Transformada de Laplace. Laplace.

Sea $f(t)$ una función de $t$ definida para todo $t>0$ . La transformada de Laplace de $f(t)$ , se denota como $\mathcal{L}[f(t)]$ y está definida como

$\displaystyle \mathcal{L}[f(t)] = \int_{0}^{\infty}{e^{-st} \, f(t) , dt}$

$\displaystyle F(s) = \int_{0}^{\infty}{e^{-st} \, f(t) , dt}$

La transforma de Laplace existe cuando la integral converge para algún valor de $s$ ; de otra manera, no existe.

Se indicará con minúscula una función de $t$ como $f(t)$ , $g(t)$ , … , mientras que la transformada de Laplace de esa función se representará una letra mayúscula como $F(s)$ , $G(s)$ , … .

A continuación, se presenta un formulario de funciones básicas con sus respectivas transformadas.

funciones-elementales-transformada-de-laplace

Demostraciones

Demostrando que $\displaystyle \mathcal{L}[1] = \frac{1}{s}$ .

$\displaystyle \mathcal{L}[f(t)] = \int_{0}^{\infty}{e^{-st} \ f(t) \ dt}$

$\displaystyle \mathcal{L}[1] = \int_{0}^{\infty}{e^{-st} \ (1) \ dt}$

$\displaystyle \mathcal{L}[1] = \int_{0}^{\infty}{e^{-st} \ dt}$

$\displaystyle \mathcal{L}[1] = \lim_{m \rightarrow \infty}{\int_{0}^{m}{e^{-st} \ dt}}$

$\displaystyle \mathcal{L}[1] = \lim_{m \rightarrow \infty}{\left. \left[-\frac{1}{s} e^{-st} \right|_{0}^{m} \right]}$

$\displaystyle \mathcal{L}[1] = \lim_{m \rightarrow \infty}{\left[ \left(-\frac{1}{s} e^{-ms} \right) - \left(- \frac{1}{s} e^{0} \right) \right]}$

$\displaystyle \mathcal{L}[1] = \lim_{m \rightarrow \infty}{\left[-\frac{1}{s} e^{-ms} + \frac{1}{s} e^{0} \right]}$

$\displaystyle \mathcal{L}[1] = -\frac{1}{s} e^{- \infty \cdot s} + \frac{1}{s} e^{0}$

$\displaystyle \mathcal{L}[1] = -\frac{1}{s} e^{- \infty} + \frac{1}{s} e^{0}$

$\displaystyle \mathcal{L}[1] = -\frac{1}{s} (0) + \frac{1}{s} (1)$

$\displaystyle \therefore \mathcal{L}[1] = \frac{1}{s}$

Donde $s>0$ .

Demostrando que $\displaystyle \mathcal{L}[t] = \frac{1}{s^2}$ .

$\displaystyle \mathcal{L}[f(t)] = \int_{0}^{\infty}{e^{-st} \ f(t) \ dt}$

$\displaystyle \mathcal{L}[t] = \int_{0}^{\infty}{e^{-st} \cdot t \ dt}$

$\displaystyle \mathcal{L}[t] = \int_{0}^{\infty}{t \ e^{-st} \ dt}$

$\displaystyle \mathcal{L}[t] = \lim_{m \rightarrow \infty}{\int_{0}^{m}{t \ e^{-st} \ dt}}$

$\displaystyle \mathcal{L}[t] = \lim_{m \rightarrow \infty}{\left[ \left. -\frac{1}{s} te^{-st} + C \right|_{0}^{m} - \int_{0}^{m}{(-\frac{1}{s} e^{-st}) \ dt} \right]}$

$\displaystyle \mathcal{L}[t] = \lim_{m \rightarrow \infty}{\left[ \left. - \frac{1}{s} te^{-st} + C \right|_{0}^{m} + \frac{1}{s} \int_{0}^{m}{e^{-st} \ dt} \right]}$

$\displaystyle \mathcal{L}[t] = \lim_{m \rightarrow \infty}{\left[ \left. - \frac{1}{s} te^{-st} + C \right|_{0}^{m} - \left. \frac{1}{s^2} e^{-st} + C \right|_{0}^{m} \right]}$

$\displaystyle \mathcal{L}[t] = \lim_{m \rightarrow \infty}{\left[ \left. - \frac{1}{s} te^{-st} - \frac{1}{s^2} e^{-st} + C \right|_{0}^{m} \right]}$

