Correct Answer - Option 2 : 10.47
Concept:
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
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
|
∞
|
∞
|
\(\frac{1}{{{K_a}}}\)
|
Calculation:
The given input, r(t) = 10 rpm
\(r\left( t \right) = \frac{{2\pi }}{{60}} \times 10 \times \frac{{180}}{\pi }t\)
= 60t degree/sec
\(G\left( s \right) = \frac{{57.3K}}{{s\left( {s + 10} \right)}}\)
Velocity error coefficient is,
\({K_v} = \mathop {\lim }\limits_{s \to 0} sG\left( s \right) = \mathop {\lim }\limits_{s \to 0} s\frac{{57.3K}}{{s\left( {s + 10} \right)}} = 5.73K\)
Steady-state error, \({e_{ss}} = \frac{{60}}{{5.73K}}\)
Given that the steady-state error = 1°
\( \Rightarrow \frac{{60}}{{5.73K}} = 1 \Rightarrow K = 10.47\)