Symbolic Laplace Transform

Circuit diagram

The Laplace Transform of the transfer $\frac{V_{s} c o p e}{V_{s}}$ is:

\begin{matrix} (1) & \frac{V_{s c o p e}}{V_{s}} = \frac{R_{b} (C_{a} R_{a} s + 1)}{(R_{a} + R_{b} + R_{s}) (\frac{C_{a} C_{b} L_{g} R_{a} R_{b} s^{3}}{R_{a} + R_{b} + R_{s}} + \frac{s^{2} (C_{a} C_{b} R_{a} R_{b} R_{s} + C_{a} L_{g} R_{a} + C_{b} L_{g} R_{b})}{R_{a} + R_{b} + R_{s}} + \frac{s (C_{a} R_{a} R_{b} + C_{a} R_{a} R_{s} + C_{b} R_{a} R_{b} + C_{b} R_{b} R_{s} + L_{g})}{R_{a} + R_{b} + R_{s}} + 1)} \end{matrix}

If the probe is correctly calibrated, we substitute:

\begin{matrix} (2) & C_{b} = \frac{C_{a} R_{a}}{R_{b}} \end{matrix}

The expression of the transfer then simplifies to:

\begin{matrix} (3) & \frac{V_{s c o p e}}{V_{s}} = \frac{R_{b} (C_{a} R_{a} s + 1)}{(R_{a} + R_{b} + R_{s}) (\frac{C_{a}^{2} L_{g} R_{a}^{2} s^{3}}{R_{a} + R_{b} + R_{s}} + \frac{s^{2} (C_{a}^{2} R_{a}^{2} R_{s} + 2 C_{a} L_{g} R_{a})}{R_{a} + R_{b} + R_{s}} + \frac{s (C_{a} R_{a}^{2} + C_{a} R_{a} R_{b} + 2 C_{a} R_{a} R_{s} + L_{g})}{R_{a} + R_{b} + R_{s}} + 1)} \end{matrix}

If we factorize this result we obtain:

\begin{matrix} (4) & \frac{V_{s c o p e}}{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{matrix}

If we normalize this result, we obtain:

\begin{matrix} (5) & \frac{V_{s c o p e}}{V_{s}} = \frac{R_{b}}{(R_{a} + R_{b} + R_{s}) (\frac{C_{a} L_{g} R_{a} s^{2}}{R_{a} + R_{b} + R_{s}} + \frac{s (C_{a} R_{a} R_{s} + L_{g})}{R_{a} + R_{b} + R_{s}} + 1)} \end{matrix}

The initial value ( $R_{s} = 0, L_{g} = 0$ ) and the final value ( $R_{s} = 0$ ) of the step response, are:

\begin{matrix} (6) & μ (0) = \frac{C_{a}}{C_{a} + C_{b}} \end{matrix}

\begin{matrix} (7) & μ (\infty) = \frac{R_{b}}{R_{a} + R_{b}} \end{matrix}

The probe is calibrated if the initial value equals the final value:

\begin{matrix} (8) & C_{b} = \frac{C_{a} R_{a}}{R_{b}} \end{matrix}

The relative overshoot $ϵ$ can be expressed in terms of the relative detuning $δ$ of $C_{b}$ :

\begin{matrix} (9) & ϵ = - \frac{9 δ}{9 δ + 10} \end{matrix}

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Last project update: 2025-03-09 21:36:43