Question

Laplace transform of \(\cos \left( {{\rm{\omega t}}} \right){\rm{is}}\frac{{\rm{s}}}{{{{\rm{s}}^2} + {{\rm{\omega }}^2}}}\). The laplace transform of e^-2t cos(4t) is
1. \(\frac{{{\rm{s}} - 2}}{{{{\left( {{\rm{s}} - 2} \right)}^2} + 16}}\)
2. \(\frac{{{\rm{s}} + 2}}{{{{\left( {{\rm{s}} - 2} \right)}^2} + 16}}\)
3. \(\frac{{{\rm{s}} - 2}}{{{{\left( {{\rm{s}} + 2} \right)}^2} + 16}}\)
4. \(\frac{{{\rm{s}} + 2}}{{{{\left( {{\rm{s}} + 2} \right)}^2} + 16}}\)

Answer 1

Correct Answer - Option 4 : \(\frac{{{\rm{s}} + 2}}{{{{\left( {{\rm{s}} + 2} \right)}^2} + 16}}\)

Concept:

If L{f(t)} = F(s), then

• L{e^at f(t)} = F(s – a)
• L{e^-at f(t)} = F(s + a)

Calculation:

\(L\left\{ {\cos 4t} \right\} = \frac{s}{{{s^2} + {4^2}}}\)

\(\therefore {\rm{L}}\left\{ {{{\rm{e}}^{ - 2{\rm{t}}}}\cos 4{\rm{t}}} \right\} = \frac{{{\rm{s}} + 2}}{{{{\left( {{\rm{s}} + 2} \right)}^2} + {4^2}}}\)

L{e^-2t cos 4t} \(= \frac{{{\rm{s}} + 2}}{{{{\left( {{\rm{s}} + 2} \right)}^2} + 16}}\)