A unity feedback system has a forward path transfer function G(s) = \(\frac{10(1+4\text{s})}{\text{s}^{2}(1+\text{s})}\). If the system is subjected to an input r(t) = 1 + t + \(\frac{\text{t}^{2}}{2}\), the steady state error of the system will be

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UPPSC AE Electrical 29 May 2022 Official Paper I

Answer 1

Concept:

The steady-state error for different inputs is given by:

Ka = 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

Calculation:

Given, \(G(s)=\frac{10(1+4\text{s})}{\text{s}^{2}(1+\text{s})}\)

\(r(t) = 1+t+{t^2\over 2}\)

Type = 2

Error = \({1\over K_a}\)

\(K_a = Lt\overset{s=0}{\rightarrow}s^2G(s)\)

\(K_a = Lt\overset{s=0}{\rightarrow}s^2\frac{10(1+4\text{s})}{\text{s}^{2}(1+\text{s})}\)

K_a = 10

Error = \({1\over 10}\)

Error = 0.1