on-call
Can anyone give me examples of the relative-determinative 'which'?
Is my test coverage up to snuff?
Why doesn't Iron Man's action affect this person in Endgame?
Will there be more tax deductions if I put the house completely under my name, versus doing a joint ownership?
Does it matter what way the tires go if no directional arrow?
How does Ctrl+c and Ctrl+v work?
Which creature is depicted in this Xanathar's Guide illustration of a war mage?
Were any toxic metals used in the International Space Station?
Single word that parallels "Recent" when discussing the near future
What was Varys trying to do at the beginning of S08E05?
Wireless headphones interfere with Wi-Fi signal on laptop
What is the effect of the Feeblemind spell on Ability Score Improvements?
Cuban Primes
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 6 hours ago
aidanh010aidanh010
12115
$endgroup$
|
show 6 more comments
$begingroup$
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 6 hours ago
aidanh010aidanh010
12115
$endgroup$
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
3 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
3 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
3 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
3 hours ago
$begingroup$
Related: Why is channel capacity a factor of bandwidth instead of frequency?
$endgroup$
– The Photon
2 hours ago
|
show 6 more comments
$begingroup$
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 6 hours ago
aidanh010aidanh010
12115
$endgroup$
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 6 hours ago
aidanh010aidanh010
12115
asked 6 hours ago
aidanh010aidanh010
12115
asked 6 hours ago
aidanh010aidanh010
12115
asked 6 hours ago
aidanh010aidanh010
12115
asked 6 hours ago
aidanh010aidanh010
12115
12115
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
3 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
3 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
3 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
3 hours ago
$begingroup$
Related: Why is channel capacity a factor of bandwidth instead of frequency?
$endgroup$
– The Photon
2 hours ago
|
show 6 more comments
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
3 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
3 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
3 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
3 hours ago
$begingroup$
Related: Why is channel capacity a factor of bandwidth instead of frequency?
$endgroup$
– The Photon
2 hours 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
3 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
3 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
3 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
3 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
3 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
3 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
3 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
3 hours ago
$begingroup$
Related: Why is channel capacity a factor of bandwidth instead of frequency?
$endgroup$
– The Photon
2 hours ago
$begingroup$
Related: Why is channel capacity a factor of bandwidth instead of frequency?
$endgroup$
– The Photon
2 hours ago
|
show 6 more comments
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 3 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 2 hours ago
JustmeJustme
2,9581413
$endgroup$
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 3 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 3 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 3 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 3 hours ago
sstobbesstobbe
2,278159
answered 3 hours ago
sstobbesstobbe
2,278159
answered 3 hours ago
sstobbesstobbe
2,278159
answered 3 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 2 hours ago
JustmeJustme
2,9581413
$endgroup$
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 2 hours ago
JustmeJustme
2,9581413
$endgroup$
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 2 hours ago
JustmeJustme
2,9581413
$endgroup$
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 2 hours ago
JustmeJustme
2,9581413
answered 2 hours ago
JustmeJustme
2,9581413
answered 2 hours ago
JustmeJustme
2,9581413
answered 2 hours ago
JustmeJustme
2,9581413
2,9581413
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
3 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
3 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
3 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
3 hours ago
$begingroup$
Related: Why is channel capacity a factor of bandwidth instead of frequency?
$endgroup$
– The Photon
2 hours ago