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Inv Laplace Tranforms

  • Thread starter fredrick08
  • Start date
  • #1
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Homework Statement


(4s^2+2)/(s^3+6s-20)


The Attempt at a Solution


ok from an earlier question, i used partial frac, to re write it as (1/(s-2))+((3s+4)/(s^2+2s+10))
which i can re write to, by completing square to (1/(s-2))+((3s+4)/((s+1)^2+9))

now i think ive done all that ok.. so my inv laplace is:

e^(2t)+invlaplace((3s+4)/((s+1)^2+9))....

now im not completely sure on how to use s and t shifts properly(i just started this stuff today), and im quite unsure what to do from now but the answer should be
e^(2t)+(1/3)e^(-t)(9cos(3t)+sin(3t)) and i have no idea how to get that from that... what happens to the 4?? can anyone help plz
 

Answers and Replies

  • #2
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anyone?
 
  • #3
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i could do..
e^(2t)+(1/3)e^(-t)invlaplace((9s+12)/(s^2+9)) but in my examples i have no idea.. how to get cos and sin inside the invlaplace brackets.. coz then i would need to
e^(2t)+(1/3)e^(-t)invlaplace(9s/(s^2+9).... which would get me 9cos(3t) but then the sin........something but then i have a 12 and would need to subtract 9 to get to 3????.. idk im confused...
 
  • #4
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plz anyone?
 
  • #5
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anyone know wat to do?
 
  • #6
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?????
 
  • #7
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surely someone knows how to do this
 
  • #8
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cmon plz someone help
 
  • #9
HallsofIvy
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Every ten minutes you complain that people are not answering you? We are not sitting here waiting for some to post a question that we will answer immediately! Have patience.

To find an inverse transform of [tex]\frac{3s+4}{(s+1)^2+ 9}[/tex], separate the two parts of the numerator: write it as
[tex]3\frac{s}{(s+1)^2+ 9}+ 4\frac{1}{(s+1)^2+ 9}[/tex]
You should be able to find both of those in any table of Laplace transforms.
 
  • #10
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ok thankyou very much.. and sorry wont happen again.
 
  • #11
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ok i know what you said to do is right, but i still dont quite understand, form what ive learnt. i have e^(2*t)+invlaplace(3*s/((s+1)^2+9)+4/((s+1)^2+9))
=e^(2 t)+(3e^(-t)*invlaplace(s/(s^(2)+9))+4e^(-t)*invlaplace(1/(s^(2)+9))
=e^(2*t)+3*e^(-t)*cos(3*t)+(4/3)*invlaplace(3/(s^2+9))
=e^(2*t)+3*e^(-t)*cos(3*t)+(4*e^(-t)*1/3)*sin(3*t)
im pretty sure this is wrong.... plz can u tell me where stuffing up
 
  • #12
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i have no idea how the 4... turns into a 9....
 
  • #13
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ok its the (3*s/(s+1)^2+9) bit im stuffing up... i know its 3e^-t*cos(3t) but i think there also has to be a -sin(3t) somewhere... but where does it come from?
 
  • #14
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plz someone?
 
  • #15
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omg did sleep thinking actually help is 3s/(s+1)^2+9=e^(-t)(3cos(3t)-sin(t) because i need 3s+1 on top so i subtract 1. then its

3*(s+1)/((s+1)^2+9)-(1/3)(3/(s+1)^2+9)???? coz i think that works
 
  • #16
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nope.... im close i think.. tho
 
  • #17
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back in couple of hrs, feel free to help = )
 
  • #18
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ok back i been thinking again and is it..
invlaplace((3(s+1))/(s+1)^2+9)-3/(s+1)^2+9))?
then it would be e^(-t)3cos(3t)-e^(-t)sin(st)

so then + (4/3)sin(3t) would be

e^(-t)3cos(3t)+e^(-t)(1/3)sin(3t)
=(1/3)e^(-t)(9cos(3t)+sin(3t) yes i think that is right, plz can someone confirm?
 
  • #19
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dang...
 
  • #20
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nope its right lol
 

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