#### PlanarMetamaterials

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The derivation is not in Paul's text, but it may be readily found in the MTL literature. I'll outline it here:Checking through equation (7.109) and earlier part of chapter 7 could not locate the proof to derive the modal terminal transformation (which is equivalent to the expression for ABCD) in post #4

Please correct me if I miss anything.

To start with, the terminal-domain ABCD matrix is defined as:

\[

\left[ \begin{array}{c} \vec{V}_T(0) \\ \vec{I}_T(0) \end{array} \right] = \left[ \begin{array}{cc} \left[A_T\right] & \left[B_T\right] \\ \left[C_T\right] & \left[D_T\right] \\ \end{array} \right] \left[ \begin{array}{c} \vec{V}_T(l) \\ \vec{I}_T(l) \end{array} \right]

\]

where \(l\) is the length of the transmission-line section and in general, each of the ABCD sub-matrices are fully populated. Alternative, we can look at the modal domain, where by definition the various modes are decoupled from each other, and therefore their properties may be described with diagonal matrices:

\[

\left[ \begin{array}{c} \vec{V}_M(0) \\ \vec{I}_M(0) \end{array} \right] = \left[ \begin{array}{cc} \left[A_M\right] & \left[B_M\right] \\ \left[C_M\right] & \left[D_M\right] \\ \end{array} \right] \left[ \begin{array}{c} \vec{V}_M(l) \\ \vec{I}_M(l) \end{array} \right]

\]

Now, we know that the voltages and current between domains are related by Paul's equations 7.8:

\[ \vec{V}_T = \left[T_V\right] \vec{V}_M \]

\[ \vec{I}_T = \left[T_I\right] \vec{I}_M \]

Such that we may take the terminal-domain expressions and write:

\[

\left[ \begin{array}{cc} \left[T_V\right] & \left[0\right] \\ \left[0\right] & \left[T_I\right] \\ \end{array} \right] \left[ \begin{array}{c} \vec{V}_M(0) \\ \vec{I}_M(0) \end{array} \right] = \left[ \begin{array}{cc} \left[A_T\right] & \left[B_T\right] \\ \left[C_T\right] & \left[D_T\right] \\ \end{array} \right] \left[ \begin{array}{cc} \left[T_V\right] & \left[0\right] \\ \left[0\right] & \left[T_I\right] \\ \end{array} \right] \left[ \begin{array}{c} \vec{V}_M(l) \\ \vec{I}_M(l) \end{array} \right]

\]

Multiplying by the inverse of the first matrix:

\[

\left[ \begin{array}{c} \vec{V}_M(0) \\ \vec{I}_M(0) \end{array} \right] = \left[ \begin{array}{cc} \left[T_V\right] & \left[0\right] \\ \left[0\right] & \left[T_I\right] \\ \end{array} \right]^{-1} \left[ \begin{array}{cc} \left[A_T\right] & \left[B_T\right] \\ \left[C_T\right] & \left[D_T\right] \\ \end{array} \right] \left[ \begin{array}{cc} \left[T_V\right] & \left[0\right] \\ \left[0\right] & \left[T_I\right] \\ \end{array} \right] \left[ \begin{array}{c} \vec{V}_M(l) \\ \vec{I}_M(l) \end{array} \right]

\]

Then, compare this with the second equation above:

\[

\left[ \begin{array}{cc} \left[A_M\right] & \left[B_M\right] \\ \left[C_M\right] & \left[D_M\right] \\ \end{array} \right] = \left[ \begin{array}{cc} \left[T_V\right] & \left[0\right] \\ \left[0\right] & \left[T_I\right] \\ \end{array} \right]^{-1} \left[ \begin{array}{cc} \left[A_T\right] & \left[B_T\right] \\ \left[C_T\right] & \left[D_T\right] \\ \end{array} \right] \left[ \begin{array}{cc} \left[T_V\right] & \left[0\right] \\ \left[0\right] & \left[T_I\right] \\ \end{array} \right]

\]

We can recast this as:

\[

\left[ \begin{array}{cc} \left[A_T\right] & \left[B_T\right] \\ \left[C_T\right] & \left[D_T\right] \\ \end{array} \right] = \left[ \begin{array}{cc} \left[T_V\right] & \left[0\right] \\ \left[0\right] & \left[T_I\right] \\ \end{array} \right] \left[ \begin{array}{cc} \left[A_M\right] & \left[B_M\right] \\ \left[C_M\right] & \left[D_M\right] \\ \end{array} \right] \left[ \begin{array}{cc} \left[T_V\right] & \left[0\right] \\ \left[0\right] & \left[T_I\right] \\ \end{array} \right]^{-1}

\]

Now, we know from Paul 7.16 that:

\[

\left[ \begin{array}{cc} \left[T_V\right] & \left[0\right] \\ \left[0\right] & \left[T_I\right] \\ \end{array} \right]^{-1} = \left[ \begin{array}{cc} \left[T_I\right] & \left[0\right] \\ \left[0\right] & \left[T_V\right] \\ \end{array} \right]^{T}

\]

And, as such:

\[

\left[ \begin{array}{cc} \left[A_T\right] & \left[B_T\right] \\ \left[C_T\right] & \left[D_T\right] \\ \end{array} \right] = \left[ \begin{array}{cc} \left[T_V\right] & \left[0\right] \\ \left[0\right] & \left[T_I\right] \\ \end{array} \right] \left[ \begin{array}{cc} \left[A_M\right] & \left[B_M\right] \\ \left[C_M\right] & \left[D_M\right] \\ \end{array} \right] \left[ \begin{array}{cc} \left[T_I\right] & \left[0\right] \\ \left[0\right] & \left[T_V\right] \\ \end{array} \right]^{T}

\]

At which point, I must apologize unreservedly for inverting the transformation matrix on the left-hand side in post #4, instead of the one on the right-hand side, which you can verify for yourself is the correct solution.