Editing Laplace transform (section)

=== Impulse response ===
Consider a linear time-invariant system with [[transfer function]]
<math display=block>H(s) = \frac{1}{(s + \alpha)(s + \beta)}.</math>

The [[impulse response]] is simply the inverse Laplace transform of this transfer function:
<math display=block>h(t) = \mathcal{L}^{-1}\{H(s)\}.</math>

;Partial fraction expansion
<!-- [[Partial fractions in Laplace transforms]] redirect here -->
To evaluate this inverse transform, we begin by expanding {{math|''H''(''s'')}} using the method of partial fraction expansion,
<math display=block>\frac{1}{(s + \alpha)(s + \beta)} = { P \over s + \alpha } + { R \over s+\beta }.</math>

The unknown constants {{math|''P''}} and {{math|''R''}} are the [[residue (complex analysis)|residues]] located at the corresponding poles of the transfer function. Each residue represents the relative contribution of that [[mathematical singularity|singularity]] to the transfer function's overall shape.

By the [[residue theorem]], the inverse Laplace transform depends only upon the poles and their residues. To find the residue {{math|''P''}}, we multiply both sides of the equation by {{math|''s'' + ''α''}} to get
<math display=block>\frac{1}{s + \beta} = P  + { R (s + \alpha) \over s + \beta }.</math>

Then by letting {{math|1=''s'' = −''α''}}, the contribution from {{math|''R''}} vanishes and all that is left is
<math display=block>P = \left.{1 \over s+\beta}\right|_{s=-\alpha} = {1 \over \beta - \alpha}.</math>

Similarly, the residue {{math|''R''}} is given by
<math display=block>R = \left.{1 \over s + \alpha}\right|_{s=-\beta} = {1 \over \alpha - \beta}.</math>

Note that
<math display=block>R = {-1 \over \beta - \alpha} = - P</math>
and so the substitution of {{math|''R''}} and {{math|''P''}} into the expanded expression for {{math|''H''(''s'')}} gives
<math display=block>H(s)  = \left(\frac{1}{\beta - \alpha} \right) \cdot \left( { 1 \over s + \alpha } - { 1  \over s + \beta }  \right).</math>

Finally, using the linearity property and the known transform for exponential decay (see ''Item'' #''3'' in the ''Table of Laplace Transforms'', above), we can take the inverse Laplace transform of {{math|''H''(''s'')}} to obtain
<math display=block>h(t) = \mathcal{L}^{-1}\{H(s)\} = \frac{1}{\beta - \alpha}\left(e^{-\alpha t} - e^{-\beta t}\right),</math>
which is the impulse response of the system.

;Convolution
The same result can be achieved using the [[Convolution theorem|convolution property]] as if the system is a series of filters with transfer functions {{math|1/(''s'' + ''α'')}} and {{math|1/(''s'' + ''β'')}}. That is, the inverse of
<math display=block>H(s) = \frac{1}{(s + \alpha)(s + \beta)} = \frac{1}{s+\alpha} \cdot \frac{1}{s + \beta}</math>
is
<math display=block> \mathcal{L}^{-1}\! \left\{ \frac{1}{s + \alpha} \right\} * \mathcal{L}^{-1}\! \left\{\frac{1}{s + \beta} \right\} = e^{-\alpha t} * e^{-\beta t} = \int_0^t e^{-\alpha x}e^{-\beta (t - x)}\, dx = \frac{e^{-\alpha t}-e^{-\beta t}}{\beta - \alpha}.</math>