Para el circuito RLC que se muestra en la figura, determine (R=3,L=2,C=K=2):
La respuesta al escalón
La respuesta al impulso
La respuesta a
x
(
t
)
=
e
−
2
t
u
(
t
)
{\displaystyle x(t)=e^{-2t}u(t)\,}
Considere un sistema LTI cuya respuesta al impulso es la función
Δ
(
t
)
=
r
(
t
+
1
)
−
2
r
(
t
)
+
r
(
t
−
1
)
{\displaystyle \Delta (t)=r(t+1)-2r(t)+r(t-1)\,}
Calcule y grafique la salida del sistema a las señales:
x
1
(
t
)
=
∑
k
=
−
∞
∞
δ
(
t
−
2
k
)
{\displaystyle x_{1}(t)=\sum _{k=-\infty }^{\infty }\delta (t-2k)\,}
x
2
(
t
)
=
∑
k
=
−
∞
∞
δ
(
t
−
4
k
)
{\displaystyle x_{2}(t)=\sum _{k=-\infty }^{\infty }\delta (t-4k)\,}
El sistema de la pregunta anterior se coloca en cascada con un segundo sistema cuya respuesta al impulso es
h
(
t
)
=
1
2
Π
(
t
−
1
2
)
{\displaystyle h(t)={\frac {1}{2}}\Pi ({\frac {t-1}{2}})\,}
Π
(
t
)
=
u
(
t
+
0.5
)
u
(
0.5
−
t
)
{\displaystyle \Pi (t)=u(t+0.5)u(0.5-t)\,}
Determine y grafique la respuesta de la cascada de sistemas a las entradas del problema anterior
Considere la cascada de dos sistemas. El primero, que llamaremos S1, comprime (operación sobre el tiempo) la señal de entrada por un factor de 2, i.e.,
y
(
t
)
=
x
(
2
t
)
{\displaystyle y(t)=x(2t)\,}
x
(
t
)
=
2
e
−
t
u
(
t
)
{\displaystyle x(t)=2e^{-t}u(t)\,}
x
(
t
)
→
S
1
→
S
2
→
y
(
t
)
{\displaystyle x(t)\rightarrow S1\rightarrow S2\rightarrow y(t)\,}
x
(
t
)
→
S
2
→
S
1
→
y
(
t
)
{\displaystyle x(t)\rightarrow S2\rightarrow S1\rightarrow y(t)\,}
¿Serán idénticas las salidas?, ¿deberían serlo?.
Considere la señal
x
1
(
t
)
=
e
−
t
u
(
t
)
{\displaystyle x_{1}(t)=e^{-t}u(t)\,}
h
(
t
)
=
u
(
t
)
u
(
1
−
t
)
{\displaystyle h(t)=u(t)u(1-t)\,}
x
2
(
t
)
=
∑
k
=
−
∞
∞
δ
(
t
−
k
T
)
{\displaystyle x_{2}(t)=\sum _{k=-\infty }^{\infty }\delta (t-kT)\,}
x
3
(
t
)
=
x
1
(
t
)
x
2
(
t
)
{\displaystyle x_{3}(t)=x_{1}(t)x_{2}(t)\,}
x
4
(
t
)
=
lim
T
→
0
x
3
(
t
)
{\displaystyle x_{4}(t)=\lim _{T\rightarrow 0}x_{3}(t)\,}
