GATE 2024 IN Q.18: solution and answer

Official answer: ASource: Official answer key, GATE 2024 IN (IISc Bengaluru)

1. Read the plot. The real part stays at $1$ for every $\omega$. The imaginary part starts at $0$ when $\omega =0$ and falls without limit as $\omega \to \infty$. So we need $\text{Re}G(j\omega )=1$ for all $\omega$ and $\text{Im}G(j\omega )$ decreasing from $0$ towards $-\infty$.

2. Option (A), $G(j\omega )=1-j\omega T$: $\text{Re}=1$ and $\text{Im}=-\omega T$. With $T>0$ it starts at $1+j0$ for $\omega =0$ and moves straight down to $-j\infty$ as $\omega \to \infty$. This matches the plot exactly.

3. Option (C), $G(j\omega )=1+j\omega T$: it lies on the same line $\text{Re}=1$, but $\text{Im}=+\omega T$ goes to $+j\infty$, so the locus would run upward. Rejected.

4. Option (B), all-pass: $|G|=\frac{\sqrt{1+{\omega }^2{T}^2}}{\sqrt{1+{\omega }^2{T}^2}}=1$ for every $\omega$, and $\angle G=-2{\tan }^{-1}\omega T$ goes from $0^{\circ }$ to $-180^{\circ }$. That traces the lower half of the unit circle from $1$ to $-1$, not a vertical line. Rejected.

5. Option (D), first-order lag: $|G|=\frac{1}{\sqrt{1+{\omega }^2{T}^2}}\to 0$ and $\angle G=-{\tan }^{-1}\omega T\to -90^{\circ }$. That traces a lower semicircle of diameter $1$ from $1$ to the origin, not a line going to infinity. Rejected.

6. So the plot is $G(j\omega )=1-j\omega T$, which is $G(s)=1-sT$: a single zero at $s=+1/T$ in the right half-plane (non-minimum phase). Its magnitude $\sqrt{1+{\omega }^2{T}^2}$ grows without limit while its phase $-{\tan }^{-1}\omega T$ goes to $-90^{\circ }$, which is why the locus heads straight down. The answer is (A).