HELP! Inverse Laplace Transform


by Patton84
Tags: inverse, laplace, transform
Patton84
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#1
Apr9-09, 10:43 PM
P: 2
I'm having some trouble getting the inverse Laplace to the following problems...I need some help

F(s)=24/s^5

F(s)= 4/[((s-2)^2)+25

F(s)= s/(s-1)(s+1)
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rock.freak667
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#2
Apr10-09, 12:24 AM
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For the first one, consider L{tn} works out to be for n>0.

For the second one what is

[tex]L^{-1} (\frac{1}{s^2+k^2)[/tex]

for the third one, split into partial fractions.
Patton84
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#3
Apr10-09, 12:44 AM
P: 2
Quote Quote by rock.freak667 View Post
For the first one, consider L{tn} works out to be for n>0.

For the second one what is

[tex]L^{-1} (\frac{1}{s^2+k^2)[/tex]

for the third one, split into partial fractions.

would this be right for the third one

A/s-1 + B/s+1 = s

djeitnstine
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#4
Apr10-09, 01:57 AM
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HELP! Inverse Laplace Transform


Quote Quote by Patton84 View Post
would this be right for the third one

A/s-1 + B/s+1 = s
Yes that's correct.
coomast
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#5
Apr10-09, 11:49 AM
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Quote Quote by djeitnstine View Post
Yes that's correct.
No that is wrong, the right formula is:

[tex]\frac{s}{(s-1)(s+1)}=\frac{A}{s-1}+\frac{B}{s+1}[/tex]

from which you need to determine A and B.

coomast
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#6
Apr10-09, 12:13 PM
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Quote Quote by coomast View Post
No that is wrong, the right formula is:

[tex]\frac{s}{(s-1)(s+1)}=\frac{A}{s-1}+\frac{B}{s+1}[/tex]

from which you need to determine A and B.

coomast

Oops I thought he multiplied it out and set it equal to s


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