Question

The inverse Laplace transform of the function \(F\left( s \right) = \frac{1}{{s\left( {s + 1} \right)}}\) is given by
1. f (t) = sin t
2. f (t) = e^-t sint
3. f(t) = e^-t
4. f(t) = 1 - e^-t

Answer 1

Correct Answer - Option 4 : f(t) = 1 - e^-t

Concept:

If L-1{F(s)} = f(t)

then

L-1F(s – a) = eat.f(t) and L-1{F(s + a)} = e-at.f(t)

Calculation:

\(F\left( s \right) = \frac{1}{{s\left( {s + 1} \right)}} = \frac{A}{s} + \frac{B}{{\left( {s + 1} \right)}}\)

\(\frac{1}{{s\left( {s + 1} \right)}} = \frac{{A\left( {s + 1} \right) + B\left( s \right)}}{{s\left( {s + 1} \right)}}\)

A(s + 1) + B(s) = 1

Put s = 0, we get A = 1

Put s = -1, we get B = -1

\(F\left( s \right) = \frac{1}{{s\left( {s + 1} \right)}} = \frac{1}{s} + \frac{{ - 1}}{{\left( {s + 1} \right)}}\)

f(t) = L-1(s)

\(f\left( t \right) = {L^{ - 1}}\left( {\frac{1}{s} - \frac{1}{{s + 1}}} \right)\)

f(t) = e0t – e-t    \(\left\{\because {{L^{ - 1}}\left( {\frac{1}{s}} \right) = 1} \right\}\)

f(t) = 1 – e-t