GATE 2025 CH – Question 60
Consider a process with transfer function $G_p = \frac{2e^{-s}}{(5s + 1)^2}$
A first-order plus dead time (FOPDT) model is to be fitted to the unit step process reaction curve (PRC) by applying the maximum slope method.
Let $\tau_m$ and $\theta_m$ denote the time constant and dead time, respectively, of the fitted FOPDT model. The value of $\frac{\tau_m}{\theta_m}$ is ____ (rounded off to 2 decimal places).
GIVEN: For $G = \frac{1}{(\tau s + 1)^2}$ the unit step output response: $y(t) = 1 - \left(1 + \frac{t}{\tau}\right)e^{-t/\tau}$, $\frac{dy}{dt} = \frac{t}{\tau^2}e^{-t/\tau}$, $\frac{d^2y}{dt^2} = \frac{1}{\tau^2}\left(1 - \frac{t}{\tau}\right)e^{-t/\tau}$
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: 5.61 to 5.67
Explanation
Work with the response normalised by the gain, so the final value is 1. The slope is greatest where $\frac{d^2y}{dt^2} = 0$, that is at $t = \tau = 5$ min (the inflection point). There $y = 1 - 2e^{-1} = 0.2642$ and the slope is $\frac{1}{5}e^{-1} = 0.07358$ per min. The tangent at this point cuts the time axis at $t_0 = 5 - \frac{0.2642}{0.07358} = 1.409$ min and reaches the final value 1 at $t_1 = 5 + \frac{1 - 0.2642}{0.07358} = 15.0$ min. The FOPDT model has the time constant $\tau_m = t_1 - t_0 = \frac{1}{0.07358} = 13.59$ min and a dead time equal to $t_0$ plus the transport delay of 1 min, $\theta_m = 1.409 + 1 = 2.409$ min. The ratio is $\frac{13.59}{2.409} = 5.64$.