Revision as of 15:25, 24 November 2008 by Sje (Talk)

definition

The Laplace transform of a function f(t), defined for all real numbers t ≥ 0, is the function F(s), defined by:

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

The lower limit of 0 is short notation to mean

$ \lim_{\varepsilon\to 0+}\int_{-\varepsilon}^\infty $

and assures the inclusion of the entire Dirac delta function δ(t) at 0 if there is such an impulse in f(t) at 0.

The parameter s is in general complex number:

$ s = \sigma + i \omega \, $

inverse laplace transform

The inverse Laplace transform is given by the following complex integral

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

Region of convergence

The Laplace transform F(s) typically exists for all complex numbers such that Re{s} > a, where a is a real constant which depends on the growth behavior of f(t), whereas the two-sided transform is defined in a range a < Re{s} < b. The subset of values of s for which the Laplace transform exists is called the region of convergence (ROC) or the domain of convergence. In the two-sided case, it is sometimes called the strip of convergence.

The integral defining the Laplace transform of a function may fail to exist for various reasons. For example, when the function has infinite discontinuities in the interval of integration, or when it increases so rapidly that exp(-pt) cannot damp it sufficiently for convergence on the interval to take place. 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 may or may not be taken.

relationship to fourier transform

Alumni Liaison

To all math majors: "Mathematics is a wonderfully rich subject."

Dr. Paul Garrett