$\displaystyle \mathcal{L}[t] = \lim_{m \rightarrow \infty}{\left\{ \left[- \frac{1}{s} (m) e^{-s \cdot m} - \frac{1}{s^2} e^{-s \cdot m} \right] - \left[- \frac{1}{s} (0) e^{-s \cdot 0} - \frac{1}{s^2} e^{-s \cdot 0} \right] \right\}}$

$\displaystyle \mathcal{L}[t] = \left[- \frac{1}{s} (\infty) e^{-s \cdot \infty} - \frac{1}{s^2} e^{-s \cdot \infty} \right] - \left[- \frac{1}{s} (0) e^{-s \cdot 0} - \frac{1}{s^2} e^{-s \cdot 0} \right]$

$\displaystyle \mathcal{L}[t] = \left[- \frac{1}{s} (\infty) e^{-\infty} - \frac{1}{s^2} e^{-\infty} \right] - \left[-(0) - \frac{1}{s^2} e^{0} \right]$

$\displaystyle \mathcal{L}[t] = \left(0 \right) - \left[- \frac{1}{s^2} (1) \right]$

$\displaystyle \mathcal{L}[t] = 0 - \left(- \frac{1}{s^2} \right)$

$\displaystyle \mathcal{L}[t] = 0 + \frac{1}{s^2}$

$\displaystyle \therefore \mathcal{L}[t] = \frac{1}{s^2}$

Donde $s>0$ .

Demostrando que $\displaystyle \mathcal{L}[e^{at}] = \frac{1}{s-a}$ .

$\displaystyle \mathcal{L}[f(t)] = \int_{0}^{\infty}{e^{-st} \ f(t) \ dt}$

$\displaystyle \mathcal{L}[e^{at}] = \int_{0}^{\infty}{e^{-st} \ e^{at} \ dt}$

$\displaystyle \mathcal{L}[e^{at}] = \int_{0}^{\infty}{e^{-st + at} \ dt}$

$\displaystyle \mathcal{L}[e^{at}] = \int_{0}^{\infty}{e^{(-s + a)t} \ dt}$

$\displaystyle \mathcal{L}[e^{at}] = \int_{0}^{\infty}{e^{-(s-a)t} \ dt}$

$\displaystyle \mathcal{L}[e^{at}] = \lim_{m \rightarrow \infty}{\int_{0}^{m}{e^{-(s-a)t} \ dt}}$

$\displaystyle \mathcal{L}[e^{at}] = \lim_{m \rightarrow \infty}{\left[ \left. -\frac{1}{s-a} e^{-(s-a)t} + C \right|_{0}^{m} \right]}$

$\displaystyle \mathcal{L}[e^{at}] = \lim_{m \rightarrow \infty}{\left\{ \left[-\frac{1}{s-a} e^{-(s-a) \cdot m} \right] - \left[-\frac{1}{s-a} e^{-(s-a) \cdot 0} \right] \right\}}$

$\displaystyle \mathcal{L}[e^{at}] = \left[-\frac{1}{s-a} e^{-(s-a) \cdot \infty} \right] - \left[-\frac{1}{s-a} e^{-(s-a) \cdot 0} \right]$

$\displaystyle \mathcal{L}[e^{at}] = \left[-\frac{1}{s-a} e^{\infty} \right] - \left[-\frac{1}{s-a} e^{0} \right]$

$\displaystyle \mathcal{L}[e^{at}] = \left[-\frac{1}{s-a} (0) \right] - \left[-\frac{1}{s-a} (1) \right]$

$\displaystyle \mathcal{L}[e^{at}] = (0) - \left[-\frac{1}{s-a} \right]$

$\displaystyle \therefore \mathcal{L}[e^{at}] = \frac{1}{s-a}$

Donde $s>a$ .

Demostrando que $\displaystyle \mathcal{L}[e^{-at}] = \frac{1}{s+a}$ .

$\displaystyle \mathcal{L}[f(t)] = \int_{0}^{\infty}{e^{-st} \ f(t) \ dt}$

$\displaystyle \mathcal{L}[e^{-at}] = \int_{0}^{\infty}{e^{-st} \ e^{-at} \ dt}$

$\displaystyle \mathcal{L}[e^{-at}] = \int_{0}^{\infty}{e^{-st - at} \ dt}$

$\displaystyle \mathcal{L}[e^{-at}] = \int_{0}^{\infty}{e^{(-s - a)t} \ dt}$

$\displaystyle \mathcal{L}[e^{-at}] = \int_{0}^{\infty}{e^{-(s+a)t} \ dt}$

$\displaystyle \mathcal{L}[e^{-at}] = \lim_{m \rightarrow \infty}{\int_{0}^{m}{e^{-(s+a)t} \ dt}}$

