Selv om vi kan bruke formelen for Laplace transformen:
F(s) = \mathcal{L}(f(t)) = \int_0^{\infty} e^{-st} f(t) \; dter det ofte raskere og enklere å bruke tabeller:
| Nr. | $f(\textcolor{blue}{t})= \mathcal{L}^{-1}(f(\textcolor{red}{s}))$ | $F(\textcolor{red}{s}) = \mathcal{L}(f(\textcolor{blue}{t}))$ | Gjelder | Utregning |
|---|---|---|---|---|
| 1 | $1$ | $\frac{1}{\textcolor{red}{s}}$ | $\textcolor{red}{s} > 0$ | årsak |
| 2 | $\textcolor{blue}{t}$ | $\frac{1}{\textcolor{red}{s}^2}$ | $\textcolor{red}{s} > 0$ | årsak |
| 3 | $\textcolor{blue}{t}^n$ | $\frac{n!}{\textcolor{red}{s}^{n+1}}$ | $\textcolor{red}{s} > 0, \quad n = \{1,2,3,\cdots\}$ | |
| 4 | $e^{a\textcolor{blue}{t}}$ | $\frac{1}{\textcolor{red}{s}-a}$ | $\textcolor{red}{s} > a$ | årsak |
| 5 | $\textcolor{blue}{t} e^{a\textcolor{blue}{t}}$ | $\frac{1}{(\textcolor{red}{s}-a)^2}$ | $\textcolor{red}{s} > a$ | |
| 6 | $\textcolor{blue}{t}^n e^{a\textcolor{blue}{t}}$ | $\frac{n!}{(\textcolor{red}{s}-a)^{n+1}}$ | $\textcolor{red}{s} > a, \quad n = \{1,2,3,\cdots\}$ | |
| 7 | $\sin(b \textcolor{blue}{t})$ | $\frac{b}{\textcolor{red}{s}^2 + b^2}$ | $\textcolor{red}{s} > 0$ | årsak |
| 8 | $\cos(b \textcolor{blue}{t})$ | $\frac{\textcolor{red}{s}}{\textcolor{red}{s}^2 + b^2}$ | $\textcolor{red}{s} > 0$ | årsak |
| 9 | $e^{a\textcolor{blue}{t}}\sin(b \textcolor{blue}{t})$ | $\frac{b}{(\textcolor{red}{s}-a)^2 + b^2}$ | $\textcolor{red}{s} > a$ | |
| 10 | $e^{a\textcolor{blue}{t}}\cos(b \textcolor{blue}{t})$ | $\frac{\textcolor{red}{s}-a}{(\textcolor{red}{s}-a)^2 + b^2}$ | $\textcolor{red}{s} > a$ | |
| 11 | $u(\textcolor{blue}{t}-a)$ | $\frac{1}{\textcolor{red}{s}} e^{-a\textcolor{red}{s}}$ | $\textcolor{red}{s} > 0, \quad a \ge 0$ | årsak |
| 12 | $\delta(\textcolor{blue}{t})$ | $1$ | $\textcolor{red}{s} > 0$ | årsak |
| 13 | $\delta(\textcolor{blue}{t-a})$ | $e^{-a\textcolor{red}{s}}$ | $\textcolor{red}{s} > 0, \quad a \ge 0$ | årsak |
| 14 | $g(\textcolor{blue}{t}) + h(\textcolor{blue}{t})$ | $\mathcal{L}(g(\textcolor{blue}{t})) + \mathcal{L}(h(\textcolor{blue}{t}))$ | Laplace-transformasjonen er lineær | |
| 15 | $ag(\textcolor{blue}{t})$ | $a\mathcal{L}(g(\textcolor{blue}{t}))$ | $a$ er konstant | Laplace-transformasjonen er lineær |
| 16 | $h(t)u(t-a)$ | $e^{-as}\mathcal{L}(h(t+a))$ | $a > 0$ | årsak |
+ Enkle eksempler
+ Kort video
Her er noen eksempler på bruk av tabellen:
\begin{array}{rclcrcl}
f(t) &=& 1 & \overset{\textnormal{Rad nr. 1}}\Longleftrightarrow & F(s) &=& \frac{1}{s}, \quad s > 0 \\
g(t) &=& t & \overset{\textnormal{Rad nr. 2}}\Longleftrightarrow & G(s) &=& \frac{1}{s^2}, \quad s > 0 \\
h(t) &=& e^{2t} & \overset{\textnormal{Rad nr. 5}}\Longleftrightarrow & H(s) &=& \frac{1}{s-2}, \quad s > 2
\end{array}Merk at vi ofte bruker små bokstaver for den opprinnelige funksjonen og store bokstaver for den Laplace-transformerte funksjonen.
