GATE 2018 EE Q.46: solution and answer

Official answer: 8.0Source: Official answer key, GATE 2018 EE (IIT Guwahati)

1. Read the steady-state value off the graph. The dashed line the response settles onto is at $y(\infty )=0.8$, not 1, so the closed loop has a steady-state error to a unit step.

2. The open loop $G(s)H(s)=\frac{K}{(s+1{)}^2(s+2)}$ has no pole at $s=0$, so the system is type 0. Its position error constant is $$K_p=\underset{s\to 0}{\lim }G(s)H(s)=\frac{K}{(1{)}^2(2)}=\frac{K}{2}$$

3. With unity feedback and a unit step input, $y(\infty )=\frac{K_p}{1+K_p}$. The final value theorem gives the same thing directly: the closed loop is $T(s)=\frac{K}{(s+1{)}^2(s+2)+K}$, so $$y(\infty )=T(0)=\frac{K}{(1{)}^2(2)+K}=\frac{K}{K+2}$$

4. Set this equal to the value read from the graph: $$\frac{K}{K+2}=0.8\Rightarrow K=0.8K+1.6\Rightarrow 0.2K=1.6\Rightarrow K=8$$

5. Check stability, because the final value theorem only holds for a stable closed loop. Expanding $(s+1{)}^2(s+2)=s^3+4s^2+5s+2$, the characteristic equation is $s^3+4s^2+5s+(2+K)=0$. For a cubic $s^3+a_2{s}^2+a_{1}s+a_0$ the Routh condition is $a_2{a}_1>a_0$ with all coefficients positive, so $4\cdot 5>2+K$, which gives $-2<K<18$. $K=8$ lies inside this range, so the loop is stable and the final value theorem applies.

6. Check against the shape of the curve. With $K=8$ the characteristic equation $s^3+4s^2+5s+10=0$ factors approximately as $(s+3.39)(s^2+0.61s+2.95)$. The complex pair is $s\approx -0.30\pm j1.69$, so ${\omega }_d\approx 1.69\text{rad/s}$ and the oscillation period is $2\pi /{\omega }_d\approx 3.7\text{s}$. That matches the spacing of the peaks at about 2.2, 5.9 and 9.6 s. The slow decay, set by the real part $0.30$, fits the ringing lasting until about 16 s.

7. Answer: $K=8.00$.