Invert mapping from physical to coupled resonator parameters

Question

If we couple two LC resonators (named "a" and "b") through a mutual inductance $M$, Kirchhoff's laws take the following form \begin{align} C_a \ddot{V}_a + \frac{V_a}{L_a'} - \frac{V_b}{M'} &= 0 \\ C_b \ddot{V}_b + \frac{V_b}{L_b'} - \frac{V_a}{M'} &= 0 \end{align} where \begin{align} L_a' &= L_a \left( 1 - \frac{M^2}{L_a L_b} \right) \\ L_b' &= L_b \left( 1 - \frac{M^2}{L_a L_b} \right) \\ \frac{1}{M'} &= \frac{M}{L_a L_b - M^2} \tag{$\star$} \, . \end{align} The primed quantities are nice because they're what we need so that Kirchhoff's laws take this nice common form for the differential equations of coupled resonators. In particular, the coupling $g$ between the resonances (and therefore the frequency splitting) is written most easily in terms of those variables $$g = \frac{1}{2} \frac{\sqrt{Z_a' Z_b'}}{M'}$$ where $Z_i' \equiv \sqrt{L_i' / C_i}$ are the impedances of each resonator. The frequencies $\omega_i' \equiv 1 / \sqrt{L_i' C_i}$ are also useful for other reasons.

Now suppose we want to design a circuit that has a certain values of $g$, $\omega_i'$ and $Z_i'$. We have to invert all of our formulas, in particular Equations ($\star$) for $L_i'$ and $M'$. How do we do this?

Answer 1

The equations ($\star$) can be inverted in the following way. Write Eq. ($\star$) as: $$ \begin{bmatrix} \frac{1}{L_a'} & \frac{1}{M'} \\ \frac{1}{M'} & \frac{1}{L_b'} \end{bmatrix} = \begin{bmatrix} L_a & -M \\ -M & L_b \end{bmatrix} ^{-1} $$

Inverting both sides, we get: $$ \begin{align} L_a &= L_a'\frac{M'^2}{M'^2-L_a'L_b'} \\ L_b &= L_b'\frac{M'^2}{M'^2-L_a'L_b'} \\ M &= M'\frac{L_a'L_b'}{M'^2-L_a'L_b'} \end{align} $$ Note that $$\frac{M'^2}{L_a'L_b'}=\frac{L_aL_b}{M^2}$$ from Eq.($\star$). Since in a realizable circuit we must have $M^2<L_aL_b$, we then also have $M'^2>L_a'L_b'$ so the quantities above are all positive as should be.

Answer 2

If I understand the question correctly, you want to determine the original parameters of the two LC circuits based on the desired frequencies, impedances and coupling parameter.

Using $Z'_i$ and $\omega_i'$ we can arrive determine $L_i'$ and $C_i$. $$ L_i'=\frac{Z_i'}{\omega_i'},\,\, C_i=\frac{1}{Z_i' \omega_i'}$$

Next you can determine $M'$ as $M'=\frac{1}{2}\frac{\sqrt{Z_a'Z_b'}}{g}$.

You now have $L_a',L_b', M'$, for which you have three equations involving three unknowns $L_a, L_b, M$ so by some algebraic manipulation you can easily recover the original circuit parameters.