First-order system impulse and step response

Given: a first-order system with time constant $\tau$:

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

(e.g. an RC low pass circuit with $\tau = RC$). Assume the system is initially at rest.

Impulse response

Take the Fourier transform, using $\frac{d}{dt} \leftrightarrow j\omega$:

$$\begin{aligned} (j\omega\tau + 1)\,Y(\omega) &= X(\omega) \\ H(\omega) = \frac{Y(\omega)}{X(\omega)} &= \frac{1}{1 + j\omega\tau} = \frac{1/\tau}{1/\tau + j\omega} \end{aligned}$$

Using the pair $e^{-at}u(t) \leftrightarrow \frac{1}{a + j\omega}$ with $a = 1/\tau$:

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

Check: for $t > 0$, $\tau h' + h = -\frac{1}{\tau}e^{-t/\tau} + \frac{1}{\tau}e^{-t/\tau} = 0$, and the jump $h(0^+) = 1/\tau$ makes $\tau\,dh/dt$ contain $\delta(t)$, matching the input.

Step response

$$\begin{aligned} s(t) &= \int_{-\infty}^{t} h(\lambda)\,d\lambda = \int_0^{t}\frac{1}{\tau}e^{-\lambda/\tau}d\lambda \\ &= \left[-e^{-\lambda/\tau}\right]_0^{t} \\ &= \left(1 - e^{-t/\tau}\right)u(t) \end{aligned}$$

Features:

• $s(\tau) = 1 - e^{-1} = 0.632$: the output reaches 63.2% of the final value in one time constant.
• It reaches about 98% after $4\tau$; final value is 1.
• Smaller $\tau$ means a faster response (wider bandwidth $1/\tau$ rad/s).

 h(t)                      s(t)
1/tau|\                    1 |      .--------
     | \                0.632|   .-'
     |  `-.                  |  /
     |     `---___           | /
   0 +--------------> t    0 +/-------------> t
     0  tau                  0  tau