Let $f ( t ) = \left\{ \begin{array} { l l }
3 & \text { if } 0 \leq t \leq 2 \
5 - t & \text { if } t 2
\end{array} \right\}$ . Then $\mathcal { L } \{ f ( t ) \}$ is
A) $3 / s ^ { 2 } - e ^ { - 2 s } / s ^ { 2 }$
B) $3 / s - e ^ { - 2 s } / s ^ { 2 }$
C) $3 / s ^ { 2 } + e ^ { - 2 s } / s ^ { 2 }$
D) $3 / s + e ^ { - 2 s } / s$
E) $3 / s ^ { 2 } + e ^ { 2 s } / s ^ { 2 }$

Question

Let $f ( t ) = \left\{ \begin{array} { l l } 
3 & \text { if } 0 \leq t \leq 2 \
5 - t & \text { if } t > 2
\end{array} \right\}$ . Then $\mathcal { L } \{ f ( t ) \}$ is
A) $3 / s ^ { 2 } - e ^ { - 2 s } / s ^ { 2 }$
B) $3 / s - e ^ { - 2 s } / s ^ { 2 }$
C) $3 / s ^ { 2 } + e ^ { - 2 s } / s ^ { 2 }$
D) $3 / s + e ^ { - 2 s } / s$
E) $3 / s ^ { 2 } + e ^ { 2 s } / s ^ { 2 }$

Quizplus · Accepted Answer

The Laplace transform of $f(t)$ is calculated by breaking it into two parts: the constant part $3$ for $0 \leq t \leq 2$, and the piecewise part $5-t$ for $t > 2$. The Laplace transform of a constant $a$ is $a/s$, and the transform of $5-t$ shifted by 2 units is $-e^{-2s}/s^2$. Adding these gives $3/s - e^{-2s}/s^2$.

Let f(t)={3 if 0≤t≤25−t if t>2}f ( t ) = \left\{ \begin{array} { l l } 3 & \text { if } 0 \leq t \leq 2 \\5 - t & \text { if } t > 2\end{array} \right\}f(t)={35−t​ if 0≤t≤2 if t>2​}

Let $f ( t ) = \left\{ \begin{array} { l l } 3 & \text { if } 0 \leq t \leq 2 \\5 - t & \text { if } t > 2\end{array} \right\}$