The Laplace transform is used frequently in
engineering and
physics; the output of a
linear time-invariant system can be calculated by convolving its unit impulse response with the input signal. Performing this calculation in Laplace space turns the convolution into a multiplication; the latter being easier to solve because of its algebraic form. For more information, see
control theory. The Laplace transform is invertible on a large class of functions. Given a mathematical or functional description of an input or output to a
system, the Laplace transform provides an alternative functional description that often simplifies the process of analyzing the behavior of the system, or in synthesizing a new system based on a set of specifications. The Laplace transform can also be used to solve differential equations and is used extensively in
mechanical engineering and
electrical engineering. The Laplace transform reduces a linear differential equation to an algebraic equation, which can then be solved by the formal rules of algebra. The original differential equation can then be solved by applying the inverse Laplace transform. English electrical engineer
Oliver Heaviside first proposed a similar scheme, although without using the Laplace transform; and the resulting operational calculus is credited as the Heaviside calculus.
Evaluating improper integrals Let {{tmath|1= \mathcal{L}\left\{f(t)\right\} = F(s) }}. Then (see the table above) \partial_s\mathcal{L} \left\{\frac{f(t)} t \right\} = \partial_s\int_0^\infty \frac{f(t)}{t}e^{-st}\, dt = -\int_0^\infty f(t)e^{-st}dt = - F(s) From which one gets: \mathcal{L} \left\{\frac{f(t)} t \right\} = \int_s^\infty F(p)\, dp. In the limit , one gets \int_0^\infty \frac{f(t)} t \, dt = \int_0^\infty F(p)\, dp, provided that the interchange of limits can be justified. This is often possible as a consequence of the
final value theorem. Even when the interchange cannot be justified the calculation can be suggestive. For example, with , proceeding formally one has \begin{align} \int_0^\infty \frac{ \cos(at) - \cos(bt) }{t} \, dt &=\int_0^\infty \left(\frac p {p^2 + a^2} - \frac{p}{p^2 + b^2}\right)\, dp \\[6pt] &=\left[ \frac{1}{2} \ln\frac{p^2 + a^2}{p^2 + b^2} \right]_0^\infty = \frac{1}{2} \ln \frac{b^2}{a^2} = \ln \left| \frac {b}{a} \right|. \end{align}
Complex impedance of a capacitor In the theory of
electrical circuits, the current flow in a
capacitor is proportional to the capacitance and rate of change in the electrical potential (with equations as for the
SI unit system). Symbolically, this is expressed by the differential equation i = C { dv \over dt} , where is the capacitance of the capacitor, is the
electric current through the capacitor as a function of time, and is the
voltage across the terminals of the capacitor, also as a function of time. Taking the Laplace transform of this equation, we obtain I(s) = C(s V(s) - V_0), where \begin{align} I(s) &= \mathcal{L} \{ i(t) \},\\ V(s) &= \mathcal{L} \{ v(t) \}, \end{align} and V_0 = v(0). Solving for we have V(s) = { I(s) \over sC } + { V_0 \over s }. The definition of the complex impedance (in
ohms) is the ratio of the complex voltage divided by the complex current while holding the initial state at zero: Z(s) = \left. { V(s) \over I(s) } \right|_{V_0 = 0}. Using this definition and the previous equation, we find: Z(s) = \frac{1}{sC}, which is the correct expression for the complex impedance of a capacitor. In addition, the Laplace transform has large applications in control theory.
Impulse response Consider a linear time-invariant system with
transfer function H(s) = \frac{1}{(s + \alpha)(s + \beta)}. The
impulse response is the inverse Laplace transform of this transfer function: h(t) = \mathcal{L}^{-1}\{H(s)\}. ; Partial fraction expansion : To evaluate this inverse transform, we begin by expanding using the method of partial fraction expansion, \frac{1}{(s + \alpha)(s + \beta)} = { P \over s + \alpha } + { R \over s+\beta }. The unknown constants and are the
residues located at the corresponding poles of the transfer function. Each residue represents the relative contribution of that
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 , we multiply both sides of the equation by to get \frac{1}{s + \beta} = P + { R (s + \alpha) \over s + \beta }. Then by letting , the contribution from vanishes and all that is left is P = \left.{1 \over s+\beta}\right|_{s=-\alpha} = {1 \over \beta - \alpha}. Similarly, the residue is given by R = \left.{1 \over s + \alpha}\right|_{s=-\beta} = {1 \over \alpha - \beta}. Note that R = {-1 \over \beta - \alpha} = - P and so the substitution of and into the expanded expression for gives H(s) = \left(\frac{1}{\beta - \alpha} \right) \cdot \left( { 1 \over s + \alpha } - { 1 \over s + \beta } \right). 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 to obtain h(t) = \mathcal{L}^{-1}\{H(s)\} = \frac{1}{\beta - \alpha}\left(e^{-\alpha t} - e^{-\beta t}\right), which is the impulse response of the system. ; Convolution : The same result can be achieved using the
convolution property as if the system is a series of filters with transfer functions and . That is, the inverse of H(s) = \frac{1}{(s + \alpha)(s + \beta)} = \frac{1}{s+\alpha} \cdot \frac{1}{s + \beta} is \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}.
