If both fast response time and good steady-state accuracy are nee

SOLUTION

Analysis:

Lag Compensator:

\(\frac{{{V_0}\left( s \right)}}{{{V_1}\left( s \right)}} = \frac{{1 + ST}}{{1 + \beta ST}}\)

Where,

\(\beta = \frac{{{R_1} + {R_2}}}{{{R_2}}},\;\beta > 1\)

In Lag Compensator, the steady-state error is reduced So, steady-state Response is increased.

In lag compensator, Transient response decreases.

Lead Compensator:

\(\frac{{{V_0}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{{\alpha \left( {1 + ST} \right)}}{{1 + \alpha ST}}\)

\(\alpha = \frac{{{R_2}}}{{{R_1} + {R_2}}}\;;\alpha < 1\)

In lead comp. steady-state error increased So, the steady-state response is decreased.

In lead comp. Transient response Improves.

Lag-lead Compensator:

Lag lead Compensator is a combined form of a lead compensator and lag comp compensator.

\(\frac{{{V_0}\left( s \right)}}{{{V_i}\left( s \right)}} = \frac{{\left( {S + {Z_1}} \right)\left( {S + {Z_2}} \right)}}{{\left( {S + {P_1}} \right)\left( {S + {P_2}} \right)}}\)

\(\beta = \underbrace {\frac{{{Z_1}}}{{{P_1}}} > 1}_{Lag}\alpha = \underbrace {\frac{{{Z_2}}}{{{P_2}}} < 1}_{Lead}\)

So, lag-lead compensator improves both steady-state accuracy and fast response time.