\$Q=\dfrac{1}{2\zeta} = \sqrt{A_{pk}^2 + 1}\$ for peak gain Apk

When Q is the quality factor of resonance gain approaches 1 or less, we switch to damping factor as resonance becomes dampening factor is more relevant.

Rule of Thumb: We approximate high Q to be just the resonant gain for Q>>1.

But in your case, Q is very low, and the peak/flat gain = 1.25

 

If gain, Apk=1.25 then Q = 1.6 , or ζ = 1/3.2 This is your answer from reading graph.

Other useful formulae for 2nd order RLC filters depend if in series or parallel resonant mode.

\$X_C(\omega_o)=\dfrac{1}{\omega_oC}\$ and \$X_L(\omega_o)=\omega_oL\$

Consider a series LP filter.

^{simulate this circuit – Schematic created using CircuitLab}

\$\omega_o=\dfrac{1}{\sqrt{LC}}=2\pi*f\$

Parallel resonant circuits for high Q, Rp >> X(ωo) the impedance of both reactances: are low and equal so the current cycles between these and very little loss current in Rp. (underdamped)

• e.g. L=10uH, C=10uF Rp=100 , thus Q=100 at ωo=1e-5 f=√2*10kHz=14.14kHz

For series resonant circuits: \$Q = \frac{1}{Rs} \sqrt{\frac{L}{C}}\$
\$Q=\dfrac{\omega_o}{BW_{-3dB}}\$ for Bandpass filters (BPF)

More math background on 2nd order systems

Normal for high Q we simplify this to just voltage or current peaking linear gain which you can convert to dB.

Another way is lookup on any "RLC Impedance Nomograph chart" and instantly find the unknown for any of the other knowns for L, C, Q, f, Z(f) using the log-log-log 3D chart like a slide rule. For examples see here.

When Q=ζ=0.707, it is the common Butterworth filter, which also has a critically damped response with no overshoot or "unity" Voltage gain as shown above.

For a more mathematical explanation read here. from U of Carlton.

Useful Q benchmarks and for fun

Q's of < 50 are easy in LC or active filters and >200 hard due to tolerances and thermal drift. However for Q > 10k it is easily possible with any piezo-crystal resonator. (MEMs, tuning fork and AT cuts) But with some Quartz SC-cut the Q => 100k only at one temperature regulated by an oven, (OCXO). Cesium clocks it is even higher, the phase noise is poor so they lock SCt OCXO to the Cesium clock.

This Q factor thus is related to noise BW, jitter, phase shift and the crystal has equivalent RLC paramters with L in the mH to Henries and C in the fempt-farads with ESR [Ohms] and a load capaitance [pF].

64k$ question

What is the Q of an atom? ( hint it is the only lossless thing in nature)

\$Q=\dfrac{1}{2\zeta} = \sqrt{A_{pk}^2 + 1}\$ for peak gain Apk

When Q is the quality factor of resonance gain approaches 1 or less, we switch to damping factor as resonance becomes dampening factor is more relevant.

Rule of Thumb: We approximate high Q to be just the resonant gain for Q>>1.

But in your case, Q is very low, and the peak/flat gain = 1.25

If gain, Apk=1.25 then Q = 1.6 , or ζ = 1/3.2 This is your answer from reading graph.

Other useful formulae for 2nd order RLC filters depend if in series or parallel resonant mode.

\$X_C(\omega_o)=\dfrac{1}{\omega_oC}\$ and \$X_L(\omega_o)=\omega_oL\$

Consider a series LP filter.

^{simulate this circuit – Schematic created using CircuitLab}

\$\omega_o=\dfrac{1}{\sqrt{LC}}=2\pi*f\$

Parallel resonant circuits for high Q, Rp >> X(ωo) the impedance of both reactances: are low and equal so the current cycles between these and very little loss current in Rp. (underdamped)

• e.g. L=10uH, C=10uF Rp=100 , thus Q=100 at ωo=1e-5 f=√2*10kHz=14.14kHz

For series resonant circuits: \$Q = \frac{1}{Rs} \sqrt{\frac{L}{C}}\$
\$Q=\dfrac{\omega_o}{BW_{-3dB}}\$ for Bandpass filters (BPF)

More math background on 2nd order systems

Normal for high Q we simplify this to just voltage or current peaking linear gain which you can convert to dB.

