Inverse Laplace: Find 2exp(t)cos(t)+3exp(t)sin(t)

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SUMMARY

The discussion focuses on finding the inverse Laplace transform of the function F(s)=(2s+1)/(s^2 -2s+2). The correct inverse transform is identified as 2exp(t)cos(t) + 3exp(t)sin(t). A participant highlights that the breakdown into two fractions is unnecessary and suggests consulting a specific Laplace transform table for clarity. The discrepancy arises from the transform table used, indicating the importance of using the correct reference material.

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footballxpaul
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find the inverse laplace transform

F(s)=(2s+1)/(s^2 -2s+2)

I have broken it down too 2s/((s-1)^2 +1) + 1/((s-1)^2 +1) , then i got...

2cos(t) + exp(t)sin(t)

the book states that

2exp(t)cos(t) + 3exp(t)sin(t) is the answer? Where did I go wrong?
 
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I'm not sure where you went wrong; I can't find a simple inverse for your first term - which is the one that doesn't match the answers - so I'm guessing that's where your problem is.

However, look at #16 on this table of transforms:

http://www.swarthmore.edu/NatSci/ec...nMethods/LaplaceZTable/LaplaceZFuncTable.html

It's exactly the form for your problem, and you don't even have to break it into two fractions.
 
thanks that transform is not on the chart my book uses.
 

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