Question

A lead compenstor used for a closed loop controller has the following transfer function
$\frac{K(1+\frac{s}{a})}{(1+\frac{s}{b})}$
For such a lead compensator

• A. $a<b$
• B. $b<a$
• C. $a>Kb$
• D. $a<Kb$

Answer 1

The correct option is A $a<b$

$Transferfunction=\frac{K(1+\frac{s}{a})}{(1+\frac{s}{b})}$
Zero of $TF=-a$
Pole of $TF=-b$
For a lead-compensator, the zero is nearer to origin as compared to pole hence the effect of zero is dominant, therefore, the lead-compensator when interduced in series with forward path of the transfer function the phase shift is increased.


So, from pole-zero configuration of the copensator $a<b$