$\displaystyle \mathcal{L}[e^{-at}] = \lim_{m \rightarrow \infty}{\left[ \left. -\frac{1}{s+a} e^{-(s+a)t} + C \right|_{0}^{m} \right]}$

$\displaystyle \mathcal{L}[e^{-at}] = \lim_{m \rightarrow \infty}{\left\{ \left[-\frac{1}{s+a} e^{-(s+a) \cdot m} \right] - \left[-\frac{1}{s+a} e^{-(s+a) \cdot 0} \right] \right\}}$

$\displaystyle \mathcal{L}[e^{-at}] = \left[-\frac{1}{s+a} e^{-(s+a) \cdot \infty} \right] - \left[-\frac{1}{s+a} e^{-(s+a) \cdot 0} \right]$

$\displaystyle \mathcal{L}[e^{-at}] = \left[-\frac{1}{s+a} e^{\infty} \right] - \left[-\frac{1}{s+a} e^{0} \right]$

$\displaystyle \mathcal{L}[e^{-at}] = \left[-\frac{1}{s+a} (0) \right] - \left[-\frac{1}{s+a} (1) \right]$

$\displaystyle \mathcal{L}[e^{-at}] = (0) - \left(-\frac{1}{s+a} \right)$

$\displaystyle \therefore \mathcal{L}[e^{-at}] = \frac{1}{s+a}$

Donde $s>a$ .

Demostrando que $\displaystyle \mathcal{L}[\sin{at}] = \frac{a}{s^2 + a^2}$ .

$\displaystyle \mathcal{L}[f(t)] = \int_{0}^{\infty}{e^{-st} \ f(t) \ dt}$

$\displaystyle \mathcal{L}[\sin{at}] = \int_{0}^{\infty}{e^{-st} \ \sin{at} \ dt}$

$\displaystyle \mathcal{L}[\sin{at}] = \lim_{m \rightarrow \infty}{\int_{0}^{m}{e^{-st} \ \sin{at} \ dt}}$

Resolviendo la integral

$\displaystyle \int{e^{-st} \ \sin{at} \ dt} = \left(- \frac{1}{s} e^{-st} \right)(\sin{at}) - \int{\left(-\frac{1}{s} e^{-st} \right) (a \cos{at}) \ dt}$

$\displaystyle \int{e^{-st} \ \sin{at} \ dt} = - \frac{1}{s} e^{-st} \sin{at} + \frac{a}{s} \int{e^{-st} \cos{at} \ dt}$

$\displaystyle \int{e^{-st} \ \sin{at} \ dt} = - \frac{1}{s} e^{-st} \sin{at} + \frac{a}{s} \left[ \left(- \frac{1}{s} e^{-st} \right) (\cos{at}) + \int{\left(- \frac{1}{s} e^{-st} \right) (a \sin{at}) \ dt} \right]$

$\displaystyle \int{e^{-st} \ \sin{at} \ dt} = - \frac{1}{s} e^{-st} \sin{at} + \frac{a}{s} \left(- \frac{1}{s} e^{-st} \cos{at} - \frac{a}{s} \int{e^{-st} \sin{at} \ dt} \right)$

$\displaystyle \int{e^{-st} \ \sin{at} \ dt} = - \frac{1}{s} e^{-st} \sin{at} - \frac{a}{s^2} e^{-st} \cos{at} - \frac{a}{s^2} \int{e^{-st} \sin{at} \ dt}$

$\displaystyle \int{e^{-st} \ \sin{at} \ dt} + \frac{a}{s^2} \int{e^{-st} \sin{at} \ dt} = - \frac{1}{s} e^{-st} \sin{at} - \frac{a^2}{s^2} e^{-st} \cos{at}$

$\displaystyle \left(1 + \frac{a^2}{s^2} \right) \int{e^{-st} \sin{at} \ dt} = - \frac{1}{s} e^{-st} \sin{at} - \frac{a}{s^2} e^{-st} \cos{at}$

$\displaystyle \left(\frac{s^2+a^2}{s^2} \right) \int{e^{-st} \sin{at} \ dt} = - \frac{1}{s} e^{-st} \sin{at} - \frac{a}{s^2} e^{-st} \cos{at}$

$\displaystyle \int{e^{-st} \sin{at} \ dt} = \frac{- \frac{1}{s} e^{-st} \sin{at} - \frac{a}{s^2} e^{-st} \cos{at} + C}{\frac{s^2+a^2}{s^2}}$