¿Puede generalizar su resultado a cualquier h(t) y x(t) ? Grafique cada una de las señales y realice las siguientes convoluciones:
e
−
t
u
(
t
)
u
(
t
−
2
)
⋆
δ
(
t
)
{\displaystyle e^{-t}u(t)u(t-2)\star \delta (t)\,}
r
(
t
)
u
(
1
−
t
)
⋆
u
(
t
−
2
)
u
(
4
−
t
)
{\displaystyle r(t)u(1-t)\star u(t-2)u(4-t)\,}
e
−
|
t
|
u
(
t
+
2
)
u
(
2
−
t
)
⋆
u
(
t
−
1
)
u
(
2
−
t
)
{\displaystyle e^{-|t|}u(t+2)u(2-t)\star u(t-1)u(2-t)\,}
sin
(
2
π
t
)
u
(
t
)
u
(
1
−
t
)
⋆
2
u
(
t
+
1
)
u
(
1
−
t
)
{\displaystyle \sin(2\pi t)u(t)u(1-t)\star 2u(t+1)u(1-t)\,}
sin
(
π
t
)
π
t
⋆
e
−
t
u
(
t
)
{\displaystyle {\frac {\sin(\pi t)}{\pi t}}\star e^{-t}u(t)\,}
Resuelto por Ender Valdivieso Carnet 06-40411
Ejercicio 1
x
1
(
t
)
=
e
−
t
u
(
t
)
u
(
t
−
2
)
{\displaystyle x_{1}(t)=e^{-t}u(t)u(t-2)\,}
h
1
(
t
)
=
δ
(
t
)
{\displaystyle h_{1}(t)=\delta (t)\,}
x
1
(
t
)
=
e
−
t
u
(
t
)
u
(
t
−
2
)
{\displaystyle x_{1}(t)=e^{-t}u(t)u(t-2)\,}
h
1
(
t
)
{\displaystyle h_{1}(t)\,}
A priori conocemos que la función delta
δ
(
t
)
{\displaystyle \delta (t)\,}
x
1
(
t
−
τ
)
=
{
e
τ
−
t
t
−
2
<
τ
<
t
0
resto
{\displaystyle x_{1}(t-\tau )=\left\{{\begin{array}{lc}e^{\tau -t}&t-2<\tau <t\\0&{\mbox{resto}}\end{array}}\right.}
Para
0
<
t
<
2
{\displaystyle 0<t<2\,}
y
1
(
t
)
=
∫
−
∞
∞
e
τ
−
t
δ
(
τ
)
d
τ
{\displaystyle y_{1}(t)=\int _{-\infty }^{\infty }e^{\tau -t}\delta (\tau )d\tau }
y
1
(
t
)
=
e
−
t
∫
−
∞
∞
δ
(
τ
)
d
τ
{\displaystyle y_{1}(t)=e^{-t}\int _{-\infty }^{\infty }\delta (\tau )d\tau }
y
1
(
t
)
=
e
−
t
u
(
t
)
u
(
2
−
t
)
{\displaystyle y_{1}(t)=e^{-t}u(t)u(2-t)\,}
Gráfica de #
y
1
(
t
)
=
x
1
(
t
)
⋆
h
1
(
t
)
{\displaystyle y_{1}(t)=x_{1}(t)\star h_{1}(t)\,}
Ejercicio 2
x
2
(
t
)
=
r
(
t
)
u
(
1
−
t
)
{\displaystyle x_{2}(t)=r(t)u(1-t)\,}
h
2
(
t
)
=
u
(
t
−
2
)
u
(
4
−
t
)
{\displaystyle h_{2}(t)=u(t-2)u(4-t)\,}
x
2
(
t
)
{\displaystyle x_{2}(t)\,}
h
2
(
t
)
{\displaystyle h_{2}(t)\,}
h
2
(
t
−
τ
)
=
{
1
t
−
4
<
τ
<
t
−
2
0
resto
