GATE 2020 EE Q.35: solution and answer

Official answer: CSource: Official answer key, GATE 2020 EE (IIT Delhi)

1. Forward path (two blocks in cascade): $G(s)=\frac{1}{s+1}\cdot \frac{1}{s^2}=\frac{1}{s^2(s+1)}$. Feedback path: $H(s)=\frac{20}{s+20}$, with negative feedback.

2. Closed-loop transfer function: $$\frac{Y(s)}{R(s)}=\frac{G(s)}{1+G(s)H(s)}=\frac{s+20}{s^2(s+1)(s+20)+20}$$

3. Characteristic equation: $s^2(s^2+21s+20)+20=0\Rightarrow s^4+21s^3+20s^2+0\cdot s+20=0$. The highest power of $s$ is 4, so the system is 4th order. No pole-zero cancellation reduces this: the only closed-loop zero is $s=-20$, and substituting it into the characteristic polynomial gives $(-20{)}^2(-19)(0)+20=20\ne 0$, so $s=-20$ is not a closed-loop pole.

4. Necessary condition for stability: every coefficient of the characteristic polynomial must be present and positive. Here the $s^1$ coefficient is zero, so the system cannot be stable. The Routh array tells us which kind of instability this is.

5. Routh array: $$\begin{array}{rl}{s}^4& :12020\\ s^3& :210\\ s^2& :\frac{21\cdot 20-1\cdot 0}{21}=20\frac{21\cdot 20-1\cdot 0}{21}=20\\ s^1& :\frac{20\cdot 0-21\cdot 20}{20}=-21\\ s^0& :20\end{array}$$

6. The first column is $1,21,20,-21,20$. The sign changes twice ($20\to -21$ and $-21\to 20$), so two closed-loop poles lie in the right half of the s-plane. The system is unstable, not merely marginally stable.

7. The system is 4th order and unstable, so the answer is option (C).