Question

The maximum phase shift that can be provided by a lead compensator with transfer function \( G(s) = \dfrac{1 + 6s}{1 + 2s} \) is:

• A. $15^\circ$
• B. $30^\circ$
• C. $45^\circ$
• D. $60^\circ$

Hint:

For lead compensators, the maximum phase lead is calculated using $\sin^{-1} \left( \frac{\alpha - 1}{\alpha + 1} \right)$. Identify $\alpha$ by comparing the transfer function with the standard form.

Answer 1

The Correct Option is B

A lead compensator has a transfer function of the form $G(s) = \dfrac{1 + \alpha T s}{1 + T s}$, where $\alpha<1$.
Given: $G(s) = \dfrac{1 + 6s}{1 + 2s}$.
Comparing with the standard form:
$\alpha T = 6$, and $T = 2 ⇒ \alpha = \dfrac{6}{2} = 3$.
The maximum phase lead $\phi_{max}$ provided by the compensator is given by:
\[ \phi_{max} = \sin^{-1} \left( \frac{\alpha - 1}{\alpha + 1} \right) \]
Substitute $\alpha = 3$:
\[ \phi_{max} = \sin^{-1} \left( \frac{3 - 1}{3 + 1} \right) = \sin^{-1} \left( \frac{2}{4} \right) = \sin^{-1}(0.5) = 30^\circ \]
Hence, the maximum phase shift is $30^\circ$.