Laplace Transform

#REDIRECT Laplace transform

Laplace transform

In mathematics and in particular, in functional analysis, the Laplace transform of a function (mathematics) ''f''(''t'') defined for all real numbers ''t'' ≥ 0 is the function ''F''(''s''), defined by: :

F(s) 
  = \left\{\mathcal{L} f\right\}(s)
  =\int_{0^-}^\infty e^{-st} f(t)\,dt.

The lower limit of 0− is short notation to mean

\lim_{\epsilon \rightarrow +0} -\epsilon \

and assures the inclusion of the entire dirac delta function

\delta (t) \

at 0 if there is such an impulse in ''f''(''t'') at 0. This integral transform has a number of properties that make it useful for analysing linear dynamical systems. The most significant advantage is that derivative and integration become multiplication and division, respectively, with

s \

. (This is similar to the way that logarithms change an operation of multiplication of numbers to addition of their logarithms.) This changes integral equations and differential equations to polynomial equations, which are much easier to solve. The inverse Laplace transform is the Bromwich integral, which is a complex number integral given by: :

f(t) = \frac{1}{2 \pi i} \int_{ \gamma - i \infty}^{ \gamma + i \infty} e^{st} F(s)\,ds.

:where

\gamma \

is a real number so that the contour path of integration is in the ''region of convergence'' of

F(s) \

normally requiring

\gamma > \operatorname{Re}(s_p) \

for every singularity

s_p \

F(s) \

. If all singularities are in the left half-plane, that is

\operatorname{Re}(s_p) < 0 \

for every

s_p \

, then

\gamma \

can be set to zero and the above inverse integral formula above becomes identical to the inverse Fourier transform. The Laplace transform can be extended to the two-sided Laplace transform or bilateral Laplace transform by setting the range of integration to be the entire real axis; if that is done the ordinary or one-sided transform becomes simply a special case consisting of those transforms making use of a Heaviside step function in the definition of the function being transformed. Moreover, the transform, both one and two-sided is sometimes defined slightly differently, by :

F(s) 
  = \left\{\mathcal{L} f\right\}(s)
  =s \int_{0^-}^\infty e^{-st} f(t)\,dt.

The Laplace transform is much used in engineering mathematics; 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: :

F(s)
  = \left\{\mathcal{L}f\right\}(s)
  =\int_{0^-}^\infty e^{-st} f(t)\,dt.

When one says "the Laplace transform" without qualification, the unilateral transform is normally intended. The bilateral transform is defined as follows: :

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

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''), whereas the two-sided transform is defined in a range ''a'' < ''s'' < ''b''. The Laplace transform can also be used to Laplace transform applied to differential equations and is used extensively in electrical engineering. There are no specific conditions that one can check a function against to know in all cases if its Laplace transform can be taken, other than to say the defining integral converges. It is however easy to give theorems on cases where it can or cannot be taken. == Relation to other transforms == ===Fourier transform=== The continuous Fourier transform is equivalent to evaluating the Laplace transform with complex argument :

\mathcal{F}f(\omega) = \mathcal{L}f(i \omega) = \int_{0^-}^\infty e^{-i \omega t} f(t)\,\mathrm{d}t.

This equivalence is usually used to determine the frequency spectrum of a signal (information theory) or dynamical system. Note that the

\frac{1}{\sqrt{2 \pi}}

constant is not included. ===Mellin transform=== The Mellin transform and its inverse are related to the two-sided Laplace transform by a simple change of variables. If in the Mellin transform :

\left\{\mathcal{M} g\right\}(s) = \int_0^\infty \theta^s g(\theta) \frac{d\theta}{\theta}

we set

\theta = \exp(-t)

we get a two-sided Laplace transform. Since an ordinary Laplace transform can be written as a special case of a two-sided transform, and since the two-sided transform can be written as the sum of two one-sided transforms, the theory of Laplace, Fourier and Mellin transforms are at bottom the same subject. However, a different point of view and different characteristic problems are associated to each of these three major integral transforms. ===Z-transform=== The Z-transform equivalence is not as straightforward as for the Fourier or Mellin transform. Take a continuous signal, its Laplace transform and its Z-transform and label them: *Continuous signal:

f(t)

