LaplaceTransform

Definition

Definition for the bilateral transform is

$$\mathcal{L}(x(t)) = X(s) \\ \mathcal{L}(x(t)) = \int_{-\infty}^{+\infty}x(t) e^{-s\cdot t} \mathrm{d} t \qquad s \in D \qquad D \subseteq \mathrm{C}$$

The set D is the set of complex numbers whose values make sense to the calculation of the integral

Principal Properties

The " ' " means total derivative

Traslation		Scaling
$$\mathcal{L} (x(t-t_0)) = X(s)\cdot e^{-t_0 s}$$		$$\mathcal{L}(x(a t)) = \frac{1}{ \mid a \mid } X( \frac{s}{a} )$$
Derivative in t		Derivative in s
$$\mathcal{L}(x'(t)) = s X( s )$$		$$\mathcal{L}(t x(t)) = - X'( s )$$
Convolution
$$\mathcal{L}( \int_{-\infty}^{+\infty}x(\tau) y(t-\tau) \mathrm{d}\tau ) = X( s ) \cdot Y(s)$$

Antitransform

The inverse of the Laplace transform is

$$x(t) = \mathcal{L}^{-1}(X(s)) \\ \mathcal{L}^{-1}(X(s)) = \frac{1}{2\pi j}\int_{r-j\infty}^{r+j\infty}X(s) e^{s\cdot t} \mathrm{d} s \qquad r \in \mathrm{R} \qquad r \pm jw \in D$$

where $j$ is the imaginary unit

Unilateral Transform

For signal defined only for positive time the transform can be defined as

$$\mathcal{L_u}(x(t)) = X(s) \\ \mathcal{L_u}(x(t)) = \int_{0}^{+\infty}x(t) e^{-s\cdot t} \mathrm{d} t \qquad s \in D \qquad D \subseteq \mathrm{C}$$

The most important modification arise in the derivative formulae where the initial conditions at

$$x(t=0)$$

enters in the formula due to the integration by parts making a difference respect to the above considered case where the function "x" is considered bounded and therefore null at the boundary

$$\mathcal{L}(x'(t)) = \int_{0}^{+\infty}x'(t) e^{-s\cdot t} \mathrm{d} t \\ \int_{0}^{+\infty}x'(t) e^{-s\cdot t} \mathrm{d} t = \biggl[x(t) e^{-s\cdot t}\biggr]_{0}^{+\infty} - \int_{0}^{+\infty}-s\cdot x(t) e^{-s\cdot t} \mathrm{d} t$$

leading to final formula

$$\mathcal{L}(x'(t)) = s X( s ) - x(0)$$

Initial Final Value Theorem

The following property can be demonstrated using the formulas for the transform of the derivative (above supplied)

$$\lim_{t \rightarrow 0^+} x(t) = \lim_{s \rightarrow \infty} s X(s)$$

$$\lim_{t \rightarrow + \infty} x(t) = \lim_{s \rightarrow 0} s X(s)$$

proof are in the attached file, less rigorously Here

-- RobertoBernetti - 23 Dec 2011