High Q peak in frequency response means what in time domain? The 2019 Stack Overflow Developer...
Keeping a retro style to sci-fi spaceships?
What aspect of planet Earth must be changed to prevent the industrial revolution?
Is every episode of "Where are my Pants?" identical?
Scientific Reports - Significant Figures
Segmentation fault output is suppressed when piping stdin into a function. Why?
Who or what is the being for whom Being is a question for Heidegger?
What was the last x86 CPU that did not have the x87 floating-point unit built in?
In horse breeding, what is the female equivalent of putting a horse out "to stud"?
How can I define good in a religion that claims no moral authority?
"... to apply for a visa" or "... and applied for a visa"?
How to test the equality of two Pearson correlation coefficients computed from the same sample?
How to stretch delimiters to envolve matrices inside of a kbordermatrix?
Can withdrawing asylum be illegal?
Windows 10: How to Lock (not sleep) laptop on lid close?
Would an alien lifeform be able to achieve space travel if lacking in vision?
Create an outline of font
How to delete random line from file using Unix command?
how can a perfect fourth interval be considered either consonant or dissonant?
Can the prologue be the backstory of your main character?
Does Parliament hold absolute power in the UK?
Searching for a differential characteristic (differential cryptanalysis)
How is simplicity better than precision and clarity in prose?
What are these Gizmos at Izaña Atmospheric Research Center in Spain?
Are my PIs rude or am I just being too sensitive?
High Q peak in frequency response means what in time domain?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)What circuit can use a falling edge to trigger this damped oscillating impulse waveform?Frequency response?How do PID controllers effect time domain and frequency domain responseMaximum Response time meaning?Questions about modelling a typical crystal radio and simulating in LTspiceWhat improved frequency response means?Some questions on a passive network's transfer function and time domain responseMid- and lowband frequency response of CEFrequency domain representationHigh frequency response of capacitors
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
$begingroup$
Reading Linear Circuit Transfer Functions and one of the graphs got me curious.
I've recreated the circuit (series RLC) and plotted the frequency response for a Q of 7.
We have a peak of ~16.3 dB when Q is 7 @ 10Khz.
Can this value be used (16.3 dB) to accurately predict something in the time domain - such as the value of Q or how long the oscillatory decay would take, the amplitude of the oscillations etc.. ?
Added in case its relevent
passive-networks frequency-response
asked 5 hours ago
efox29efox29
8,06953481
$endgroup$
add a comment |
$begingroup$
Reading Linear Circuit Transfer Functions and one of the graphs got me curious.
I've recreated the circuit (series RLC) and plotted the frequency response for a Q of 7.
We have a peak of ~16.3 dB when Q is 7 @ 10Khz.
Can this value be used (16.3 dB) to accurately predict something in the time domain - such as the value of Q or how long the oscillatory decay would take, the amplitude of the oscillations etc.. ?
Added in case its relevent
passive-networks frequency-response
asked 5 hours ago
efox29efox29
8,06953481
$endgroup$
$begingroup$
How did you measure the decay and value vs Q on this example?"
$endgroup$
– Sunnyskyguy EE75
3 hours ago
$begingroup$
@SunnyskyguyEE75 I don't fully understand the question. I caculated the values for my R L and C to give me a Q = 0.5 and Q = 7 (green and blue respectively). In this case, I know ahead of time, the Q and f because its what I used to calculate R, L and C
$endgroup$
– efox29
3 hours ago
$begingroup$
because the Zreal=Zreactive for Q=1 the apparent voltage amplitude from phasor current is sqrt (1+1) = sqrt(2) so for Q>>1 it equals gain , try Q=1
$endgroup$
– Sunnyskyguy EE75
3 hours ago
$begingroup$
Did you get an ringing T asymptote of about 300us for 7 ?. So if T=300us = 1/(2πΔf) or Δf= then 530Hz yet Δf=fo/Q = 10k/7=1.43k
$endgroup$
– Sunnyskyguy EE75
3 hours ago
add a comment |
$begingroup$
Reading Linear Circuit Transfer Functions and one of the graphs got me curious.
