For a system with the given open loop transfer function, the steady state error when the input f (t) = 1 + 2t applied is .....

Open loop transfer function G (s) H (s) = \(\frac{100(s+2)}{s(s+3)(s+4)}\) 

Concept:

To compute the steady-state error, use the Final Value Theorem for the error signal:

\( e_{ss} = \lim_{s \to 0} s \cdot \frac{1}{1 + G(s)H(s)} \cdot R(s) \)

Given:

\( G(s)H(s) = \frac{100(s+2)}{s(s+3)(s+4)} \)

Input: \( f(t) = 1 + 2t \Rightarrow R(s) = \frac{1}{s} + \frac{2}{s^2} \)

Calculation:

This is a Type-1 system (one pole at the origin), so for a ramp input, the steady-state error is given by:

\( e_{ss} = \frac{2}{K_v} \) where \( K_v = \lim_{s \to 0} s \cdot G(s)H(s) \)

\( K_v = \lim_{s \to 0} s \cdot \frac{100(s+2)}{s(s+3)(s+4)} = \frac{100 \cdot 2}{3 \cdot 4} = \frac{200}{12} = \frac{50}{3} \)

\( e_{ss} = \frac{2}{K_v} = \frac{2}{\frac{50}{3}} = \frac{6}{50} = \frac{3}{25} \)

Hence, the correct answer is option 2