Impedance with two-port elements

The two types of energy transducing elements, transformers and gyrators, “reflect” or “transmit” impedance through themselves, such that they are “felt” on the other side.

For a transformer, the elemental equations are \[\begin{aligned} \mathcal{V}_2(t) = \mathcal{V}_1(t)/TF \quad \text{and} \quad \mathcal{F}_2(t) = -TF \mathcal{F}_1(t), \end{aligned}\] the Laplace transforms of which are \[\begin{aligned} \label{eq:transformer_laplace} \mathcal{V}_2(s) = \mathcal{V}_1(s)/TF \quad \text{and} \quad \mathcal{F}_2(s) = -TF \mathcal{F}_1(s). \end{aligned}\]

If, on the \(2\)-side, the input impedance is \(Z_3\), as in , the equations of are subject to the continuity and compatibility equations \[\begin{aligned} \mathcal{V}_2 = \mathcal{V}_3 \quad \text{and} \quad \mathcal{F}_2 = -\mathcal{F}_3. \end{aligned}\] Substituting these into and solving for \(\mathcal{V}_1\) and \(\mathcal{F}_1\), \[\begin{aligned} \mathcal{V}_1 = TF \mathcal{V}_3 \quad \text{and} \quad \mathcal{F}_1 = \mathcal{F}_3/TF. \end{aligned}\] The elemental equation for element \(3\) is \(\mathcal{V}_3 = \mathcal{F}_3 Z_3\), which can be substituted into the through-variable equation to yield \[\begin{aligned} \mathcal{F}_1 = \frac{1} {Z_3 TF} \mathcal{V}_3.\end{aligned}\] Working our way back from \(\mathcal{V}_3\) to \(\mathcal{V}_1\), we apply the compatibility equation \(\mathcal{V}_2 = \mathcal{V}_3\) and the elemental equation \(\mathcal{V}_2 = \mathcal{V}_1/TF\), as follows: \[\begin{aligned} \mathcal{F}_1 &= \frac{1} {Z_3 TF} \mathcal{V}_2 \\ &= \frac{1} {Z_3 TF^2} \mathcal{V}_1.\end{aligned}\] Solving for the effective input impedance \(Z_1\), \[\begin{aligned} Z_1 &\equiv \frac{\mathcal{V}_1(s)}{\mathcal{F}_1(s)} \\ &= TF^2 Z_3. \end{aligned}\]

For a gyrator with gyrator modulus \(GY\), in the configuration shown in , a similar derivation yields the effective input impedance \(Z_1\), \[\begin{aligned} Z_1 &= GY^2/Z_3. \end{aligned}\]