I've recreated the circuit (series RLC) and plotted the frequency response for a Q of 7.
We have a peak of ~16.3 dB when Q is 7 @ 10Khz.
Can this value be used (16.3 dB) to accurately predict something in the time domain - such as the value of Q or how long the oscillatory decay would take, the amplitude of the oscillations etc.. ?
Added in case its relevent
passive-networks frequency-response
asked 5 hours ago
efox29efox29
8,06953481
$endgroup$
Reading Linear Circuit Transfer Functions and one of the graphs got me curious.
I've recreated the circuit (series RLC) and plotted the frequency response for a Q of 7.
We have a peak of ~16.3 dB when Q is 7 @ 10Khz.
Can this value be used (16.3 dB) to accurately predict something in the time domain - such as the value of Q or how long the oscillatory decay would take, the amplitude of the oscillations etc.. ?
Added in case its relevent
passive-networks frequency-response
passive-networks frequency-response
asked 5 hours ago
efox29efox29
8,06953481
asked 5 hours ago
efox29efox29
8,06953481
edited 3 hours ago
efox29
asked 5 hours ago
efox29efox29
8,06953481
asked 5 hours ago
efox29efox29
8,06953481
asked 5 hours ago
efox29efox29
8,06953481
8,06953481
$begingroup$
How did you measure the decay and value vs Q on this example?"
$endgroup$
– Sunnyskyguy EE75
3 hours ago
$begingroup$
@SunnyskyguyEE75 I don't fully understand the question. I caculated the values for my R L and C to give me a Q = 0.5 and Q = 7 (green and blue respectively). In this case, I know ahead of time, the Q and f because its what I used to calculate R, L and C
$endgroup$
– efox29
3 hours ago
$begingroup$
because the Zreal=Zreactive for Q=1 the apparent voltage amplitude from phasor current is sqrt (1+1) = sqrt(2) so for Q>>1 it equals gain , try Q=1
$endgroup$
– Sunnyskyguy EE75
3 hours ago
$begingroup$
Did you get an ringing T asymptote of about 300us for 7 ?. So if T=300us = 1/(2πΔf) or Δf= then 530Hz yet Δf=fo/Q = 10k/7=1.43k
$endgroup$
– Sunnyskyguy EE75
3 hours ago
add a comment |
$begingroup$
How did you measure the decay and value vs Q on this example?"
$endgroup$
– Sunnyskyguy EE75
3 hours ago
$begingroup$
@SunnyskyguyEE75 I don't fully understand the question. I caculated the values for my R L and C to give me a Q = 0.5 and Q = 7 (green and blue respectively). In this case, I know ahead of time, the Q and f because its what I used to calculate R, L and C
$endgroup$
– efox29
3 hours ago
$begingroup$
because the Zreal=Zreactive for Q=1 the apparent voltage amplitude from phasor current is sqrt (1+1) = sqrt(2) so for Q>>1 it equals gain , try Q=1
$endgroup$
– Sunnyskyguy EE75
3 hours ago
$begingroup$
Did you get an ringing T asymptote of about 300us for 7 ?. So if T=300us = 1/(2πΔf) or Δf= then 530Hz yet Δf=fo/Q = 10k/7=1.43k
$endgroup$
– Sunnyskyguy EE75
3 hours ago
$begingroup$
How did you measure the decay and value vs Q on this example?"
$endgroup$
– Sunnyskyguy EE75
3 hours ago
$begingroup$
How did you measure the decay and value vs Q on this example?"
