I don't understand why the laplace transform of some function, say f(t), has to be "piecewise continuous" and not "continuous". Is "piecewise continuous" sort of like the minimum requirement? This troubles me because I don't think f(t)=t is piecewise continuous, it's simply continuous…
Also, I don't understand what's so special about a function being of "exponential order". Can someone explain to me why this property is so special and apparently makes the Laplace Transform exist?
Also, my teacher says that f(t) has to be of exponential order from [0,infinity] and other sources say that it at least has to be of exponential order from [T,infinity] where T>0.
Thanks.
Best Answer
The Laplace transform is defined as:
$F(s)=\int _0 ^{+\infty} e ^{-st} f(t) dt$
Your first question: As one can see the limit of the integral is from $0$ to $\infty$. So, it is inherently assumed that $f(t)$ is zero for $t<0$. As a result, when we talk about $f(t)=t$, it is actually $f(t)=t, t\geq0$, which is a piecewise continuous function.
Second question: A function $f(t)$ is said of exponential order if there exists a constant $a$ and positive constants $t_0$ and $M$ such that $|f(t)|<Me^{at}$, for all $t>t_0$ at which $f(t)$ is defined. This condition should hold because otherwise the Laplace transform will not exist. To prove it, lets first split the integral into two parts:
$F(s)=\int _0 ^{+\infty} e ^{-st} f(t) dt = \int _0 ^{t_0} e ^{-st} f(t) dt+\int _{t_0} ^{\infty} e ^{-st} f(t) dt$
It is easy to show that the first part exists. Now, we need to show that the existence of the laplace transform depends on the convergence of the second part:
$|\int _{t_0} ^{\infty} e ^{-st} f(t) dt|\leq \int _{t_0} ^{\infty} |e ^{-st} f(t)| dt \leq \int _{t_0} ^{\infty} Me ^{-st} e^{at} dt = M\int _{t_0} ^{\infty} e^{-(s-a)t} dt$
This integral converges if $s>a$. So, we conclude that if $f(x)$ is exponentially ordered, there exists a constant $a$ in which $F(s)$ exists for $s>a$.