Question

A first order linear system is initially relaxed for a unit step signal u(t), the response is v(t) = (1 – e^-3t), for t > 0. If a signal 3u(t) + δ(t) is applied to the same system, the response is
1. (3 – 6e^-3t) u(t)
2. (3 – 3e^-3t) u(t)
3. 3u(t)
4. (3 + 3e^3t) u(t)

Answer 1

Correct Answer - Option 3 : 3u(t)

Given that a first-order system.

Its unit step response v(t) = (1 – e^-3t), t > 0

Input r(t) = u(t)

By applying the Laplace transform,

\(V\left( s \right) = \frac{1}{s} - \frac{1}{{s + 3}}\left[ {L\left[ {{e^{ - 3t}}} \right] = \frac{1}{{s + 3}}} \right]\)

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

The transfer function of the system,

\(H\left( s \right) = \frac{{V\left( s \right)}}{{R\left( s \right)}} = \frac{{\frac{1}{s} - \frac{1}{{s + 3}}}}{{\frac{1}{s}}} = 1 - \frac{s}{{s + 3}}\)

For a system, the transfer function is always the same for any combination of input, output.

Therefore, for the input r₂(t) = 3u(t) + δ(t)

\({R_2}\left( s \right) = \frac{3}{s} + 1 = \frac{{3 + s}}{s}\)

Then the output is,

\(H\left( s \right) = \frac{{{V_2}\left( s \right)}}{{{R_2}\left( s \right)}} \Rightarrow {V_2}\left( s \right) = H\left( s \right) \cdot {R_2}\left( s \right)\)

\( \Rightarrow {V_2}\left( s \right) = \left( {1 - \frac{s}{{s + 3}}} \right)\left( {\frac{{3 + s}}{s}} \right)\)

\(= \frac{{3 + s}}{s} - 1 = \frac{3}{s} + 1 - 1 = \frac{3}{s}\)

By taking inverse Laplace transform

The output V₂(t) = 3u(t)