Convolución entre dos señales no constantes
1) Elegimos que señal vamos a reflejar
Reflejamos la señal menos compleja, que es \(\color{orange}x(t)\) \[ \color{orange}x(t) \color{black} \rightarrow \color{orange}x(-t) \]
2) Resolver las integrales de los diferentes intervalos
Intervalo a
Cuando \(t < -1\)
\[
y(t) = 0
\]
Intervalo b
Cuando \(-1 < t < 0\)
\[
y(t) = \int_{-1}^{t}(t - \tau) \cdot 2\tau d\tau
\]
\[
y(t) = \int_{-1}^{t} [t2\tau - 2\tau^2] d\tau
\]
Sacamos afuera lo que no depende de \(\tau\).
\[
y(t) = -t\int_{-1}^{t} 2\tau d\tau + \int_{-1}^{t} 2\tau^2 d\tau
\]
\[
y(t) = -t\tau^2 \Big|^{\tau}_{-1} + \frac{2}{3}\tau^3 \Big|^{\tau}_{-1}
\]
\[
y(t) = -t(t^2 - (-1)^2) + \frac{2}{3}(t^3 - (-1)^3)
\]
\[
y(t) = -t(t^2 - 1) + \frac{2}{3}(t^3 + 1)
\]
\[
y(t) = -t^3 + t + \frac{2}{3}t^3 + \frac{2}{3}
\]
\[
\color{limegreen}\boxed{
\color{black} y(t) = -\frac{1}{3}t^3 + t + \frac{2}{3}
}
\]
Intervalo c
Cuando \(0 < t < 1\)
\[
y(t) = \int_{t - 1}^{t} (t - \tau) \cdot (-2\tau) d\tau + \int^{t-1}_{-1} 1 \cdot (-2\tau) d\tau
\]
\[
y(t) =
- t \tau^2 \Big|^{t}_{t-1}
- \frac{2}{3} \tau^3 \Big|^{t^{-1}}_{t-1}
- \tau^2 \Big|^{t-1}_{-1}
\]
\[
y(t) =
- t(t^2 - (t - 1)^2)
- \frac{2}{3} (t^2 - (t-1)^3)
- [(t - 1)^2 - (-1)^2]
\]
\[
y(t) =
- t^3 + t(t - 1)^2
- \frac{2}{3}t^3 + \frac{2}{3}(t - 1)^3
- (t - 1)^2 + 1
\]
Intervalo d
Cuando \(1 < t < 2\)
\[
y(t) =
\int^{1}_{t - 1} (t - \tau) - 2\tau d\tau +
\int^{t - 1}_{1} 1 \cdot (2\tau) d\tau
\]
\[
y(y) =
- t\tau^2 \Big|^{1}_{t - 1}
- frac{2}{3} \tau^3 \Big|^{1}_{t - 1}
- \tau^2 \Big|^{t - 1}_{1}
\]
\[
y(t) =
- t(1^2 - (t - 1)^2)
- \frac{2}{3}(1^3 - (t - 1)^3)
- [(t - 1)^2 - (-1)^2]
\]
\[
y(t) =
- t + t(t-1)^2
- \frac{2}{3} + \frac{2}{3}(t-1)^3
- (t - 1)^2 + 1
\]
Intervalo e
Cuando 2 < t < 4
\[
y(t) = \int^{1}_{t - 3} 1 \cdot (-2\tau) d\tau
\]
\[
y(t) = \tau^2 \Big|^{1}_{t - 3}
\]
\[
y(t) = - [1^2 - (t - 3)^2]
\]
\[
y(t) = -1 - (t - 3)^2
\]
Intervalo f
Cuando 4 < t
\[
y(t) = 0
\]