I am trying to take the laplace transform of \cos(t)u(t-\pi).

Bailee Ortiz 2022-04-27 Answered
I am trying to take the laplace transform of cos(t)u(tπ). Is it valid for me to treat it as ((cos(t)+π)π)u(tπ) and treat cos(t)π as f(t) and use the 2nd shifting property, or is this not the correct procedure?
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Answers (2)

utloverej
Answered 2022-04-28 Author has 15 answers
Here is one approach:
L(cos(t))=ss2+1
Lcos(t)u(tπ)=eπsL(cos(tπ))=eπsL(cos(t))=eπsss2+1
Recall from the sum formula:
cos(tπ)=costcosπ+sintsinπ=cost
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Salvador Ayala
Answered 2022-04-29 Author has 10 answers
Note that cos(t)=cos((tπ)+π)=cos(tπ)
Because cos(α+β)=cosαcosβsinαsinβ  where  α=tπ  and  β=π
Hence,
L{cos(t)u(tπ)}=L{cos(tπ)u(tπ)}=
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