Step response of RC filter

The step response $s(t)$ of the RC low-pass filter is its output when $x(t) = u(t)$ and the capacitor is initially uncharged.

        R
 x(t) o--/\/\/--+--------o
  +             |        +
               === C    y(t)
  -             |        -
 ---o-----------+--------o

Impulse response

By KVL, with $i(t) = C\,\dfrac{dy}{dt}$:

$$x(t) = RC\frac{dy(t)}{dt} + y(t)$$

Taking the Fourier transform gives $H(j\omega) = \dfrac{1}{1 + j\omega RC}$, whose inverse transform (using $e^{-at}u(t) \leftrightarrow \frac{1}{a + j\omega}$) is

$$h(t) = \frac{1}{RC}e^{-t/RC}u(t)$$

Step response

The step response is the running integral of the impulse response:

$$\begin{aligned} s(t) &= u(t) * h(t) = \int_{-\infty}^{t} h(\tau)\,d\tau \\ &= \int_{0}^{t} \frac{1}{RC}e^{-\tau/RC}\,d\tau \quad (t > 0) \\ &= \frac{1}{RC}\left[-RC\,e^{-\tau/RC}\right]_0^t \\ &= \left(1 - e^{-t/RC}\right)u(t) \end{aligned}$$

(Check: solving $RC\,\dot y + y = 1$ with $y(0) = 0$ gives the same.)

s(t)
 1 |- - - - - - - - - - - - -
   |           _______-----
.632|- - - -.-'
   |      .'
   |    /
   |  /
   +-/-----+---------------> t
   0      RC

Discussion

• $\tau = RC$ is the time constant: $s(RC) = 1 - e^{-1} = 0.632$.
• $s(t)$ reaches about 98% at $4RC$ and 99.3% at $5RC$.
• Rise time (10% to 90%) $= RC\ln 9 \approx 2.2RC$. Since bandwidth $\omega_c = 1/RC$, rise time $\approx 2.2/\omega_c$: a wider bandwidth gives a faster rise.
• There is no overshoot (unlike the ideal LPF), and the response is causal because $h(t) = 0$ for $t < 0$.