Question

Consider the system described by following state space equations

\(\left[ {\begin{array}{*{20}{c}} {{{\dot x}_1}}\\ {{{\dot x}_2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&1\\ { - 1}&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0\\ 1 \end{array}} \right]u;y = \left[ {\begin{array}{*{20}{c}} 1&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}} \end{array}} \right]\)

If u is unit step input, then the steady state error of the system is

Answer 1

Correct Answer - Option 1 : 0

Concept:

The transfer function from the state space representation is given by:

G(s) = C(SI – A)-1 B + D

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:

From the given state-space representation,

\(A = \left[ {\begin{array}{*{20}{c}} 0&1\\ { - 1}&{ - 1} \end{array}} \right],B = \left[ {\begin{array}{*{20}{c}} 0\\ 1 \end{array}} \right],C = \left[ {\begin{array}{*{20}{c}} 1&0 \end{array}} \right]\)

\(\left[ {sI - A} \right] = \left[ {\begin{array}{*{20}{c}} s&0\\ 0&s \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 0&1\\ { - 1}&{ - 1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} s&{ - 1}\\ 1&{s + 1} \end{array}} \right]\)

\({\left[ {sI - A} \right]^{ - 1}} = \frac{1}{{{s^2} + s + 1}}\left[ {\begin{array}{*{20}{c}} {s + 1}&1\\ { - 1}&s \end{array}} \right]\)

Transfer function \( = \left[ {\begin{array}{*{20}{c}} 1&0 \end{array}} \right]\frac{1}{{{s^2} + s + 1}}\left[ {\begin{array}{*{20}{c}} {s + 1}&1\\ { - 1}&s \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0\\ 1 \end{array}} \right]\)

\( = \frac{1}{{{s^2} + s + 1}}\)

Open-loop transfer function \( = \frac{{\frac{1}{{{s^2} + s + 1}}}}{{1 - \frac{1}{{{s^2} + s + 1}}}}\)

\( = \frac{1}{{s\left( {s + 1} \right)}}\)

It is a type 1 system. The steady-state error for a step input is zero.