Topic: Control Systems
Transfer function of a lead compensator aimed at achieving gain crossover frequency of 0.5rad/sec and phase margin of 30 deg is
- A. $5[\frac{\sqrt{3}s+0.5}{s+0.5\sqrt{3}}]$
- B. $5[\frac{\sqrt{3}s-0.5}{s-0.5\sqrt{3}}]$
- C. $[\frac{\sqrt{3}s+0.5}{s+0.5\sqrt{3}}]$
- D. $[\frac{0.5s+\sqrt{3}}{s+0.5\sqrt{3}}]$
Answer: A. $5[\frac{\sqrt{3}s+0.5}{s+0.5\sqrt{3}}]$
Explanation: For a maximum phase margin $\phi_m = 30^\circ$, the attenuation factor $\alpha = \frac{1-\sin(30^\circ)}{1+\sin(30^\circ)} = \frac{1-0.5}{1+0.5} = 1/3$. To center this maximum phase at the crossover frequency $\omega_m = 0.5$ rad/sec, the time constant $T$ must satisfy $\omega_m = \frac{1}{T\sqrt{\alpha}} \implies T = \frac{1}{0.5 \sqrt{1/3}} = 2\sqrt{3}$. The lead compensator structure is $K_c \frac{1+sT}{1+s\alpha T}$. The pole and zero locations match the form presented in option (a), which correctly provides the $30^\circ$ phase lead at $0.5$ rad/sec.