GATE 2020 EC Q.40: solution and answer

Official answer: ASource: Official answer key, GATE 2020 EC (IIT Delhi)

1. Put the equation in root-locus form $1+KG(s)H(s)=0$ by collecting the $K$ terms: $s^3+3s^2+2s+K(s+3)=0$. This gives $$G(s)H(s)=\frac{K(s+3)}{s(s+1)(s+2)}$$ The open-loop poles are at $0,-1,-2$ and the only zero is at $-3$.

2. Real-axis rule: a point on the real axis is on the locus (for $K>0$) only if the number of real poles and zeros to its right is odd. That puts $(-1,0)$ (one to the right) and $(-3,-2)$ (three to the right) on the locus. $(-2,-1)$ (two to the right) and $(-\infty ,-3)$ (four to the right) are not on it.

3. What to expect: on $(-1,0)$ two branches start at the poles $0$ and $-1$ and move towards each other, so they must meet and break away somewhere inside $(-1,0)$. On $(-3,-2)$ one branch runs from the pole at $-2$ to the zero at $-3$, so no break point is needed there. The break-point equation below confirms both.

4. Asymptotes: $P-Z=3-1=2$, so the angles are $\pm 90^{\circ }$ and the centroid is ${\sigma }_a=\frac{(0-1-2)-(-3)}{2}=0$. The two branches that break away from $(-1,0)$ go off towards these vertical asymptotes.

5. Break-point condition $dK/ds=0$ with $K=-\frac{s^3+3s^2+2s}{s+3}$: $$(3s^2+6s+2)(s+3)-(s^3+3s^2+2s)=0$$ $$\Rightarrow 2s^3+12s^2+18s+6=0\Rightarrow s^3+6s^2+9s+3=0$$

6. Find where the roots lie by sign changes of $f(s)=s^3+6s^2+9s+3$: $f(0)=3$, $f(-1)=-1$, $f(-2)=1$, $f(-3)=3$, $f(-4)=-1$. So there is one root in $(-1,0)$, one in $(-2,-1)$ and one in $(-4,-3)$. That accounts for all three roots of the cubic. (Numerically they are $s\approx -0.47,-1.65,-3.88$.)

7. Keep only the roots that lie on the locus. The roots near $-1.65$ and $-3.88$ fall in $(-2,-1)$ and $(-\infty ,-3)$, which are not part of the locus. Check: $K=-\frac{s(s+1)(s+2)}{s+3}$ is negative at both, so they belong to the $K<0$ locus.

8. Check the root in $(-1,0)$: at $s\approx -0.47$, $$K=\frac{0.47\times 0.53\times 1.53}{2.53}\approx 0.15>0,$$ so it is a genuine break-away point. No valid root lies in $(-3,-2)$, which confirms there is no break point on that segment. So the only break-away or break-in point lies in $(-1,0)$, which is option (A).