Laplace Transform

Definition

$\mathcal{L}:\{f(t)\}\to \{F(s)\}$

for $f(t)$, function defined on $t\ge 0$ ( on $[0,\infty )$)

Laplace transform of $f$

$F(s)=\mathcal{L}(f(t)):={\int }_0^{\infty }{e}^{-st}f(t)dt$

Inverse transform of $F$

${\mathcal{L}}^{-1}(F(s))=f(t)$

Solution

For given equation $y^{\prime \prime }+a{y}^{\prime }+by=r(t),y(0),y^{\prime }(0)$

1. make a subsidiary equation Let $Y(s)=\mathcal{L}(y(t)),R(s)=\mathcal{L}(r(t))$ $\Rightarrow (s^{2}Y-Sy(0)-y^{\prime }(0))+\alpha (sY-y(0))+bY=R$ $\Rightarrow (s^2+as+b)Y=(s+a)y(0)+y^{\prime }(0)+R$^[subsidiary equation]

2. find a solution of a subsidiary equation by algebra Let $Q(s):=\frac{1}{s^2+as+b}=(\frac{1}{(s+\frac{1}{2}a{)}^2+b^2-\frac{1}{4}{a}^2})$^[Transfer function] $\Rightarrow Y(s)=((s+a)y(0)+y^{\prime }(0))Q(s)+R(s)Q(s)$

3. find Inversion of $Y(s)$ $y(t)={\mathcal{L}}^{-1}(Y(s))$

Facts

Linearity of the Laplace Transform

$\mathcal{L}(af(t)+bg(t))=a\mathcal{L}(f(t))+b\mathcal{L}(g(t)),\forall a,b\in \mathbb{R}$

First shifting (s-shifting) theorem

If $\mathcal{L}(f)=F(s)$, then $\mathcal{L}(e^{at}f)=F(s-a)$

Existence and uniqueness of Laplace Transform

If there is $M,k$ with $|f(t)|\le M{e}^{kt}\forall t\ge 0$^[growth restriction, growth of exponential order] $f(t)$ has Laplace Transform

If $\exists t_1,\dots ,t_n$ s.t. $f$ is continuous on $(t_{i-1},t_i)$ and has a finite limits as $t\to t_i^{+}$ or $t\to t_i^{-}$ $f$ is piecewise continuous on $a\le t\le b$

If $f(t):[0,\infty )\to \mathbb{R}$ is piecewise continuous and satisfies growth-restriction,
then there exists $\mathcal{L}(f(t))\forall s>k$

Laplace transform of derivatives

If $\mathcal{L}(f(t))=F(s)$ and $f$ satisfies the growth restriction, $\mathcal{L}(f^{\prime }(t))=s\mathcal{L}(f)-f(0)=sF(s)-f(0)$ $\mathcal{L}(f^{\prime \prime }(t))=s\mathcal{L}(f^{\prime })-f^{\prime }(0)=s^{2}F(s)-sf(0)-f^{\prime }(0)$ $\mathcal{L}(f^{(n)}(t))=s^{n}F(s)-s^{n-1}f(0)-s^{n-2}{f}^{\prime }(0)-\cdots -f^{(n-1)}(0)$

Laplace Transform of Integrals

If $\mathcal{L}(f(t))=F(s)$, then $\mathcal{L}({\int }_0^{t}f(\tau )d\tau )=\frac{1}{S}F(s)$, thus ${\int }_0^{t}f(\tau )d\tau ={\mathcal{L}}^{-1}(\frac{1}{S}F(s))$

Laplace transform of Unit step function

$L(u(t-a))=\frac{e^{-as}}{s},s>0$

Second shifting theorem (Time-shifting / t-shifting)

If $\mathcal{L}(f(t))=F(s)$,
then $\mathcal{L}(f(t-a)u(t-a))=e^{-as}F(s)$ and $\mathcal{L}(f(t)u(t-a))=e^{-as}\mathcal{L}(f(t+a))$ where $u$ is unit step function

Laplace transform of Dirac’s delta function

$\mathcal{L}(\delta (t-a))=e^{-as}$

Laplace transform and Convolution

If $\mathcal{L}(f(t))=F(s),\mathcal{L}(g(t))=G(s)$,
then $H(s)=\mathcal{L}(h(t))=\mathcal{L}((f\ast g)(t))=F(s)G(s)$