Question

Determine Laplace transform of f(t)=10 sin(2t-30degrees).
1.F(s)=20/(s+2)^2+4
2.20cos30deg./s^2+4
3.-5s+17.32/s^2+4
4.8.66/s^2+4

Answer 1

whts laplace?

Answer 2

its gauassians function

Answer 3

gauassians?

Answer 4

Lt(f(t) 10sin(2t-30 deg) F(s)
=(20/(s+2^2)+4)(s)
=-10(s-2sqrt3)F(s)/2s^2+8
=20/(s+2^2)+4/s

Answer 5

\[L \left\{ \sin(2t - \pi/6) \right\} = \int\limits_{0}^{\infty}e ^{-st}\sin(2t - \pi/6) dt.\] Let's call that integral L and apply integration by parts twice to set up a linear equation in L. Here I let u be e^-st and dv be the trig function dt. I come up with
\[L = \sqrt{3}/4 - s/8 - (s ^{2}/4)L.\]

Answer 6

Solving for L , then multiplying by 10, gives us \[10 L = (10 \sqrt{3} - 5s)/(s ^{2} + 4).\] for the answer, at least according to my calculations.