\(G\left( s \right) = \frac{{ - s + 1}}{{s + 1}}\)
r(t) = u(t)
\(\Rightarrow r\left( s \right) = \frac{1}{s}\)
Output, C(s) = R(s) G(s)
\(= \frac{1}{s}\left( {\frac{{1 - s}}{{1 + s}}} \right)\)
\(\Rightarrow c\left( s \right) = \frac{1}{s} - \frac{2}{{1 + s}}\)
By applying inverse Laplace transform,
⇒ c(t) = u(t) – 2e-t u(t)
At t = 1.5 sec,
c(t = 1.5s) = 1 – 2e
-1.5 = 0.553