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GATE 2023 CH – Question 44

Instrumentation and Process Control · Cascade and feedforward control · 2 marks · Multiple choice

A cascade control strategy is shown in the figure below. The transfer function between the output ($y$) and the secondary disturbance ($d_2$) is defined as $G_{d_2}(s) = \frac{y(s)}{d_2(s)}$.
Which one of the following is the CORRECT expression for the transfer function $G_{d_2}(s)$ ?

$y_{set}$ goes to a summer (with the feedback of $-y$), then to the primary controller of gain 1, then to a second summer (with the feedback of the secondary variable, $-$), then to the secondary controller of gain 10, then to a third summer where $d_2$ enters through $\frac{1}{0.1s + 1}$. The sum drives the process $\frac{1}{s + 1}$ and its output is added to $d_1$ (through $\frac{1}{2s + 1}$) to give $y$.
  1. $\frac{1}{(11s + 21)(0.1s + 1)}$
  2. $\frac{1}{(s + 1)(0.1s + 1)}$
  3. $\frac{s + 1}{(s + 2)(0.1s + 1)}$
  4. $\frac{s + 1}{(s + 1)(0.1s + 1)}$

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Show answer and explanation

Correct answer: (A) $\frac{1}{(11s + 21)(0.1s + 1)}$

Explanation

Set $y_{set} = 0$ and $d_1 = 0$. Let $g = \frac{1}{0.1s + 1}$. The secondary variable is $y_2 = u + gd_2$, where $u = 10(u_1 - y_2)$ and the primary controller output is $u_1 = -y$. Then $y_2 = -10y - 10y_2 + gd_2$, so $11y_2 = -10y + gd_2$. The process gives $y = \frac{y_2}{s + 1}$, so $(s + 1)y = \frac{gd_2 - 10y}{11}$, that is $(11s + 21)y = gd_2$ and $G_{d_2} = \frac{1}{(11s + 21)(0.1s + 1)}$.