Question

The response h(t) of a linear time invariant system to an impulse δ (t) , under initially relaxed condition is h(t)  = e^-t + e^-2t . The response of this system for a unit step input u(t) is
1. u (t) + e^-t + e^-2t
2. (e^-t + e^-2t) u (t)
3. (1.5 – e^-t – 0.5e^-2t) u (t)
4. e^-t δ (t) + e^-2t u (t)

Answer 1

Correct Answer - Option 3 : (1.5 – e^-t – 0.5e^-2t) u (t)

h (t) = e^-t + e^-2t

Taking Laplace transform,

\(H\left( s \right) = \frac{1}{{s + 1}} + \frac{1}{{s + 2}}\)

For unit step input, r (t) = u (t)

∴ R (s) = 1/s

\( Y\left( s \right) = R\left( s \right)H\left( s \right) = \frac{1}{s}\left[ {\frac{1}{{s + 1}} + \frac{1}{{s + 2}}} \right]\)

\( Y\left( s \right) = \frac{A}{s}+\frac{B}{s+1}+\frac{C}{s}+\frac{D}{s+2}\)

A = 1, B = -1, C = 0.5, D = -0.5

Now, Y(s) becomes, 

\(\\ Y\left( s \right) = \frac{3}{{2s}} - \frac{1}{{s + 1}} - \frac{1}{2}\left( {\frac{1}{{s + 2}}} \right)\)

Taking inverse Laplace,

Y (t) = u (t) [ 1.5 – e^-t – 0.5 e ^–2t ]