Correct Answer - Option 1 : a < b
Concept:
The general expression for a lead compensator is;
\(G\left( s \right)=\frac{\left( \alpha \right)\left( 1+Ts \right)}{\left( 1+\alpha Ts \right)}\)
And for a lead compensator α < 1
Analysis:
Given lead compensator expression is;
\({{G}_{c}}\left( s \right)=\frac{k\left( s+a \right)}{\left( s+b \right)}\)
On comparing it with the standard expression we get;
\(T=\frac{1}{a}~,~~\alpha T=\frac{1}{b}\)
\(\Rightarrow \frac{\alpha }{a}=\frac{1}{b}\Rightarrow \alpha =\frac{a}{b}\)
⇒ a < b
Hence option – 1 is correct.