GATE 2023 EE Q.14: solution and answer

Official answer: DSource: Official answer key, GATE 2023 EE (IIT Kanpur)

1. PI controller: $C(s)=K_p+\frac{K_i}{s}=3+\frac{1}{s}=\frac{3s+1}{s}$, so the loop gain is $L(s)=C(s)G(s)=\frac{3s+1}{s(s-1)}$.

2. Check stability before using the final value theorem. The characteristic equation $1+L(s)=0$ gives $$s(s-1)+(3s+1)=s^2+2s+1=(s+1{)}^2=0$$ Both closed-loop poles are at $s=-1$, so the closed loop is stable even though the plant has a pole at $s=+1$.

3. Plant output: $\frac{Y(s)}{R(s)}=\frac{L}{1+L}=\frac{3s+1}{(s+1{)}^2}$. With $R(s)=\frac{1}{s}$, the final value theorem gives $$y(\infty )=\underset{s\to 0}{\lim }s\cdot \frac{3s+1}{(s+1{)}^2}\cdot \frac{1}{s}=1$$ This fits the loop being type 1 (the integrator in $C$), which gives zero steady-state error to a step.

4. Controller output: $U(s)=C(s)E(s)$, where $\frac{E(s)}{R(s)}=\frac{1}{1+L}=\frac{s(s-1)}{(s+1{)}^2}$. So $$\frac{U(s)}{R(s)}=\frac{3s+1}{s}\cdot \frac{s(s-1)}{(s+1{)}^2}=\frac{(3s+1)(s-1)}{(s+1{)}^2}$$

5. Final value theorem, valid because every pole of $sU(s)$ is at $s=-1$: $$u(\infty )=\underset{s\to 0}{\lim }s\cdot \frac{(3s+1)(s-1)}{(s+1{)}^2}\cdot \frac{1}{s}=\frac{(1)(-1)}{1}=-1$$

6. Consistency check at steady state: the plant DC gain is $G(0)=\frac{1}{0-1}=-1$, so $y(\infty )=G(0)u(\infty )=(-1)(-1)=1$. This agrees with step 3.

7. The controller output settles at $-1$ and the plant output at $1$, in the order the question asks for (controller first). The answer is option (D).