Circuit simplification containing transmission lines, assuming a high quality factor

Question

I am trying to digest the paper of M.S Khalil et al. 2012(1) on asymmetric resonator transmission (in superconducting devices). In it, early on, a resonator circuit coupled to a transmission line is drawn, as in fig.a. Now, assuming a high internal quality factor \$Q_i >> 1\$ where \$R=Q_i/(\omega_0 C)\$, and \$\omega_0=1/\sqrt{LC}\$ being the resonance frequency of the uncoupled resonator circuit, the circuit is redrawn as in fig. b.

I can't seem to figure out this step, neither why assuming a high internal quality factor is necessary, nor how this simplification is made. Further, no additional information is given in the paper. Are there people who understand this simplification and its motivation? Thank you in advance!

(1) M. S. Khalil, M. J. A. Stoutimore, F. C. Wellstood, and K. D. Osborn , "An analysis method for asymmetric resonator transmission applied to superconducting devices", Journal of Applied Physics 111, 054510 (2012) https://doi.org/10.1063/1.3692073

Answer 1

First consider a conventional capacitor with loss. The loss could be due to ohmic loss in the metal, best represented by a small series resistance, or it could be due to slightly conducting dielectric, best represented by a large parallel resistance, or it could be some more complex combination. If considered over a wide range of frequencies (in other words, low Q), these models give very different frequency responses.

However, in the case of a high Q resonator, the shape of the frequency response near resonance is approximately the same for a small series Rs or a large parallel Rp, with the equivalence Rs = L/(Rp*C). You can see this by writing the impedance of the parallel combination of L and C^ for both cases. With a series Rs in the capacitor

Z = jwL(1-w^2LCRs)/(1-w^2LC+jwCRs)

with a parallel Rp in the capacitor

Z = jWL/(1-w^2LC+jwL/Rp)

If the Q is high, w^2LCRs << 1, and then the two equations have the same variation with frequency if L/Rp = CRs.

Therefore the loss, regardless of the actual mechanism or combinations of mechanisms, can be modeled equivalently as a parallel Rp.

That's the reason why high Q is required. Once you have that, the formula for the parallel resistor just comes from the standard formula for the Q of a capacitor with parallel resistance: Q = wCR. The original model (a) in the figure implies the inductors are assumed lossless.