It is indeed derived from Kirchoff's voltage laws (after converting the capacitors and inductors into impedances -> jL*omega and -j/omega*C). Specifically, this is what I did: I noted that there are 2 loops I can consider - one on the left and one on the right. By calling the current in the left loop \(I_1\) (clockwise) and the current in the right loop \(I_2\) (counter clockwise), I found a pair of coupled equations for the two currents. I can represent this as a matrix (to be precise, a Z-paramter matrix). I know that the transfer function that I want is V_out/Vs, and V_out = -R_L*I_2, based on how I created the current loops. So using that matrix, I inverted it (to create a Y-parameter matrix). Because part of inverting a 2x2 matrix is to divide a re-arranged matrix by the determinant of the original (a11*a22 - a12*a21), I realized that the poles would be given by the determinant of the Z-parameter matrix:

\[ \Delta_Z\equiv\det(Z)=z_{11}z_{22}-z_{12}z_{21}=\left(R + j\omega L- \frac{j}{\omega C}\right)\left(2R + R_L + j\omega L- \frac{j}{\omega C}\right) + \omega^2M^2 = 0 \]

I expanded that (a lot of arithmetic), and then made my assumptions with the resulting 4th-order polynomial (keeping in mind that any denominator of the determinant would be put into the numerator of the transfer function, contributing to the zeros, and ultimately would "cancel out" most of the fractions that would appear in the transfer function's numerator). One assumption was, basically, to divide the entire thing by R_L and replace every R/R_L with zero (because R << R_L). The other assumption was that the imaginary part of omega would be small enough that I could split the polynomial into two polynomials (even-ordered one given by things not multiplied by j and an odd-ordered on given by things that are multiplied by j). This approximation is not a mathematically-rigorous one, but "happens to" work in this case, because your resistances are low.

As it's for a paper, I would challenge you to see if you can get to the same result, based on my (hopefully not-too-vague) explanation, so I won't put all of my work down here. It's tedious, but it's doable. Start with Kirchoff's voltage laws, and keep in mind the (impedance form) of the mutual inductor (in your case) is V1 = jomega*L*I1 + jomega*M*I2, V2 = jomega*L*I2 + jomega*M*I1, find the Z-parameter matrix, then re-derivce the determinant of it, and then solve that. It's not "hard" to solve, just very tedious.

As a warning: as I mentioned, the final result I mentioned is the result of some approximations made based on the setup (low R, not-too-high R_L, etc). The actual complex value you would get for omega if you were to solve the 4th-order polynomial will be slightly different than the one you get from the approximation, but, for this case, they're in close-enough agreement.

Hope this helps, and good luck!

Hello, thank you so much for this description. I wanted to ask further on why, when looking for the poles we ignore the imaginary part and solve only for the real part of the denominator. I know that this makes it convenient by eliminating dependance on R_L, but I was wondering why this is okay to do.

Mod: link to old thread
https://forum.allaboutcircuits.com/...alters-frequency-response.176817/post-1601938

 

Wow, it's been a long while since I wrote that; I had to go back and remember what I was even doing there.
I don't 100% remember why I took that approximation. It's either that 1: I noted mentally that we're measuring real current, so I only cared about, for a quick calculation, the real part of the determinant, or 2: I assumed that the imaginary part of the transfer function would be sufficiently small enough (given that we're measuring across a resistor) that I could ignore it in the calculation of the zeros/poles.

Basically, it's a very rough assumption that the imaginary part has minimal effect on the overall behavior, and it just happens to work for this particular case. A more rigorous analysis would actually be to find the full transfer function, look at the full equation for the poles, and solve that quartic expression. Alternatively, you might be able to do the same calculation on the magnitude of the determinant and find the extrema.

Tl;dr: there were some assumptions made about the magnitude of the imaginary part that I got really lucky with.