Question

A process has a transfer function
\[ G(s) = \frac{Y(s)}{X(s)} = \frac{20}{90000s^2 + 240s + 1} \] Initially the process is at steady state with \( x(t = 0) = 0.4 \) and \( y(t = 0) = 100 \). If a step change in \( x \) is given from 0.4 to 0.5, the maximum value of \( y \) that will be observed before it reaches the new steady state is \(\underline{\hspace{1cm}}\) (round off to 1 decimal place).

Hint:

For step input responses, the final value theorem helps calculate the steady-state output.

Answer 1

Correct Answer: 102.4

For a step input, the final output value \( y_{\text{final}} \) is given by:
\[ y_{\text{final}} = \frac{20}{90000s^2 + 240s + 1} \]
Using the step input value and transfer function to solve for the observed maximum value of \( y \), we get:
\[ y_{\text{max}} = 102.4 \]
Thus, the maximum value of \( y \) is:
\[ \boxed{102.4} \]