다음 함수 $f \left (t \right)$ 를 단위계단함수(unit step function)로 나타내고 $f \left (t \right)$ 의 라플라스 변환을 구하면?

$f \left (t \right) = \begin{cases} 1, 0 \leq t<3 \\ 0, 3 \leq t<5 \\ -1, t \geq 5 \end{cases}$

• ①$f \left (t \right) =1+u \left (t-3 \right) -u \left (t-5 \right)$, $\displaystyle \mathscr{L} \left \{f \left (t \right) \right \} = \frac{1} {s} + \frac{e ^ {3s}} {s} - \frac{e ^ {5s}} {s}$
• ②$f \left (t \right) =1+u \left (t-3 \right) -u \left (t-5 \right)$, $\displaystyle \mathscr{L} \left \{f \left (t \right) \right \} = \frac{1} {s} + \frac{e ^ {-3s}} {s} - \frac{e ^ {-5s}} {s}$
• ③$f \left (t \right) =1-u \left (t-3 \right) -u \left (t-5 \right)$, $\displaystyle \mathscr{L} \left \{f \left (t \right) \right \} = \frac{1} {s} - \frac{e ^ {3s}} {s} - \frac{e ^ {5s}} {s}$
• ④$f \left (t \right) =1-u \left (t-3 \right) -u \left (t-5 \right)$, $\displaystyle \mathscr{L} \left \{f \left (t \right) \right \} = \frac{1} {s} - \frac{e ^ {-3s}} {s} - \frac{e ^ {-5s}} {s}$