GATE 2026 EE Q.29: solution and answer

Official answer: ASource: Official answer key, GATE 2026 EE (IIT Guwahati)

1. Read the plot. From the lowest frequencies the magnitude falls at a steady $-20\text{dB/decade}$, so there is one pole at the origin (a $1/s$ factor). At ${\omega }_0$ the slope changes by $+20\text{dB/decade}$, from $-20$ to $0$. That means a simple zero with corner frequency ${\omega }_0$, i.e. a factor $\left(1+\frac{s}{{\omega }_0}\right)$.

2. Fix the gain from the low-frequency asymptote. Write $G(s)=\frac{K\left(1+\frac{s}{{\omega }_0}\right)}{s}$. For $\omega <{\omega }_0$ the magnitude asymptote is $K/\omega$, and the figure shows this line crossing $0\text{dB}$ exactly at $\omega ={\omega }_0$. So $K/{\omega }_0=1$, which gives $K={\omega }_0$. The low-frequency term is then $\frac{{\omega }_0}{s}=\frac{1}{s/{\omega }_0}$.

3. Put the pieces together: $$G(s)=\frac{1+\frac{s}{{\omega }_0}}{\frac{s}{{\omega }_0}}$$ This is option (A).

4. Check option (A) with asymptotes. For $\omega \ll {\omega }_0$: $|G(j\omega )|\approx \frac{{\omega }_0}{\omega }$, which falls at $-20\text{dB/dec}$ and reaches $0\text{dB}$ at ${\omega }_0$. For $\omega \gg {\omega }_0$: $|G(j\omega )|\approx \frac{\omega /{\omega }_0}{\omega /{\omega }_0}=1$, which is $0\text{dB}$ and flat. This matches the figure.

5. Rule out (B). Its low-frequency term is $s/{\omega }_0$, so the plot would rise at $+20\text{dB/dec}$ up to $0\text{dB}$ at ${\omega }_0$ and then stay flat. That is a high-pass shape, the mirror image of the figure.

6. Rule out (C) and (D). In both, $\left|1\pm \frac{j\omega }{{\omega }_0}\right|=\sqrt{1+(\omega /{\omega }_0{)}^2}$, so numerator and denominator have the same magnitude and $|G(j\omega )|=1$ at every frequency. Their magnitude plot is a flat $0\text{dB}$ line (an all-pass response) with no $-20\text{dB/dec}$ segment. The sign of the $s/{\omega }_0$ term changes only the phase.