$endgroup$
– Sunnyskyguy EE75
3 hours ago
$begingroup$
@SunnyskyguyEE75 I don't fully understand the question. I caculated the values for my R L and C to give me a Q = 0.5 and Q = 7 (green and blue respectively). In this case, I know ahead of time, the Q and f because its what I used to calculate R, L and C
$endgroup$
– efox29
3 hours ago
$begingroup$
@SunnyskyguyEE75 I don't fully understand the question. I caculated the values for my R L and C to give me a Q = 0.5 and Q = 7 (green and blue respectively). In this case, I know ahead of time, the Q and f because its what I used to calculate R, L and C
$endgroup$
– efox29
3 hours ago
$begingroup$
because the Zreal=Zreactive for Q=1 the apparent voltage amplitude from phasor current is sqrt (1+1) = sqrt(2) so for Q>>1 it equals gain , try Q=1
$endgroup$
– Sunnyskyguy EE75
3 hours ago
$begingroup$
because the Zreal=Zreactive for Q=1 the apparent voltage amplitude from phasor current is sqrt (1+1) = sqrt(2) so for Q>>1 it equals gain , try Q=1
$endgroup$
– Sunnyskyguy EE75
3 hours ago
$begingroup$
Did you get an ringing T asymptote of about 300us for 7 ?. So if T=300us = 1/(2πΔf) or Δf= then 530Hz yet Δf=fo/Q = 10k/7=1.43k
$endgroup$
– Sunnyskyguy EE75
3 hours ago
$begingroup$
Did you get an ringing T asymptote of about 300us for 7 ?. So if T=300us = 1/(2πΔf) or Δf= then 530Hz yet Δf=fo/Q = 10k/7=1.43k
$endgroup$
– Sunnyskyguy EE75
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Q is (among other definitions) the voltage gain at resonance, and a voltage gain of 7 times is $$20 * log(7) = 16.9dB$$ which seems close enough as your cursor is clearly not actually on resonance (phase would be -90 not -93). So dB of resonant gain is trivially converted to or from Q.
Q gives you risetime and whether the circuit is over/under or critically damped in the time domain, as well as how well damped the ringing in an under damped circuit is.
answered 4 hours ago
Dan MillsDan Mills
12.2k11225
$endgroup$
$begingroup$
It's always something simple. This has given me a items to explore deeper into.
$endgroup$
– efox29
4 hours ago
$begingroup$
I though Av= √{1+Q²} so when Q=1 Av=1.414 or +3dB and for LPF the f-3dB breakpoints are not symmetrical about peak unlike a simple BPF so your Q=6.6 ( close enuf)
$endgroup$
– Sunnyskyguy EE75
4 hours ago
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%2f432436%2fhigh-q-peak-in-frequency-response-means-what-in-time-domain%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
$begingroup$
Q is (among other definitions) the voltage gain at resonance, and a voltage gain of 7 times is $$20 * log(7) = 16.9dB$$ which seems close enough as your cursor is clearly not actually on resonance (phase would be -90 not -93). So dB of resonant gain is trivially converted to or from Q.
Q gives you risetime and whether the circuit is over/under or critically damped in the time domain, as well as how well damped the ringing in an under damped circuit is.
answered 4 hours ago
Dan MillsDan Mills
12.2k11225
$endgroup$
$begingroup$
It's always something simple. This has given me a items to explore deeper into.
$endgroup$
– efox29
4 hours ago
$begingroup$
I though Av= √{1+Q²} so when Q=1 Av=1.414 or +3dB and for LPF the f-3dB breakpoints are not symmetrical about peak unlike a simple BPF so your Q=6.6 ( close enuf)
$endgroup$
– Sunnyskyguy EE75
4 hours ago
add a comment |
$begingroup$
Q is (among other definitions) the voltage gain at resonance, and a voltage gain of 7 times is $$20 * log(7) = 16.9dB$$ which seems close enough as your cursor is clearly not actually on resonance (phase would be -90 not -93). So dB of resonant gain is trivially converted to or from Q.
Q gives you risetime and whether the circuit is over/under or critically damped in the time domain, as well as how well damped the ringing in an under damped circuit is.
answered 4 hours ago
Dan MillsDan Mills
12.2k11225
$endgroup$
$begingroup$
It's always something simple. This has given me a items to explore deeper into.
