ISRO EC 2025 Q.56: solution and answer

Official answer: DSource: ISRO's final answer key of 15 May 2026

1. The initial slope is $-20\text{dB/decade}$, so there is one pole at the origin: $G(s)=\frac{K}{s}\cdots$.

2. At $\omega =1$ the slope steepens from $-20$ to $-40\text{dB/decade}$, so there is a simple pole at $s=-1$. At $\omega =20$ it steepens from $-40$ to $-60\text{dB/decade}$, so there is a simple pole at $s=-20$. The slope never rises, so there are no zeros.

3. In Bode form: $G(s)=\frac{K}{s(1+s)\left(1+\frac{s}{20}\right)}$.

4. For $\omega \le 1$ only the $K/\omega$ term matters. The plot reads $40\text{dB}$ at $\omega =1$, so $20{\log }_{10}K=40$ and $K=100$. At $\omega =0.1$ this gives $20{\log }_{10}(100/0.1)=60\text{dB}$, matching the $60\text{dB}$ level on the plot.

5. Hence $$G(s)=\frac{100}{s(s+1)\left(1+\frac{s}{20}\right)}=\frac{2000}{s(s+1)(s+20)}.$$

6. Check: between $\omega =1$ and $\omega =10$ the slope is $-40\text{dB/decade}$, so the level falls by $40\text{dB}$, from $40\text{dB}$ to $0\text{dB}$ at $\omega =10$. This matches the plot.

7. A magnitude plot cannot show the sign of a pole, so the stable minimum-phase form is taken, with poles at $s=-1$ and $s=-20$. Options (a), (b) and (c) contain $(s-1)$ or $(s-20)$, which are right-half-plane poles. Option (c) also has the wrong gain: its low-frequency constant is $1000/20=50$, not $100$. Only option (d) matches.