GATE Chemical 2025 | Question: 50

Question

Consider a process with transfer function

$$G_{p}=\frac{2 e^{-s}}{(5 s+1)^{2}}$$

A first-order plus dead time $\text{(FOPDT)}$ 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:
\[
\begin{array}{l}
y_{(t)}=1-\left(1+\frac{t}{\tau}\right) e^{-t / \tau} \\
\frac{d y_{(t)}}{d t}=\frac{t}{\tau^{2}} e^{-t / \tau} \\
\frac{d^{2} y_{(t)}}{d t^{2}}=\frac{1}{\tau^{2}}\left(1-\frac{t}{\tau}\right) e^{-t / \tau}
\end{array}
\]

Answer 1

In this question i used the concept of maximum slope method as stated in the question. here we are just trying to convert that complex transfer function to simple first order. for that first i saw at what time we are getting maximum slope. that is t=5 sec (actually 6 because there is already a time delay of 6 secs). so t=6sec. means if we plot y(t) graph then we will get maximum slope at t=6sec. 

 

after that we want to know the coordinates of that point on the graph. which is coming to be (y(6),6). y(6) can be calculated easily but that given equation

after getting the coordinates. we found the slope of the tangent at those coordinates. which is coming to be around 2/5e.

after that we will write the eqn of the tangent as we now know both slope and coordinates. after writing the equation we put y=0 to get the required time . i mean where the tangent is cutting the x axis. we will get t=2.4sec.( which is the dead time i.e thetaM)

now ,from this equation . we will get tauP=K/thetaP( here we can write "m" instead of p because "m" is given in the question. we will get tauP(taum)==13.59

so final ans==[dead time(t)/time constant(taum)]===5.663