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GATE 2025 EC – Question 26

Control Systems · Bode and Root-Locus Plots · 1 mark · Multiple choice

Consider the unity-negative-feedback system shown in Figure (i) below, where gain $K\ge0$. The root locus of this system is shown in Figure (ii) below.

For what value(s) of $K$ will the system in Figure (i) have a pole at $-1+j1$?

Diagram for GATE 2025 EC question 26
  1. $K=5$
  2. $K=\frac15$
  3. For no positive value of $K$
  4. For all positive values of $K$

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Correct answer: (C) For no positive value of $K$

Explanation

From Figure (ii) the open-loop zeros are at $-3,-4$ and the poles at $-1,-2$, so the characteristic equation is $(s+1)(s+2)+K(s+3)(s+4)=0$, i.e. $(1+K)s^2+(3+7K)s+(2+12K)=0$. Substituting $s=-1+j$ ($s^2=-2j$): the real part gives $5K-1=0$ and the imaginary part gives $5K+1=0$. These cannot both hold, so $-1+j1$ is not on the root locus for any positive $K$.