High Q peak in frequency response means what in time domain? The 2019 Stack Overflow Developer...
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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.
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
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
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 4 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
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
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
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
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
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
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 5 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 |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
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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 5 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 5 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 5 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 5 hours ago
Dan MillsDan Mills
12.2k11225
answered 5 hours ago
Dan MillsDan Mills
12.2k11225
answered 5 hours ago
Dan MillsDan Mills
12.2k11225
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
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 |
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$begingroup$
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
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