Finding the schematic of a black box with unknown impedances
I can give you a possible approach. For any two-port containing passive elements you can measure its admittance matrix Y, at a given frequency.







simulate this circuit &ndash; Schematic created using CircuitLab

This matrix Y is such that:
$$\begin{bmatrix} i_1 \\ 
i_2 \end{bmatrix}=
\begin{bmatrix}
Y_{11}       &amp; Y_{12}  \\
Y_{21}       &amp; Y_{22}
\end{bmatrix} \cdot \begin{bmatrix} V_1 \\ V_2 \end{bmatrix}$$

Actually \$Y_{12}=Y_{21}\$ for two-port containing linear passive elements, and therefore you only need to perform 3 measurements to your "black box". You would then have 3 complex numbers.

Then you need to calculate the admittance matrix of your "target" two-port:



Its admittance matrix is rather easy to find out in terms of the admittances of the components, and you'll end up with:
$$\begin{bmatrix}
Y_{1}+Y_{2}       &amp; -Y_{2}  \\
-Y_{2}       &amp; Y_{3}+Y_{2}
\end{bmatrix}$$
where \$Y_{1}\$, \$Y_{2}\$ and \$Y_3\$ are the admittances of the components \$Y_1=\frac 1 {Z_1}\$, \$Y_2=\frac 1 {Z_2}\$, etc.

You now simply match the coefficients of this matrix with the values measured from the "black box" and solve for \$Y_{1}\$, \$Y_{2}\$ and \$Y_{3}\$. Finally \$Z_1\$, \$Z_2\$ and \$Z_3\$ are obtained doing the inverse.

The admittance matrix was chosen because of your particular "target" two port topology.