The Laplace Transform of the transfer $\frac{V_scope}{V_s}$ is:
\begin{equation} \frac{V_{scope}}{V_{s}}=\frac{R_{b} \left(C_{a} R_{a} s + 1\right)}{C_{a} C_{b} L_{g} R_{a} R_{b} s^{3} + R_{a} + R_{b} + R_{s} + s^{2} \left(C_{a} C_{b} R_{a} R_{b} R_{s} + C_{a} L_{g} R_{a} + C_{b} L_{g} R_{b}\right) + s \left(L_{g} + R_{a} R_{b} \left(C_{a} + C_{b}\right) + R_{s} \left(C_{a} R_{a} + C_{b} R_{b}\right)\right)} \end{equation}If the probe is correctly calibrated, we substitute:
\begin{equation} C_{b}=\frac{C_{a} R_{a}}{R_{b}} \end{equation}The expression of the transfer then simplifies to:
\begin{equation} \frac{V_{scope}}{V_{s}}=\frac{R_{b} \left(C_{a} R_{a} s + 1\right)}{C_{a}^{2} L_{g} R_{a}^{2} s^{3} + R_{a} + R_{b} + R_{s} + s^{2} \left(C_{a}^{2} R_{a}^{2} R_{s} + 2 C_{a} L_{g} R_{a}\right) + s \left(2 C_{a} R_{a} R_{s} + L_{g} + R_{a} R_{b} \left(\frac{C_{a} R_{a}}{R_{b}} + C_{a}\right)\right)} \end{equation}If we factorize this result we obtain:
\begin{equation} \frac{V_{scope}}{V_{s}}=\frac{R_{b}}{C_{a} L_{g} R_{a} s^{2} + C_{a} R_{a} R_{s} s + L_{g} s + R_{a} + R_{b} + R_{s}} \end{equation}If we normalize this result, we obtain:
\begin{equation} \frac{V_{scope}}{V_{s}}=\frac{R_{b}}{\left(R_{a} + R_{b} + R_{s}\right) \left(\frac{C_{a} L_{g} R_{a} s^{2}}{R_{a} + R_{b} + R_{s}} + \frac{s \left(C_{a} R_{a} R_{s} + L_{g}\right)}{R_{a} + R_{b} + R_{s}} + 1\right)} \end{equation}The initial value ($R_s=0, \, L_g=0$) and the final value ($R_s=0$) of the step response, are:
\begin{equation} \mu{\left(0 \right)}=\frac{C_{a}}{C_{a} + C_{b}} \end{equation} \begin{equation} \mu{\left(\infty \right)}=\frac{R_{b}}{R_{a} + R_{b}} \end{equation}The probe is calibrated if the initial value equals the final value:
\begin{equation} C_{b}=\frac{C_{a} R_{a}}{R_{b}} \end{equation}The relative overshoot $\epsilon$ can be expressed in terms of the relative detuning $\delta$ of $C_b$:
\begin{equation} \epsilon=- \frac{9 \delta}{9 \delta + 10} \end{equation}
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Last project update: 2022-02-16 10:55:49