Question

Consider a single-input-single-output (SISO) system with the transfer function \[ G_p(s)=\frac{2(s+1)}{\left(\frac{1}{2}s+1\right)\left(\frac{1}{4}s+1\right)} \] where the time constants are in minutes. The system is forced by a unit step input at time $t=0$. The time at which the output response reaches its maximum is \(\underline{\hspace{2cm}}\) minutes (rounded off to two decimal places).

Hint:

For overdamped systems, the maximum in step response appears where the first derivative of the inverse Laplace expression becomes zero.

Answer 1

Correct Answer: 0.54 - 0.56

The transfer function is: \[ G(s)=2(s+1)\Big/\left[(0.5s+1)(0.25s+1)\right] \] For a unit step input, \[ Y(s)=\frac{2(s+1)}{s(0.5s+1)(0.25s+1)} \] The time of maximum output occurs when \[ \frac{dy(t)}{dt}=0 \] This requires using inverse Laplace and solving the derivative expression for its root. After performing partial fractions and differentiating the time-domain response, the first positive solution of \[ \frac{dy}{dt}=0 \] occurs at: \[ t \approx 0.55\ \text{min} \] Thus the required time is within: \[ \boxed{0.54\text{ to }0.56\text{ min}} \]