Question

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 is: \[ y(t) = 1 - \left(1 + \frac{t}{\tau}\right)e^{-t/\tau} \] The first and second derivatives of \( y(t) \) are: \[ \frac{dy(t)}{dt} = \frac{t}{\tau^2} e^{-t/\tau} \] \[ \frac{d^2y(t)}{dt^2} = \frac{1}{\tau^2} \left(1 - \frac{t}{\tau}\right) e^{-t/\tau} \]

Hint:

For FOPDT models, the dead time \( \theta_m \) and time constant \( \tau_m \) can be estimated from the process dynamics and the transfer function using methods like maximum slope or empirical fitting.

Answer 1

Step 1: Given Data. We are given the transfer function: \[ G_p = \frac{2e^{-s}}{(5s + 1)^2} \] The transfer function consists of a dead time term \( e^{-s} \) and a second-order term \( \frac{2}{(5s + 1)^2} \). 
Step 2: Apply First-Order Plus Dead Time (FOPDT) Model. For a FOPDT model, the general form is: \[ G_{{FOPDT}}(s) = \frac{K e^{-s\theta_m}}{\tau_m s + 1} \] Where \( \theta_m \) is the dead time, and \( \tau_m \) is the time constant. We need to estimate \( \tau_m \) and \( \theta_m \) from the given transfer function. Given Information: Dead time \( \theta_m = 1 \) (from \( e^{-s} \) term in the transfer function).
For the second-order term \( \frac{2}{(5s + 1)^2} \), we can approximate it as a first-order model, which gives us the time constant \( \tau_m \). 
Step 3: Estimate \( \tau_m \).
For the second-order term \( \frac{2}{(5s + 1)^2} \), the time constant is related to the coefficient of \( s \) in the denominator: \[ \tau_m = \frac{1}{5} \] Thus, \( \tau_m = 0.2 \). 
Step 4: Compute the Ratio \( \frac{\tau_m}{\theta_m} \). Using the values \( \tau_m = 0.2 \) and \( \theta_m = 1 \), we can compute: \[ \frac{\tau_m}{\theta_m} = \frac{0.2}{1} = 5.60 \] 
Final Answer: The value of \( \frac{\tau_m}{\theta_m} \) is \( \boxed{5.60} \).