GATE 2025 IN Q.28: solution and answer

Official answer: BSource: Official answer key, GATE 2025 IN (IIT Roorkee)

1. Closed-loop characteristic function: $F(s)=1+G(s)H(s)$. Its poles are the open-loop poles, i.e. the poles of $G(s)H(s)$. Its zeros are the closed-loop poles.

2. Data: the contour of infinite radius encloses the whole right half-plane and contains $P=2$ poles of $G(s)H(s)$, so $F(s)$ also has $P=2$ poles inside it. No poles lie on the $j\omega$-axis, so the contour needs no indentation.

3. Argument principle (the direction matters): as $s$ goes once round a closed contour, the image $F(s)$ encircles the origin $N=Z-P$ times in the same direction the contour is traversed. Here $Z$ and $P$ are the zeros and poles of $F$ inside the contour.

4. The contour in Figure (b) runs up the semicircle and down the $j\omega$-axis, which keeps the enclosed right half-plane on its left, so it is counterclockwise. Therefore the net number of counterclockwise encirclements of the origin by $1+G(s)H(s)$ is $N_{\text{ccw}}=Z-P$.

5. Closed-loop stability means no closed-loop poles in the right half-plane, i.e. $Z=0$: $$N_{\text{ccw}}=0-2=-2$$

6. A count of $-2$ counterclockwise encirclements is 2 clockwise encirclements of the origin by $1+G(s)H(s)$. Since $G(s)H(s)=F(s)-1$ is just $F(s)$ shifted left by 1, this is the same as $G(s)H(s)$ encircling $-1+j0$ twice clockwise.

7. Check the options: (A) pairs $G(s)H(s)$ with the origin and (D) pairs $1+G(s)H(s)$ with $-1+j0$, so both use the wrong critical point. (C) is the textbook rule for the usual clockwise Nyquist contour, and it gets the direction backwards for the counterclockwise contour drawn here. Only (B) is correct.