GATE 2018 IN Q.13: solution and answer

Official answer: ASource: Official answer key, GATE 2018 IN (IIT Guwahati)

1. Split the phase into two parts. When factors are multiplied, their phases add. The constant $100^2$ in the numerator adds nothing. $$\angle G(j\omega )=\angle e^{-j0.01\omega }+\angle \frac{100^2}{(j\omega {)}^2+0.2j\omega +100^2}$$

2. Match the quadratic to the standard form $s^2+2\zeta {\omega }_{n}s+{\omega }_n^2$. This gives ${\omega }_n=100\text{rad/s}$ and $2\zeta {\omega }_n=0.2$, so $\zeta =\frac{0.2}{2\times 100}=0.001$. The poles are almost undamped.

3. Phase of the quadratic. At $s=j\omega$ the denominator is $(100^2-{\omega }^2)+j0.2\omega$. Its imaginary part is always positive. Its real part changes sign at $\omega =100$. So $$\begin{array}{rl}\omega <100& :{\varphi }_2=-{\tan }^{-1}\frac{0.2\omega }{100^2-{\omega }^2}\approx 0^{\circ }\\ \omega =100& :{\varphi }_2=-90^{\circ }\\ \omega >100& :{\varphi }_2=-180^{\circ }+{\tan }^{-1}\frac{0.2\omega }{{\omega }^2-100^2}\approx -180^{\circ }\end{array}$$ The phase passes $-45^{\circ }$ at $\omega \approx 99.9\text{rad/s}$ and $-135^{\circ }$ at $\omega \approx 100.1\text{rad/s}$. The whole $180^{\circ }$ fall therefore happens within about $0.2\text{rad/s}$. On a 0 to 200 rad/s axis that is a vertical downward step at ${\omega }_n=100\text{rad/s}$.

4. Phase of the pure delay. $|e^{-j\omega T}|=1$, but $\angle e^{-j\omega T}=-\omega T$. With $T=0.01\text{s}$, ${\varphi }_1=-0.01\omega \text{rad}=-0.573\omega$ degrees. This is a straight line through the origin with a negative slope, and it continues on both sides of 100 rad/s.

5. Add the two parts at the key frequencies: $$\begin{array}{rl}\omega \to 0& :\varphi =0^{\circ }+0^{\circ }=0^{\circ }\\ \omega =100^{-}& :\varphi =-57.3^{\circ }+0^{\circ }=-57.3^{\circ }\\ \omega =100^{+}& :\varphi =-57.3^{\circ }-180^{\circ }=-237.3^{\circ }\\ \omega =200& :\varphi =-114.6^{\circ }-180^{\circ }=-294.6^{\circ }\end{array}$$

6. So the phase has three parts: a falling straight line, a $180^{\circ }$ downward drop at 100 rad/s, and then a falling line with the same slope. Option (A) is the only plot with this shape. Its axis labels are not evenly spaced, so it is a sketch and its corner values cannot be read off it. The shape identifies it.

7. Rule out the others. (B) and (D) are flat except for the step, so they leave out the delay's linear phase lag. (D) also steps up to $+180^{\circ }$. Tracked continuously, the phase of a pair of left-half-plane poles moves from $0^{\circ }$ down to $-180^{\circ }$, never up. (C) has a rising slope, which would need a lead term $e^{+0.01s}$ instead of a delay.