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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:

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

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 differentiation 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 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 \$ of $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.

Contents

1 Engineering/physics notation

2 Relation to other transforms

2.1 Fourier transform
2.2 Mellin transform
2.3 Z-transform

3 Properties and theorems

4 Common transforms

5 External links

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 solve 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 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 $θ = exp( - t)$ 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

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

Differentiation

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

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

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

Frequency division

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

Frequency integration

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

Integration

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

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 decompositions 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

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

nth 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 period $T$

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

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

External links

Online Computation of the transform or inverse transform, wims.unice.fr

Categories: Integral transforms

Science Fair Project Encyclopedia Contents Page

03-10-2013 05:06:04
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