Problema 6 02 08
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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)\,}
Problema 7 02 08
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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)\,} Problema 8 02 08
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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
Problema 9 02 08
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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?.
Problema 10 02 08
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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) ? Problema 11 02 08
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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)\,} Subsección 1 Problema 1
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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|}}
Subsección 2 Problema 1
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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}}}
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