Realizar la convolución de una señal por si misma
\[
x(t) = e^{-3t}u(t - 1)
\]
\[
y(t) = x(t) * x(t)
\]
Desarrollo
2) Reflejamos y comenzamos a resolver las integrales
Intervalo 1
Cuando \(t - 1 < 1\)
\(t < 2\)
\[
y(t) =
\int_{t - 1}^{1}
\color{purple} x(\tau) \color{black}
\color{orange} x(t - \tau) \color{black}
d\tau
\]
\[
\color{limegreen} \boxed{ \color{black}
y(t) = 0
}
\]
Intervalo 2
Cuando
\(t - 1 > 1\)
\(t > 2\)
\[
y(t) =
\int_{1}^{t - 1}
\color{purple} x(\tau) \color{black}
\color{orange} x(t - \tau) \color{black}
d\tau
\]
\[
y(t) =
\int_{1}^{t - 1}
\color{purple} e^{-3\tau}u(\tau - 1) \color{black}
\color{orange} e^{-3(t - \tau)}u(t - \tau - 1) \color{black}
d\tau
\]
\[
y(t) =
\int_{1}^{t - 1}
\color{purple} e^{-3\tau}1 \color{black}
\color{orange} e^{(-3t + 3\tau)}1 \color{black}
d\tau
\]
\[
y(t) =
\int_{1}^{t - 1}
\color{purple} e^{-3\tau} \color{black}
\color{orange} e^{(-3t + 3\tau)} \color{black}
d\tau
\]
\[
y(t) =
\int_{1}^{t - 1}
\color{purple} e^{-3\tau} \color{black}
\color{orange} e^{-3t} e^{3\tau}\color{black}
d\tau
\]
\[
y(t) =
e^{-3t}
\int_{1}^{t - 1}
\color{purple} e^{-3\tau} \color{black}
\color{orange} e^{3\tau}\color{black}
d\tau
\]
\[
y(t) =
e^{-3t}
\int_{1}^{t - 1}
e^{0}
d\tau
\]
\[
y(t) =
e^{-3t}
\int_{1}^{t - 1}
1
d\tau
\]
\[
y(t) =
e^{-3t}
\tau \Big|^{t - 1}_{1}
\]
\[
y(t) =
e^{-3t}
((t - 1) - 1)
\]
\[
\color{limegreen} \boxed{ \color{black}
y(t) = e^{-3t}(t - 2)
}
\]
3) Finalmente escribimos la salida \(y(t)\)
\[
\color{limegreen} \boxed{ \color{black}
y(t) =
\begin{cases}
0 & t < 2 \\
e^{-3t}(t - 2) & t > 2 \\
\end{cases}
}
\]