{\displaystyle h_{2}(t-\tau )=\left\{{\begin{array}{lc}1&t-4<\tau <t-2\\0&{\mbox{resto}}\end{array}}\right.}
x
2
(
τ
)
=
{
τ
0
<
τ
<
1
0
resto
{\displaystyle x_{2}(\tau )=\left\{{\begin{array}{lc}\tau &0<\tau <1\\0&{\mbox{resto}}\end{array}}\right.}
Para
2
<
t
<
3
{\displaystyle 2<t<3\,}
y
2
(
t
)
=
∫
0
t
−
2
τ
d
τ
{\displaystyle y_{2}(t)=\int _{0}^{t-2}\tau d\tau }
y
2
(
t
)
=
(
t
−
2
)
2
2
{\displaystyle y_{2}(t)={\frac {(t-2)^{2}}{2}}}
Para
3
<
t
<
4
{\displaystyle 3<t<4\,}
y
2
(
t
)
=
∫
0
1
τ
d
τ
{\displaystyle y_{2}(t)=\int _{0}^{1}\tau d\tau }
y
2
(
t
)
=
1
2
{\displaystyle y_{2}(t)={\frac {1}{2}}}
4
<
t
<
5
{\displaystyle 4<t<5\,}
y
2
(
t
)
=
∫
t
−
4
1
τ
d
τ
{\displaystyle y_{2}(t)=\int _{t-4}^{1}\tau d\tau }
y
2
(
t
)
=
1
2
−
(
t
−
4
)
2
2
{\displaystyle y_{2}(t)={\frac {1}{2}}-{\frac {(t-4)^{2}}{2}}}
Enotonces la función
y
2
(
t
)
{\displaystyle y_{2}(t)\,}
y
2
(
t
)
=
{
(
t
−
2
)
2
2
2
<
t
<
3
1
2
3
<
t
<
4
1
2
−
(
t
−
4
)
2
2
4
<
t
<
5
0
resto
{\displaystyle y_{2}(t)=\left\{{\begin{array}{lc}{\frac {(t-2)^{2}}{2}}&2<t<3\\{\frac {1}{2}}&3<t<4\\{\frac {1}{2}}-{\frac {(t-4)^{2}}{2}}&4<t<5\\0&{\mbox{resto}}\end{array}}\right.}
Gráfica de #
y
2
(
t
)
=
x
2
(
t
)
⋆
h
2
(
t
)
{\displaystyle y_{2}(t)=x_{2}(t)\star h_{2}(t)\,}
Ejercicio 3
x
3
(
t
)
=
e
−
|
t
|
u
(
t
+
2
)
u
(
2
−
t
)
{\displaystyle x_{3}(t)=e^{-|t|}u(t+2)u(2-t)\,}
h
3
(
t
)
=
u
(
t
−
1
)
u
(
2
−
t
)
{\displaystyle h_{3}(t)=u(t-1)u(2-t)\,}
Gráfica de
x
3
(
t
)
{\displaystyle x_{3}(t)\,}
h
3
(
t
)
{\displaystyle h_{3}(t)\,}
x
3
(
τ
)
=
{
e
τ
−
2
<
τ
<
0
e
−
τ
0
<
τ
<
2
0
resto
{\displaystyle x_{3}(\tau )=\left\{{\begin{array}{lc}e^{\tau }&-2<\tau <0\\e^{-\tau }&0<\tau <2\\0&{\mbox{resto}}\end{array}}\right.\,}
h
3
(
t
−
τ
)
=
{
1
t
−
2
<
τ
<
t
−
1
0
resto
{\displaystyle h_{3}(t-\tau )=\left\{{\begin{array}{lc}1&t-2<\tau <t-1\\0&{\mbox{resto}}\end{array}}\right.\,}
−
1
<
t
<
0
{\displaystyle -1<t<0\,}
y
3
(
t
)
=
∫
−
2
t
−
1
e
τ
d
τ
{\displaystyle y_{3}(t)=\int _{-2}^{t-1}e^{\tau }d\tau \,}
y
3
