Matematicas Especiales

Matematicas especiales

Ejercicios de transformada inversa de Laplace

1)

L^(-1) {(4s-18)/(9-5^2 )-(3s-12)/(s^2+18) }

Solución

Se aplica la propiedad de linealidad

L^(-1) {(4s-18)/(-(s^2-9) )}- L^(-1) {(3s-12)/(s^2+18)}

=-4L^(-1) {s/s^(2-9) }+18L^(-1) {1/s^(2-9) }-3L^(-1) {s/(s^2 (√18)2)}+12 L^(-1) {1/s^2 +(√18)2}

=-4cosh3t + 18 (senh(3t)/3)-3cos (√18t)+12 (sen(√18t)/√18)

=-4cosh3t + 6senh(3t) - 3cos(√18t)+4/√2 sen√18 t

=-4cosh3t + 6senh(3t) – 3cos(√18t)+2√2 sen√18t R/

2)

L^(-1) {(s^2-3)/(s^2-s-6)(s^2+2s+5) }

Solucion

Se expresa la fraccion en terminos de fracciones parciales

(s^2-3)/(s^2-s-6)(s^2+2s+5) =(s^2-3)/(s-3)(s+2)(s^2+2s+5)

=A/(s-3)+B/(s+2)+(Cs+D)/(s^2+2s+5)

s^2-3=A(s+2)(s^2+2s+5)+B(s-3)(s^2+2s+5)+(Cs+D)(s-3)(s+2)

s^2-3=A(s^3+2s^2+5s+2s^2+4s+10)+B(s^3+〖2s〗^2+5s-〖3s〗^2-6s-15)+(Cs+D)(s^2-5-6)

s^2-3=〖As〗^3+4As^2+9As+10A+Bs^3-Bs^3-〖Bs〗^2-Bs-15B+Cs^3-〖Cs〗^2-6Cs+Ds^2-Ds-6

s^2-3=(A+B+C) S^3+(4A-B-C-D) S^2+(9A-B-6C-D)S+(10A-15B-6D)

A+B+C=0 (1)

4A-B-C+D=1 (2)

9A-B-6C-D=0 (3)

10A-15B-6D=-3 (4)

A= (s^2-3)/((s+2)(s^2+2s+5))=6/((5)(9+6+5))=6/100=3/50 A=3/50

A= (s^2-3)/((s-3)(s^2+2s+5))=1/((-5)(4-4+5))=-1/25 B=1/25

En la ecuacion (1) se reemplazan los valores de A y B para obtener C.

3/50-1/25+C=0 = C=1/25-3/50= C=-1/50

En la ecuación (3) reemplazamos y se obtiene D

9A-B-6C-D=0 = 9(3/50)-(-1/25)-6(-1/50)-D=0

27/50+1/25+6/50=D=35/50=D=7/10

Por tanto la fracción algebraica queda:

(s^2-3)/(s-3)(s+2)(s^2+2s+5) =(3/50)/(s-3)+(-1/25)/(s+2)+(-1/50 s+7/10)/(s^2+2s+5)

Por tanto

L^(-1) {(s^2-3)/(s-3)(s+2)(s^2+2s+5) }=L^(-1) {(3/50)/(s-3) - (1/25)/(s+2)+(-1/50 s+7/10)/(s^2+2s+5)}

Se aplica la propiedad de linealidad

=3/50 L^(-1) {1/(s-3)}-1/25 L^(-1) {1/(s+2)}-1/50 L^(-1) {s/((s^2+2s+1)+5-1)}+7/10 L^(-1) {1/((s+1)2+4)}

=3/50 ℮^3t-1/25 ℮^(-2t)-1/50 L^(-1) {((s+1)-1)/((s+1)2+4)}+7/10 ℮^(-t) ((sen 2t)/2)

=3/50 ℮^3t-1/25 ℮^(-2t)+7/20 ℮^(-t) sen2t-1/50 L^(-1) {(s+1)/((s+1)2+4)}+1/50 L^(-1) {1/((s+1)2+4)}

=3/50 ℮^3t-1/25 ℮^(-2t)+7/20 ℮^(-t) sen2t-1/50 ℮^(-t) cos2t+1/50 (sen2t/2).℮^(-t)

=3/50 ℮^3t-1/25 ℮^(-2t)+℮^(-t) [7/20 sen2t-1/50 cos2t+1/100 sen2t]

=3/50 ℮^3t-1/25 ℮^(-2t)+℮^(-t) [7/20 sen2t-1/50 cos2t]

3)

L^(-1) {(5s-2)/(〖3s〗^2+4s+8)}

Solución

L^(-1) {(5s-2)/(〖3s〗^2+4s+8)}=L^(-1) {5s/(〖3s〗^2+4s+8)}-2L^(-1) {1/(〖3s〗^2+4s+8)} =se aplica la propiedad de linealidad

=5L^(-1) {s/(3(s^2+4/3 s+8/3)}-2L^(-1) {1/(3(s^(2+4/3 s+8/3)) )}

=5/3 L^(-1) {s/(〖(s〗^2+4/3 s+((4/3)/2)2)8/3-4/9)}-2/3 L^(-1) {1/((s+2/3)2+20/9)}

=5/3 L^(-1) {s/((s+2/3)2+20/9)}-2/3 L^(-1) {1/((s+2/3)2+20/9)}

=5/3 L^(-1) {s/((s+2/3)2+((2√5)/3)2)}-2/3 L^(-1) {1/((s+2/3)2+((2√5)/3)2)}

=5/3 L^(-1) {((s+2/3)-2/3)/((s+2/3)2+((2√5)/3)2)}-2/3 ℮^(-2/3 t) (sen((2√5)/3)t/(((2√5)/3) ))

=5/3 L^(-1) {(s+2/3)/((s+2/3)2+((2√5)/3)2)}-5/3.2/3 L^(-1) {1/((s+2/3)2+((2√5)/3)2)}-2/3.3/(2√5) ℮^(-2/3 t).sen (2√5)/3

=5/3 ℮^(-2/3

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