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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 dynamical 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.
Engineering/physics notation
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.
Relation to other transforms
Fourier transform
The Continuous Fourier transform is equivalent to evaluating <math>s = i \omega<math>
- <math>F(\omega) = \left. F(s) \right|_{s = i \omega}
=\int_{0^-}^\infty e^{-i \omega t} f(t)\,dt.<math>
This equivalence is usually used to determine the frequency spectrum of a signal or dynamical system.
Note that the <math>\frac{1}{\sqrt{2 \pi}}<math> constant is not included.
Z-transform
The Z-transform equivalence is not as straight forward as for the Fourier transform.
Take a continuous signal, its Laplace transform and its Z-transform and label them:
- Continous signal: <math>f(t)<math>
- Laplace transform: <math>F(s)<math>
- Z-transform: <math>F(z)<math>
Multiply <math>f(t)<math> by a Dirac comb and name the result <math>f^{*}(t)<math>
- <math>f^{*}(t) = f(t) \delta_T(t) = \sum_{n=0}^{\infty} f(n T) \delta(t - n T)<math>
and taking the Laplace transform results in
- <math>F^{*}(s) = \int_{0^-}^{\infty} f^{*}(t) e^{-s t}\,dt. = \int_{0^-}^{\infty} f(t) \delta(t - n T) e^{-s t}\,dt. = \sum_{n=0}^{\infty} f(n T) e^{-n T s}<math>
Then the equivalence can be stated:
- <math>F(z) = \left. F^{*}(s) \right|_{s = z}<math>
This equation relates the sampled values of a continuous signal to the discrete sequence resulting from the Z-transform.
Properties and Theorems
- <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>
- <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>
- <math>\mathcal{L}\left\{ \int_0^t f(\tau)\, d\tau \right\}
= \mathcal{L}\left\{ 1 * f(t)\right\} = {1 \over s} \mathcal{L}\{f\}<math>
- <math>f(0^+)=\lim_{s\to \infty}{sF(s)}<math>
- <math>f(\infty)=\lim_{s\to 0}{sF(s)}<math>, all poles in left-hand plane.
- The final value theorem is useful because it gives the long-term bahaviour without having to perform partial fraction decompositions or other difficult algebra. If a functions poles are in the right hand plane (e.g. <math>e^t<math> or <math>\sin(t)<math>) the behaviour of this formula is undefined.
- <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>
- <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.
- <math>n<math>th-power shifting
- <math>\mathcal{L}\{\,t^nf(t)\} = (-1)^nD_s^n[F(s)]<math>
- <math>\mathcal{L}\{f * g\}
= \mathcal{L}\{ f \} \mathcal{L}\{ g \}<math>
Common Transforms
- <math>\mathcal{L}\{\,t^n\} = \frac {n!}{s^{n+1}}<math>
- <math>\mathcal{L}\{\,e^{-at}\} = \frac {1}{s+a}<math>
- <math>\mathcal{L}\{\,\sin(\omega t)\} = \frac {\omega}{s^2 + \omega^2}<math>
- <math>\mathcal{L}\{\,\cos(\omega t)\} = \frac {s}{s^2 + \omega^2}<math>
- <math>\mathcal{L}\{\,\sinh(bt)\} = \frac {b}{s^2-b^2}<math>
- <math>\mathcal{L}\{\,\cosh(bt)\} = \frac {s}{s^2 - b^2}<math>
- <math>\mathcal{L}\{\,\ln(t)\} = - \frac{\ln(s)+\gamma}{s}<math>
- <math>\mathcal{L}\{\,\sqrt[n]{t}\} = s^{-\frac{n+1}{n}} \cdot \Gamma\left(1+\frac{1}{n}\right)<math>
- 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>
- 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>
- <math>\mathcal{L}\{\,\operatorname{erf}(t)\} = {e^{s^2/4} \operatorname{erfc} \left(s/2\right) \over s}<math>
- Periodic Function period <math>T<math>
- <math>\mathcal{L}\{ f \}
= {1 \over 1 - e^{-Ts}} \int_0^T e^{-st} f(t)\,dt<math>
| Laplace transform | Time function |
<math>1<math>
| <math>\delta(t)<math>, unit impulse
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| <math>\frac{1}{s}<math>
| <math>u(t)<math>, unit step |
| <math>\frac{1}{(s+a)^n}<math>
| <math>\frac{t^{n-1}}{(n-1)!}e^{-at}<math> |
| <math>\frac{a}{s(s+a)}<math>
| <math>1-e^{-at}<math> |
| <math>\frac{1}{(s+a)(s+b)}<math>
| <math>\frac{1}{b-a}\left(e^{-at}-e^{-bt}\right)<math> |
| <math>\frac{s+c}{(s+a)^2+b^2}<math>
| <math>e^{-at}\left(\cos{(bt)}+\left(\frac{c-a}{b}\right)\sin{(bt)}\right)<math> |
| <math>\frac{s\sin\varphi+a\cos\varphi}{s^2+a^2}<math>
| <math>\sin{(at+\varphi)}<math> |
External links
- Online Computation (http://wims.unice.fr/wims/wims.cgi?lang=en&+module=tool%2Fanalysis%2Ffourierlaplace.en) of the transform or inverse transform, wims.unice.fr
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