$endgroup$
– efox29
4 hours ago
$begingroup$
I though Av= √{1+Q²} so when Q=1 Av=1.414 or +3dB and for LPF the f-3dB breakpoints are not symmetrical about peak unlike a simple BPF so your Q=6.6 ( close enuf)
$endgroup$
– Sunnyskyguy EE75
4 hours ago
add a comment |
$begingroup$
Q is (among other definitions) the voltage gain at resonance, and a voltage gain of 7 times is $$20 * log(7) = 16.9dB$$ which seems close enough as your cursor is clearly not actually on resonance (phase would be -90 not -93). So dB of resonant gain is trivially converted to or from Q.
Q gives you risetime and whether the circuit is over/under or critically damped in the time domain, as well as how well damped the ringing in an under damped circuit is.
answered 4 hours ago
Dan MillsDan Mills
12.2k11225
$endgroup$
Q is (among other definitions) the voltage gain at resonance, and a voltage gain of 7 times is $$20 * log(7) = 16.9dB$$ which seems close enough as your cursor is clearly not actually on resonance (phase would be -90 not -93). So dB of resonant gain is trivially converted to or from Q.
Q gives you risetime and whether the circuit is over/under or critically damped in the time domain, as well as how well damped the ringing in an under damped circuit is.
answered 4 hours ago
Dan MillsDan Mills
12.2k11225
answered 4 hours ago
Dan MillsDan Mills
12.2k11225
answered 4 hours ago
Dan MillsDan Mills
12.2k11225
answered 4 hours ago
Dan MillsDan Mills
12.2k11225
12.2k11225
$begingroup$
It's always something simple. This has given me a items to explore deeper into.
$endgroup$
– efox29
4 hours ago
$begingroup$
I though Av= √{1+Q²} so when Q=1 Av=1.414 or +3dB and for LPF the f-3dB breakpoints are not symmetrical about peak unlike a simple BPF so your Q=6.6 ( close enuf)
$endgroup$
– Sunnyskyguy EE75
4 hours ago
add a comment |
$begingroup$
It's always something simple. This has given me a items to explore deeper into.
$endgroup$
– efox29
4 hours ago
$begingroup$
I though Av= √{1+Q²} so when Q=1 Av=1.414 or +3dB and for LPF the f-3dB breakpoints are not symmetrical about peak unlike a simple BPF so your Q=6.6 ( close enuf)
$endgroup$
– Sunnyskyguy EE75
4 hours ago
$begingroup$
It's always something simple. This has given me a items to explore deeper into.
$endgroup$
– efox29
4 hours ago
$begingroup$
It's always something simple. This has given me a items to explore deeper into.
$endgroup$
– efox29
4 hours ago
$begingroup$
I though Av= √{1+Q²} so when Q=1 Av=1.414 or +3dB and for LPF the f-3dB breakpoints are not symmetrical about peak unlike a simple BPF so your Q=6.6 ( close enuf)
$endgroup$
– Sunnyskyguy EE75
4 hours ago
$begingroup$
I though Av= √{1+Q²} so when Q=1 Av=1.414 or +3dB and for LPF the f-3dB breakpoints are not symmetrical about peak unlike a simple BPF so your Q=6.6 ( close enuf)
$endgroup$
– Sunnyskyguy EE75
4 hours ago
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%2f432436%2fhigh-q-peak-in-frequency-response-means-what-in-time-domain%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
$begingroup$
How did you measure the decay and value vs Q on this example?"
$endgroup$
– Sunnyskyguy EE75
3 hours ago
$begingroup$
@SunnyskyguyEE75 I don't fully understand the question. I caculated the values for my R L and C to give me a Q = 0.5 and Q = 7 (green and blue respectively). In this case, I know ahead of time, the Q and f because its what I used to calculate R, L and C
$endgroup$
– efox29
3 hours ago
$begingroup$
because the Zreal=Zreactive for Q=1 the apparent voltage amplitude from phasor current is sqrt (1+1) = sqrt(2) so for Q>>1 it equals gain , try Q=1
$endgroup$
– Sunnyskyguy EE75
3 hours ago
$begingroup$
Did you get an ringing T asymptote of about 300us for 7 ?. So if T=300us = 1/(2πΔf) or Δf= then 530Hz yet Δf=fo/Q = 10k/7=1.43k
$endgroup$
– Sunnyskyguy EE75
3 hours ago