Find the Laplace transform of displaystyle f{{left({t}right)}}={t}{e}^{{-{t}}} sin{{left({2}{t}right)}} Then you obtain displaystyle{F}{left({s}right)}=frac{{{4}{s}+{a}}}{{left({left({s}+{1}right)}^{2}+{4}right)}^{2}} Please type in a = ?

Haven

Haven

Answered question

2020-11-29

Find the Laplace transform of f(t)=tetsin(2t)
Then you obtain F(s)=4s+a((s+1)2+4)2
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Answer & Explanation

davonliefI

davonliefI

Skilled2020-11-30Added 79 answers

Step 1
Given that f(t)=tetsin(2t)
We have to Laplace transform of f(t)=tetsin(2t)  that we have to find  L{tetsin(2t)}
Use Laplace transform formula
L{tkf(t)}=(1)kdkdsk(L{f(t)})
For, tetsin(2t),f(t)=etsin(2t)
L{tetsin(2t)}=ddx(L{etsin(2t)})(1)
Now, use formula L{ebtsin(at)}=a(sb)2+a2
Now, L{etsin(2t)}=2(s+1)2+4
Substitute L{etsin(2t)}=2(s+1)2+4 in equation (1)
L{tetsin(2t)}=ddx(2(s+1)2+4)
Step 2
Now,
L{tetsin(2t)}=ddx(2(s+1)2+4)
=(4(s+1)((s+1)2+4)2)
=4s+4((s+1)2+4)2
Therefore, F(s)=4s+4((s+1)2+4)2
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