Laplace transform list of properties

Posted December 8th, 2007 by Structure

Consider the Laplace transform of a original function f.

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

We give some useful property if this transform.

Change of scale.
Let $a>0$ then g(t)=f(at) is function which admits Laplace transform and

$\mathcal{L}(g(t))(s)=\int_{0}^{+\infty}f(at)e^{-st}dt=\frac{1}{a}\int_{0}^{+\infty}f(\tau)e^{-\frac{s}{a}\tau}d\tau=\frac{1}{a}F(\frac{s}{a})$

Multiply by exponential
Let $\lambda\in C$ then
$g(t)=e^{\lambda t}f(t)$ has Laplace transform

$\mathcal{L}(g(t))(s)=\mathcal{L}(e^{\lambda t}f(t))(s)=\int_{0}^{+\infty}e^{\lambda t}f(t)e^{-st}dt=\int_{0}^{+\infty}f(t)e^{-(s-\lambda )t}dt=F(s-\lambda)$

Product with t variable.
If f is an original function, $f\in\mathcal{O}_{\sigma_0}$ then for $\forall \epsilon>0,g_k(t)=t^kf(t)\in\mathcal{O}_{\sigma_0+\epsilon}$ and

$\mathcal{L}(tf(t))(s)=\int_{0}^{+\infty}tf(t)e^{-st}dt=-F'(s)$

$\mathcal{L}(t^2f(t))(s)=\int_{0}^{+\infty}t^2f(t)e^{-st}dt=(-1)^2F"(s)$

$\mathcal{L}(t^nf(t))(s)=\int_{0}^{+\infty}t^nf(t)e^{-st}dt=(-1)^nF^{(n)}(s)$

Laplace transform for derivative
If f and f' are original function then

$\mathcal{L}(f'(t))(s)=\int_{0}^{+\infty}f'(t)e^{-st}dt=f(t)e^{-st}|_0^{+\infty}+s\int_0^{+\infty}f(t)e^{-st}dt=sF(s)-f(0+)$

Where

$f(0+)=\lim_{t\to 0, t>0}f(t)$

If higher order derivatives exists and are originals,

$\mathcal{L}(f"(t))(s)=\int_{0}^{+\infty}f"(t)e^{-st}dt=s^2F(s)-f(0+)s-f'(0+)$

$\mathcal{L}(f^{(n)}(t))(s)=\int_{0}^{+\infty}f^{(n)}(t)e^{-st}dt=s^nF(s)-f(0+)s^{n-1}-f'(0+)s^{n-2}-....-f^{(n-1)}(0+)=$

Convolution of originals.
Let f,g be original function.
Then

$f*g(t)=\int_{0}^{+\infty}f(\tau)g(t-\tau)d\tau=\int_{0}^{t}f(\tau)g(t-\tau)d\tau$

is an original and

$\mathcal{L}(f*g(t))(s)=\int_{0}^{+\infty}f*g(t)e^{-st}dt=\int_{0}^{+\infty}\int_{0}^{t}f(\tau)g(t-\tau)d\tau e^{-st}dt=$

$=\int_{0}^{+\infty}f(\tau)e^{-s\tau}(\int_{\tau}^{+\infty}g(t-\tau) e^{-s(t-\tau)}dt)d\tau=\int_{0}^{+\infty}f(\tau)e^{-s\tau}d\tau\int_{\tau}^{+\infty}g(t-\tau)e^{-s(t-\tau)}dt=$