Impulse Response

In the same way that the impulse response of a digital filter is given by the inverse z transform of its transfer function, the impulse response of an analog filter is given by the inverse Laplace transform of its transfer function, viz.,

$$h(t) = {\cal L}_t^{-1}\{H(s)\} = \tau e^{-\frac{t}{\tau}} u(t)$$

where $u(t)$ denotes the Heaviside unit step function

$$u(t) \isdef \left\{\begin{array}{ll} 1, & t\geq 0 \\ [5pt] 0, & t<0. \\ \end{array}\right.$$

This result is most easily checked by taking the Laplace transform of an exponential decay with time-constant $\tau>0$:

$$\begin{eqnarray*} {\cal L}_s\{e^{-\frac{t}{\tau}}\} &\isdef & \int_0^{\infty}e^... ... _0^\infty\\ &=& \frac{1}{s+\frac{1}{\tau}} = \frac{RC}{RCs+1}. \end{eqnarray*}$$

In more complicated situations, any rational $H(s)$ (ratio of polynomials in $s$) may be expanded into first-order terms by means of a partial fraction expansion (see §6.8) and each term in the expansion inverted by inspection as above.