Question

The transfer function of a compensator is given as
$G_c\left(s\right)=\frac{s+a}{s+b}$

$G_c\left(s\right)$ is lead compensator if

• A. a = 1, b = 2
• B. a = 3, b = 2
• C. a = -3, b = -1
• D. a = 3, b = 1

Answer 1

The correct option is A a = 1, b = 2

$G_c\left(s\right)=\frac{s+a}{s+b}$

$G_c\left(j\omega \right)=\frac{j\omega +a}{j\omega +b}$
$\angle G_c\left(j\omega \right)=ta{n}^{-1}\frac{\omega }{a}-ta{n}^{-1}\frac{\omega }{b}$

$=ta{n}^{-1}\left(\frac{\frac{\omega }{a}-\frac{\omega }{b}}{1+\frac{{\omega }^2}{ab}}\right)$

For $G_c\left(s\right)$ to be a lead compensator
$\angle G_c\left(j\omega \right)>0$

$\frac{\omega }{a}>\frac{\omega }{b}\Rightarrow b>a$

Option (a) satisfies the above equation.