\[\mathscr{L}\{f(t)\}=F(s)=\int_0^\infty e^{-st} f(t)\,dt\]
\[\mathscr{L}\{e^{at}\}=\]
\[\mathscr{L}\{e^{at}\}=\int_0^\infty e^{-st}e^{at}\,dt\]
\[\mathscr{L}\{e^{at}\}=\int_0^\infty e^{(a-s)t}\,dt\]
\[\mathscr{L}\{e^{at}\}=\left[\frac{e^{(a-s)t}}{(a-s)}\right]_0^\infty\]
\[(\text{If }s>a)\Rightarrow \lim_{t\rightarrow \infty} e^{(a-s)t}= 0\]
\[\Rightarrow\mathscr{L}\{e^{at}\}=\left[\frac{e^{(a-s)t}}{(a-s)}\right]_0^\infty=0-\frac{1}{(a-s)}=\frac{1}{s-a},\qquad s>a\]
\[\therefore \boxed{\mathscr{L}\{e^{at}\}=\frac{1}{s-a},\qquad s>a}\]
par(mfrow=c(1,2))
t <- seq(0,3,length.out=100)
s <- seq(0,3,length.out=100)
plot(t,exp(2*t),type = "l",ylim = c(0,100), col="blue", lwd=2)
plot(s,1/(s-2), type = "l",ylim = c(1,30), col="blue", lwd=2)+abline(v=2, lty =3, col="red", lwd=2)integer(0)mtext("Laplace Transformation", side = 3, line = -3, outer = TRUE)
define(f(t),%e^(2*t));
ys: laplace(f(t),t,s);
plot2d(ys,[s,0,3],[y,0,50]);| \[f(t)\] | \[F(s)\] | ||
|---|---|---|---|
| \[1\] | \[\frac{1}{s}\] | \[s>0\] | |
| \[e^{at}\] | \[\frac{1}{s-a}\] | \[s>a\] | |
| \[t^n\] | \[\frac{n!}{s^{n+1}}\] | \[s>0,\quad n\in \mathbb{N}\] | |
| \[\sin kt\] | \[\frac{k}{s^2+k^2}\] | \[s>0\] | |
| \[\cos kt\] | \[\frac{s}{s^2+k^2}\] | \[s>0\] | |
| \[e^{at}t^n\] | \[\frac{n!}{(s-a)^{n+1}}\] | \[s>0,\quad n\in \mathbb{N}\] | |
| \[e^{at}\sin kt\] | \[\frac{k}{(s-a)^2+k^2}\] | \[s>a\] | |
| \[e^{at}\cos kt\] | \[\frac{s-a}{(s-a)^2+k^2}\] | \[s>a\] |
Theorem 5.1 (Translation) \[\boxed{\mathscr{L}\{e^{at}f(t)\}=F(s-a)}\] where \(F(s)=\mathscr{L}\{f(t)\}\)
\[\mathscr{L}\{e^{at}f(t)\}=\int_0^\infty e^{-st}e^{at}f(t)\,dt\] \[=\int_0^\infty e^{-(s-a)t}f(t)\,dt\]
\[=F(s-a)\]
\[\square\]
\[\mathscr{L}\{e^{-2t}t^3\}=\]
\[\mathscr{L}\{e^{-2t}t^3\}=\mathscr{L}\{t^3\}\Big|_{s\rightarrow s-(-2)}\] Evaluate and then Shift:
\[=\frac{3!}{s^4}\Big|_{s\rightarrow s+2}\] \[=\frac{3!}{(s+2)^4}\]
A function \(f(t)\) is said to be of exponential order if \(\exists M,C,T>0\) such that \[|f(t)|\leq Me^{ct},\qquad \forall t\geq T\]
A function \(f(t)\) is a piecewise continuous for \(t\geq 0\) if all of the following are true.
Theorem 5.2 (Existance of Laplace Transform) If \(f(t)\) is piecewise continuous for \(t\geq 0\) and of exponential order for some positive constant \(C\) and for \(t\geq T\), then \[\mathscr{L}\{f(t)\}\] exists for \(s>C\).