$\displaystyle \int{e^{-st} \sin{at} \ dt} = - \frac{s^2}{s^2+a^2} \cdot \frac{1}{s} e^{-st} \sin{at} - \frac{s^2}{s^2+a^2} \cdot \frac{a}{s^2} e^{-st} \cos{at} + C$

$\displaystyle \int{e^{-st} \sin{at} \ dt} = - \frac{s}{s^2+a^2} e^{-st} \sin{at} - \frac{a}{s^2+a^2} e^{-st} \cos{at} + C$

Regresando

$\displaystyle \mathcal{L}[\cos{at}] = \lim_{m \rightarrow \infty}{\int_{0}^{m}{e^{-st} \ \cos{at} \ dt}}$

$\displaystyle \mathcal{L}[\sin{at}] = \lim_{m \rightarrow \infty}{\left[ \left. - \frac{s}{s^2+a^2} e^{-st} \sin{at} - \frac{a}{s^2+a^2} e^{-st} \cos{at} + C \right|_0^m \right]}$

$\displaystyle \mathcal{L}[\cos{at}] = \lim_{m \rightarrow \infty}{\left\{\left[ - \frac{s}{s^2+a^2} e^{-s \cdot m} \sin{a \cdot m} - \frac{a}{s^2+a^2} e^{-s \cdot m} \cos{a \cdot m} \right] - \left[- \frac{s}{s^2+a^2} e^{-s \cdot 0} \sin{a \cdot 0} - \frac{a}{s^2+a^2} e^{-s \cdot 0} \cos{a \cdot 0} \right] \right\}}$

$\displaystyle \mathcal{L}[\cos{at}] = \left[ - \frac{s}{s^2+a^2} e^{-s \cdot \infty} \sin{a \cdot \infty} - \frac{a}{s^2+a^2} e^{-s \cdot \infty} \cos{a \cdot \infty} \right] - \left[- \frac{s}{s^2+a^2} e^{-s \cdot 0} \sin{a \cdot 0} - \frac{a}{s^2+a^2} e^{-s \cdot 0} \cos{a \cdot 0} \right]$

$\displaystyle \mathcal{L}[\sin{at}] = \left[ - \frac{s}{s^2+a^2} e^{-\infty} \sin{\infty} - \frac{a}{s^2+a^2} e^{-\infty} \cos{\infty} \right] - \left[- \frac{s}{s^2+a^2} e^{0} \sin{0} - \frac{a}{s^2+a^2} e^{0} \cos{0} \right]$

$\displaystyle \mathcal{L}[\sin{at}] = \left[ - \frac{s}{s^2+a^2} (0)(\infty) - \frac{a}{s^2+a^2} (0)(\infty) \right] - \left[- \frac{s}{s^2+a^2} (1)(0) - \frac{a}{s^2+a^2} (1)(1) \right]$

$\displaystyle \mathcal{L}[\sin{at}] = (-0 - 0) - \left(-0 - \frac{a}{s^2+a^2} \right)$

$\displaystyle \therefore \mathcal{L}[\sin{at}] = \frac{a}{s^2+a^2}$

Donde $s>0$ .

Demostrando que $\displaystyle \mathcal{L}[\cos{at}] = \frac{s}{s^2 + a^2}$ .

$\displaystyle \mathcal{L}[f(t)] = \int_{0}^{\infty}{e^{-st} \ f(t) \ dt}$

$\displaystyle \mathcal{L}[\cos{at}] = \int_{0}^{\infty}{e^{-st} \ \cos{at} \ dt}$

$\displaystyle \mathcal{L}[\cos{at}] = \lim_{m \rightarrow \infty}{\int_{0}^{m}{e^{-st} \ \cos{at} \ dt}}$

Resolviendo la integral

$\displaystyle \int{e^{-st} \ \cos{at} \ dt} = \left(- \frac{1}{s} e^{-st} \right)(\cos{at}) - \int{\left(-\frac{1}{s} e^{-st} \right) (-a \sin{at}) \ dt}$

$\displaystyle \int{e^{-st} \ \cos{at} \ dt} = - \frac{1}{s} e^{-st} \cos{at} - \frac{a}{s} \int{e^{-st} \sin{at} \ dt}$

$\displaystyle \int{e^{-st} \ \cos{at} \ dt} = - \frac{1}{s} e^{-st} \cos{at} - \frac{a}{s} \left[ \left(- \frac{1}{s} e^{-st} \right) (\sin{at}) - \int{\left(- \frac{1}{s} e^{-st} \right) (a \cos{at}) \ dt} \right]$

