Laplace Transform (Chapter 10)

10.1 Definition of the Laplace Transform

For a function $f(t)$ defined on the interval $t\in[0,\infin)$, the Laplace transform of $f(t)$ is defined by

$$ F(s)=\int_0^\infin f(t)e^{-st}dt=L[f] $$

provided that the integral converges

Giving a new transformed function in terms of a new parameter $s$.

<aside> 💡 Common Transforms

• $f(t)=t^n$

  $$ L[t^n]=\frac{n!}{s^{n+1}},s>0 $$

• $f(t)=e^{at}$

  $$ L[e^{at}]=\frac{1}{s-a},s>a $$

• $f(t)=\cos(bt)$

  $$ L[\cos(bt)]=\frac s{s^2+b^2},s>0 $$

• $f(t)=\sin(bt)$

  $$ L[\sin(bt)]=\frac b{s^2+b^2},s>0 $$

</aside>

Linearity of the Laplace Transform

$$ L[f+g]=L[f]+L[g] $$

$$ L[f+g]=\int_0^\infin(f(t)+g(t))e^{-st}dt=\int_0^\infin f(t)e^{-st}dt+\int_0^\infin f(s)e^{-st}ds $$

$$ L[cf]=cL[f] $$

$$ L[cf]=\int_0^\infin cf(t)e^{-st}dt=c\int_0^\infin f(t)e^{-st}dt $$

Thus the Laplace Transform is a Linear Transform!

→ this is based on the properties of convergent improper integrals

Piecewise Continuous Functions

<aside> 💡 A function, $f$, is piecewise continous on the inteval $[a,b]$ if we can divide $[a,b]$ into a finte number of subintervals such that

• $f$ is continuous on each sub interval
• $f$ approaches a finite limit as the endpoints of each interval are approached from within

If $f$ is piecewise continuous on every interval of the form $[0,b]$ where $b$ is a constant, then $f$ is piecewise continuous on $[0,\infin]$. It does not have to be continous at every point as long as you can split the function into subintervals such that each of them are continuous

</aside>

Ex. Consider the function $f(t)$ defined by

$$ f(t)=\begin{cases}t&0\le t<1\\ -1&t\ge 1 \end{cases} $$

$$ L[f]=\int_0^\infin e^{-st}f(t)dt=\int_0^1e^{-st}tdt+\int_1^\infin-e^{-st}dt\\ =\frac1{s^2}[1-e^{-s}(2s+1)]~~~~s>0

$$