The steady-state error for a system is 0.1. Steady-state error for the previously mentioned system being closed loop with a unity negative feedback and pulse input for 1 sec having a magnitude of 10 is?

Concept:

The steady-state error for a system is defined as:

\({e_{ss}} = \mathop {\lim }\limits_{s \to 0} \frac{{sR\left( s \right)}}{{1 + G\left( s \right)H\left( s \right)}}\)

R(s) = Input

G(s) = open loop transfer function

H(s) = feedback gain = 1 for unity feedback system

Analysis:

For a unit step input, we have:

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

The steady-state error becomes:

\({e_{ss}} = \mathop {\lim }\limits_{s \to 0} \frac{{s\left( {\frac{1}{s}} \right)}}{{1 + G\left( s \right)}} = 0.1\)

1 + G(0) = 10

G(0) = 9

Again we have the input as:

r(t) = 10 [u(t) – u(t - 1)]

\(R\left( s \right) = 10\left[ {\frac{1}{s} - \frac{1}{s}{e^{ - s}}} \right]\)

= \(10\left( {\frac{{1 - {e^{ - s}}}}{s}} \right)\)

∴ The steady-state error for the input will be:

\(e_{ss}' = \mathop {\lim }\limits_{s \to 0} \frac{{s \times 10\frac{{\left( {1 - {e^{ - s}}} \right)}}{s}}}{{1 + G\left( s \right)}}\)

= \(\frac{{10\left( {1 - {e^0}} \right)}}{{10}}\)

\(e_{ss}' = 0\)

Important Points

The shifting in input signal does not affect the steady-state error.