Laplace transform



         


In mathematics and in particular, in functional analysis, the Laplace transform of a function f(t) defined for all real numbers t ≥ 0 is the function F(s), defined by:

<math>F(s)
= \left\{\mathcal{L} f\right\}(s) =\int_0^\infty e^{-st} f(t)\,dt.<math>

This integral transform has a number of properties that make it useful for analysing linear dynamic systems. The most significant advantage is that integration and differentiation become multiplication and division. (This is similar to the way that logarithms change multiplication of numbers to addition.) This changes integral equations and differential equations to polynomial equations, which are much easier to solve. The inverse is the Bromwich integral, which is a complex integral.

Also, the output of a linear dynamic 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, which often makes matters easier. For more information, see control theory.

The Laplace transform is named in honor of Pierre-Simon Laplace.

A sometimes convenient abuse of notation, prevailing especially among engineers and physicists, writes this in the following form:

<math>F(s)
= \mathcal{L} \left\{f(t)\right\} =\int_0^\infty e^{-st} f(t)\,dt.<math>

When one talks about the Laplace transform, one is generally referring to the unilateral version. There also exists a bilateral Laplace transform, which is defined as follows:

<math>F_B(s)
= \left\{\mathcal{L} f\right\}(s) =\int_{-\infty}^{\infty} e^{-st} f(t)\,dt.<math>

The Laplace transform F(s) typically exists for all real numbers s > a, where a is a constant which depends on the growth behavior of f(t).

The Laplace transform can also be used to solve differential equations and is used extensively in electrical engineering.

An interesting aspect of Laplace transforms is that mathematicians to this day do not know its domain. In other words, there is no specific set of rules that one can check a function against to know if its Laplace transform can be taken.

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Properties

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Linearity

<math>\mathcal{L}\left\{a f(t) + b g(t) \right\}
= a \mathcal{L}\left\{ f(t) \right\} + b \mathcal{L}\left\{ g(t) \right\}<math>
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nth power

<math>\mathcal{L}\{\,t^n\} = \frac {n!}{s^{n+1}}<math>
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Exponential

<math>\mathcal{L}\{\,e^{-at}\} = \frac {1}{s+a}<math>
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Sine

<math>\mathcal{L}\{\,\sin(bt)\} = \frac {b}{s^2 + b^2}<math>
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Cosine

<math>\mathcal{L}\{\,\cos(bt)\} = \frac {s}{s^2 + b^2}<math>
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Hyperbolic sine

<math>\mathcal{L}\{\,\sinh(bt)\} = \frac {b}{s^2-b^2}<math>
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Hyperbolic cosine

<math>\mathcal{L}\{\,\cosh(bt)\} = \frac {s}{s^2 - b^2}<math>
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Natural logarithm

<math>\mathcal{L}\{\,\ln(t)\} = - \frac{\ln(s)+\gamma}{s}<math>
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nth root

<math>\mathcal{L}\{\,\sqrt[n]{t}\} = s^{-\frac{n+1}{n}} \cdot \Gamma\left(1+\frac{1}{n}\right)<math>
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Bessel function of the first kind

<math>\mathcal{L}\{\,J_n(t)\} = \frac{\left(s+\sqrt{1+s^2}\right)^{-n}}{\sqrt{1+s^2}}<math>
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Modified Bessel function of the first kind

<math>\mathcal{L}\{\,I_n(t)\} = \frac{\left(s+\sqrt{-1+s^2}\right)^{-n}}{\sqrt{-1+s^2}}<math>
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Error function

<math>\mathcal{L}\{\,\operatorname{erf}(t)\} = {e^{s^2/4} \operatorname{erfc} \left(s/2\right) \over s}<math>
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Differentiation

<math>\mathcal{L}\{f'\}
= s \mathcal{L}\{f\} - f(0)<math>
<math>\mathcal{L}\{f''\}
= s^2 \mathcal{L}\{f\} - s f(0) - f'(0)<math>
<math>\mathcal{L}\left\{ f^{(n)} \right\}
= s^n \mathcal{L}\{f\} - s^{n - 1} f(0) - \cdots - f^{(n - 1)}(0)<math>
<math>\mathcal{L}\{ t f(t)\}
= -F'(s)<math>
<math>\mathcal{L}\left\{ \frac{f(t)}{t} \right\} = \int_s^\infty F(\sigma)\, d\sigma<math>
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Integration

<math>\mathcal{L}\left\{ \int_0^t f(\tau)\, d\tau \right\}
= {1 \over s} \mathcal{L}\{f\}<math>
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s shifting

<math>\mathcal{L}\left\{ e^{at} f(t) \right\}
= F(s - a)<math>
<math>\mathcal{L}^{-1} \left\{ F(s - a) \right\}
= e^{at} f(t)<math>
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t shifting

<math>\mathcal{L}\left\{ f(t - a) u(t - a) \right\}
= e^{-as} F(s)<math>
<math>\mathcal{L}^{-1} \left\{ e^{-as} F(s) \right\}
= f(t - a) u(t - a)<math>

Note: <math>u(t)<math> is the step function.

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nth-power shifting

<math>\mathcal{L}\{\,t^nf(t)\} = (-1)^nD_s^n[F(s)]<math>
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Convolution

<math>\mathcal{L}\{f * g\}
= \mathcal{L}\{ f \} \mathcal{L}\{ g \}<math>
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Laplace transform of a function with period p

<math>\mathcal{L}\{ f \}
= {1 \over 1 - e^{-ps}} \int_0^p e^{-st} f(t)\,dt<math>






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