The GATE Grind

GATE 2024 CH – Question 48

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

For the block diagram shown in the figure, the correct expression for the transfer function $G_d = \frac{y_1(s)}{d(s)}$ is

a cascade of two loops. The setpoint $y_1^{sp}$ goes to a summer (with the feedback of $-y_1$), then $G_{c1}$, which gives $y_2^{sp}$ to a second summer (with the feedback of $-y_2$), then $G_{c2}$ which gives the manipulated input $u$. The input $u$ drives $G_{p2}$, whose output is added to the disturbance $d$ to give $y_2$, and also drives $G_{p1}$, whose output is $y_1$.
  1. $\frac{-G_{p1}G_{c2}}{(1 + G_{c1}G_{c2}G_{p1})(1 + G_{c2}G_{p2})}$
  2. $\frac{-G_{p1}G_{c2}}{1 + G_{c2}G_{p2} + G_{c1}G_{c2}G_{p1}G_{p2}}$
  3. $\frac{-G_{p1}G_{c2}}{1 + G_{c2}G_{p2} + G_{c1}G_{c2}G_{p1}}$
  4. $\frac{1}{1 + G_{c2}G_{p2} + G_{c1}G_{c2}G_{p1}G_{p2}}$

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Correct answer: (C) $\frac{-G_{p1}G_{c2}}{1 + G_{c2}G_{p2} + G_{c1}G_{c2}G_{p1}}$

Explanation

Set $y_1^{sp} = 0$. Then $y_2^{sp} = -G_{c1}y_1 = -G_{c1}G_{p1}u$. The manipulated input is $u = G_{c2}(y_2^{sp} - y_2)$ with $y_2 = G_{p2}u + d$, so $u = G_{c2}(-G_{c1}G_{p1}u - G_{p2}u - d)$ and $u(1 + G_{c1}G_{c2}G_{p1} + G_{c2}G_{p2}) = -G_{c2}d$. With $y_1 = G_{p1}u$: $G_d = \frac{-G_{p1}G_{c2}}{1 + G_{c2}G_{p2} + G_{c1}G_{c2}G_{p1}}$.