Question

laplace

Answer 1

\[Y(s)=\frac{ e^{-4s} }{(s-2)^2+9 }\]

Answer 2

inverse laplace

Answer 3

http://www.terpconnect.umd.edu/~lvrmr/2013-2014-F/Classes/MATH246/EXAMS/TableLaplace.pdf

Answer 4

Looks like the one for sin times the one for the heaviside

Answer 5

\[Y(s)=\frac{ 1 }{ (s-2)^2+(3)^2 }*\frac{ 1 }{ e^{4t} }\]

Answer 6

I mean the one for exp sin

Answer 7

ya, but what can we do e^-4t

Answer 8

We have: \[\mathcal L ^{-1}
\bigg[
e^{-4s}J(s) \bigg](t) = u(t-4) \mathcal L ^{-1}
\bigg[J(s) \bigg](t-4)
\]

Answer 9

We have: \[
\mathcal L ^{-1}
\bigg[\frac{1}{(s-2)^2+3^2}\bigg](t-4) =
\frac 13 \mathcal L ^{-1}
\bigg[\frac{3}{(s-2)^2+3^2}\bigg](t-4)
\]

Answer 10

\[
=\frac 13 e^{2(t-4)}\sin(3(t-4))
\]

Answer 11

I think all together:\[
\frac 13 u(t-4)e^{2(t-4)}\sin(3(t-4))
\]

Answer 12

how can we graph it?>