+ Eksempel 1: $f(t) = 5t – 7$
+ Kort video
Laplace-transformasjonen til $f(t) = 5t – 7$:
\begin{aligned}
F(s) & = \mathcal{L}\Big(f(t)\Big) \\
\textnormal{Setter inn funksjonen} \quad \Rightarrow \quad F(s) & = \mathcal{L}(5t - 7) \\
\textnormal{Bruker rad 14 i tabellen} \quad \Rightarrow \quad F(s) & = \mathcal{L}(5t) - \mathcal{L}(7) \\
\textnormal{Bruker rad 15 i tabellen} \quad \Rightarrow \quad F(s) & = 5\mathcal{L}(t) - 7\mathcal{L}(1) \\
\textnormal{Bruker rad 1 og 2 i tabellen} \quad \Rightarrow \quad F(s) & = 5 \cdot \frac{1}{s^2} - 7 \cdot \frac{1}{s} \\
\textnormal{Forenkler uttrykket} \quad \Rightarrow \quad F(s) & = \frac{5}{s^2} - \frac{7}{s} \\
\Rightarrow \quad F(s) & = \frac{5 - 7s}{s^2}
\end{aligned}+ Eksempel 2: $f(t) = 5e^t + 2\sin(4t)$
+ Kort video
Laplace-transformasjonen til $f(t) = 5e^t + 2\sin(4t)$:
\begin{aligned}
F(s) & = \mathcal{L}\Big(f(t)\Big) \\
\textnormal{Setter inn funksjonen} \quad \Rightarrow \quad F(s) & = \mathcal{L}(5e^t + 2\sin(4t)) \\
\textnormal{Bruker rad 14 i tabellen} \quad \Rightarrow \quad F(s) & = \mathcal{L}(5e^t) + \mathcal{L}(2\sin(4t)) \\
\textnormal{Bruker rad 15 i tabellen} \quad \Rightarrow \quad F(s) & = 5\mathcal{L}(e^t) + 2\mathcal{L}(\sin(4t)) \\
\textnormal{Bruker rad 4 og 7 i tabellen} \quad \Rightarrow \quad F(s) & = 5 \cdot \frac{1}{s-1} + 2 \cdot \frac{4}{s^2 + 4^2} \\
\textnormal{Forenkler uttrykket} \quad \Rightarrow \quad F(s) & = \frac{5}{s-1} + \frac{8}{s^2 + 16}
\end{aligned}+ Eksempel 3: $f(t) = (5 + 2\sin(3t))e^{4t}$
+ Kort video
Laplace-transformasjonen til $f(t) = (5 + 2\sin(3t))e^{4t}$:
\begin{aligned}
F(s) & = \mathcal{L}\Big(f(t)\Big) \\
\textnormal{Setter inn funksjonen} \quad \Rightarrow \quad F(s) & = \mathcal{L}\Big((5 + 2\sin(3t)) e^{4t} \Big) \\
\textnormal{Multipliserer } e^{4t} \textnormal{ inn i parentesen} \quad \Rightarrow \quad F(s) & = \mathcal{L}\Big(5 e^{4t} + 2 \sin(3t) e^{4t} \Big) \\
\textnormal{Bruker rad 14 i tabellen} \quad \Rightarrow \quad F(s) & = \mathcal{L}\Big(5 e^{4t}\Big) + \mathcal{L}\Big(2\sin(3t) e^{4t} \Big) \\
\textnormal{Bruker rad 15 i tabellen} \quad \Rightarrow \quad F(s) & = 5\mathcal{L}\Big(e^{4t}\Big) + 2\mathcal{L}\Big(\sin(3t) e^{4t}\Big) \\
\textnormal{Bruker rad 4 og 9 i tabellen} \quad \Rightarrow \quad F(s) & = 5 \cdot \frac{1}{s-4} + 2 \cdot \frac{3}{(s-4)^2 + 3^2} \\
\textnormal{Forenkler uttrykket} \quad \Rightarrow \quad F(s) & = \frac{5}{s-4} + \frac{6}{(s-4)^2 + 9}
\end{aligned}