(
t
)
=
e
t
−
1
−
e
−
2
{\displaystyle y_{3}(t)=e^{t-1}-e^{-2}\,}
Para
0
<
t
<
1
{\displaystyle 0<t<1\,}
y
3
(
t
)
=
∫
t
−
2
t
−
1
e
τ
d
τ
{\displaystyle y_{3}(t)=\int _{t-2}^{t-1}e^{\tau }d\tau \,}
y
3
(
t
)
=
e
t
−
1
−
e
t
−
2
{\displaystyle y_{3}(t)=e^{t-1}-e^{t-2}\,}
1
<
t
<
2
{\displaystyle 1<t<2\,}
y
3
(
t
)
=
∫
t
−
2
0
e
τ
d
τ
+
∫
0
t
−
1
e
−
τ
d
τ
{\displaystyle y_{3}(t)=\int _{t-2}^{0}e^{\tau }d\tau +\int _{0}^{t-1}e^{-\tau }d\tau \,}
y
3
(
t
)
=
−
e
−
t
+
1
−
e
t
−
2
+
2
{\displaystyle y_{3}(t)=-e^{-t+1}-e^{t-2}+2\,}
Para
2
<
t
<
3
{\displaystyle 2<t<3\,}
y
3
(
t
)
=
∫
t
−
2
t
−
1
e
−
τ
d
τ
{\displaystyle y_{3}(t)=\int _{t-2}^{t-1}e^{-\tau }d\tau \,}
y
3
(
t
)
=
e
−
t
+
1
−
e
−
t
+
2
{\displaystyle y_{3}(t)=e^{-t+1}-e^{-t+2}\,}
Para
3
<
t
<
4
{\displaystyle 3<t<4\,}
y
3
(
t
)
=
∫
t
−
2
2
e
−
τ
d
τ
{\displaystyle y_{3}(t)=\int _{t-2}^{2}e^{-\tau }d\tau \,}
y
3
(
t
)
=
e
−
t
+
2
−
e
−
2
{\displaystyle y_{3}(t)=e^{-t+2}-e^{-2}\,}
y
3
(
t
)
=
0
{\displaystyle y_{3}(t)=0\,}
t
{\displaystyle t\,}
En síntesis, la función sería de la forma
y
3
(
t
)
=
{
e
t
−
1
−
e
−
2
0
<
t
<
1
e
t
−
1
−
e
t
−
2
0
<
t
<
1
−
e
−
t
+
1
−
e
t
−
2
+
2
1
<
t
<
2
e
−
t
+
1
−
e
−
t
+
2
2
<
t
<
3
e
−
t
+
2
−
e
−
2
3
<
t
<
4
0
resto
{\displaystyle y_{3}(t)=\left\{{\begin{array}{lc}e^{t-1}-e^{-2}&0<t<1\\e^{t-1}-e^{t-2}&0<t<1\\-e^{-t+1}-e^{t-2}+2&1<t<2\\e^{-t+1}-e^{-t+2}&2<t<3\\e^{-t+2}-e^{-2}&3<t<4\\0&{\mbox{resto}}\end{array}}\right.\,}
y
3
(
t
)
=
x
3
(
t
)
⋆
h
3
(
t
)
{\displaystyle y_{3}(t)=x_{3}(t)\star h_{3}(t)\,}
Ejercicio 4
x
4
(
t
)
=
sin
(
2
π
t
)
u
(
t
)
u
(
1
−
t
)
{\displaystyle x_{4}(t)=\sin(2\pi t)u(t)u(1-t)\,}
h
4
(
t
)
=
2
u
(
t
+
1
)
u
(
1
−
t
)
{\displaystyle h_{4}(t)=2u(t+1)u(1-t)\,}
x
4
(
t
)
{\displaystyle x_{4}(t)\,}
h
4
(
t
)
{\displaystyle h_{4}(t)\,}
x
4
(
τ
)
=
{
(
2
π
t
)
0
<
τ
<
1
0
resto
{\displaystyle x_{4}(\tau )=\left\{{\begin{array}{lc}(2\pi t)&0<\tau <1\\0&{\mbox{resto}}\end{array}}\right.\,}
h
4
(
t
−
τ
)
=
{
