High Q peak in frequency response means what in time domain?





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$begingroup$


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.



enter image description here



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
enter image description here










share|improve this question











$endgroup$












  • $begingroup$
    How did you measure the decay and value vs Q on this example?"
    $endgroup$
    – Sunnyskyguy EE75
    4 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
    4 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
    4 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


















1












$begingroup$


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.



enter image description here



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
enter image description here










share|improve this question











$endgroup$












  • $begingroup$
    How did you measure the decay and value vs Q on this example?"
    $endgroup$
    – Sunnyskyguy EE75
    4 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
    4 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
    4 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














1












1








1





$begingroup$


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.



enter image description here



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
enter image description here










share|improve this question











$endgroup$




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.



enter image description here



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
enter image description here







passive-networks frequency-response






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 4 hours ago







efox29

















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
    4 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
    4 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
    4 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
    4 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
    4 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
    4 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
4 hours ago






$begingroup$
How did you measure the decay and value vs Q on this example?"
$endgroup$
– Sunnyskyguy EE75
4 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
4 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
4 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
4 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
4 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










1 Answer
1






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oldest

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3












$begingroup$

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.






share|improve this answer









$endgroup$













  • $begingroup$
    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














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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

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.






share|improve this answer









$endgroup$













  • $begingroup$
    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


















3












$begingroup$

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.






share|improve this answer









$endgroup$













  • $begingroup$
    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
















3












3








3





$begingroup$

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.






share|improve this answer









$endgroup$



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.







share|improve this answer












share|improve this answer



share|improve this answer










answered 5 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




















  • $begingroup$
    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




















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