Another way is lookup on any "RLC Impedance Nomograph chart" and instantly find the unknown for any of the other knowns for L, C, Q, f, Z(f) using the log-log-log 3D chart like a slide rule. For examples see here.

When Q=ζ=0.707, it is the common Butterworth filter, which also has a critically damped response with no overshoot or "unity" Voltage gain as shown above.

For a more mathematical explanation read here. from U of Carlton.

Useful Q benchmarks and for fun

Q's of < 50 are easy in LC or active filters and >200 hard due to tolerances and thermal drift. However for Q > 10k it is easily possible with any piezo-crystal resonator. (MEMs, tuning fork and AT cuts) But with some Quartz SC-cut the Q => 100k only at one temperature regulated by an oven, (OCXO). Cesium clocks it is even higher, the phase noise is poor so they lock SCt OCXO to the Cesium clock.

This Q factor thus is related to noise BW, jitter, phase shift and the crystal has equivalent RLC paramters with L in the mH to Henries and C in the fempt-farads with ESR [Ohms] and a load capaitance [pF].

64k$ question

What is the Q of an atom? ( hint it is the only lossless thing in nature)

\$Q=\dfrac{1}{2\zeta} = \sqrt{A_{pk}^2 + 1}\$ for peak gain Apk
Normal for high Q we simplify this to just voltage or current peaking linear gain which you can convert to dB.

But in your case the gain is 1.25 thus very low Q.

If Apk=1.25 Q = 1.6 , ζ = 1/3.2

Other useful formulae are; \$\zeta = \frac{R_s}{2} \sqrt{\frac{C}{L}}\$ and \$Q = R_s \sqrt{\frac{C}{L}}\$

More background info

Normal for high Q we simplify this to just voltage or current peaking linear gain which you can convert to dB.

Another way is lookup on any "RLC Impedance Nomograph chart" and instantly find the unknown for any of the other knowns for L, C, Q, f, Z(f) using the log-log-log 3D chart like a slide rule. For examples see here.

When Q=ζ=0.707, it is the common Butterworth filter, which also has a critically damped response with no overshoot or "unity" Voltage gain as shown above.

Useful Q benchmarks and for fun

Q's of < 50 are easy in LC or active filters and >200 hard due to tolerances and thermal drift. However for Q > 10k it is easily possible with any piezo-crystal resonator. (MEMs, tuning fork and AT cuts) But with some Quartz SC-cut the Q => 100k only at one temperature regulated by an oven, (OCXO). Cesium clocks it is even higher, the phase noise is poor so they lock SCt OCXO to the Cesium clock.

This Q factor thus is related to noise BW, jitter, phase shift and the crystal has equivalent RLC paramters with L in the mH to Henries and C in the fempt-farads with ESR [Ohms] and a load capaitance [pF].

64k$ question

What is the Q of an atom? ( hint it is the only lossless thing in nature)

\$Q=\dfrac{1}{2\zeta} = \sqrt{A_{pk}^2 + 1}\$ for peak gain Apk

When Q is the quality factor of resonance gain approaches 1 or less, we switch to damping factor as resonance becomes dampening factor is more relevant.

Rule of Thumb: We approximate high Q to be just the resonant gain for Q>>1.

But in your case, Q is very low, and the peak/flat gain = 1.25

If gain, Apk=1.25 then Q = 1.6 , or ζ = 1/3.2 This is your answer from reading graph.

Other useful formulae for 2nd order RLC filters depend if in series or parallel resonant mode.

\$X_C(\omega_o)=\dfrac{1}{\omega_oC}\$ and \$X_L(\omega_o)=\omega_oL\$

Consider a series LP filter.

^{simulate this circuit – Schematic created using CircuitLab}

\$\omega_o=\dfrac{1}{\sqrt{LC}}=2\pi*f\$

Parallel resonant circuits for high Q, Rp >> X(ωo) the impedance of both reactances: are low and equal so the current cycles between these and very little loss current in Rp. (underdamped)

• e.g. L=10uH, C=10uF Rp=100 , thus Q=100 at ωo=1e-5 f=√2*10kHz=14.14kHz

For series resonant circuits: \$Q = \frac{1}{Rs} \sqrt{\frac{L}{C}}\$
\$Q=\dfrac{\omega_o}{BW_{-3dB}}\$ for Bandpass filters (BPF)

More math background on 2nd order systems

Normal for high Q we simplify this to just voltage or current peaking linear gain which you can convert to dB.

