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If you have accurate circuit model for your structure, then Q factor can be directly calculated. If your structure is complicated, there is no direct formula for Q, you have to do a full-wave simulation, and calculate Q factor based on its fundamental definition. https://en.wikipedia.org/wiki/Q_factor.
In your microstrip case, you have to calculate the energy stored in the microstrip stub from integration of E & H fields and find the dissipated energy per cycle.
I don't know if any full-wave simulation software can directly do this for you. Any one else help?
if it is a lumped port, then it only calculates Q for that port. For a Q factor of the entire structure, I think integration describes better. It is up to your interest.
Assuming that Q is relatively high (say >10), you can determine Q based on the complex impedance versus frequency curve.
For a circuit that behaves around its resonant frequency as an RLC circuit (for example quarter wave sections), the -3 dB impedance points are at the positions where |Im(Z)|=Re(Z).
When plotted on a smith chart the curve should cross the horizontal centerline vertically. If not, you should rotate the graph (for example by adding some loss free transmission line section).
You may also match your resonator (lossfree components) to a convenient real impedance so that at the center frequency RC = 0. Where |RC| = 0.33, you have the VSWR=2 frequencies. When they are (for examle) 10 MHz apart, the -3 dB frequencies are 10*rt(2) = 14 MHz apart. Now you can calculate Q based on fcenter/(-3dB band width).
Assuming that Q is relatively high (say >10), you can determine Q based on the complex impedance versus frequency curve.
For a circuit that behaves around its resonant frequency as an RLC circuit (for example quarter wave sections), the -3 dB impedance points are at the positions where |Im(Z)|=Re(Z).
When plotted on a smith chart the curve should cross the horizontal centerline vertically. If not, you should rotate the graph (for example by adding some loss free transmission line section).
You may also match your resonator (lossfree components) to a convenient real impedance so that at the center frequency RC = 0. Where |RC| = 0.33, you have the VSWR=2 frequencies. When they are (for examle) 10 MHz apart, the -3 dB frequencies are 10*sqrt(2) = 14 MHz apart. Now you can calculate Q based on fcenter/(-3dB band width).
The 45 degrees is only because it is easy to determine (Re(Z) = |Im(Z)| ). You may also use the resonant impedance and d(im(Z))/df (series circuit behavior) or d(Im(Y))/df (parallel circuit behavior) to determine resonator Q. For these methods you have to make sure that the S11 curve goes vertically through the horizontal center line of the Smith Chart.
RC = reflection coefficient = (Z-Z0)/(Z+Z0). Where Z0 is a real reference value (for example 50 Ohms for telecom or 75 Ohms in video). In simulation you can take whatever value that is convenient for you. The method based on reflection coefficient can be used in all cases, as long as you have RC=0 at the center frequency.
Assuming that Q is relatively high (say >10), you can determine Q based on the complex impedance versus frequency curve.
For a circuit that behaves around its resonant frequency as an RLC circuit (for example quarter wave sections), the -3 dB impedance points are at the positions where |Im(Z)|=Re(Z).
When plotted on a smith chart the curve should cross the horizontal centerline vertically. If not, you should rotate the graph (for example by adding some loss free transmission line section).
I thought adding loss free line section will change the resonator frequency. The Q derived is for the new resonance frequency, rather than at the orginal line section's resonance frequency?
You may also match your resonator (lossfree components) to a convenient real impedance so that at the center frequency RC = 0. Where |RC| = 0.33, you have the VSWR=2 frequencies. When they are (for examle) 10 MHz apart, the -3 dB frequencies are 10*rt(2) = 14 MHz apart. Now you can calculate Q based on fcenter/(-3dB band width).
To get RC=0, the real impedance has to be equal to the external impedance (for example 50 ohm for meas. system). Why sqrt(2) has to be used to multiply the 10MHz?
Thank you very much for your explanation.
You are correct on the addition of a line section (So I am wrong here).
A theorectical LCR series or parallel circuit (assuming relative high Q) has a S11 curve that is symmetrical around the horizontal center line of the Smith chart. When simulating real resonators, you may see that the S11 curve doesn't go vertically across the horizontal center line of the Smith chart. Adding an inductive or capacitive parallel component at the feed can correct this so that you can calculate Q factor on d(Im(Z))/df or based on |Im(Z)| = Re(Z). Of course you can also use admittance. Sorry for the confusion about the addition of some transmission line length.
matched resonators
RC=0.33 is used frequently as this equals SWR=2 (in for example antenna systems). You can take whatever value for change in reflection coefficient, but you need a correction to get the - 3dB impedance points of your resonator.
You may also use |RC|=0.45 (return loss = 7dB, SWR=2.63) to find the -3 dB points directly (then you don't need the 0.71 correction factor). Of course at resonance you need RC=0 (so you need to match your resonator to a convenient impedance). Matching can be done via a small capacitor between feed and high impedance point of the resonator. When Q is very high, an air gap or simple capacitive probe towards the resonator may have sufficient capacitance. You may also connect the feed line on a tap (like a tapped LC circuit).
You may simulate an RLC circuit (with known Q) in your simulator and determine the Q factor based on the S11 or impedance curve. This to know that you are on the right track.
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