Symbolic time-domain analysis

Circuit diagram

The 1-rst order ( $R_{s} = 0$ ) detuned unit step response is found as:

\begin{matrix} (1) & μ_{t} = \frac{R_{b} (\frac{(C_{a} R_{a}^{2} - C_{b} R_{a} R_{b}) e^{- \frac{t (R_{a} + R_{b})}{C_{a} R_{a} R_{b} + C_{b} R_{a} R_{b}}}}{C_{a} R_{a} R_{b} + C_{b} R_{a} R_{b}} + 1)}{R_{a} + R_{b}} \end{matrix}

The 1-rst order ( $R_{s} \neq 0, C_{b} = C_{a} \frac{R_{a}}{R_{b}}$ ) tuned unit step response is found as:

\begin{matrix} (2) & μ_{t} = \frac{R_{b} (1 - e^{- \frac{t (R_{a} + R_{b} + R_{s})}{C_{a} R_{a} R_{s}}})}{R_{a} + R_{b} + R_{s}} \end{matrix}

The settling error to the final value of a wrongly tuned probe can be written as a function of time:

\begin{matrix} (3) & ϵ_{t} = \frac{(C_{a} R_{a} - C_{b} R_{b}) e^{- \frac{t (R_{a} + R_{b})}{R_{a} R_{b} (C_{a} + C_{b})}}}{(C_{a} + C_{b}) (R_{a} + R_{b})} \end{matrix}

After equating this value with a 1 LSB error, we find the settling time as function of the relative positivedetuning $δ$ of $C_{b}$ :

\begin{matrix} (4) & τ_{s} = 0.9 C_{a} R_{b} (9 δ + 10) \log (\frac{9 \cdot 2^{n} δ}{9 δ + 10}) \end{matrix}

The settling error to the final value of a wrongly tuned probe can be written as a function of time:

\begin{matrix} (5) & ϵ_{t} = \frac{(C_{a} R_{a} - C_{b} R_{b}) e^{- \frac{t (R_{a} + R_{b})}{R_{a} R_{b} (C_{a} + C_{b})}}}{(C_{a} + C_{b}) (R_{a} + R_{b})} \end{matrix}

After equating this value with a 1 LSB error, we find the settling time as function of the relative negativedetuning $δ$ of $C_{b}$ :

\begin{matrix} (6) & τ_{s} = 0.9 C_{a} R_{b} (9 δ + 10) \log (- \frac{9 \cdot 2^{n} δ}{9 δ + 10}) \end{matrix}

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