2
t
−
1
<
τ
<
t
+
1
0
resto
{\displaystyle h_{4}(t-\tau )=\left\{{\begin{array}{lc}2&t-1<\tau <t+1\\0&{\mbox{resto}}\end{array}}\right.\,}
Para
−
1
<
t
<
0
{\displaystyle -1<t<0\,}
y
4
(
t
)
=
∫
0
t
+
1
2
sin
(
2
∗
π
τ
)
d
τ
{\displaystyle y_{4}(t)=\int _{0}^{t+1}2\sin(2*\pi \tau )d\tau \,}
y
4
(
t
)
=
1
−
cos
(
2
π
t
)
{\displaystyle y_{4}(t)=1-\cos(2\pi t)\,}
Para
0
<
t
<
1
{\displaystyle 0<t<1\,}
y
4
(
t
)
=
∫
0
1
2
sin
(
2
∗
π
τ
)
d
τ
{\displaystyle y_{4}(t)=\int _{0}^{1}2\sin(2*\pi \tau )d\tau \,}
y
4
(
t
)
=
0
{\displaystyle y_{4}(t)=0\,}
1
<
t
<
2
{\displaystyle 1<t<2\,}
y
4
(
t
)
=
∫
t
−
1
1
2
sin
(
2
∗
π
τ
)
d
τ
{\displaystyle y_{4}(t)=\int _{t-1}^{1}2\sin(2*\pi \tau )d\tau \,}
y
4
(
t
)
=
cos
(
2
π
t
)
−
1
{\displaystyle y_{4}(t)=\cos(2\pi t)-1\,}
y
4
(
t
)
=
{
1
−
cos
(
2
π
t
)
−
1
<
t
<
0
cos
(
2
π
t
)
−
1
1
<
t
<
2
0
resto
{\displaystyle y_{4}(t)=\left\{{\begin{array}{lc}1-\cos(2\pi t)&-1<t<0\\\cos(2\pi t)-1&1<t<2\\0&{\mbox{resto}}\end{array}}\right.\,}
Gráfica de #
y
4
(
t
)
=
x
4
(
t
)
⋆
h
4
(
t
)
{\displaystyle y_{4}(t)=x_{4}(t)\star h_{4}(t)\,}
Ejercicio 5
x
5
(
t
)
=
sin
(
π
t
)
π
t
{\displaystyle x_{5}(t)={\frac {\sin(\pi t)}{\pi t}}\,}
h
5
(
t
)
=
e
−
t
u
(
t
)
{\displaystyle h_{5}(t)=e^{-t}u(t)\,}
x
5
(
t
)
{\displaystyle x_{5}(t)\,}
h
5
(
t
)
{\displaystyle h_{5}(t)\,}
x
5
(
τ
)
=
sin
(
π
τ
)
(
π
τ
)
{\displaystyle x_{5}(\tau )={\frac {\sin(\pi \tau )}{(\pi \tau )}}\,}
h
5
(
t
−
τ
)
=
{
e
τ
−
t
τ
<
t
0
resto
{\displaystyle h_{5}(t-\tau )=\left\{{\begin{array}{lc}e^{\tau -t}&\tau <t\\0&{\mbox{resto}}\end{array}}\right.\,}
y
5
(
t
)
=
∫
−
∞
t
sin
(
π
τ
)
(
π
τ
)
e
τ
−
t
d
τ
{\displaystyle y_{5}(t)=\int _{-\infty }^{t}{\frac {\sin(\pi \tau )}{(\pi \tau )}}e^{\tau -t}d\tau \,}
Una versión imprimible se encuntra en el siguiente archivoArchivo
Sean,
x
1
(
t
)
=
e
−
|
t
|
u
(
t
)
u
(
2
−
t
)
{\displaystyle x_{1}(t)=e^{-|t|}u(t)u(2-t)\,}
x
2
(
t
)
=
∑
k
=
−
∞
∞
δ
(
t
−
4
k
)
