Help with inverse Laplace transforms

  • Thread starter rusty009
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  • #1
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Homework Statement



I have three inverse laplace transforms I can't solve, they are,

i)
(s-1)/(s^2 + 8s + 17)

ii)
(s+3)/(s^2 + 4s)

iii)
2/[(s+1)*(s^2 + 1)]


Homework Equations



The laplace transform table,

http://en.wikipedia.org/wiki/Laplace_transform#Table_of_selected_Laplace_transforms

The Attempt at a Solution




i) I completed the square and got,

(s-1)/(s^2 + 8s + 17)

-->(s-1)/[(s+4)^2 -1]

then split up into, s/[(s+4)^2 -1] and -1/[(s+4)^2 -1]

my answer was e^-4t *cos(t) - e^-4t *sin(t)
but the answer is e^-4t *cos(t) - 5e^-4t *sin(t) , I do not know where the 5 has come from ?


ii) I'm not sure where to start on this one

(s+3)/(s^2 + 4s) , I have taken out a factor of 1/s but im not sure were to go from there.


iii) For this one, I'm not having problem with the actual laplace but rather partial fractions, what I have done,

2/[(s+1)*(s^2 + 1)] = A/(s+1) + B/(s^2 + 1)

A(s^2 + 1) + B(s+1) = 2

for s=-1 I found A= 1

but I'm stuck at finding B, how do I make (s^2 + 1) = 0

Thanks a lot in advance !
 

Answers and Replies

  • #2
25
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Part 3:

It should be A/(s+1) + (Bs + c)/(s2 + 1)

Hence

2 = A.(s2 + 1) + (Bs +c).(s + 1)

You can then find all the values using s = -1, 0 and 1.
 
  • #3
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i) I completed the square and got,

(s-1)/(s^2 + 8s + 17)

-->(s-1)/[(s+4)^2 -1]

No, (s+4)^2 - 1 = s^2 + 8s + 16 - 1 = s^2 + 8s + 15, you have 17

ii) I'm not sure where to start on this one

(s+3)/(s^2 + 4s) , I have taken out a factor of 1/s but im not sure were to go from there

Note that s^2 + 4s = (s+2)^2 - 4. Does that help?

2/[(s+1)*(s^2 + 1)]

Your setup is wrong, s^2 + 1 has no real roots therefore the proper way to try to split it up is to rewrite it as A/(s+1) + (Bs + C)/(s^2 + 1). Work with that.
 
  • #4
412
4
(s-1)/(s^2 + 8s + 17)

-->(s-1)/[(s+4)^2 -1]

then split up into, s/[(s+4)^2 -1] and -1/[(s+4)^2 -1]

1) (s+4)^2 -1 = s^2+8s+16-1 .. but that's not 17?
2) Is s/[(s+4)^2 -1] = sin .. ? You should look at your numerator

NoMoreExams beat me so only look at 2 ..
 
  • #5
412
4
iii) For this one, I'm not having problem with the actual laplace but rather partial fractions, what I have done,

2/[(s+1)*(s^2 + 1)] = A/(s+1) + B/(s^2 + 1)

A(s^2 + 1) + B(s+1) = 2

for s=-1 I found A= 1

but I'm stuck at finding B, how do I make (s^2 + 1) = 0

Thanks a lot in advance !

First it is just wrong as NotStine mentioned. Second, you might want to review some partial fractions. You not always make things equal to 0 .. Use linear algebra
 
  • #6
623
0
Also if you ARE allowed to use a table, you can use this: http://www.vibrationdata.com/Laplace.htm

i) (s-1)/(s^2 + 8s + 17) = (s-1)/((s + 4)^2 + 1). By equation 2.27, lambda = -1, alpha = 4, beta = 1. So plugging that in we get e^(-4t)[cos(t) - 5sin(t)]

etc.
 

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