Unit step response of a first-order system with transfer function \(G\left( s \right) = \frac{1}{{1 + \tau s}}\) is

Concept:

Y(s) = X(s) . G(s)

X(s): Applied input.

G(s): transfer function

Y(s): Output response

A unit step input is defined as u(t). Its Laplace transform is given by:

\(x\left( t \right) \leftrightarrow X\left( s \right) = \frac{1}{s}\)

Calculation:

The transfer function is given as:

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

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

The unit step response is the response when the input is a unit step function, i.e.
\(Y\left( s \right) = \frac{1}{s} \cdot \frac{1}{{\left( {1 + \tau s} \right)}}\)

Using Partial fraction, we get:

\(\frac{1}{s} \cdot \frac{1}{{1 + \tau s}} = \frac{A}{s} + \frac{B}{{1 + \tau s}}\)

On solving it, we’ll get:

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

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

Taking the Inverse Laplace Transform, we get:

\(y\left( t \right) = \left( {1 - {e^{ - \frac{t}{\tau }}}} \right)u\left( t \right)\)