GATE 2025 EE Q.29: solution and answer

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

1. The loop has unity negative feedback (the take-off after $G(s)$ returns directly to the summing junction with a $-$ sign), so the characteristic equation is $1+G(s)=0$ and the standard root-locus rules for $K\ge 0$ apply to $G(s)=\frac{Ks(s+2)}{(s+5)(s+7)}$.

2. Open-loop zeros: $s=0,-2$. Open-loop poles: $s=-5,-7$. All four are real, so they split the real axis into segments.

3. Real-axis rule for $K\ge 0$: a real point is on the locus if the total number of real open-loop poles and zeros to its right is odd.

4. $-2<s<0$: one point ($0$) to the right, which is odd, so this segment is on the locus. $-1$ lies here, so A is correct.

5. $-5<s<-2$: two points ($0,-2$) to the right, which is even, so this segment is not on the locus. $-4$ is not on it, so B is wrong.

6. $-7<s<-5$: three points ($0,-2,-5$) to the right, which is odd, so this segment is on the locus. $-6$ lies here, so C is correct.

7. $s<-7$: four points to the right, which is even, so this segment is not on the locus. $-10$ is not on it, so D is wrong.

8. Direct check with the magnitude/sign condition. A real point is on the locus for $K\ge 0$ only if $G(s)=-1$ gives a non-negative $K$:
   $$\begin{array}{rl}s=-1:& G=\frac{K(-1)(1)}{(4)(6)}=-\frac{K}{24}\Rightarrow K=24\ge 0\\ s=-4:& G=\frac{K(-4)(-2)}{(1)(3)}=\frac{8K}{3}\Rightarrow K=-\frac{3}{8}<0\\ s=-6:& G=\frac{K(-6)(-4)}{(-1)(1)}=-24K\Rightarrow K=\frac{1}{24}\ge 0\\ s=-10:& G=\frac{K(-10)(-8)}{(-5)(-3)}=\frac{16K}{3}\Rightarrow K=-\frac{3}{16}<0\end{array}$$
   Only $-1$ and $-6$ need a non-negative $K$, which confirms A and C.

9. Shape check: there are two poles and two zeros, so there are no asymptotes and both branches end at the zeros $0$ and $-2$. The two branches that start at $-7$ and $-5$ meet in $(-7,-5)$ and break away into the complex plane as a conjugate pair; they cannot reach $(-2,0)$ along the real axis because $(-5,-2)$ is not on the locus, so they re-enter the real axis in $(-2,0)$ and end at the zeros. Setting $\frac{dK}{ds}=0$ with $K=-\frac{(s+5)(s+7)}{s(s+2)}$ gives $s^2+7s+7=0$, so $s=\frac{-7\pm \sqrt{21}}{2}\approx -5.79,-1.21$: a breakaway point in $(-7,-5)$ and a break-in point in $(-2,0)$, consistent with the segments found above.