$\displaystyle \int{e^{-st} \ \cos{at} \ dt} = - \frac{1}{s} e^{-st} \cos{at} - \frac{a}{s} \left(- \frac{1}{s} e^{-st} \sin{at} + \frac{a}{s} \int{e^{-st} \cos{at} \ dt} \right)$

$\displaystyle \int{e^{-st} \ \cos{at} \ dt} = - \frac{1}{s} e^{-st} \cos{at} + \frac{a}{s^2} e^{-st} \sin{at} - \frac{a}{s^2} \int{e^{-st} \cos{at} \ dt}$

$\displaystyle \int{e^{-st} \ \cos{at} \ dt} + \frac{a}{s^2} \int{e^{-st} \cos{at} \ dt} = - \frac{1}{s} e^{-st} \cos{at} + \frac{a^2}{s^2} e^{-st} \sin{at}$

$\displaystyle \left(1 + \frac{a^2}{s^2} \right) \int{e^{-st} \cos{at} \ dt} = - \frac{1}{s} e^{-st} \cos{at} + \frac{a}{s^2} e^{-st} \sin{at}$

$\displaystyle \left(\frac{s^2+a^2}{s^2} \right) \int{e^{-st} \cos{at} \ dt} = - \frac{1}{s} e^{-st} \cos{at} + \frac{a}{s^2} e^{-st} \sin{at}$

$\displaystyle \int{e^{-st} \cos{at} \ dt} = \frac{- \frac{1}{s} e^{-st} \cos{at} + \frac{a}{s^2} e^{-st} \sin{at} + C}{\frac{s^2+a^2}{s^2}}$

$\displaystyle \int{e^{-st} \cos{at} \ dt} = - \frac{s^2}{s^2+a^2} \cdot \frac{1}{s} e^{-st} \cos{at} + \frac{s^2}{s^2+a^2} \cdot \frac{a}{s^2} e^{-st} \sin{at} + C$

$\displaystyle \int{e^{-st} \cos{at} \ dt} = - \frac{s}{s^2+a^2} e^{-st} \cos{at} + \frac{a}{s^2+a^2} e^{-st} \sin{at} + C$

Regresando

$\displaystyle \mathcal{L}[\cos{at}] = \lim_{m \rightarrow \infty}{\int_{0}^{m}{e^{-st} \ \cos{at} \ dt}}$

$\displaystyle \mathcal{L}[\cos{at}] = \lim_{m \rightarrow \infty}{\left[ \left. - \frac{s}{s^2+a^2} e^{-st} \cos{at} + \frac{a}{s^2+a^2} e^{-st} \sin{at} \right|_0^m \right]}$

$\displaystyle \mathcal{L}[\cos{at}] = \lim_{m \rightarrow \infty}{\left\{\left[ - \frac{s}{s^2+a^2} e^{-s \cdot m} \cos{a \cdot m} + \frac{a}{s^2+a^2} e^{-s \cdot m} \sin{a \cdot m} \right] - \left[- \frac{s}{s^2+a^2} e^{-s \cdot 0} \cos{a \cdot 0} + \frac{a}{s^2+a^2} e^{-s \cdot 0} \sin{a \cdot 0} \right] \right\}}$

$\displaystyle \mathcal{L}[\cos{at}] = \left[ - \frac{s}{s^2+a^2} e^{-s \cdot \infty} \cos{a \cdot \infty} + \frac{a}{s^2+a^2} e^{-s \cdot \infty} \sin{a \cdot \infty} \right] - \left[- \frac{s}{s^2+a^2} e^{-s \cdot 0} \cos{a \cdot 0} + \frac{a}{s^2+a^2} e^{-s \cdot 0} \sin{a \cdot 0} \right]$

$\displaystyle \mathcal{L}[\cos{at}] = \left[ - \frac{s}{s^2+a^2} e^{-\infty} \cos{\infty} + \frac{a}{s^2+a^2} e^{-\infty} \sin{\infty} \right] - \left[- \frac{s}{s^2+a^2} e^{0} \cos{0} - \frac{a}{s^2+a^2} e^{0} \sin{0} \right]$

$\displaystyle \mathcal{L}[\cos{at}] = \left[ - \frac{s}{s^2+a^2} (0)(\infty) + \frac{a}{s^2+a^2} (0)(\infty) \right] - \left[- \frac{s}{s^2+a^2} (1)(1) + \frac{a}{s^2+a^2} (1)(0) \right]$