Question

The transfer function relating the input x(t) to the output y(t) of a system is given by G(s) = 1/(s + 3). A unit-step input is applied to the system at time t = 0. Assuming that y(0) = 3, the value of y(t) at time t = 1 is ______ (Answer should be rounded off to two decimal places)

Answer 1

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

\(\Rightarrow \frac{{Y\left( s \right)}}{{X\left( s \right)}} = \frac{1}{{s + 3}}\) 

⇒ (s + 3) Y(s) = X(s)

⇒ sY(s) + 3Y(s) = X(s)

\(\Rightarrow \frac{{dy}}{{dt}} + 3y = x\left( t \right)\) 

⇒ sY(s) - Y(0) + 3Y(s) = X(s)

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

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

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

\(\Rightarrow y\left( t \right) = \frac{1}{3} + \frac{8}{3}{e^{ - 3t}}\) 

At t = 1,

\(y\left( t \right) = \frac{1}{3} + \frac{8}{3}{e^{ - 3}} = 0.462.\)