Question

A unity feedback control system has forward path transfer function \(G\left( s \right) = \frac{K}{{s\left( {s + 2} \right)}}.\) If the design specification is that the steady state error due to unit ramp input is 0.05. The value of gain K will be.
1. 10
2. 20
3. 40
4. 80

Answer 1

Correct Answer - Option 3 : 40

Concept:

KP = position error constant = \(\mathop {\lim }\limits_{s \to 0} G\left( s \right)H\left( s \right)\)

Kv = velocity error constant = \(\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)\)

Ka = acceleration error constant = \(\mathop {\lim }\limits_{s \to 0} {s^2}G\left( s \right)H\left( s \right)\)

Steady state error for different inputs is given by

Input	Type -0	Type - 1	Type -2
Unit step	\(\frac{1}{{1 + {K_p}}}\)	0	0
Unit ramp	∞	\(\frac{1}{{{K_v}}}\)	0
Unit parabolic	∞	∞	\(\frac{1}{{{K_a}}}\)

 

From the above table, it is clear that for type – 1 system, a system shows zero steady-state error for step-input, finite steady-state error for Ramp-input and \(\infty \) steady-state error for parabolic-input.

Calculation:

The given forward path transfer function is \(G\left( s \right) = \frac{K}{{s\left( {s + 2} \right)}}\)

Velocity error constant, \({K_v} = \mathop {\lim }\limits_{s \to 0} s \times \frac{K}{{s\left( {s + 2} \right)}} = \frac{K}{2}\)

Steady-state error \( = \frac{2}{K} = 0.05\)

⇒ K = 40