Question

The output y(t) of a system is related to its input x(t) as \(y\left( t \right) = \mathop \smallint \limits_0^t x\left( {\tau - 2} \right)d\tau ,\)

where, x (t) = 0 and y(t) = 0 for t ≤ 0. The transfer function of the system is
1. \(\frac{1}{s}\)
2. \(\frac{{\left( {1 - {e^{ - 2s}}} \right)}}{s}\)
3. \(\frac{{{e^{ - 2s}}}}{s}\)
4. \(\frac{1}{s} - {e^{ - 2s}}\)

Answer 1

Correct Answer - Option 3 : \(\frac{{{e^{ - 2s}}}}{s}\)

\(y\left( t \right) = \mathop \smallint \limits_0^t x\left( {\tau - 2} \right)\;d\tau \)

Transfer function is the impulse response So, the input is impulse function, i.e. δ(t)

\(h\left( t \right) = \mathop \smallint \limits_0^t \delta \left( {\tau - 2} \right)d\tau\)

⇒ h(t) = u(t-2)

By applying Laplace transform

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