ma20010770021.pdf

L sen7 t u 2 π t e 2 π s l sen7 t t t 2 π e 2 π s

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L { sen(7 t ) U 2 π ( t ) } = e - 2 π s · L sen(7 t ) | t = t +2 π = e - 2 π s · L { sen(7 t + 7 · 2 π ) } = e - 2 π s · L { sen(7 t ) } = e - 2 π s · 7 s 2 +49
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Es importante recordar que sen( θ ± n · 2 π ) = sen( θ ) y que sen( θ ± π ) = - sen( θ ). Nuestro avance en la ED va ahora: ( s 2 + 25 ) L { y } - s · 2 - 0 = L { sen(7 t ) U 2 π ( t ) } ( s 2 + 25 ) L { y } - 2 s = e - 2 π s · 7 s 2 +49 As´ ı L { y } = 2 s s 2 +25 + e - 2 π s · 7 ( s 2 +49) · ( s 2 +25) Por lo tanto, la soluci´ on a la ED cumple: y ( t ) = L - 1 { L { y }} = L - 1 2 s s 2 + 25 + e - 2 π s · 7 ( s 2 + 49) · ( s 2 + 25) Por la linealidad de la transformada inversa: y ( t ) = L - 1 2 s s 2 + 25 + L - 1 e - 2 π s · 7 ( s 2 + 49) · ( s 2 + 25) Calculemos por separado estas dos inversas. La primera de ellas tiene que ver con las condiciones iniciales que aparec´ ıan en el miembro izquierdo y que pasa- ron al derecho: Por la linealidad de la ED la interpretaremos como la respuesta del sistema a entrada cero y dadas las condiciones internas (condiciones inicia- les) del problema: y 0 ( t ) = L - 1 2 s s 2 + 25 = 2 sen(5 t ) La segunda inversa de Laplace a calcular se relaciona con la respuesta del sis- tema a la entrada r ( t ) = sen(7 t ) U 2 π ( t ). Para calcularla utilicemos el segundo teorema de traslaci´ on en su versi´ on de la inversa: L - 1 F ( s ) · e - a s = L - 1 { F ( s ) } t = t - a · U a ( t ) Es decir, 1) Omitimos el factor exponencial, 2) Obtenemos la inversa de la expresi´ on resultante, 3) En el resultado hacemos la sustituci´ on de t por t - a ,
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  • Spring '11
  • Mariag
  • 25 l, Inverso multiplicativo, Pierre-Simon Laplace, Multiplicación, eje t, ·L {f

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