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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.,

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

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

$\displaystyle 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^{-t/\tau}\}
&\isdef & \int_0^{\infty}e^{-t/\tau...
...ght\vert _0^\infty\\
&=& \frac{1}{s+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.


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Next: The Continuous-Time Impulse

written by Julius Orion Smith III
Julius Smith's background is in electrical engineering (BS Rice 1975, PhD Stanford 1983). He is presently Professor of Music and Associate Professor (by courtesy) of Electrical Engineering at Stanford's Center for Computer Research in Music and Acoustics (CCRMA), teaching courses and pursuing research related to signal processing applied to music and audio systems. See http://ccrma.stanford.edu/~jos/ for details.


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