= RC low-pass step response
{c}
{title2=$v_{\rm out}(t)=V_s+(v_0-V_s)e^{-t/\tau}$}
The <step response> describes a sudden persistent input change. Suppose the input becomes a constant $V_s$ at $t=0$, and the initial <capacitor> voltage is $v_0$. Finite <electric current> leaves that voltage continuous. Subtract $V_s$ from the <RC low-pass filter> equation: the remaining error satisfies $\tau\,de/dt+e=0$. Separating variables gives
$$
v_{\rm out}(t)=V_s+(v_0-V_s)e^{-t/\tau}\qquad(t\geq0).
$$
For an initially uncharged <capacitor> and a $0$ to $3\,\mathrm V$ step in the constructed example,
$$
v_{\rm out}(t)=3\left(1-e^{-t/(103.4\,\mu\mathrm{s})}\right)\,\mathrm V.
$$
At one <time constant>, $t=103.4\,\mu\mathrm{s}$, the voltage is $1.8964\,\mathrm V$, about $63.2\%$ of its final value. The voltage error then continues its <exponential decay>. This is different from a pure delay: the output starts changing immediately, yet takes time to approach the target.
\Image[/rc-low-pass-bode-and-step-response.png]
{title=<RC low-pass frequency response> and <RC low-pass step response> from one <time constant>}
{description=Original analytical curves for $R=2.2\,\mathrm{k}\Omega$ and $C=47\,\mathrm{nF}$; no experimental points. Left: a <Bode plot> of voltage gain and output-minus-input <phase difference>, with <cutoff frequency> $1.5392\,\mathrm{kHz}$. Right: the response to a $0$ to $3\,\mathrm V$ step, compared with doubling $C$ at fixed $R$. Doubling the <time constant> halves the <cutoff frequency> and doubles the time to reach the same fraction of the final voltage.}
Define <settling time> relative to the total voltage change: require $|v_{\rm out}-V_s|\leq\epsilon|v_0-V_s|$ thereafter, with $0<\epsilon<1$. The <exponential decay> is monotone, so
$$
t_\epsilon=\tau\ln\frac1\epsilon,\qquad
f_ct_\epsilon=\frac{\ln(1/\epsilon)}{2\pi}.
$$
For $1\%$ settling, $t_{0.01}=4.6052\tau\approx476.17\,\mu\mathrm{s}$; the $10\%$ to $90\%$ rise takes $\tau\ln9\approx227.19\,\mu\mathrm{s}$. Reducing the <cutoff frequency> by doubling $C$ doubles both times. This reciprocal relation is the useful connection between <signal bandwidth> and response speed: stronger suppression of rapid <noise> comes with slower tracking of wanted changes.
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