Question

A forcing function (t² – 2t) u (t – 1) is applied to a linear system. The \({\cal L}\)- transform of the forcing function is
1. \(\frac{{2 - s}}{{{s^3}}}\epsilon{^{ - 2s}}\)
2. \(\left( {\frac{{1 - {s^2}}}{S}} \right)\epsilon{^{ - s}}\)
3. \(\frac{1}{s}{e^{ - s}} - \frac{1}{{{s^2}}}\epsilon{^{ - 2s}}\)
4. \(\left( {\frac{{2 - {s^2}}}{{{s^3}}}} \right)\epsilon{^{ - S}}\)

Answer 1

Correct Answer - Option 4 : \(\left( {\frac{{2 - {s^2}}}{{{s^3}}}} \right)\epsilon{^{ - S}}\)

Concept:

The definition of unilateral Laplace transform is:

\(X\left( s \right) = \mathop \smallint \nolimits_0^\infty x\left( t \right){e^{ - st}}dt\)

Laplace transform of a function of f(t) is shown by:

\(\cal L\) [f(t)] = F(s)

Differentiation property

\(u\left( t \right) \leftrightarrow \frac{1}{s}\)

\(u\left( t-1 \right) \leftrightarrow \frac{e^{-s}}{s}\)

\(tf\left( t \right) \leftrightarrow - \frac{d}{{ds}}\left( {F\left( s \right)} \right)\)

Calculation:

given that forcing function

f(t) = (t2 – 2t) u (t – 1)

f(t) = (t – 1)² u (t – 1) - u (t – 1)

Now Laplace transform of the given function

\(X\left( s \right) = \mathop \smallint \nolimits_0^\infty x\left( t \right){e^{ - st}}dt\)

F(s) = \({\cal L}\) [f(t)] = \(\cal L\) [(t – 1)2 u (t – 1)] - \(\cal L\) [u (t – 1)]

\(u\left( t-1 \right) \leftrightarrow \frac{e^{-s}}{s}\)

\((t-1)^2u\left( t-1 \right) \leftrightarrow \frac{2e^{-s}}{s^3}\)

Now combine those transform

\(F\left( s \right) = \frac{{{2e^{ - s}}}}{{{s^3}}} - \frac{{{e^{ - s}}}}{s}\)

\(F\left( s \right) = \left( {\frac{{2 - {s^2}}}{{{s^3}}}} \right){e^{ - s}}\)