GATE 2025 EE Q.23: solution and answer

Official answer: DSource: Official answer key, GATE 2025 EE (IIT Roorkee)

1. Open-loop RHP poles. $G(s)$ is strictly stable, so all its poles lie in the open left half-plane and none lie on the $j\omega$ axis. The poles of $1+G(s)$ are the poles of $G(s)$, so the number of open-loop RHP poles is $$P=0$$

2. The zero at $s=3$ does not enter the count. $(s-3)$ puts a zero of $G(s)$ in the RHP. The Nyquist criterion counts the zeros of $1+G(s)$. For this unity negative feedback loop those zeros are the closed-loop poles, because $T(s)=\frac{G(s)}{1+G(s)}$. A zero of $G(s)$ is neither a pole nor a zero of $1+G(s)$, so it plays no part in the count.

3. Argument principle, with no sign convention assumed. Let the Nyquist contour enclose the whole RHP. As $s$ goes once round it, $G(s)$ traces the Nyquist plot. The net number of times that plot encircles $-1$, counted in the same sense as the contour is traversed, is $$N=Z-P$$ Here $Z$ is the number of zeros of $1+G(s)$ in the RHP, which is the number of closed-loop RHP poles.

4. What $P=0$ implies. With $P=0$ we get $N=Z\ge 0$. So the plot of a stable $G(s)$ can encircle $-1$ only in the direction the contour is traversed, and every net encirclement is one closed-loop RHP pole. Increasing $\omega$ (clockwise contour) gives clockwise encirclements. Decreasing $\omega$ (anticlockwise contour) gives anticlockwise ones. The question does not say which way the plot is traced, and its 'anticlockwise' fits the second reading. Either way, the number of encirclements is what decides stability.

5. Put in the data. The plot encircles $-1$ exactly once, so $$Z=N+P=1+0=1$$ The closed loop has one pole in the right half-plane.

6. Ruling out marginal stability. The point $-1$ is encircled, so it does not lie on the plot. That means $1+G(j\omega )\ne 0$ for every $\omega$, so no closed-loop pole sits on the $j\omega$ axis. The system is therefore not marginally stable. It has one RHP pole, so it is unstable: option (D).

7. A concrete check. Take $G(s)=\frac{s-3}{(s+1)(s+2)}$. It is strictly stable and its numerator is $(s-3)$. The closed-loop characteristic equation is $$(s+1)(s+2)+(s-3)=s^2+4s-1=0\Rightarrow s=-2\pm \sqrt{5}$$ so one root, $s\approx 0.236$, is in the RHP and $Z=1$. A numerical trace of this plot, with $G(0)=-\frac{3}{2}$ to the left of $-1$, gives exactly one encirclement of $-1$. It is clockwise for $\omega$ from $-\infty$ to $\infty$ and anticlockwise when traced the other way. This is the situation the question describes, and the loop is unstable.