Question

 A unity feedback system has \(G\left( s \right) = \frac{{K\left( {s + 12} \right)}}{{\left( {s + 14} \right)\left( {s + 18} \right)}}\)What is the value of K to yield 10% error in steady state for unit step input
1. 672
2. 189
3. 100
4. 21

Answer 1

Correct Answer - Option 2 : 189

Concept:

Steady-state error is defined as the difference between the input (command) and the output of a system in the limit as time goes to infinity (i.e. when the response has reached a steady-state).

Steady-state error is dependent on the type of input as well as on the type of the system.

Type	Step input	Ramp input	Parabolic input
Type – 0	\(\frac{A}{{1 + {k_p}}}\)	∞	∞
Type – 1	0	\(\frac{A}{{{K_v}}}\)	∞
Type – 2	0	0	\(\frac{A}{K_a}\)

Where:

k_p is positional error constant

\({k_p} = \mathop {\lim }\limits_{s \to 0} G(s)H(s)\)

k_v is velocity error constant

\({k_v} = \mathop {\lim }\limits_{s \to 0} s.G(s)H(s)\)

k_a is acceleration error constant

\({k_a} = \mathop {\lim }\limits_{s \to 0} = {s^2}G(s)H(s)\)

Calculation:

Given:

ess = 10% = 0.1

\(G\left( s \right) = \frac{{K\left( {s + 12} \right)}}{{\left( {s + 14} \right)\left( {s + 18} \right)}}\)

Since the type of system is zero and the type of input is a unit step, the steady-state error (ess) will be:

ess = \(\frac{A}{{1 + {k_p}}}\)

\(\frac{1}{{1 + {k_p}}}\) = 0.1

kp = 9

\({k_p} = \mathop {\lim }\limits_{s \to 0} G(s)H(s)\)

\(\mathop {\lim }\limits_{s \to 0} \frac{{K\left( {s + 12} \right)}}{{\left( {s + 14} \right)\left( {s + 18} \right)}}\) = 9

K = 189

Note: All the above formulae are valid for unity feedback systems only.