Ejercicio resueltos trasformada de la place

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TRANSFORMADA DE LAPLACE

f(t) = t²e^(-αt)
L{ t²}s s+α =2!/s^3 =2/〖(s + α〗^3

f(t) = 3e^(-2t) - 9e^(-t) + 6

3L{ e^(-2t) } - 9L{ e^(-t) } + L{ 6}

f(t) = (2t-1)³=2t³-3(2t)²+6t-1

=2t³ - 12t² + 6t – 1

2L{ t^3 } - 12L{ t^2 } + 6L{ t} - L{ 1}

12/s^4 +24/s³+ 6/s²+1/s

TRANSFORMADA INVERSA DE LAPLACE
F(S)=((S+5))/((S+1)(S+3)) = A/((S+1))+B/((S+3))
S+ 5= AS + 3A + BS + B
5=3A+B A=2
1=AS+BS B=-1
= l^(-1){ 2/((S+1)) } + l^(-1){ (-1)/((S+3)) }
2e^(-t) - e^(-3t)

F(S)=(3(S+4))/(S(S+1)(S+2)) = A/S+B/((S+1))+C/((S+2))

3(S+4)= A(S+1)(S+2)+ B(S+2)S+C(S)(S+1)
3S +12 = (AS²+BS²+CS²) + (2AS+AS+2BS+CS) + 2A


0=A + B + C A = 63=3A + 2B + C B = -9
12=2A C = -3

= l^(-1){ 6/S } + l^(-1){ 9/((S+1)) } + l^(-1){ (-3)/((S+1)) }
=6l^(-1){ 1/S } -9l^(-1){ 1/((S+1)) } -9 l^(-1){ 1/((S+1)) }
6 – 9e^(-t) + 3e^(-2t)


F(S)=(6S+3)/S^2 = A/S+B/S^2
6S+2=AS+B
6/S+3/S^2 = 6l^(-1){ 1/S } + 3l^(-1){ 1/S^2 }
6 + 3t³

F(S)=(5s+2)/((S+1)(S+2)²) = A/((S+1))+B/((S+2))+C/((S+2)²)

(5s+2)/((S+1)(S+2)²) = (A(S+2)^2+ B(S+1)(S+2)+C (S+1))/((S+1)(S+2)²)

5s+2=AS²+2AS+4A+BS²+2BS+BS+2B+CS+C

0= A + BA=-1
5= 2A + 3B +C B=1
2= 4A + 2B +C C=4

= l^(-1){ (-1)/((S+1)) } + l^(-1){ 1/((S+2)) } + l^(-1){ 4/((S+2)) }
= 〖-l〗^(-1){ 1/((S+1)) }+ l^(-1){ 1/((S+2)) } + 4l^(-1){ 1/((S+2)) }
-e^(-t) – e^(-2t) + 4e^2t




F(S)= (2s²+4S+5)/(S(S+1)) = A/S+B/((S+1))
2s²+4S+5 = AS + A + BS
2=0A=-1
4=A +B B=1
5= A C=4

= l^(-1){ 5/S } + l^(-1){ (-1)/((S+1)) }
= 5l^(-1){ 1/S } - l^(-1){...
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