Question

The bilateral Laplace transform of a function \({\rm{f}}\left( {\rm{t}} \right) = \left\{ {\begin{array}{*{20}{c}} {1{\rm{\;if\;a}} \le {\rm{t}} \le {\rm{b}}}\\ {0{\rm{\;otherwise}}} \end{array}} \right.\)
1. \(\frac{{{\rm{a}} - {\rm{b}}}}{{\rm{s}}}\)
2. \({\rm{}}\frac{{{{\rm{e}}^2}\left( {{\rm{a}} - {\rm{b}}} \right)}}{{\rm{s}}}\)
3. \(\frac{{{{\rm{e}}^{ - {\rm{as}}}} - {{\rm{e}}^{ - {\rm{bs}}}}}}{{\rm{s}}}\)
4. \(\frac{{{{\rm{e}}^{{\rm{s}}\left( {{\rm{a}} - {\rm{b}}} \right)}}}}{{\rm{s}}}\)

Answer 1

Correct Answer - Option 3 : \(\frac{{{{\rm{e}}^{ - {\rm{as}}}} - {{\rm{e}}^{ - {\rm{bs}}}}}}{{\rm{s}}}\)

Given \({\rm{f}}\left( {\rm{t}} \right) = \begin{array}{*{20}{c}} 1\\ 0 \end{array}{\rm{\;}}\begin{array}{*{20}{c}} {{\rm{\;}}:{\rm{\;a}} \le {\rm{t}} \le {\rm{b}}}\\ {{\rm{\;\;}}:{\rm{otherwise}}} \end{array}\)

\(\begin{array}{l} {\rm{L}}\left\{ {{\rm{f}}\left( {\rm{t}} \right)} \right\} = \mathop \smallint \limits_{ - \infty }^\infty {{\rm{e}}^{ - {\rm{st}}}}{\rm{f}}\left( {\rm{t}} \right){\rm{dt}}\\ = \mathop \smallint \limits_{ - \infty }^{\rm{a}} {{\rm{e}}^{ - {\rm{st}}}}{\rm{f}}\left( {\rm{t}} \right) + \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {{\rm{e}}^{ - {\rm{st}}}}{\rm{f}}\left( {\rm{t}} \right){\rm{dt}} + \mathop \smallint \limits_{\rm{b}}^\infty {{\rm{e}}^{ - {\rm{st}}}}{\rm{f}}\left( {\rm{t}} \right){\rm{dt}}\\ = 0 + \mathop \smallint \limits_{\rm{a}}^{\rm{b}} {{\rm{e}}^{ - {\rm{st}}}}{\rm{dt}} + 0\\ = \left. {\frac{{{{\rm{e}}^{ - {\rm{st}}}}}}{{ - {\rm{s}}}}} \right|_{\rm{a}}^{\rm{b}} = \frac{{ - 1}}{{\rm{s}}}\left[ {{{\rm{e}}^{ - {\rm{bs}}}} - {{\rm{e}}^{ - {\rm{as}}}}} \right]\\ = \frac{{{{\rm{e}}^{ - {\rm{as}}}} - {{\rm{e}}^{ - {\rm{bs}}}}}}{{\rm{s}}} \end{array}\)