GATE 2024 EE Q.58: solution and answer

Official answer: 0.09 to 0.11Source: Official answer key, GATE 2024 EE (IISc Bengaluru)

1. Read the low-frequency asymptote. The asymptotic magnitude has one constant slope of $-20$ dB/decade from low frequency up to at least $100$ rad/s, and the asymptotic phase stays at $-90^{\circ }$ up to at least $10$ rad/s. So $G(s)$ has exactly one pole at the origin and no corner frequency below $100$ rad/s. In this band $G(j\omega )\approx \frac{K}{j\omega }$, which makes it a type-1 system.

2. Use the gain crossover to find $K$. The crossover ${\omega }_{gc}=10$ rad/s lies inside the $-20$ dB/dec band, so $$\left|\frac{K}{j{\omega }_{gc}}\right|=1\Rightarrow \frac{K}{10}=1\Rightarrow K=10$$

3. Velocity error constant. For a type-1 system, $K_v={\lim }_{s\to 0}sG(s)=K=10{\text{s}}^{-1}$. Graphically, $K_v$ is the frequency where the initial $-20$ dB/dec line (or its extension) crosses $0$ dB. Here that line is still the actual asymptote at $10$ rad/s, so $K_v$ equals the gain crossover frequency, $10$.

4. Ramp error for unity feedback. The question says the closed loop is stable, so the final value theorem applies: $$e_{ss}=\underset{s\to 0}{\lim }s\cdot \frac{1}{1+G(s)}\cdot \frac{1}{s^2}=\frac{1}{{\lim }_{s\to 0}sG(s)}=\frac{1}{K_v}=\frac{1}{10}=0.10$$