Question

The unilateral Laplace transform of f(t) is\(\frac{1}{{{s^2} + s + 1}}\) . Which one of the following is the unilateral Laplace transform of g(t) = t⋅f(t)?

1. \(\frac{{ - s}}{{\left( {{s^2} + s + 1} \right)}}\)
2. \(\frac{{ - \left( {2s + 1} \right)}}{{{{\left( {{s^2} + s + 1} \right)}^2}}}\)
3. \(\frac{s}{{{{\left( {{s^2} + s + 1} \right)}^2}}}\)
4. \(\frac{{2s + 1}}{{{{\left( {{s^2} + s + 1} \right)}^2}}}\)

Answer 1

Correct Answer - Option 4 : \(\frac{{2s + 1}}{{{{\left( {{s^2} + s + 1} \right)}^2}}}\)

Concept

Multiplication of function in one domain corresponds to the differentiation in another domain

If f(t) having the Laplace transform F(s) then t ⋅ f(t) will have the transform as

f(t) ↔\( - \frac{{dF\left( s \right)}}{{ds}}\)

Calculation:

Given function f(t) is . And  g(t) = t ⋅ f(t)

Laplace transform of g(t) is:

\(L[g(t)] = - \frac{d}{{ds}}\left( {\frac{1}{{{s^2}s + 1}}} \right)\)

\( = - \frac{{\frac{{d\left( 1 \right)}}{{ds}}\left( {{s^2} + s + 1} \right) - 1 \times \frac{{d\left( {{S^2} + S + 1} \right)}}{{ds}}}}{{{{\left( {{S^2} + S + 1} \right)}^2}}}\)

\( = - \left( {\frac{{0 - 1\left( {2s + 1} \right)}}{{{{\left( {{s^2} + s + 1} \right)}^2}}}} \right)\)

\( = \frac{{2s + 1}}{{{{\left( {{s^2} + s + 1} \right)}^2}}}\)