Phase delay Starting with the Laplace transform, X(s) = \frac{s\sin(\varphi) + \omega \cos(\varphi)}{s^2 + \omega^2} we find the inverse by first rearranging terms in the fraction: \begin{align} X(s) &= \frac{s \sin(\varphi)}{s^2 + \omega^2} + \frac{\omega \cos(\varphi)}{s^2 + \omega^2} \\ &= \sin(\varphi) \left(\frac{s}{s^2 + \omega^2} \right) + \cos(\varphi) \left(\frac{\omega}{s^2 + \omega^2} \right). \end{align} We are now able to take the inverse Laplace transform of our terms: \begin{align} x(t) &= \sin(\varphi) \mathcal{L}^{-1}\left\{\frac{s}{s^2 + \omega^2} \right\} + \cos(\varphi) \mathcal{L}^{-1}\left\{\frac{\omega}{s^2 + \omega^2} \right\} \\ &= \sin(\varphi)\cos(\omega t) + \cos(\varphi)\sin(\omega t). \end{align} This is just the
sine of the sum of the arguments, yielding: x(t) = \sin (\omega t + \varphi). We can apply similar logic to find that \mathcal{L}^{-1} \left\{ \frac{s\cos\varphi - \omega \sin\varphi}{s^2 + \omega^2} \right\} = \cos{(\omega t + \varphi)}.
Statistical mechanics In
statistical mechanics, the Laplace transform of the density of states g(E) defines the
partition function. That is, the canonical partition function Z(\beta) is given by Z(\beta) = \int_0^\infty e^{-\beta E}g(E)\,dE and the inverse is given by g(E) = \frac{1}{2\pi i} \int_{\beta_0-i\infty}^{\beta_0+i\infty} e^{\beta E}Z(\beta) \, d\beta
Spatial (not time) structure from astronomical spectrum The wide and general applicability of the Laplace transform and its inverse is illustrated by an application in astronomy which provides some information on the
spatial distribution of matter of an
astronomical source of
radiofrequency thermal radiation too distant to
resolve as more than a point, given its
flux density spectrum, rather than relating the
time domain with the spectrum (frequency domain). Assuming certain properties of the object, e.g. spherical shape and constant temperature, calculations based on carrying out an inverse Laplace transformation on the spectrum of the object can produce the only possible
model of the distribution of matter in it (density as a function of distance from the center) consistent with the spectrum. When independent information on the structure of an object is available, the inverse Laplace transform method has been found to be in good agreement.
Birth and death processes Consider a
random walk, with steps \{+1,-1\} occurring with probabilities . Suppose also that the time step is a
Poisson process, with parameter . Then the probability of the walk being at the lattice point n at time t is P_n(t) = \int_0^t\lambda e^{-\lambda(t-s)}(pP_{n-1}(s) + qP_{n+1}(s))\,ds\quad (+e^{-\lambda t}\quad\text{when}\ n=0). This leads to a system of
integral equations (or equivalently a system of differential equations). However, because it is a system of convolution equations, the Laplace transform converts it into a
system of linear equations for \pi_n(s) = \mathcal L(P_n)(s), namely: \pi_n(s) = \frac{\lambda}{\lambda+s}(p\pi_{n-1}(s) + q\pi_{n+1}(s))\quad (+\frac1{\lambda + s}\quad \text{when}\ n=0) which may now be solved by standard methods.
Tauberian theory The Laplace transform of the measure \mu on [0,\infty) is given by \mathcal L\mu(s) = \int_0^\infty e^{-st}d\mu(t). It is intuitively clear that, for small , the exponentially decaying integrand will become more sensitive to the concentration of the measure \mu on larger subsets of the domain. To make this more precise, introduce the distribution function: M(t) = \mu([0,t)). Formally, we expect a limit of the following kind: \lim_{s\to 0^+}\mathcal L\mu(s) = \lim_{t\to\infty} M(t).
Tauberian theorems are theorems relating the asymptotics of the Laplace transform, as , to those of the distribution of \mu as . They are thus of importance in asymptotic formulae of
probability and
statistics, where often the spectral side has asymptotics that are simpler to infer. == See also ==