Question

L{sin2t*sin2t}

Answer 1

wouldn't that be \[\mathcal L \{ \sin^2 (2t) \}\]
??

Answer 2

does * here denote convolution?

Answer 3

i thought that was multiplication lol

Answer 4

solution plz

Answer 5

thats a Laplace transform right?

Answer 6

u know L(sin2t) ?

Answer 7

if u know it,L{sin 2t * sin 2t} is just L(sin2t).L(sin2t). ok?

Answer 8

im guessig its convolution

Answer 9

and L(sin 2t)=2/(s^2+4)
so whats the final answer?

Answer 10

its convolution thats why i can separate into two laplace transforms........

Answer 11

yeah just saw that

Answer 12

\[\mathcal L \{ \sin^2 (2t) \}=\frac{1}{2}\mathcal L \{ 1-\cos (4t) \}=\frac{1}{2}(\frac{1}{s}-\frac{s}{s^2+16})=\frac{8}{s(s^2+16)}\]

Answer 13

\[\begin{equation} \mbox{L}\left\{ f_{1} (t) \, f_{2} (t) \right\} = \frac{1}{2 \pi {\rm j}} \int\limits\limits^{c + {\rm j}\infty}_{c - {\rm j}\infty} F_{1} (p) \, F_{2} (s - p) \, {\rm d}p\ . \end{equation}\]

Answer 14

its convolution in time domain

Answer 15

Ah - OK then the dual of this is the more known case.