Question

The output of a linear system for step input is t²e^-t. Then the transfer function is
1. \(\frac{s}{\left(s + 1\right)^2}\)
2. \(\frac{2s}{\left(s + 1\right)^3}\)
3. \(\frac{s}{s^2\left(s + 1\right)}\)
4. \(\frac{1}{\left(s + 1\right)^3}\)

Answer 1

Correct Answer - Option 2 : \(\frac{2s}{\left(s + 1\right)^3}\)

Concept:

A transfer function is defined as the ratio of the Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

TF = L[output]/L[input]

\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}}\)

For unit impulse input i.e. r(t) = δ(t)

⇒ R(s) = δ(s) = 1

Now transfer function = C(s)

Therefore, the transfer function is also known as the impulse response of the system.

Transfer function = L[IR]

IR = L-1 [TF]

Analysis:-

Given,

r(t) = u(t)

c(t) = t² e^-t

\(R\left( s \right) = \frac{1}{s}\)

\(C\left( s \right) = \frac{2}{{{{\left( {s + 1} \right)}^3}}}\)

Now,

\(H\left( s \right) = \frac{{C\left( s \right)}}{{R\left( s \right)}}\)

\(= \frac{{2s}}{{{{\left( {s + 1} \right)}^3}}}\)

Note:-

\(\frac{{{t^{n - 1}}}}{{\left( {n - 1} \right)!}}{e^{ - at}}\;u\left( t \right)\) L.T à  \(\frac{1}{{{{\left( {s + a} \right)}^n}}}\)