*Laplace transform:

F(s)

*Z-transform:

F(z)

Multiply

f(t)

by a Dirac comb and name the result

f^{*}(t)

f^{*}(t) = f(t) \delta_T(t) = \sum_{n=0}^{\infty} f(n T) \delta(t - n T)

and taking the Laplace transform results in :

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}

Then the equivalence can be stated: :

F^{*}(s) = \left. F(z) \right|_{z=e^{sT}}

This equation relates the sampled values of a continuous signal to the discrete sequence resulting from the Z-transform. == Properties and theorems== *Linearity :

\mathcal{L}\left\{a f(t) + b g(t) \right\}
  = a \mathcal{L}\left\{ f(t) \right\} +
    b \mathcal{L}\left\{ g(t) \right\}

*Derivative :

\mathcal{L}\{f'\}
  = s \mathcal{L}\{f\} - f(0)

\mathcal{L}\{f''\}
  = s^2 \mathcal{L}\{f\} - s f(0) - f'(0)

\mathcal{L}\left\{ f^{(n)} \right\}
  = s^n \mathcal{L}\{f\} - s^{n - 1} f(0) - \cdots - f^{(n - 1)}(0)

* Frequency :

\mathcal{L}\{ t f(t)\}
  = -F'(s)

* Frequency :

\mathcal{L}\left\{ \frac{f(t)}{t} \right\} = \int_s^\infty F(\sigma)\, d\sigma

* Integral :

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

* Initial value theorem :

f(0^+)=\lim_{s\to \infty}{sF(s)}

* Final value theorem :

f(\infty)=\lim_{s\to 0}{sF(s)}

, all poles in left-hand plane. : The final value theorem is useful because it gives the long-term behaviour without having to perform Partial_fraction or other difficult algebra. If a functions poles are in the right hand plane (e.g.

e^t

\sin(t)

) the behaviour of this formula is undefined. *

s

shifting :

\mathcal{L}\left\{ e^{at} f(t) \right\}
  = F(s - a)

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

t

shifting :

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

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

:Note:

u(t)

is the step function. *

n

th-power shifting :

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

*Convolution :

\mathcal{L}\{f * g\}
  = \mathcal{L}\{ f \} \mathcal{L}\{ g \}

==Common transforms== *

n

th power :

\mathcal{L}\{\,t^n\} = \frac {n!}{s^{n+1}}

*Exponential :

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

*Sine :

\mathcal{L}\{\,\sin(\omega t)\} = \frac {\omega}{s^2 + \omega^2}

*Cosine :

\mathcal{L}\{\,\cos(\omega t)\} = \frac {s}{s^2 + \omega^2}

*Hyperbolic sine :

\mathcal{L}\{\,\sinh(bt)\} = \frac {b}{s^2-b^2}

*Hyperbolic cosine :

\mathcal{L}\{\,\cosh(bt)\} = \frac {s}{s^2 - b^2}

*Natural logarithm :

\mathcal{L}\{\,\ln(t)\} = - \frac{\ln(s)+\gamma}{s}

*''n''th root :

\mathcal{L}\{\,\sqrt[n]{t}\} = s^{-\frac{n+1}{n}} \cdot \Gamma\left(1+\frac{1}{n}\right)

*Bessel function of the first kind :

\mathcal{L}\{\,J_n(t)\} = \frac{\left(s+\sqrt{1+s^2}\right)^{-n}}{\sqrt{1+s^2}}

*Modified Bessel function of the first kind :

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

*Error function :

