If the input signal x(t) and impulse response h(t) of a continuous-time system are described as

x(t) =e-3t u(t) and h(t) = u(t – 1), the output y(t) will be

1
\(\frac{1}{3}\left[ {1 - {e^{ - 3\left( {t - 1} \right)}}} \right]\)u(t - 1)
2
\(\frac{1}{3}\left[ {1 - {e^{ - 3t\;}}} \right]\)u(t - 1)
3
\(\frac{1}{3}\left[ {1 + {e^{ - 3\left( {t - 1} \right)}}} \right]\)u(t - 1)
4
\(\frac{1}{3}\left[ {1 + {e^{ - 3t}}} \right]\)u(t - 1)

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