{\displaystyle x_{2}(t)=\sum _{k=-\infty }^{\infty }\delta (t-4k)}
x
3
(
t
)
=
∑
k
=
−
∞
∞
Π
1
(
t
−
8
k
)
{\displaystyle x_{3}(t)=\sum _{k=-\infty }^{\infty }\Pi _{1}(t-8k)}
Determine:
y
1
(
t
)
=
x
1
(
t
)
∗
x
2
(
t
)
{\displaystyle y_{1}(t)=x_{1}(t)*x_{2}(t)\,}
y
2
(
t
)
=
x
1
(
t
)
∗
x
3
(
t
)
{\displaystyle y_{2}(t)=x_{1}(t)*x_{3}(t)\,}
editar
Realizado Por: Jesús Querales #05-38758
y
1
(
t
)
=
x
1
(
t
)
∗
x
2
(
t
)
=
x
2
(
t
)
∗
x
1
(
t
)
{\displaystyle y_{1}(t)=x_{1}(t)*x_{2}(t)=x_{2}(t)*x_{1}(t)\,}
x
2
(
t
)
=
x
(
t
)
{\displaystyle x_{2}(t)=x(t)\,}
x
1
(
t
)
=
h
(
t
)
{\displaystyle x_{1}(t)=h(t)\,}
y
1
(
t
)
=
∫
−
∞
∞
x
(
τ
)
h
(
t
−
τ
)
d
τ
{\displaystyle y_{1}(t)=\int _{-\infty }^{\infty }x(\tau )h(t-\tau )\,d\tau }
x
(
τ
)
=
∑
k
=
−
∞
∞
δ
(
τ
−
4
k
)
{\displaystyle x(\tau )=\sum _{k=-\infty }^{\infty }\delta (\tau -4k)}
h
(
t
−
τ
)
=
e
−
|
t
−
τ
|
u
(
t
−
τ
)
u
(
2
−
t
+
τ
)
{\displaystyle h(t-\tau )=e^{-|t-\tau |}u(t-\tau )u(2-t+\tau )\,}
y
1
(
t
)
=
∫
−
∞
∞
∑
k
=
−
∞
∞
δ
(
τ
−
4
k
)
e
−
|
t
−
τ
|
u
(
t
−
τ
)
u
(
2
−
t
+
τ
)
d
τ
{\displaystyle y_{1}(t)=\int _{-\infty }^{\infty }\sum _{k=-\infty }^{\infty }\delta (\tau -4k)e^{-|t-\tau |}u(t-\tau )u(2-t+\tau )\,d\tau }
y
1
(
t
)
=
∫
t
t
−
2
∑
k
=
−
∞
∞
δ
(
τ
−
4
k
)
e
−
|
t
−
τ
|
{\displaystyle y_{1}(t)=\int _{t}^{t-2}\sum _{k=-\infty }^{\infty }\delta (\tau -4k)e^{-|t-\tau |}}
y
1
(
t
)
=
∑
k
=
−
∞
∞
e
−
|
t
−
4
k
|
∫
t
t
−
2
δ
(
τ
−
4
k
)
=
∑
k
=
−
∞
∞
e
−
|
t
−
4
k
|
{\displaystyle y_{1}(t)=\sum _{k=-\infty }^{\infty }e^{-|t-4k|}\int _{t}^{t-2}\delta (\tau -4k)=\sum _{k=-\infty }^{\infty }e^{-|t-4k|}}
editar
Realizado Por: Alexander Gamero #05-38196
x
1
(
t
)
=
e
−
|
t
|
u
(
t
)
u
(
2
−
t
)
{\displaystyle x_{1}(t)=e^{-|t|}\ u(t)\ u(2-t)\,}
[
0
,
2
]
{\displaystyle [0,2]\,}
t
≥
0
{\displaystyle t\geq 0}
x
1
(
t
)
=
e
−
t
u
(
t
)
u
(
2
−
t
)
{\displaystyle x_{1}(t)=e^{-t}\ u(t)\ u(2-t)\,}
x
3
(
t
)
=
∑
k
=
−
∞
∞
Π
1
(
t
−
8
k
)
{\displaystyle x_{3}(t)=\sum _{k=-\infty }^{\infty }\Pi _{1}(t-8k)}