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

* Periodic Function periodic function

T

\mathcal{L}\{ f \}
  = {1 \over 1 - e^{-Ts}} \int_0^T e^{-st} f(t)\,dt

Laplace transform	Time function
$1$	$\delta(t)$ , unit impulse
$\frac{1}{s}$	$u(t)$ , unit step
$\frac{1}{(s+a)^n}$	$\frac{t^{n-1}}{(n-1)!}e^{-at}$
$\frac{a}{s(s+a)}$	$1-e^{-at}$
$\frac{1}{(s+a)(s+b)}$	$\frac{1}{b-a}\left(e^{-at}-e^{-bt}\right)$
$\frac{s+c}{(s+a)^2+b^2}$	$e^{-at}\left(\cos{(bt)}+\left(\frac{c-a}{b}\right)\sin{(bt)}\right)$
$\frac{s\sin\varphi+a\cos\varphi}{s^2+a^2}$	$\sin{(at+\varphi)}$

==External links== *[http://wims.unice.fr/wims/wims.cgi?lang=en&+module=tool%2Fanalysis%2Ffourierlaplace.en Online Computation] of the transform or inverse transform, wims.unice.fr * [http://eqworld.ipmnet.ru/en/auxiliary/aux-inttrans.htm Tables of Integral Transforms] at EqWorld: The World of Mathematical Equations. == Bibliography == * A. D. Polyanin and A. V. Manzhirov, ''Handbook of Integral Equations'', CRC Press, Boca Raton, 1998. Integral transforms Differential_equations

Laplace transform

L{ty'}= L{t�y} = -Ds{L{y'}} = -Ds{sY-y0} = -(Y + s�Ds{Y}) = -Y - s�Y' User:Euyyn 11:24, 13 Sep 2004 (UTC) == What is the point of LaPlace? == Say I'm not familiar with him, and am looking him up in the Wikipedia for that reason. I would then like to find out what is so important about this Transform. When can it be used, and how? What is, again, the point of the LaPlace transform? The article does a poor job of informing 'me'.-- User:Ec5618 11:32, Apr 21, 2005 (UTC) : First of all, it is ''Laplace'', with small p. Its uses are explained in the second paragraph in the article. I am not sure its signifance can be explained in non-mathematical terms, as it is a sofisticated tool used for solving differential equations and such. :Let me know if you have further questions. User:Oleg Alexandrov 15:34, 21 Apr 2005 (UTC) :: Also, this is an encyclopedia, not a textbook. For a non-specialist, this might be tricky to understand, but then, to understand it well, you might need to read a book or take a course, as again, this is a complex tool. User:Oleg Alexandrov 15:36, 21 Apr 2005 (UTC) ::: I understand that an encyclopedia cannot possibly contain all knowledge in a way that any person could understamd it, however, if you look at the CPR article, for example, you'll see that it's quite possible for wikipedia to be textbook-like. Perhaps this article should be simplified, and expanded somewhat. -- User:Gerriegijsen 15:58, May 13, 2005 (UTC) :::: I think the CPR article is a bad example. Firstly, you need to know a bit of mathematics to understand why the Laplace transform is useful; at least differential equations and integration. This is rather different from CPR, where not much beyond common sense is necessary to understand the gist of the procedure. Secondly, in my opinion the CPR article goes too much into detail for an encyclopedia (this is justified for that specific example, I think). :::: On the other hand, the Laplace transform article can certainly be improved. It shouldn't have a technical definition in the first sentence and it should include an example of solving a differential equation with Laplace transforms. Apparently, no editor has yet found the time and motivation to do this. -- User:Jitse Niesen 00:11, 14 May 2005 (UTC) == Engineering/physics notation == How is this equation different from the one at the top? User:Taral 16:13, 27 May 2005 (UTC)

See other meanings of words starting from letter:

L

LA | LB | LC | LD | LE | LF | LG | LH | LI | LJ | LK | LM | LN | LO | LP | LR | LS | LT | LU | LW | LX | LY | LZ |

Words begining with Laplace_transform:

Laplace_Transform
Laplace_transform
Laplace_transform
Laplace_Transformation
Laplace_transform_applied_to_differential_equations
Laplace_transform_applied_to_differential_equations