Laplace transform involving step function \(\displaystyle{f{{\left({t}\right)}}}={\frac{{{\sin{{\left({2}{t}\right)}}}}}{{{e}^{{{2}{t}}}}}}+{t}{\left\lbrace\cdot\right\rbrace}{u}{\left({t}-{4}\right)}\)

Brielle James 2022-04-13 Answered

Laplace transform involving step function
f(t)=sin(2t)e2t+t·u(t4)

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Answers (2)

srasloavfv
Answered 2022-04-14 Author has 6 answers
F(s)=Lt[sin(2t)e2t+tθ(t4)](s)=0est(sin(2t)e2t+tθ(t4))dt
Where θ(x) is the Heaviside step function.
Now:
F(s)=0estsin(2t)e2tdt+0testθ(t4)dt=
0et(2+s)sin(2t)dt+4testdt
So, we get:
1. When R[s]>2
0et(2+s)sin(2t)dt=2s2+4(2+s)
2. When R[s]>0
4testdt=e4s(1+4s)s2
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lildeutsch11xq2j
Answered 2022-04-15 Author has 13 answers

You can use the formula for finding the laplace transform of secondary part-
L{f(t)u(ta)}=L{f(t+a)}×e-ass

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