Question

The response of the system \({\rm{G}}\left( {\rm{s}} \right) = \frac{{{\rm{s}} - 2}}{{\left( {{\rm{s}} + 1} \right)\left( {{\rm{s}} + 3} \right)}}\) to the unit step input \(\rm u(t)\) is \(\rm y(t)\).

The value of \(\frac{{{\rm{dy}}}}{{{\rm{dt}}}}\) at \(\rm t = 0^+\) is _________

Answer 1

We have, \({\rm{G}}\left( {\rm{s}} \right) = \frac{{{\rm{s}} - 2}}{{\left( {{\rm{s}} + 1} \right)\left( {{\rm{s}} + 3} \right)}}\)

Now, \(\rm Y(s) = G(s). X(s)\)

We have, \(\rm X(s) = 1/s\)

\(\Rightarrow {\rm{Y}}\left( {\rm{s}} \right) = \frac{{\left( {{\rm{s}} - 2} \right)}}{{{\rm{s}}\left( {{\rm{s}} + 1} \right)\left( {{\rm{s}} + 3} \right)}}\)

Using properties of Laplace transform we have

\(\frac{{{\rm{dy}}\left( {\rm{t}} \right)}}{{{\rm{dt}}}}\mathop \leftrightarrow \limits^{{\rm{LT}}} {\rm{s}}.{\rm{Y}}\left( {\rm{s}} \right)\)

Thus, \(\frac{{{\rm{dy}}\left( {\rm{t}} \right)}}{{{\rm{dt}}}}\mathop \leftrightarrow \limits^{{\rm{LT}}} \frac{{{\rm{s}} - 2}}{{\left( {{\rm{s}} + 1} \right)\left( {{\rm{s}} + 3} \right)}}\)

Taking inverse Laplace of \(\frac{{{\rm{s}} - 2}}{{\left( {{\rm{s}} + 1} \right)\left( {{\rm{s}} + 3} \right)}}\) assuming right sided ROC, as we need to calculate \(\frac{{{\rm{dy}}\left( {\rm{t}} \right)}}{{{\rm{dt}}}}\) at \(\rm t = 0^+\), we have

\(\frac{{\left( {{\rm{s}} - 2} \right)}}{{\left( {{\rm{s}} + 1} \right)\left( {{\rm{s}} + 3} \right)}} = \frac{{ - \frac{3}{2}}}{{\left( {{\rm{s}} + 1} \right)}} + \frac{{\frac{5}{2}}}{{\left( {{\rm{s}} + 3} \right)}}\)

Thus, \(\frac{{{\rm{dy}}\left( {\rm{t}} \right)}}{{{\rm{dt}}}} = - \frac{3}{2}{{\rm{e}}^{ - {\rm{t}}}}{\rm{u}}\left( {\rm{t}} \right) + \frac{5}{2}{{\rm{e}}^{ - 3{\rm{t}}}}{\rm{u}}\left( {\rm{t}} \right)\)

\(\Rightarrow {\left. {\frac{{{\rm{dy}}\left( {\rm{t}} \right)}}{{{\rm{dt}}}}} \right|_{{\rm{t}} = {0^ + }}} = - \frac{3}{2} + \frac{5}{2} = \frac{2}{2} = 1\)