The T1 matrix of the device under test is found as:
\begin{equation} T_{1}=\left[\begin{matrix}\frac{L_{s} s + R_{s}}{L_{r}^{0.5} L_{s}^{0.5} k_{c} s} & \frac{- L_{r} L_{s} k_{c}^{2} s^{2} + L_{r} L_{s} s^{2} + L_{r} R_{s} s + L_{s} R_{r} s + R_{r} R_{s}}{L_{r}^{0.5} L_{s}^{0.5} k_{c} s}\\\frac{1}{L_{r}^{0.5} L_{s}^{0.5} k_{c} s} & \frac{L_{r} s + R_{r}}{L_{r}^{0.5} L_{s}^{0.5} k_{c} s}\end{matrix}\right] \end{equation}The matrix equation for the two-port (DUT) is found as:
$$\left[\begin{matrix}V_{i}\\I_{i}\end{matrix}\right]=\left[\begin{matrix}\frac{L_{s} s + R_{s}}{\sqrt{L_{r}} \sqrt{L_{s}} k_{c} s} & \frac{- L_{r} L_{s} k_{c}^{2} s^{2} + L_{r} L_{s} s^{2} + L_{r} R_{s} s + L_{s} R_{r} s + R_{r} R_{s}}{\sqrt{L_{r}} \sqrt{L_{s}} k_{c} s}\\\frac{1}{\sqrt{L_{r}} \sqrt{L_{s}} k_{c} s} & \frac{L_{r} s + R_{r}}{\sqrt{L_{r}} \sqrt{L_{s}} k_{c} s}\end{matrix}\right]\cdot\left[\begin{matrix}V_{o}\\I_{o}\end{matrix}\right]$$
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Last project update: 2024-03-04 22:11:05