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March 4th, 2013, 08:23 AM   #1
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Inverse Laplace of: (1) / ( (1+s)^2 +1 ) = ?

Will you please tell which conversions from the table did you use, and how did you used them for solving it? THanks
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March 4th, 2013, 12:07 PM   #2
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Re: Inverse Laplace of: (1) / ( (1+s)^2 +1 ) = ?

First notice that



But we there we have (s+1) instead of s so :


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March 5th, 2013, 12:04 AM   #3
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Re: Inverse Laplace of: (1) / ( (1+s)^2 +1 ) = ?

Quote:
Originally Posted by zaidalyafey
First notice that



But we there we have (s+1) instead of s so :


Thanks for answering Ok now can I see a reference or documentation for this? I mean where you add +1 to s and it equals ? Perhaps a web page or a book?
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March 5th, 2013, 03:34 AM   #4
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Re: Inverse Laplace of: (1) / ( (1+s)^2 +1 ) = ?

Quote:
Originally Posted by baylar10

Ok now can I see a reference or documentation for this? I mean where you add +1 to s and it equals ? Perhaps a web page or a book?
you always can prove it yourself using :



When I first learned about Laplace I looked at this PDF file, it is very helpful :
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March 5th, 2013, 06:31 AM   #5
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Re: Inverse Laplace of: (1) / ( (1+s)^2 +1 ) = ?

Quote:
Originally Posted by zaidalyafey
Quote:
Originally Posted by baylar10

Ok now can I see a reference or documentation for this? I mean where you add +1 to s and it equals ? Perhaps a web page or a book?
you always can prove it yourself using :



When I first learned about Laplace I looked at this PDF file, it is very helpful :
Thanks very much for the book Turns out you have to use First shifting rule!
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March 5th, 2013, 06:38 AM   #6
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Re: Inverse Laplace of: (1) / ( (1+s)^2 +1 ) = ?

yep, excellent.

Have you read it all ?
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March 5th, 2013, 11:13 AM   #7
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Re: Inverse Laplace of: (1) / ( (1+s)^2 +1 ) = ?

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Originally Posted by zaidalyafey
yep, excellent.

Have you read it all ?
No way!!!
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