Another way is lookup on any "RLC Impedance Nomograph chart" and instantly find the unknown for any of the other knowns for L, C, Q, f, Z(f) using the log-log-log 3D chart like a slide rule. For examples see here.

When Q=ζ=0.707, it is the common Butterworth filter, which also has a critically damped response with no overshoot or "unity" Voltage gain as shown above.

For a more mathematical explanation read here. from U of Carlton.

Useful Q benchmarks and for fun

Q's of < 50 are easy in LC or active filters and >200 hard due to tolerances and thermal drift. However for Q > 10k it is easily possible with any piezo-crystal resonator. (MEMs, tuning fork and AT cuts) But with some Quartz SC-cut the Q => 100k only at one temperature regulated by an oven, (OCXO). Cesium clocks it is even higher, the phase noise is poor so they lock SCt OCXO to the Cesium clock.

This Q factor thus is related to noise BW, jitter, phase shift and the crystal has equivalent RLC paramters with L in the mH to Henries and C in the fempt-farads with ESR [Ohms] and a load capaitance [pF].

64k$ question

What is the Q of an atom? ( hint it is the only lossless thing in nature)

\$Q=\dfrac{1}{2\zeta} = \sqrt{A_{pk}^2 + 1}\$ for peak gain Apk
Normal for high Q we simplify this to just voltage or current peaking linear gain which you can convert to dB.

But in your case the gain is 1.25 thus very low Q.

If Apk=1.25 Q = 1.6 , ζ = 1/3.2

Normal for high Q we simplify this to just voltage or current peaking linear gain which you can convert to dB.

Another way is lookup on any "RLC Impedance Nomograph chart" and instantly find the unknown for any of the other knowns for L, C, Q, f, Z(f) using the log-log-log 3D chart like a slide rule. For examples see here.

When Q=ζ=0.707, it is the common Butterworth filter, which also has a critically damped response with no overshoot or "unity" Voltage gain as shown above.

For fun

Q's of < 50 are easy in LC or active filters and >200 hard due to tolerances and thermal drift and > 10k possible with piezo-crystal resonators and with some Quartz SC-cut Q> 100k only at one temperature regulated by an oven, (OCXO). This Q factor thus is related to noise BW, jitter, phase shift and the crystal has equivalent RLC paramters with L in the mH to Henries and C in the fempt-farads with ESR [Ohms] and a load capaitance [pF].

64k$ question

What is the Q of an atom? ( hint it is the only lossless thing in nature)

\$Q=\dfrac{1}{2\zeta} = \sqrt{A_{pk}^2 + 1}\$ for peak gain Apk
Normal for high Q we simplify this to just voltage or current peaking linear gain which you can convert to dB.

But in your case the gain is 1.25 thus very low Q.

If Apk=1.25 Q = 1.6 , ζ = 1/3.2

Other useful formulae are; \$\zeta = \frac{R_s}{2} \sqrt{\frac{C}{L}}\$ and \$Q = R_s \sqrt{\frac{C}{L}}\$

More background info

Normal for high Q we simplify this to just voltage or current peaking linear gain which you can convert to dB.

Another way is lookup on any "RLC Impedance Nomograph chart" and instantly find the unknown for any of the other knowns for L, C, Q, f, Z(f) using the log-log-log 3D chart like a slide rule. For examples see here.

When Q=ζ=0.707, it is the common Butterworth filter, which also has a critically damped response with no overshoot or "unity" Voltage gain as shown above.

Useful Q benchmarks and for fun

Q's of < 50 are easy in LC or active filters and >200 hard due to tolerances and thermal drift. However for Q > 10k it is easily possible with any piezo-crystal resonator. (MEMs, tuning fork and AT cuts) But with some Quartz SC-cut the Q => 100k only at one temperature regulated by an oven, (OCXO). Cesium clocks it is even higher, the phase noise is poor so they lock SCt OCXO to the Cesium clock. 

This Q factor thus is related to noise BW, jitter, phase shift and the crystal has equivalent RLC paramters with L in the mH to Henries and C in the fempt-farads with ESR [Ohms] and a load capaitance [pF].

64k$ question

What is the Q of an atom? ( hint it is the only lossless thing in nature)