Laplace Transforms

Definition

Let $x(t)$$x(t)$ be a continuous-time signal and let $s$$s$ be a complex number such that $s=\sigma +j\omega$$s = \sigma + j\omega$. Then,

where $\mathcal{L}\{x(t)\}=X(s)$$\mathcal{L}\{x(t)\} = X(s)$ is the Laplace Transform of $x(t)$$x(t)$.

Properties

Linearity

• $\mathcal{L}\{c_1{x}_1(t)+c_2{x}_2(t)\}=c_1\mathcal{L}\{x_1(t)\}+c_2\mathcal{L}\{x_2(t)\}$$\Large \mathcal{L}\{c_1x_1(t) + c_2x_2(t)\} = c_1\mathcal{L}\{x_1(t)\} + c_2\mathcal{L}\{x_2(t)\}$

Exponential Shift

• $\mathcal{L}\{e^{\alpha t}x(t)\}=X(s-\alpha )$$\Large \mathcal{L}\{e^{\alpha t}x(t)\} = X(s-\alpha)$

Derivatives

• $\mathcal{L}\{\frac{d}{dx}(x(t))\}=sX(s)-x(0)$$\Large \mathcal{L}\{\frac{d}{dx}(x(t))\} = sX(s) - x(0)$

Integrals

• $\mathcal{L}\{{\int }_0^{t}x(\tau )d\tau \}=\frac{X(s)}{s}$$\Large \mathcal{L}\{\int_{0}^{t}x(\tau)\:d\tau\} = \frac{X(s)}{s}$

Convolution

• $\mathcal{L}\{(x\ast y)(t)\}=X(s)Y(s)$$\Large \mathcal{L}\{(x*y)(t)\} = X(s)Y(s)$

Initial Value Theorem

• $x(0)=\underset{s\to \mathrm{\infty }}{lim}sX(s)$$\Large x(0) = \lim\limits_{s \to \infty}sX(s)$

Final Value Theorem

• $\underset{t\to \mathrm{\infty }}{lim}x(t)=\underset{s\to 0}{lim}sX(s)$$\Large \lim\limits_{t \to \infty} x(t) = \lim\limits_{s \to 0} sX(s)$

Condition: all poles of $sX(s)$$sX(s)$ must have strictly negative real parts.

Laplace Transform Table