Question

A process described by the transfer function \[ G_p(s) = \frac{10s + 1}{5s + 1} \] is forced by a unit step input at time $t = 0$. The output value immediately after the step input (at $t = 0^+$) is \(\underline{\hspace{2cm}}\) (rounded off to the nearest integer).

Hint:

Use the initial value theorem to find the instantaneous output of systems with step inputs.

Answer 1

Correct Answer: 2

For a unit step input, \[ U(s) = \frac{1}{s} \] The output is: \[ Y(s) = G_p(s) \cdot U(s) = \frac{10s + 1}{(5s+1)s} \] To find the output at $t = 0^+$, we use the initial value theorem: \[ y(0^+) = \lim_{s\to\infty} sY(s) \] \[ y(0^+) = \lim_{s\to\infty} s \cdot \frac{10s+1}{(5s+1)s} = \lim_{s\to\infty} \frac{10s + 1}{5s + 1} \] Divide numerator and denominator by \(s\): \[ y(0^+) = \frac{10 + \frac{1}{s}}{5 + \frac{1}{s}} \xrightarrow{s\to\infty} \frac{10}{5} = 2 \] Thus the immediate response after the step input is 2.