Question

Comprehension

The transfer function of a compensator is given as

\(G_c(S)=\frac{s+a}{s+b}\)

1. a = 1, b = 2
2. a = 3, b = 2
3. a = -3, b = -1
4. a = 3, b = 1
5.

Answer 1

Correct Answer - Option 1 : a = 1, b = 2

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

Only two options are valid as per the above condition, options 1 and 3

But we can't go with option 3, because for lead compensator zero is near to the origin compared to pole.

Hence, a = 1, b = 2