Se pide encontrar la salida \(y(t)\) d
el sistema LTI siendo \(x(t)\) y
\(h(t)\) las siguientes señales:
\( \color{purple}x(t) \color{black}= e^{-t}u(t-1)\)
\( \color{orange} h(t) \color{black}= -2\delta(t+1) + \delta(t-1)+e^{-4t}u(t-2)\)
Intervalo 1
Cuando \(t - 1 < -1\)
Si despejamos nos queda que \(t < 0\)
\[
\boxed{
y(t) = 0
}
\]
Intervalo 2
Cuando \(-1 < t - 1 < 1\)
Si despejamos nos queda que
\(-1 + 1 < t < 1 + 1\)
\(0 < t < 2\)
\[
y(t) = \int^{}_{-1}
\color{purple} x(t - \tau) \color{black}
\color{orange} h(\tau) \color{black}
d\tau
\]
\[
y(t) = \int^{}_{-1}
\color{purple} e^{-(t - \tau)} u(t - \tau - 1) \color{black}
\color{orange} (-2\delta(\tau + 1)) \color{black}
d\tau
\]
\[
y(t) = \int^{}_{-1}
\color{purple} e^{-(t - \tau)} 1 \color{black}
\color{orange} (-2) \color{black}
d\tau
\]
\[
y(t) = -2\int^{}_{-1}
\color{purple} e^{-(t - \tau)} \color{black}
d\tau
\]
\[
y(t) = -2e^{-(t - \tau)}\Big|^{}_{-1}
\]
\[
y(t) = -2e^{-(t - (-1))}
\]
\[
\boxed{
y(t) = -2e^{-(t + 1)}
}
\]
Intervalo 3
Cuando \( 1 < t - 1 < 2 \)
Si despejamos nos queda que
\(1 + 1 < t < 2 + 1\)
\(2 < t < 3\)
\[
y(t) =
\int^{}_{1}
\color{purple} x(t - \tau) \color{black}
\color{orange} h(\tau) \color{black}
d\tau
+
\int^{}_{-1}
\color{purple} x(t - \tau) \color{black}
\color{orange} h(\tau) \color{black}
d\tau
\]
\[
y(t) =
\int^{}_{1}
\color{purple} e^{-(t - \tau)} u(t - \tau - 1) \color{black}
\color{orange} \delta(\tau - 1) \color{black}
d\tau
+
\int^{}_{-1}
\color{purple} e^{-(t - \tau)} u(t - \tau - 1) \color{black}
\color{orange} (-2\delta(\tau + 1)) \color{black}
d\tau
\]
\[
y(t) =
\int^{}_{1}
\color{purple} e^{-(t - \tau)} u(t - \tau - 1) \color{black}
\color{orange} 1 \color{black}
d\tau
+
\int^{}_{-1}
\color{purple} e^{-(t - \tau)} u(t - \tau - 1) \color{black}
\color{orange} (-2) \color{black}
d\tau
\]
\[
y(t) =
\int^{}_{1}
\color{purple} e^{-(t - \tau)}1 \color{black}
d\tau
-2\int^{}_{-1}
\color{purple} e^{-(t - \tau)}1 \color{black}
d\tau
\]
\[
y(t) =
\int^{}_{1}
\color{purple} e^{-(t - \tau)} \color{black}
d\tau
-2\int^{}_{-1}
\color{purple} e^{-(t - \tau)} \color{black}
d\tau
\]
\[
y(t) =
e^{-(t - \tau)} \Big|^{b}_{1}
-2 e^{-(t - \tau)} \Big|^{b}_{-1}
\]
\[
y(t) =
e^{-(t - 1)}
-2 e^{-(t - (-1))}
\]
\[
\boxed{
y(t) = e^{-(t - 1)} -2 e^{-(t + 1)}
}
\]
Intervalo 4
Cuando \(t - 1 > 2\)
Si despejamos nos queda que
\(t > 2 + 1\)
\(t > 3\)
\[
y(t) =
\int^{t - 1}_{2}
\color{purple} x(t - \tau) \color{black}
\color{orange} h(\tau) \color{black}
d\tau
+
\int^{}_{1}
\color{purple} x(t - \tau) \color{black}
\color{orange} h(\tau) \color{black}
d\tau
+
\int^{}_{-1}
\color{purple} x(t - \tau) \color{black}
\color{orange} h(\tau) \color{black}
d\tau
\]
\[
y(t) =
\int^{t - 1}_{2}
\color{purple} e^{-(t - \tau)} u(t - \tau - 1) \color{black}
\color{orange} e^{-4\tau} u(\tau - 2) \color{black}
d\tau
+
\int^{}_{1}
\color{purple} e^{-(t - \tau)} u(t - \tau - 1) \color{black}
\color{orange} \delta(\tau - 1) \color{black}
d\tau
+
\int^{}_{-1}
\color{purple} e^{-(t - \tau)} u(t - \tau - 1) \color{black}
\color{orange} (-2\delta(\tau + 1)) \color{black}
d\tau
\]
\[
y(t) =
\int^{t - 1}_{2}
\color{purple} e^{-(t - \tau)}1 \color{black}
\color{orange} e^{-4\tau}1 \color{black}
d\tau
+
\int^{}_{1}
\color{purple} e^{-(t - \tau)}1 \color{black}
\color{orange} (1)\color{black}
d\tau
+
\int^{}_{-1}
\color{purple} e^{-(t - \tau)}1\color{black}
\color{orange} (-2) \color{black}
d\tau
\]
\[
y(t) =
\int^{t - 1}_{2}
\color{purple} e^{-(t - \tau)} \color{black}
\color{orange} e^{-4\tau} \color{black}
d\tau
+
\int^{}_{1}
\color{purple} e^{-(t - \tau)} \color{black}
d\tau
-
2\int^{}_{-1}
\color{purple} e^{-(t - \tau)}\color{black}
d\tau
\]
\[
y(t) =
\color{purple} e^{-(t - \tau)} \color{black}
\color{orange} e^{-4\tau} \color{black}
\Big|^{t - 1}_{2}
+
e^{-(t - \tau)} \Big|^{}_{1}
-2 e^{-(t - \tau)} \Big|^{}_{-1}
\]
\[
y(t) =
(e^{-(t - 2)}e^{-4(2)}) - (e^{-(t - (t - 1))}e^{-4(t - 1)})
+
e^{-(t - 1)}
-
2 e^{-(t - (-1))}
\]
\[
y(t) =
(e^{-(t - 2)}e^{-8}) - (e^{-(t - t + 1)}e^{-4(t - 1)})
+
e^{-(t - 1)}
-
2 e^{-(t + 1)}
\]
\[
y(t) =
e^{-8 - (t - 2)} - e^{- 1 - 4(t - 1)}
+
e^{-(t - 1)}
-
2 e^{-(t + 1)}
\]
\[
y(t) =
e^{(-8 - t + 2)} - e^{(- 1 - 4t + 4)}
+
e^{-(t - 1)}
-
2 e^{-(t + 1)}
\]
\[
\boxed{
y(t) =
e^{(-6 - t)} - e^{(- 4t + 3)}
+
e^{-(t - 1)}
-
2 e^{-(t + 1)}
}
\]
\[
\color{limegreen} \boxed{ \color{black}
y(t) =
\begin{cases}
0 & t < 0 \\
-2e^{-(t+1)} & 0 < t < 2 \\
e^{-(t-1)}-2e^{-(t+1)} & 2 < t < 3 \\
e^{(-6-t)}-e^{(-4t+3)}+e^{-(t-1)}-2e^{-(t+1)} & t > 3
\end{cases}
}
\]
