RC step response

What comes out of an RC lowpass filter when the input is a unit step. A smooth exponential approach to a final value, with time constant $\tau =RC$. One of the canonical first-order responses.

First-order RC charging: $v_C(t)=V(1-e^{-t/\tau })$. At $t=\tau$ the response has reached ~63.2% of the final value.

For an RC lowpass with input $v_{\text{in}}(t)=V_{b}u(t)$ (a battery $V_b$ switched on at $t=0$), the capacitor voltage is

$v_C(t)=V_b(1-e^{-t/RC})u(t),$

and the current through the resistor is

$i(t)=\frac{V_b}{R}{e}^{-t/RC}u(t).$

Before $t=0$ everything is zero; after $t=0$, the capacitor charges asymptotically toward $V_b$, and the current decays exponentially from its initial peak.

Derivation via convolution

The RC lowpass has impulse response $h(t)=(1/RC)e^{-t/RC}u(t)$. The input is $x(t)=u(t)$. The output is the convolution:

$y(t)={\int }_{-\infty }^{\infty }u(\tau )\frac{e^{-(t-\tau )/RC}}{RC}u(t-\tau )d\tau .$

The two step factors restrict the integrand: $u(\tau )$ is zero for $\tau <0$, $u(t-\tau )$ is zero for $\tau >t$. So the integrand is nonzero only for $0<\tau <t$, requiring $t>0$. For $t<0$, $y(t)=0$.

For $t>0$:

$y(t)={\int }_0^t\frac{e^{-(t-\tau )/RC}}{RC}d\tau =\frac{e^{-t/RC}}{RC}{\int }_0^t{e}^{\tau /RC}d\tau =\frac{e^{-t/RC}}{RC}\cdot RC\cdot (e^{t/RC}-1)=1-e^{-t/RC}.$

So $y(t)=(1-e^{-t/RC})u(t)$, scaling by $V_b$ for the battery-charging case.

What it looks like

• At $t=0$: $v_C=0$, $i=V_b/R$ (maximum current, capacitor acts like a wire when uncharged).
• At $t=RC$: $v_C\approx 0.632V_b$, $i\approx 0.368V_b/R$.
• At $t=5RC$: $v_C\approx 0.993V_b$, capacitor is fully charged for practical purposes.

The “five time constants to settle” rule for first-order systems is the $e^{-5}\approx 0.007$ that’s left in the exponential.

A useful pattern

Every convolution of two causal signals (signals with a $u(t)$ factor) reduces to an integral from $0$ to $t$. The lower limit comes from $u(\tau )$, the upper limit from $u(t-\tau )$.

Frequency-domain shortcut

In the s-domain, the same calculation is one line. $X(s)=1/s$ (step) and $H(s)=1/(RCs+1)=(1/RC)/(s+1/RC)$ (RC lowpass transfer function). Output:

$Y(s)=X(s)H(s)=\frac{1}{s(RCs+1)}.$

Partial fractions: $1/[s(RCs+1)]=1/s-RC/(RCs+1)=1/s-1/(s+1/RC)$. Inverse-transforming gives $y(t)=u(t)-e^{-t/RC}u(t)=(1-e^{-t/RC})u(t)$, same answer with much less algebra. That’s the practical reason for developing the Laplace transform.