T
=
8
{\displaystyle T=8\,}
x
1
(
t
)
{\displaystyle x_{1}(t)\,}
x
3
(
t
)
{\displaystyle x_{3}(t)\,}
k
=
0
{\displaystyle k=0\,}
x
3
(
t
)
=
u
(
−
t
+
1
2
)
u
(
t
+
1
2
)
{\displaystyle x_{3}(t)=u(-t+{\frac {1}{2}})\ u(t+{\frac {1}{2}})\,}
y
(
t
)
=
∫
−
∞
∞
x
3
(
τ
)
x
1
(
t
−
τ
)
d
τ
{\displaystyle y(t)=\int _{-\infty }^{\infty }x_{3}(\tau )x_{1}(t-\tau )\,d\tau }
−
1
2
≤
t
≤
1
2
{\displaystyle -{\frac {1}{2}}\leq t\leq {\frac {1}{2}}}
y
(
t
)
=
∫
−
1
2
t
e
−
t
+
τ
d
τ
=
1
−
e
−
t
+
1
2
{\displaystyle y(t)=\int _{-{\frac {1}{2}}}^{t}e^{-t+\tau }\,d\tau =1-e^{-t+{\frac {1}{2}}}}
1
2
≤
t
≤
3
2
{\displaystyle {\frac {1}{2}}\leq t\leq {\frac {3}{2}}}
y
(
t
)
=
∫
−
1
2
1
2
e
−
t
+
τ
d
τ
=
e
−
t
+
1
2
−
e
−
(
t
+
1
2
)
{\displaystyle y(t)=\int _{-{\frac {1}{2}}}^{\frac {1}{2}}e^{-t+\tau }\,d\tau =e^{-t+{\frac {1}{2}}}-e^{-(t+{\frac {1}{2}})}}
3
2
≤
t
≤
5
2
{\displaystyle {\frac {3}{2}}\leq t\leq {\frac {5}{2}}}
y
(
t
)
=
∫
t
−
2
1
2
e
−
t
+
τ
d
τ
=
e
−
t
+
1
2
−
e
−
2
{\displaystyle y(t)=\int _{t-2}^{\frac {1}{2}}e^{-t+\tau }\,d\tau =e^{-t+{\frac {1}{2}}}-e^{-2}}
y
(
t
)
=
0
{\displaystyle y(t)=0\,}
y
(
t
)
{\displaystyle y(t)\,}
t
por
t
−
8
k
{\displaystyle t\ {\mbox{por}}\ t-8k\,}
y
(
t
)
=
∑
k
=
−
∞
∞
{
1
−
e
−
t
−
1
2
+
8
k
,
si
−
1
2
+
8
k
≤
t
≤
1
2
+
8
k
e
−
(
t
−
1
2
−
8
k
)
−
e
−
(
t
+
1
2
−
8
k
)
,
si
1
2
+
8
k
≤
t
≤
3
2
+
8
k
e
−
t
+
1
2
+
8
k
−
e
−
2
,
si
3
2
+
8
k
≤
t
≤
5
2
+
8
k
0
,
para todo lo demas
{\displaystyle y(t)=\sum _{k=-\infty }^{\infty }{\begin{cases}1-e^{-t-{\frac {1}{2}}+8k},&{\mbox{si }}-{\frac {1}{2}}+8k\leq t\leq {\frac {1}{2}}+8k\\e^{-(t-{\frac {1}{2}}-8k)}-e^{-(t+{\frac {1}{2}}-8k)},&{\mbox{si }}{\frac {1}{2}}+8k\leq t\leq {\frac {3}{2}}+8k\\e^{-t+{\frac {1}{2}}+8k}-e^{-2},&{\mbox{si }}{\frac {3}{2}}+8k\leq t\leq {\frac {5}{2}}+8k\\0,&{\mbox{ para todo lo demas}}\end{cases}}}
![{\displaystyle [0,2]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c930c6b148a1afeee1a559f22d76ee3262a73a56)
