# A unity feedback control system has forward path transfer function $G\left( s \right) = \frac{K}{{s\left( {s + 2} \right)}}.$ If the design specific

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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

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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 = $\mathop {\lim }\limits_{s \to 0} sG\left( s \right)H\left( s \right)$

K= 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